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\begin{document}
\input{preamble.tex}
\lecture{2}{Mar 2, 2009}{Jayalal Sarma M.N.}{Jayalal Sarma M.N.}
In the previous lecture we saw that constant factors don't matter when
define class of functions accepted by a Turing machine in time $t$ or
space $s$. Hence the classes defined as invariant under $\bigO$
notation. We can define the following complexity classes based on
this.
\begin{definition}
\[ \P = \bigcup_{k \ge 0} \DTIME(n^k) \]
\[ \PSPACE = \bigcup_{k \ge 0} \DSPACE(n^k) \]
\[ \L = \DSPACE(\log n) \]
\end{definition}
From the simulations we already saw it is clear that $\L \subseteq \P
\subseteq \PSPACE$.
\section{Diagonalisation to prove Heirarachy Theorems}
In this lecture we want to introduce one of the important technique to
prove lower bounds that has been used in the classical complexity
theory. The method is diagonalisation and is a simple adaptation of
the method you have seen in previous courses. We will do a warm up
proof to review diagonalisation.
A standard example is the claim : the set of functions from $\{0,1\}^*
\to \{0,1\}$ is not countable. Diagonalisation is used to prove this
by contradiction. Suppose they are countable via an enumeration of the
functions $f_1, f_2, \ldots $. It is easy to construct a function $f$
which on input $i$ is the exact negation of $f_i$. This function is
clearly from $\{0,1\}^* \to \{0,1\}$ and is different from each
function in the enumeration. Hence the contradiction. The proof we
want present uses exactly the same idea, but over an enumeration of
Turing machines.
We need to fix a technical notion before we state the theorem. This is
called constructibility and is a sort of ``niceness'' for the resource
function that we consider. A function $f: \N \to \N$ is called a fully
space-constructible function if there exists a DTM that on any input
$x$ of length $n$ halts visiting exactly $s(n)$ squares of the
work-tape. For any fully space constructable function $s$ we can
construct a 2-tape machine that on input $x$ marks $\#$-symbols on
exactly $s(n)$ squares on the work-tape. We will call this the {\em
space-marking machine}. Most of the common functions, like $\lceil
\log n \rceil$, $cn$, $n^c$ are all space constructible functions.
\begin{theorem}[Space Hierarchy Theorem]
Let $s_2(n)$ be a fully constructible function. Suppose $s_1(n) =
o(s_2(n))$ then,
\[ \DSPACE(s_1(n)) \subsetneq \DSPACE(s_2(n)) \]
\end{theorem}
\begin{proof}
We are going to demonstrate a language $L$ that is accepted by a
Turing machine $M$ which runs in space $s_2(n)$ and differs from every
language in $\DSPACE(s_1(n))$. We are going to define $L$ by giving
the description for the machine $M$. We use the standard enumeration
of one-tape Turing machines $M_1, M_2, \ldots$. This is sufficient
because Tape reduction (we proved this in last lecture) does not
increase space beyond a constant factor. In fact the machine $M$ is
going to be a 2-tape machine.
Given input $x$ the machine $M$ is set to do the following:
\begin{enumerate}
\item {\em Initialisation:} $M$ marks off $s_2(n)$ cells on the firs
tape. This is possible by the assumption that $s_2(n)$ is {\em
space constructable}. $M$ also writes the integer $t_2(n) =
2^{s_2(n)}$. This is just writing off $s_2(n)$ zeros and a 1 on
the left. Again, space constructability of $s_2(n)$ is being
used. This could be thought of as a counter for the simulation
process.
\item Now $M$ processes input $x$ as follows : It simulates $M_x$ in
the enumeration on input $x$ with its first tape as $M_x$'s
work-tape. If at any point of the simulation, a cell which is not
marked is touched, $M_x$ aborts the simulation and
rejects. Otherwise it decrements the counter and proceeds to the
next step of $M_x$ on $x$. If the counter reaches down to zero,
$M_x$ halts and rejects.
\item If $M_x$ on input $x$ halts before counter being zero, and
without touching any unmarked cells of tape 1, then this means $M$
was able to correctly simulate $M$, and it needs to differ on
this input. $M$ accepts on $x$ if and only if $M_x$ rejects on $x$.
\end{enumerate}
Space bound for the simulation is clearly $O(s_2(n))$. Hence the
language $L$ accepted by $M$ is inside $\DSPACE(s_2(n))$.
Suppose language accepted by $M$ is in $\DSPACE(s_1(n))$. Since
$s_1(n) = o(s_2(n))$, there is an input $x$, $L(M_x) = L(M)$. Let us
see how $M$ simulates $M_x$ on input $x$. Let $n=|x|$ and $t_1(n)$ be
the number of possible configuration of $M_x$. We can also assume that
$t_1(n) \le t_2(n)$ by choice of $x$ (we skip this calculation).
If $M_x$ halts in time $t_2(n)$, then there is a contradiction in step
3, because $M$ does just opposite of what $M_x$ does on input $x$. If
$M_x$ does not halt in time $t_2(n)$. But by choice of $x$ there can't
be more than $t_2(n)$ configurations. Hence $M_x$ must be running into
an infinite loop where as $M$ never does. Thus a contradiction.
This proves the theorem.
\end{proof}
Note that the above proof diagonalised against an enumeration of only
one-tape Turing machines. In addition $M$ was a 2-tape Turing
machine. But in the case of space as the resource, this did not make a
difference (up to constant factors). However, we saw that we will have
to lose a quadratic factor in terms of time as the
resource. Essentially the same proof works to prove a weaker form
heirarchy theorem for time as well. We will skip the details.
\begin{theorem}[Time Hierarchy Theorem]
Let $t_2(n)$ be a fully constructible function. Suppose $(t_1(n))^2 =
o(t_2(n))$ then,
\[ \DTIME(t_1(n)) \subsetneq \DTIME(t_2(n)) \]
\end{theorem}
We did not prove this in complete detail in class. However, the
details are similar to the Space Hierarchy Theorem. A stronger form of
this theorem can in fact be proved by the stronger version of tape
reduction that we stated in the last lecture. This version essentially
says that if $t_1(n) \log t_1(n) = o(t_2(n))$, then $\DTIME(t_1(n))
\subsetneq \DTIME(t_2(n))$.
There are many corollaries to these two fundamental heirarchy
theorems. However, most of them follow from the definition itself.
\begin{corollary}
\begin{itemize}
\item $\L \subsetneq \PSPACE$.
\item If $\L \subseteq \DTIME(n^{k})$ then $\L \ne \P$.
\end{itemize}
\end{corollary}
We did not state the second corollary above explicitly in class. But
it may be interesting to note that it says, if the number of
configurations of any log-space bounded TM $M$ is bounded by
$2^{c.s(n)}$ for an absolute constant $c$(independent of $M$) then,
$\L \ne \P$.
\section{Nondeterminism}
Nondeterministic Turing machines are deterministic Turing machines
with an additional {\em guessing power}. That is at each configuration
of the computation, there are multiple values for the successor
function, and the Turing machine tries to simultaneously compute in
each of those choices, and accepts if atleast one of the choices
accept. To make it a little more formal, the non-deterministic Turing
machine has an additional {\em guess bit} in each state which gets
associated with each configuration, and there are at most two (without
loss of generality) outgoing transitions from each state for a given
input and worktape contents. If the guess bit 1, the Turing machine
takes one transition and if not the other. Finally, if for any input
$x$, there is atleast one choice of the guess bits such that the
computation accepts, then the machine is said to accept input $x$.
Similar to the deterministic complexity classes we can define their
non-deterministic counterparts too.
\[ \NP = \bigcup_{k \ge 0} \NTIME(n^k) \]
\[ \NPSPACE = \bigcup_{k \ge 0} \NSPACE(n^k) \]
\[ \NL = \NSPACE(\log n) \]
\subsection{Simulating Non-determinism with Determinism}
Now we will see some deterministic simulations of Non-deterministic
Complexity classes. The most important one of this is the Savitch's
theorem and we did this in class. However we will list down other ones
too. A simple simulation of the nondeterministic machine by a
deterministic Turing machine explores all possibilities of the guess
bit one-by-one and checks if the given non-deterministic Turing
machine accepts with that guess bits. This gives $\NTIME(t)
\subseteq \DSPACE(t)$ and $\NSPACE(t) \subseteq
\cup_{c>0} \DSPACE(2^{ct})$. The second inclusion also uses the
elementary theorem about configurations that we proved in the first
lecture. This gives $\NL \subseteq \P$ and $\NP \subseteq \PSPACE$.
Now we can prove the third main theorem of this lecture, which is the
simulation due to Savitch.
\begin{theorem}[Savitch's Theorem]
If $s(n)$ is space constractible, then,
\[ \NSPACE(s(n)) \subseteq \DSPACE((s(n))^2) \]
\end{theorem}
\begin{proof}
To prove the theorem, let $L \in \NSPACE(s(n))$ via a Turing machine
$M$. Since the space used is bounded by $s(n)$ there can be atmost
$2^{s(n)}$ configurations in the configuration graph of $M$. $M$
accepts if and only if there is atleast one path from the start
configuration to the accepting configuration (we can ensure that there
is at most one accepting configuration).
Now we define the following predicate for each configuration $\alpha$,
$\beta$, $\reach(\alpha,\beta,k)$ is 1 if and only if there is a
computation path starting from configuration $\alpha$ to configuration
$\beta$ of length at most $k$.
Basically we will give a way to test if
$\reach(\alpha,\beta,2^{O(s(n))})$. We will keep this $2^{O(s(n))}$
implicitly so that we will not need to find the explicit constant. The basic idea is a simple search for a middle node. That is,
\[ \reach(\alpha,\beta,k) \iff \exists \gamma
\left(\reach \left(\alpha,\gamma, \left\lceil \frac{k}{2} \right\rceil \right)
\land \reach\left(\gamma, \beta, \left\lfloor \frac{k}{2} \right\rfloor \right) \right) \]
This naturally gives rise to a deterministic recursive algorithm which
runs over all possible $\gamma$ that could be the middle node. It is
clear that the depth of the recursion is bounded by $\log k$. At each
step of the recursion, the argument includes $\gamma$ which is a $\log
(2^{(O(s(n)))}) = O(s(n))$ sized number. Thus the above recursive
algorithm can be simulated by a stack which uses atmost $\log
k.O(s(n))$ space. This gives the $O((s(n))^2)$ space bound.
\end{proof}
An easy corollary of Savitch's theorem is that for $\PSPACE$
non-determinism does not really help. That is $\PSPACE = \NPSPACE$.
There is a version of heirarchy theorem for non-deterministic Turing
machines. We will leave some of it to the excerices.
\end{document}