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\begin{document}

\centerline{\bf Homework 8 Morally Due April 16 at 3:30PM}

\begin{enumerate}
\item 
(30 points)
Show that NP is closed under intersection. 

\newpage 

\item 
(40 points) 
In this problem we will look at instructions a Turing Machine could have
and how they would be modeled by Boolean Formulas (by modifying
the proof of the Cook-Levin Theorem)

We will DO one such problem and then assign two others. 

\begin{enumerate}
\item 
(0 points- I give the answer so that you can do the other parts more easily) 
Assume that a Turing Machine has an instruction of the following type:

$$\delta(q,a) = (p,RR)$$

which means that if the head is looking at the symbol $a$,
and the state is $q$, then the head of the Turing Machine moves TWO steps to the right
and the state changes to $p$. 
Nothing on the tape changes, though the configuration will change since
the head has moved.

Give the formula that models this instruction. 

ANSWER:

We first look at what happens 
if the configuration is $(q,a)bb$? Here is the
sequence of parts of the configurations. 

\[
\begin{array}{|c|c|c|}
\hline
  (q,a) & b & b  \cr
\hline
  a     & b & (p,b)  \cr
\hline
\end{array} 
\]

The formula is

$$
(z_{i,j,(q,a)} \wedge z_{i,j+1,b} \wedge z_{i,j+2,b}) \rightarrow (z_{i+1,j,a} \wedge z_{i+1,j+1,b} \wedge z_{i+1,j+2,(p,b)})$$

This is NOT the final answer since the configuration could have other
symbols where I have the $bb$. 
Here is what happens if the config is $(q,a)\sigma_1\sigma_2$. 

\[
\begin{array}{|c|c|c|}
\hline
  (q,a) & \sigma_1 & \sigma_2  \cr
\hline
  a     & \sigma_1 & (p,\sigma_2)  \cr
\hline
\end{array} 
\]

Hence the formula is 

$$\bigwedge_{(\sigma_1,\sigma_2)\in\Sigma\times\Sigma}$$

$$(z_{i,j,(q,a)} \wedge z_{i,j+1,\sigma_1}\wedge z_{i,j+2,\sigma_2}) \rightarrow (z_{i+1,j,a} \wedge z_{i+1,j+1,\sigma_1} \wedge z_{i+1,j+2,(p,\sigma_2)})$$

When you answer the questions below you NEED to have both a diagram
like this one:

\[
\begin{array}{|c|c|c|}
\hline
  (q,a) & \sigma_1 & \sigma_2  \cr
\hline
  a     & \sigma_1 & (p,\sigma_2)  \cr
\hline
\end{array} 
\]

and the formula like this one:

$$\bigwedge_{(\sigma_1,\sigma_2)\in\Sigma\times\Sigma}$$

$$(z_{i,j,(q,a)} \wedge z_{i,j+1,\sigma_1} \wedge (z_{i,j+2,\sigma_2}) 
\rightarrow 
(z_{i+1,j,a} \wedge z_{i+1,j+1,\sigma_1} \wedge z_{i+1,j+2,(p,\sigma_2)})$$

And NOW for the problems YOU need to do.

\item
(20 points) Do what I did above for the transition

$\delta(q,a)=(p,b,L)$

which means that if the head is looking at an $a$ and the machine is in state $q$
then it will overwrite the $a$ it is looking with by a $b$ {\bf AND} change state to $p$ {\bf AND} move Left. 

\item
(20 points) Do what I did above for the transition

$\delta(q,a)=(p,L,b)$ 

which means that if the head is looking at an $a$ and the machine is in state $q$
then it will move Left {\bf AND THEN} write a $b$ in the square (overwriting whatever
was there) {\bf AND THEN} change state to $p$. 

\end{enumerate}

\newpage

\item 
(30 points) 
Let $a,b\in\N$, $a,b\ge 2$.
Let $B$ be solvable in time $2^{n^b}$ time where $n$ is the length of the input to $B$. 
(NOTE- in this problem the input to $B$ will be an ordered pair $(x,y)$ where $|x|=n$
and $|y|=n^a$ (as you will see soon). since $n \ll n^a$ we will consider the length of the
input to $B$ to be $O(n^a)$.)
Let 

$$A = \{ x \st (\exists y, |y|=|x|^a)[(x,y)\in B].$$

Give an algorithm that determines if $x\in A$. 
Give $T(n)$, the time bound on the algorithm for inputs of length $n$. 
$T(n)$ should be of the form $2^{O(n^c)}$ for a $c$ that depends on $a,b$. 

\end{enumerate}

\end{document}