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\begin{document}
\lecture{5}{9/22/2011}{Time bounded Non-determinism}{Gautam Prakriya}
%\draft % put this here only if you want to indicate this is a draft
\sloppy
%\section{Environments}
%\begin{theorem}
%This is an example of the theorem environment.
%Other theorem-like environments are: claim and exercise.
%\end{theorem}
%\begin{example}
%This is an example of the example environment.
%\end{example}
In the last lecture we introduced the notion of $\NP$-completeness. We showed
that Satisfiability is $\NP$-complete under polytime mapping reductions that
run in time $n \cdot (\poly(\log n )$ (where each bit in the reduction can be computed simultaneously in logspace and
$\polylog$ time.). The same holds for many other $\NP$-complete problems such as VERTEX-COVER,
$3$-SAT, INDEPENDENT-SET etc. Infact, most of the natural problems in $\NP$ which are not known to be
in $\P$ are NQLIN-complete. We begin this lecture by illustrating this fact - we show that
$3$-SAT and VC are NQLIN-complete.
In Sections 2 and 3, we answer two natural questions - (a) if $\P$ differs from $\NP$, are there problems in $\NP$ that are not
$\NP$-complete?, (b) Are there any $\NP$-complete problems which are not NQLIN-complete?
We answer the question (a) in the affirmitive by proving the conditional existence of $\NP$-intermediate languages
and address (b) by proving a time hierarchy theorem for non-deterministic machines.
We end the lecture by introducing the concept of Relativisation.
\section{Completeness for NQLIN (Contd.)}
\begin{theorem}
$3$-SAT is complete for NQLIN under $\leq^{QLIN}_{m}$.
\end{theorem}
\begin{proof}
Given a CNF, we need to convert it to a CNF with clauses of size $ \leq 3$.
We do this by splitting each clause that has more than $3$ literals into a number of clauses with
atmost $3$ literals each. While doing so, we introduce additional variables to preserve satisfiability.
For instance we replace $(l_1 \vee l_2 \vee l_3 \vee l_4)$ by
$(l_1 \vee l_2 \vee z) \wedge (l_3 \vee l_4 \vee \overline{z})$. Where $z$ is the new-literal introduced.
In general a clause with $m > 3$ literals can be replaced by $m-2$ clauses each containing $3$ literals.
$$(l_1 \vee l_2 \vee \ldots l_m)$$
can be written as
$$(l_1 \vee l_2 \vee z_1) \wedge (\overline{z_1} \vee l_3 \vee z_2) \wedge (\overline{z_2}
\vee l_4 \vee z_3) \wedge \ldots \wedge (\overline{z_{m-2}} \vee l_{m-1} \vee l_m)$$
\end{proof}
\begin{theorem}
VC is complete for NQLIN under $\leq^{QLIN}_{m}$.
\end{theorem}
\begin{figure}[h]
\centering
%\input{05_01.pstex_t}
\scalebox{0.5}{\input{05_02.pdf_t}}
\caption{$x_2 \wedge (x_1 \vee \overline{x_3}) \wedge
(\overline{x_2} \vee x_3 \vee x_4)$ has a satisfying assignment iff this graph has a vertex cover of size $\leq 4 + 0 + 1 + 3$}
\label{VC}
\end{figure}
\begin{proof}
Given a $3$-SAT formula $\phi$, we define a reduction that returns a graph $G$ and a number $k$,
s.t. $G$ has a vertex cover of size $k$ iff $\phi$ is satisfiable.
For each boolean variable $x$, we produce an edge which connects nodes $x$ and $\overline{x}$ in the graph.
The gadget for clauses is as follows - For each clause, we introduce a node for each
literal in the clause. We connect these nodes to each other and to the nodes in the variable gadgets with an identical label.
Figure $1$ contains the graph produced by the reduction for the formula $x_2 \wedge (x_1 \vee \overline{x_3}) \wedge
(\overline{x_2} \vee x_3 \vee x_4)$.
$k$ is defined to be $n + \sum (|C_j| - 1)$ where $n$ is the number of variables and $|C_j|$ is the size of clause $j$.
We now show that $\phi$ is satisfiable iff $G$ has a vertex cover of size $k_{\phi}$.
Note that any vertex cover must contain atleast one literal from each of the vertex gadgets and
atleast $|C_j| - 1$ from the gadget for clause $j$ ($\forall j$).
Now, Suppose $G$ has a vertex cover of size $n + \sum (|C_j| - 1)$, it is clear that exactly one vertex from each
clause is not picked. This implies that atleast one outgoing edge from each clause is covered by a vertex in some vertex gadget
(Each clause has $|C_j|$ outgoing edges.).
The literals from the vertex gadgets which are in the vertex cover define a
assignment in an obvious way. The argument above shows that each clause must contain atleast one of
these literals. Therefore, the assignment thus defined satisfies $\phi$.
If $\phi$ has a satisfying assignment, pick the literals in the vertex gadgets that are set to true
by the assignment. Since each clause must contain one of these literals, atleast one outgoing edge from each
clause gadget is covered. Choosing the remaining vertices from each clause gives us a vertex cover of the required
size.
\end{proof}
\section{Nondeterministic-Time Hierarchy Theorem}
Recall that in an earlier lecture, we looked at the deterministic time-hierarchy.
We now show a similar result for non-deterministic machines.%The Non-deterministic-time hierarchy
%is much finer than the deterministic-time hierarchy. This is because simulation in the non-deterministic setting involves only
%a constant factor overhead.
The argument we used in the deterministic case doesn't extend in an obvious way. This is because we don't know
how to complement a $\NTIME(t(n))$-language in $\NTIME(t'(n))$\footnote{The issue is that one might
have to run through all non-deterministic choices of a machine on an input to to flip the answer of the machine.}.
However, we can complement a $\NTIME(t(n))$-language in $2^{t(n)}$ deterministic steps.
We use this fact along with a modified diagonalization argument to prove the following theorem.
\begin{theorem}Let $t,t' : \mathbb{N} \mapsto \mathbb{N}$ s.t, $t'$
is time constructible and $t'(n) = \omega(t(n+1))$, then $\NTIME(t(n)) \subsetneq \NTIME(t'(n))$
\end{theorem}
\begin{proof}
In the deterministic case, machine $M$(the universal machine we construct) diagonalizes the $i^{th}$ input against
Machine $M_i$. Here we associate an interval of inputs $I_i$ with each machine $M_i$ and ensure that
for all $i$, $M$ disagrees with $M_i$ on atleast one element of $I_i$(The $I_i$'s are all disjoint).
The idea is that if the interval $I_i$ is large enough then $M$ when given one of the larger strings in $I_i$ as input
can deterministically complement $M_i$ on one of the smaller elements of the interval.
This technique is known as ``Delayed Diagonalization''.
As in the deterministic case, we demand that each machine occurs infinitely often,
so that the asymptotic behaviour of $t$ and $t'$ is captured.
The intervals are defined in such a way that the length of each element is one
more than that of the previous one. Let the elements of $I_i$ be $I_{i,1}, I_{i,2} \ldots I_{i,m}$,
we require $|I_{i,k}| = |I_{i,k-1}| + 1$ for $2 \leq k \leq m$.
%It is infact sufficient to consider just unary inputs.
For instance $I_i$ could be the set $\{0^l,0^{l+1}, \ldots 0^{l+m-1}\}$.
We now describe how $M$ simulates $M_i$ on $I_i$. The details of this simulation will
also tell us what $m$(= $|I_i|$) should be.
\begin{enumerate}
\item For input $I_{i,k}$, $M$ simulates $M_i$ on $I_{i,k+1}$ $(k < m)$ . Note that
$t'(|I_{i,k}|)$ is enough time to simulate $t(|I_{i,k+1}|)$ computation steps of $M_i$.
This follows from the following facts: (a) $t'(|I_{i,k}|) = \omega(t(|I_{i,k+1}|))$ and
(b) Non-deterministic simulation can be done with a constant factor overhead in time.
\item On input $I_{i,m}$, $M$ simulates all runs of $M_i$ on $I_{i,1}$ deterministically
and flips the result. This tells us that the interval $I_i$ has to be large enough, so that
$t'(|I_{i,m}|) \geq 2^{\Omega(t'(|I_{i,1})|)}$ holds.
\end{enumerate}
We won't get into details of how $M$ computes these intervals. For a detailed illustration
of how this can be done see page 70 of the book by Arora and Barak.
Now, suppose for the sake of contradiction that $M(x) = M_i(x)$ for all $x \in I_i$.
This would imply by (1) that $M(I_{i,m}) = M_i(I_{i,1})$. This is a contradiction,
since in (2) we set $M(I_{i,m}) = \overline{M_i(I_{i,1})}$. Therefore, we have shown that $M$ disagrees
with $M_i$ on atleast one element of $I_i$.
\end{proof}
\begin{figure}[h]
\centering
%\input{05_01.pstex_t}
\scalebox{0.6}{\input{05_01.pdf_t}}
\caption{The Delayed Diagonalization of $M$ against $M_i$ on interval $I_i$}
\label{the-label-for-cross-referencing}
\end{figure}
\section{NP-Intermediate Problems}
Let NPC represent the set of $\NP$ complete problems under poly-time many-one reductions.
We define the set of $\NP$-intermediate problems as follows.
\begin{definition}
$\NPI = \NP \setminus (\P \cup \NPC )$
\end{definition}
\begin{theorem}
If $\P \neq \NP$ then $\NPI \neq \phi$.
\end{theorem}
\begin{proof}
To prove this theorem, we need to present a language $ A \in \NP \setminus \P$
s.t. SAT is not $ \leq^{p}_{m}$ reducible to $A$. We will infact do more - we show that
$\exists A \in \NP \setminus \P$ s.t. SAT is not $\leq^{p}_{o}$ to $A$.
%In other words, we need to find a language $A \in \NP$ s.t (1) for any poly-time $DTM$ $M$, $A \neq L(M)$ and
%(2) for any poly-time $DOTM$ $N$ with oracle access to $A$, $SAT \neq L(N^A)$.
Firstly, let $M_1, M_2, M_3 \ldots$ be an enumeration of all
deterministic Turing machines and $N_1, N_2, N_3 \ldots $ be an enumeration of all
deterministic oracle Turing machines. We clock machines $M_i$ and $N_i$ by $n^i$ $\forall i$.
(Since we cannot enumerate the set of all poly-time Turing machines, we
instead enumerate all Turing machines and clock them with polynomials.)
Our goal is to construct a set $A \in \NP$ s.t $\forall i$:
\begin{enumerate}
\item $\bf C_{2i-1}$: $A \neq L(M_i)$ .
\item $\bf C_{2i}$: $L({N_i}^A) \neq$ SAT .
\end{enumerate}
It is clear that by satisfying $C_{2i-1}$, $\forall i$, we realize $A \notin \P$, and by satisfying $C_{2i}$, $\forall i$,
we ensure that there are no polynomial time oracle reductions from SAT to $A$, i.e, $A$ is not NP-hard .
Before getting into the details of the construction, we make a simple observation which we use
crucially in our construction. It follows from our hypothesis($\P \neq \NP$) that SAT differs
from all polynomial time computable languages at infinitely many points.
Therefore, given any language time $L \in P$ and $n \in \mathbb{N}$, $\exists N \geq n$
s.t. $L$ disagrees with SAT on $x_N$ (where by $x_k$ we refer to $k^{th}$ input in lexicographic order.).
As in the previous theorem, we associate disjoint intervals of inputs with the machines.
($M_i$'s and $N_i$'s). Suppose $I$ is the interval we associate with some $M_i$. Then by forcing
$A$ to behave like SAT on $I$ and picking a ''large enough`` $I$, we ensure that $A$ disagrees
with $L(M_i)$ on $I$, therefore satisfying $C_{2i-1}$. Similarly, on an interval $J$ corresponding to some $N_i$, we force $A$ to
behave like $\emptyset$(the empty set).\footnote{We could replace $\emptyset$ by any language in P.} .
If $J$ is large enough, $L(N_i^A)$ will disagree with SAT on some element of $J$, thus ensuring that $A$ satisfies $C_{2i}$.
%\footnote{Any language in P would do}
%Let $I$ be the interval on which we attempt to diagonalize $A$ against $M_i$. we force $A$ to behave like $SAT$ on $I$.
%Formally $I \cap A$ = $I \cap$ SAT (where $I$ is the interval under consideration).If the interval $I$ is
%large enough, then clearly, $L(M_i)$ will disagree with $A$ on some element of $I$.
%Similarly, suppose we are attempting to diagonalize $A$ against $N_i$ on an interval $J$.
We now give a formal description of our construction. %We will later modify this to ensure that
%the language $A$ we construct is in $NP$.
$A$ is constructed in phases. In phase $i$ we satisfy conditions $C_{2i-1}$ and $C_{2i}$.
\begin{algorithm}
\begin{algtab}
Initially we set $A =$ SAT and $y = \epsilon$\\
\algforeach{phase $i = 0, 1, 2, ...$}
Note that except for a finite set of inputs, $A$ is identical to SAT. Therefore, $A$ is $\NP$-hard.
It follows that $A$ disagrees with $L(M_i)$ on infinitely many inputs. Let $\omega$ be the lexicographically
smallest string that follows $y$ s.t, $M_i$ disagrees with $A$ on $\omega$.\\
$A \longleftarrow (A \cap \Sigma^{\leq |\omega|}) \cup (\emptyset \cap \Sigma^{ > |\omega|})$, i.e, $A$ contains no
strings of size greater than $|\omega|$. Since $A$ is now finite $A \in \P$.\\
Since $A \in \P$, $L(N_i^A)$ disagrees with SAT at infinitely many points. Let $u$ be the lexicographically smallest
that follows $\omega$ s.t, $L(N_i^A)$ disagrees with SAT at $u$. The size of the largest oracle query $N_i^A$
could have made to $A$ on input $u$ is $|u|^i$. (Since we clock $N_i^A$ with $n^i$.) We set $y$ to be the
lexicographically smallest string of size $|u|^i + 1$. This ensures that when $A$ is modified in future iterations,
the computation of $N_i^A$ on $u$ remains unaffected.\\
$A \longleftarrow (A \cap \Sigma^{ < |y|}) \cup ($SAT$ \cap \Sigma^{ \geq |y|})$, i.e, $A$ agrees with
SAT for all strings of size $\geq |y|$.\\
\end{algtab}
\end{algorithm}
The $A$ obtained clearly satisfies $C_{2i-1}$ and $C_{2i}$ $\forall i$. However, it is not clear that
the $A$ is in $\NP$. The issue is given $x$ we need to first find which of SAT and $\emptyset$ $A$ behaves like
at $x$. This can be achieved by running the above algorithm till we hit $x$, however, this may take exponential time.
The problem lies in steps $(3)$ and $(5)$ - Checking whether $M_i$(or $N_i^A$) disagree with SAT at some input
can potentially take time exponential in $x$.
We get around this by using the technique of delayed diagonalization to modify the above
construction. Recall that the idea behind delayed diagonalization is to stretch the intervals out,
so that the elements at the end of the interval are much larger than the ones in the beginning. This
allows us to brute-force check an otherwise hard to compute condition on the initial elements.
In our setting, this translates to modifying steps $(3)$ and $(5)$ by ''waiting long enough``
so that it becomes easy to detect a disagreement with SAT for the earlier elements.
We formalize this by defining a Polynomial-time computable function COND which on input $x$
tells whether $A$ behaves like SAT or $\emptyset$ at $x$.
If COND$(x)$ is odd, $A$ agrees with SAT on $x$, otherwise, $A$ agrees with $\emptyset$.
\begin{algorithm}
PROCEDURE COND$(n \in \mathbb{N})$.\\
\begin{algtab}
\algif{$n = 0$}
\algreturn{$1$}\\
COMPUTE COND$(n-1)$ \\
Check first $n$ strings in lexicographic order and see if they
witness $C_{COND(n-1)}$. Clock each check for $n$ steps. If undecided in given time bound,
conclude that the witness was not found.\\
\algif{Witness found}
\algreturn{COND$(n-1) + 1$ }\\
\algelse
\algreturn{COND$(n-1)$}
\end{algtab}
\end{algorithm}
In Step $4$, we run a deterministic algorithm for SAT on the first $n$ strings. %Since each of these strings has
%length atmost $O(log(n))$
%We can run the deterministic brute-force algorithm for SAT.
Suppose that $c =$ COND$(n-1)$ is even. Then in step $4$ we check if $N_{c/2}^A$
disagrees with SAT on the first $n$ strings. Since we clock our computation on each of
these strings by $n$, any oracle query to the language $A$ has size atmost $n$.
This ensures that the computation of $N_{c/2}^A$ on the first $n$ inputs remains
unaffected regardless of future changes to $A$. This is because, any changes made
to $A$ in subequent iterations will effect only the membership of strings with size $ > n$.
It is easy to check that this procedure runs in polynomial time. We have thus shown that $A \in \NP$
and $A \notin \P \cup$ NPC.
\end{proof}
Problems which are believed to be $\NP$-intermediate include Graph Isomorphism and Factoring.
\section{Relativisation}
\begin{definition}
A Statement is said to relativize if it continues to hold if all Turing Machines are
given access to the same oracle.
\end{definition}
\begin{example}
The Non-deterministic Time hierarchy theorem relativises:
$\forall $ oracles A $\NTIME^A(t(n)) \subsetneq \NTIME^A(t'(n))$
where, $t'(n) = \omega(t(n+1))$.
\end{example}
\begin{example}
Recall that $K_{N} = \{ |$ NTM $M$ accepts $x$ in time $ \leq t\}$ was shown to be
complete for $\NP$ under $\leq ^{log}_m$.
This statement relativises. $K_{N}^A = \{ |$ NTM $M^A$ accepts $x$ in time $ \leq t\}$
is complete for $\NP^A$ under $\leq ^{log}_m$, $\forall$ oracles A.
\end{example}
\section{Next Time}
In the next lecture, We continue studying the concept of relativisation.
We show that the statements $\P = \NP$ and $\P \neq \NP$ do not relativize. Which means any proof resolving
$\P$ vs. $\NP$ must contain a statement which doesn't relativize. Note that most of the statements and
Proofs we have studied so far do relativize.
%which implies that we require techniques different from the ones we have studied so far to resolve the $\P vs. \NP$ question.
We also investigate non-determinism with bounds on space. We will study results which suggest
that space bounded non-determinism is better understood, e.g. non-deterministic space is closed under complementation.
\section*{Acknowledgements}
In writing the notes for this lecture, I perused the notes by Baris Aydinlioglu for
lecture 3,4 from the Spring 2007 offering of CS~810, and the notes by
Dmitri Svetlov for lectures 3,4 from the Spring 2010 offering of CS~710.
\section*{References}
Sanjeev Arora and Boaz Barak. \textit{Computational Complexity: A Modern Approach.} 2009.
\end{document}