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\fancyhead[L,LO]{
\textit{\small Math 387 Group problems, Monday Week 4}
}
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\begin{document}
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\bigskip

\begin{prob}
  Let $A=\{\langle R,S\rangle : \text{$R$ and $S$ are regular expressions and
  $L(R)\subseteq L(S)$}\}$.  Show that $A$ is decidable.  
\end{prob}

\begin{prob}
  A \emph{useless state} in a PDA is a state that is never entered on any input
  string. Let $L=\{<P> : \text{$P$ is a PDA with a useless state}\}$.  Prove
  that~$L$ is decidable.
\end{prob}

\begin{prob}
  Let $A$ and $B$ be DFAs.  Let $A$ have states~$Q_{A}$ and $B$ have states
  $Q_{B}$.  Show that if $L(A)\neq L(B)$, then there exists a word~$w$ of length
  less than or equal to~$|Q_{A}||Q_{B}|$ that is accepted by one machine and not
  the other.
\end{prob}

\begin{prob}
  Prove that if~$G$ is a CFG is in Chomsky normal form, then for any string~$w$
  of length $|w|=n\geq1$, exactly $2n-1$ steps are required for any derivation of $w$.
  (For the relevance of this problem, recall that our text uses it to prove that
  the acceptance question for CFG is decidable, i.e. the language
  $A_{\textrm{CFG}}=\{\langle G, w\rangle: \text{$G$ is a CFG that generates
  $w$}\}$ is decidable.)  Your proof will judged according to the clarity of
  its exposition---it should not require too much ingenuity on the part of the
  reader.
\end{prob}

\begin{prob} (bonus)
  Let 
  \[
    E=\{\langle M\rangle: \text{$M$ is a DFA that accepts some string with
  more $1$s than $0$s}\}.
\]
  Is $E$ decidable?
\end{prob}
\end{document}
