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\begin{document}
\header{6.841/18.405J: Advanced Complexity}{Monday, February 24, 2003}
{5}{Madhu Sudan}{Daniel Zaharopol}
\section{Fortnow's time/space lower bound for \lang{SAT}}
Fortnow gave a lower bound for either time or space for an algorithm
to solve \lang{SAT}, though the proof doesn't say which actually has
the bound \cite{Fortnow2000}.
To be more specific, consider what we ``believe'' about complexity
classes. We believe PH does not collapse, which in turn would imply
that NP $\neq$ P. This, in turn, would imply both that \lang{SAT}
does not have a linear-time algorithm, and that it does not have a log
space algorithm.
We will prove that, given two TMs, one cannot solve \lang{SAT} in
linear time while the other solves it in logarithmic space. This
result is not trivial; it was open until 1997, when proven by
Fortnow. The main idea of the proof, surprisingly, is to use
alternation to derive a contradiction.
Let's state the theorem more formally.
\begin{theorem}
\label{thm:fortnow}
$\lang{SAT} \in L \Rightarrow \exists \epsilon > 0 \ {\rm such\ that\ }
\lang{SAT} \notin {\rm TIME}(n^{1 + \epsilon})$
\end{theorem}
We prove this, as usual, by contradiction: assume $\lang{SAT} \in L$
and $\lang{SAT} \in {\rm TIME}(n^{1 + \epsilon})$.
Intuitively, we can see where the contradiction comes from by looking
at the ``power'' of alternation given either alternative. Suppose
$\lang{SAT} \in {\rm DTIME}(n^{1 + \epsilon})$. This would imply, in
a way, that alternation was not powerful. For example, it would give
us that $\Sigma_i^P \subset P \forall i$.
On the other hand, Savitch's theorem tells us that alternation is
powerful for small space computation. But since $\lang{SAT} \in L$,
this would tell us that alternation was powerful.
In principle, this is a contradiction. We now quantify it.
The contradiction will come from the Time Hierarchy Theorem (we could
also derive it, with some more work, from the Space Hierarchy
Theorem). Essentially, we will derive that ${\rm TIME}(T(n)) \subseteq
{\rm TIME}(T^{1 - \delta})$, for some $\delta > 0$. We will do this
by finding some intermediary inclusions.
\subsection{Cook's Strong Theorem}
There is a stronger version of Cook's Theorem which says that
languages in ${\rm NTIME}(T(n))$ reduce to \lang{SAT} on formulae of
length $T(n) \log T(n)$. We don't know that this bound is tight
(after all, if $P = NP$, then they reduce to formulae of length $1$).
\begin{theorem}[Cook's Strong Theorem\cite{Cook1988}] let $M$ be a
nondeterministic Turing machine running in time $t(n)$. There is a
reduction running in $O(t(n)\log t(n))$ time and $O(\log t(n))$ space
that maps inputs $x$ of length $n$ to formulae $\phi$ of size
$O(t(n)\log t(n))$ such that $$x\in L \iff \phi \in SAT.$$
\end{theorem}
We can, of course, also apply this result to deterministic time to get
that ${\rm TIME}(T(n)) \subseteq \lang{SAT}$ on length $T \log T$.
Then, given our assumption about \lang{SAT}, ${\rm TIME}(T(n))
\subseteq {\rm SPACE}(O(\log T))$. This is our first inclusion to
derive our contradiction.
\begin{lemma}
${\rm SPACE}(s) \subseteq {\rm ATIME}[2i, s 2^{\frac{s}{i}} i]$
\end{lemma}
\proof{We prove this in a similar manner to Savitch's theorem.
Consider the computation of a ${\rm SPACE}(s)$ TM; we can write the
states of its tape(s) in a table with width $s(n)$ and height $2^s$.
In the proof of Savitch's Theorem, we continually divided the
computation in half, and examined each half to see if it was a valid
computation.
In this proof, we divide each time into $w$ pieces and guess the $w$
intermediate points. Each time we divide out, we perform two
alternations.
Each time we branch out, we guess the next computation universally.
This requires writing down each computation (of length $s$), and doing
this $i$ times if we perform $2i$ alternations in total. Then the
total time taken is $wsi$.
Let $w = 2^{\frac{s}{i}}$, so that we continue to split the table into
$w$ pieces each time we go down a level.
Then the total time taken is $wsi = s 2^{\frac{s}{i}} i$, with $2i$
alternations.}
This gets us another inclusion, so we've built up to ${\rm TIME}(T(n))
\subseteq {\rm SPACE}(c_1 \log T) \subseteq \Sigma_{2i}^{{\rm
TIME}(T^{(c_2 / i)})}$, for constants $c_1$ and $c_2$.
\subsection{The Final Step}
\begin{lemma}
$i \exists \lang{TQBF} \in {\rm TIME}(n^{(1 + \epsilon)^{3i}})$
\end{lemma}
\proof{We proceed by induction.
Consider $2 \exists \lang{TQBF}$. A problem is of the form $\exists
x_1 \ldots x_n \forall y_1 \ldots y_m \phi(x_1, \ldots, x_n, y_1,
\ldots, y_m)$. But $\forall y_1 \ldots y_m \phi$ is in ${\rm
co-NTIME}(n)$ and therefore in ${\rm TIME}(n^{1 + \epsilon})$.
Then our entire formula is in ${\rm NTIME}(n^{1 + \epsilon})$; from
Cook's Strong Theorem we know that this reduces to a satisfiability
problem in \lang{SAT} of length $n^{1 + \epsilon} \log n$, which is in
${\rm TIME}((n^{1 + \epsilon} \log n) ^ {1 + \epsilon})$. This is
clearly in ${\rm TIME}(n^{(1 + \epsilon)^3})$.
We now proceed by induction, using the same argument, and get that $i
\exists \lang{TQBF} \in {\rm TIME}(n^{(1 + \epsilon)^{3i}})$.
}
This gives us that $\Sigma_{2i}^{{\rm TIME}(T^{(c_2 / i)})} \subseteq
{\rm TIME}((T^{(c_2 / i)})^{(1 + \epsilon)^{6i}})$.
Then ${\rm TIME}(T^{(c_2 / i) (1 + \epsilon)^{6i}}) \supset {\rm
TIME}(T)$ by our chain of inclusions.
This is a contradiction for proper choices of $i$ and $\epsilon$; for
example, if we choose $i = 10c_2$ and $\epsilon = 1/(60c_2)$.
Then we conclude that initial theorem about \lang{SAT} was correct.
\section{Randomness}
Here we consider randomness as a resource, and look at the eight
complexity classes of interest when considering randomness.
\subsection{Augmented Turing Machines}
We've already seen various ``augmented Turing machines,'' of the form
$M(\cdot , \cdot)$. The idea is that each one takes a ``real input''
--- a string we wish to accept or reject --- and an ``auxilliary
input,'' a string that somehow helps the calculation. We generally
assume $M(x,y)$ operates in polynomial time, and that $y$ is
polynomial in the length of $x$.
For example, nondeterminism asks if, given $x$, there exists $y$ such
that $M(x, y)$ accepts.
Non-uniformity says that, for all $n$, there exists $y_n$ such that,
for all $x$ with $|x| = n$, $x \in L$ if and only if $M(x, y)$
accepts.
When considering randomness, $y$ is not related to $x$ at all, but
captures some physical process. We ask if randomness can make
computations faster.
\subsection{Randomness}
We'd like a machine $M$ for a language $L$ to accept $x$ with a high
probability if $x \in L$, for the range of auxilliary inputs $y$.
There are several kinds of errors we can accept. A ``false positive''
occurs when $x \notin L$ but $M$ accepts on some $y$. A ``false
negative'' occurs when $x \in L$ but is rejected on some $y$.
We can require either that there are no false positives, no false
negatives, none of either, or we can accept either. This gives us
four natural complexity classes.
We get the other four by considering machines that operate either in
polynomial time or logarithmic space.
For logarithmic machines, we stipulate that they can read $y$ only
once (they must store what information they need in their own memory).
\subsection{Complexity Classes}
In all of the following, the final ``P'' stands for ``polynomial.''
\begin{itemize}
\item ZPP --- zero-error probabilistic
\item RP --- Randomized
\item coRP --- Complement-randomized
\item BPP --- Bounded-error probabilistic
\end{itemize}
We say that $L \in {\rm ZPP}$ if there exists a TM $M$ such that ${\rm
Pr}_y [M {\rm accepts } x] = 1$ if $x \in L$, ${\rm Pr}_y [M {\rm
accepts } x] = 0$ if $x \notin L$. This seems to give no benefit from
being probabilistic, so we allow one more output for this class only:
``I don't know.''
RP is in a sense analogous to NP. With NP, we know that $x \in L$ if
there exists a $y$ such that $M(x, y)$ accepts. A language is in RP,
meanwhile, if there exists a TM $M$ such that ${\rm Pr}_y[M {\rm
accepts } x] \geq \frac{1}{2}$ if $x \in L$, and ${\rm Pr}_y[M {\rm
accepts } x] = 0$ if $x \notin L$.
coRP is just the complementary class, defined as the complement of
languages in RP. It's easy to see that this is equivalent to there
existing $M$ such that ${\rm Pr}_y[M {\rm accepts } x] = 1$ if $x \in
L$, and ${\rm Pr}_y[M {\rm accepts } x] \leq \frac{1}{2}$ if $x \notin
L$.
As an exercise, one can prove ${\rm ZPP} = {\rm RP} = {\rm coRP}$.
There more interesting class for study is BPP. Here, we say that $L
\in BPP$ if there exists a TM $M$ such that ${\rm Pr}_y[M {\rm accepts
} x] \geq \frac{2}{3}$ if $x \in L$, and ${\rm Pr}_y[M {\rm accepts }
x] \leq \frac{1}{3}$ if $x \notin L$. It turns out that these bounds
can be anything as long as they differ appreciably; for example, we
could even have $\frac{2}{3} + \frac{1}{|x|}$ and $\frac{2}{3}$ as our
bounds.
\begin{thebibliography}{10}
\bibitem[C88]{Cook1988} S. Cook. Short propositional formulae represent
non-deterministic computation. {\em Information Processing Letters},
26:269-270, 1988.
\bibitem[F00]{Fortnow2000} L. Fortnow. Time-space tradeoffs for
satisfiability. {\em Journal of Computer and System Sciences},
60(2):337-353, April 2000.
\end{thebibliography}
\end{document}