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\begin{document}


\centerline{\bf Homework 9 Morally Due April 23 at 3:30PM}

\begin{enumerate}


\item
(30 points)
In this problem sets are sets of natural numbers.

Recall that 

$A\in \Sigma_1$ if there exists $B$ decidable such that

$$A = \{ x \st (\exists y)[ B(x,y) ] \}.$$

{\bf Definition} $A$ is an ADAM SET if 
there exists a Turing Machine $M$ with the following behaviour:

\begin{itemize}
\item
If $x\in A$ then $M(x)$ halts.
\item
If $x\notin A$ then $M(x)$ does not halt.
\end{itemize} 

And NOW for the problem: 

\begin{enumerate}
\item
Show that if $A\in \Sigma_1$ then $A$ is an ADAM set.
\item
Show that if $A$ is an ADAM set then $A\in\Sigma_1$.
\end{enumerate} 

\newpage

\item 
(30 points) 
{\bf Definition} Let $w\in\Sigma^*$. Then $\IS(w)$ is the set of words that can be
formed by removing any set of symbols from $w$. For example

$$\IS(abab) = \{e, a, b, aa, ab, ba, bb, aab, aba, abb, bab, abab \}$$

If $L$ is a language (a subset of $\Sigma^*$) then 

$$\IS(L) = \bigcup_{w\in L} \IS(w).$$

For example if $A=\{abab,bbbb\}$ then

$$\IS(A) = \{e, a, b, aa, ab, ba, bb, aab, aba, abb, bab, bbb, abab, bbbb \}$$

\begin{enumerate}
\item 
(25 points) 
Show that if $L\in\Sigma_1$ then $\IS(L)\in \Sigma_1$.
(You may use the quantifier definition of $\Sigma_1$ or the ADAM definition of $\Sigma_1$.
Either one will work.) 
\item
(5 points) 

VOTE one of the following (Note: You do not need to vote correctly to receive points):

\begin{itemize}
\item 
If $L$ is decidable then $\IS(L)$ is decidable.
Fire and Brimstone Speech to Follow.
\item 
There exists an $L$ that is decidable such that $\IS(L)$ is NOT decidable.
\item 
The question is UNKNOWN TO SCIENCE.
\end{itemize} 

\end{enumerate} 

\newpage 


\item 
(40- 8 points each)
You are designing an algorithm for CNFSAT. 
I will incompletely describe some short cuts you can take.
Fill in the BLANK

The input is of the form 

$C_1 \wedge \cdots \wedge C_m$

where each $C_i$ is an OR of literals (a literal is a var or its negation).

\begin{enumerate}
\item
If $C_1= (x_3)$ then you can do $BLANK_1$.

\item 
If $x_4$ appears in the formula but $\neg x_4$ never appears then you can do $BLANK_2$. 

\item
If $C_2 = (x_8)$ and $C_3 = (x_9)$ and $C_4=(\neg x_8 \vee \neg x_9)$ then you can do $BLANK_3$. 

\item
If $C_4 = (x_{10} \vee\neg  x_{11} \vee x_{12} \vee \neg x_{12})$ then you can do $BLANK_4$. 

\item
If there are no negation signs in the formula then you can do $BLANK_5$

\end{enumerate} 


\end{enumerate}

\end{document}