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\hbox to 5.78in {\bf 6.841/18.405J: Advanced Complexity \hfill Wednesday, Feburary 5th, 2003}
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\hbox to 5.78in { {\Large \hfill Lecture 1\hfill} }
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\hbox to 5.78in { {\it Instructor: Madhu Sudan \hfill Scribe: Steve Weis} }
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\begin{document}
\header
\section{Administrivia}
The course instructor is Madhu Sudan, {\tt madhu@mit.edu}. E-mail him
if you have not received a class e-mail. The TA is Prahladh Harsha,
whose e-mail is {\tt prahladh@theory.lcs.mit.edu} . The course web
page is htpp://theory.lcs.mit.edu/\~{}madhu/ST03. Each student is
responsible for scribing one lecture. E-mail Prof. Sudan to sign up.
The first homework has been posted on the web page and is due in one
week on February 12th. There will be 4-5 problem sets over the
course. They will typically be due two weeks after they are assigned.
\section{Course Contents}
This course examines what kind of computational resources are
interesting and how they affect our ability to solve mathematical
problems. The 6.840 course dealt with classical concepts of resource:
time, space and nondeterminism. This course will consider resources
including alternation, non-uniformity, randomness and quantum
physics. We will also explore notions of what are proofs, how
interaction affects a computation and how knowledge affects computation.
\section{Classical Computational Complexity}
Traditionally, complexity theory has involved identifying a particular
resource, such as time, space or nondeterminism, and quantifying how
much of that resource a particular problem requires. By representing a
problem as a point on a plane, we can depict its hardness. To indicate
that a problem, $A$, is no harder than another, $B$, we draw an arrow
from $A$ to $B$. So, $A \rightarrow B$ would indicate that the problem
$A$ is easier than $B$, i.e. $A \leq B$. If paths exist in both
directions between two problems, then those problems are equivalent
difficulty. Clustering sets of equivalent problems yields phenomena
such as complexity classes, which we are familiar with. This is
depicted below:
\begin{center}
\includegraphics[scale=.65]{l01-classes.eps}
\end{center}
\section{Complexity Classes}
There are many time complexity classes and we would like to be able to
find which ones are ``interesting'' by ruling out as many arrows
between them as possible. To start with, we can define classes by
applying crude boundaries on resources. These will be logarithmic,
polynomial and exponential boundaries on time and space with or
without nondeterminism:
\vspace{1mm}
\begin{tabular}{|l|l|l|l|l|l|l|l|}\hline
& L & NL & P & NP & PSPACE & EXP \\ \hline
\hline
Restriction & Space & Space & Time & Time & Space & Time \\ \hline
Boundary & Log & Log & Poly & Poly & Poly & Exp \\ \hline
Nondeterminism & N & Y & N & Y & N & N \\ \hline
Example & Addition & Path & Primes & SAT & TQBF & Chess \\ \hline
\end{tabular}\\
\vspace{1mm}
Each one of these classes is contained in the class to its right,
i.e. $L \subseteq NL \subseteq P \subseteq \ldots, \subseteq
EXP$. Note that $PSPACE=NPSPACE$, an important result from 6.840. In
fact by Savitch's Theorem, $NSPACE(s(n)) \subseteq SPACE(s^2(n))$ for
any $s(n) \geq \log n$. See
Section
\ref{savitch}.
\section{Reductions}
Reductions allow for a formal definition of relative hardness between
problems. Recall the definition of a language, $L \subseteq \{0,1\}^*$. We
are concerned with the computation problem of when given $x \in
\{0,1\}^*$, deciding whether $x \in L$. There turn out to be two
useful definitions of a reduction:
\begin{itemize}
\item Karp Reduction: $L_1 \leq L_2$ if there exists an {\em
efficient} function $f$ such that $x \in L_1 \iff f(x) \in L_2$. This
is also called a many-to-one reduction.
\item Turing Reduction : $L_1 \leq L_2$ if given access to an oracle
deciding $L_2$, we can decide $L_1$ {\em efficiently}. This is also
called many-to-many reduction.
\end{itemize}
Turing reductions are clearly more general since Karp reductions work
only between languages while Turing reductions are well-defined for
any computational problem. Moreover, a Karp reduction implies a Turing
reduction. However, Karp reductions are generally easier to find than
Turing reductions. Clearly, a Karp reduction implies a Turing
reduction. However, for some problems there are known Turing
reductions, but no known Karp reductions. Also, we believe that Karp
reductions allow for a finer distinction between classes than what
Turing reductions provide us. For example, consider the computational
problem $\overline{\rm 3SAT}$ determining whether a 3-CNF formula
$\phi$ is unsatisfiable. Clearly, there exists a Turing reduction
${\rm 3SAT} \leq \overline{\rm 3SAT}$, that simply negates the output
of the oracle. Thus, if we had just Turing reductions, we would be let
to believe that $\overline{\rm 3SAT}$ is ``no harder'' than 3SAT and
this would imply that NP $=$ coNP. However Karp reductions allow us to
distinguish between NP and coNP as unfortunately, no one has
constructed a Karp reduction ${\rm 3SAT} \leq \overline{\rm 3SAT}$.
\section{Hierarchy Theorems}
The Time and Space Hierarchy Theories define strict partitions between
complexity classes. The theories are analogous to each other, except
that the space bound is tighter:
{\theorem For any time constructible function $t$, ${\rm TIME}(t(n))
\subsetneq {\rm TIME}\left(\omega(t(n) \log t(n))\right)$.}
{\theorem For any space constructible function $s$, ${\rm SPACE}(s(n))
\subsetneq {\rm SPACE}\left(\omega(s(n))\right)$.}\\
Both of these theorems apply to deterministic classes and are proved
through diagonalization arguments covered in 6.840. The basic idea is
as follows: Suppose we want to show some language $L$ is decidable in
time $t_2(n)$ but not in time $t_1(n)$. We do this by defining a
language $L$ that is decidable in time $t_2(n)$ but different from all
languages decidable in $t_1(n)$. $L$ is made to differ from every
language in ${\rm TIME}(t_1(n))$ by simulating all the machines that
run in time $t_1(n)$ and flipping the answer on some input. Thus,
there cannot be any equivalent language in ${\rm TIME}(t_1(n))$, since
$L$ explicitly differs from each of them on some input.
Finally, Blum's Speedup Theorem says that any problem solvable in time
$cn$ can be solved in time $c'n$ where $0 < c' < c$, i.e.:
{\theorem $L \in {\rm TIME}(t(n)) \Rightarrow L \in {\rm
TIME}\left(\frac{t(n)}{2}\right)$}
This tells us that the actual constant hidden in the running time of a
problem is irrelevant.
\section{Open Questions}
One of the biggest complexity class questions is whether
P$=$NP. Suppose that P$\neq$ NP, then is it true that we can decide
the exact complexity of every problem? For instance, if a language $L$
is described by a non-deterministic Turing machine, then can we decide
if $L \in $P? Unfortunately, this problem is undecidable. In fact,
under our assumption P$\neq$ NP, there is an infinite chain of
problems between P and NP.
Other open problems are whether coNP $=$ NP, L $=$ P or P $=$
PSPACE. Because a turning machine cannot use more space than it has
time, ${\rm TIME}(t(n)) \subseteq {\rm SPACE}(t(n))$ implying that P
$\subseteq$ PSPACE. Similarly, ${\rm SPACE}(s(n)) \subseteq {\rm
TIME}\left(2^{O(s(n))}\right)$ which implies that L $\subseteq$
P. However, by the Space Hierarchy theorem, L $\subsetneq$
PSPACE. Thus, we know that one of the containments L$\subset$P, P
$\subset$ PSPACE, must be tight.
\section{Savitch's Theorem}\label{savitch}
Savitch's theorem states that something computable in $s(n)$
non-deterministic space is computable in $s(n)^2$ deterministic
space. We'll now give a different proof (rather sketch) of this
theorem from what is commonly seen in textbooks.
We first use the following claim
{\claim If $L_1 \leq L_2$ in ${\rm SPACE}(S_1)$ and $L_2 \in {\rm
SPACE}(S_2)$, then $L_1$ is decidable in ${\rm SPACE}(2S_1+S_2)$.}\\
\textbf{Proof Sketch:} This misleadingly appears intuitive. One could
first compute the reduction $L_1\leq L_2$ and then call the decider
for $L_2$. Unfortunately, we are operating under a three-tape Turing
machine model: one tape is read-only, one is read/write and the third
is write-only. The space restrictions apply only to the read/write
tape. The write-only tape, hence the output of the reduction may be of
any size. To get around this, we will run the decider for $L_2$ and
then perform the reduction each time the decider needs to read a
particular bit of its input. Note that the only space needed for this
is $2S_1+S_2$: $S_1, S_2$ for the reduction and the decider
respectively, while another $S_1$ is required to indicate which bit is
needed.
Now on to Savitch's theorem:
{\theorem For any $s(n) \geq \log n$, ${\rm NSPACE}(s(n)) \subset
{\rm SPACE}(s^2(n))$}
\textbf{Proof Sketch:} We did not reach this in class, however, we'll
provide a brief sketch below.
We'll use Claim~1 repeatedly to prove this Theorem. Let
$N=2^{s(n)}$. Consider the language $${\rm PATH}_{N,i}=\{\langle
G,s,t\rangle : |V(G)| = N, \mbox{ there exists a path of length at
most $i$ from $s$ to $t$.}\}$$ Since ${\rm PATH}_{N,N}$ is a complete
problem for ${\rm NSPACE}(s(n))$, it is sufficient if we show that
${\rm PATH}_{N,N} \in {\rm SPACE}(s^2(n))$. For this, we reduce ${\rm
PATH}_{N,i}$ to ${\rm PATH}_{N,i/2}$ in the following manner and then
use Claim~1. From an instance $\langle G,s,t\rangle$ of ${\rm
PATH}_{N,i}$, we construct the new graph $G'$ which has exactly the
same set of vertices in $G$ but two vertices in $G'$ are connected if
there is a path of length at most 2 in the original graph $G$. Thus,
$\langle G,s,t\rangle \in {\rm PATH}_{N,i} \iff \langle G',s,t\rangle
\in {\rm PATH}_{N,i/2}$. Clearly this reduction takes space $O(\log
N)$. Suppose, $S_i$ is the space required to decide ${\rm
PATH}_{N,i}$, then by Claim~1, we have that recurrence $S_i \leq S_{i/2}+O(\log
N)$. Also, we note that $S_1 = O(1)$. Solving the recurrence we have
$S_i = (\log i)(\log N)$. Hence, $S_N = (\log N)^2 = s^2(n)$, which is
what we wanted to prove.
\end{document}
% LocalWords: Administrivia Madhu Prahladh Harsha madhu PSPACE Poly TQBF CNF
% LocalWords: NPSPACE Savitch's NSPACE coNP diagonalization c'n NDTM