\documentclass{article}

\usepackage{../style}

\usepackage{../grammar}

\usepackage{../graphics}

\begin{document}

\section*{Homework 7}

%\HEADER

\begin{enumerate}

\item Suppose the context-sensitive grammar

\begin{center}

\begin{tabular}{lcl}

$S$ & $::=$ &  $bSAA\;|\; \epsilon$\\

$A$ & $::=$ & $a$\\

$bA$ & $::=$ & $Ab$\\

\end{tabular}

\end{center}

where $S$ is the starting symbol of the grammar. 

Give a derivation  of the string $"\!aaabaaabb"$. 

What can you say about the number of as and bs in the

strings recognised by this grammar.

\solution{

  S -> bSAA ->  bbSAAAA ->

  bbbSAAAAAA ->

  bbbAAAAAA ->

  bbAbAAAAA -> .. ->

  bbAAAAAAb -> .. -> AAAbAAAbb -> .. -> aaabaaabb

  }

\item Consider the following grammar 

\begin{plstx}[margin=1cm]

  : \meta{S\/} ::= \meta{N\/}\cdot \meta{P\/}\\

  : \meta{P\/} ::= \meta{V\/}\cdot \meta{N\/}\\

  : \meta{N\/} ::= \meta{N\/}\cdot \meta{N\/}\\

  : \meta{N\/} ::= \meta{A\/}\cdot \meta{N\/}\\

  : \meta{N\/} ::= \texttt{student} \mid \texttt{trainer} \mid \texttt{team} \mid \texttt{trains}\\

  : \meta{V\/} ::= \texttt{trains} \mid \texttt{team}\\

  : \meta{A\/} ::= \texttt{The} \mid \texttt{the}\\

\end{plstx}

where $S$ is the start symbol and $S$, $P$, $N$, $V$ and $A$ are non-terminals.

Using the CYK-algorithm, check whether or not the following string can be parsed

by the grammar:

\begin{center}

\texttt{The trainer trains the student team}

\end{center}

\solution{

\begin{center}

  \begin{tikzpicture}[scale=0.7,line width=0.8mm]

  \draw (-2,0) -- (4,0);

  \draw (-2,1) -- (4,1);

  \draw (-2,2) -- (3,2);

  \draw (-2,3) -- (2,3);

  \draw (-2,4) -- (1,4);

  \draw (-2,5) -- (0,5);

  \draw (-2,6) -- (-1,6);

  \draw (0,0) -- (0, 5);

  \draw (1,0) -- (1, 4);

  \draw (2,0) -- (2, 3);

  \draw (3,0) -- (3, 2);

  \draw (4,0) -- (4, 1);

  \draw (-1,0) -- (-1, 6);

  \draw (-2,0) -- (-2, 6);

  \draw (-1.5,-0.5) node {\footnotesize{}\texttt{The}}; 

  \draw (-0.5,-1.0) node {\footnotesize{}\texttt{trainer}}; 

  \draw ( 0.5,-0.5) node {\footnotesize{}\texttt{trains}}; 

  \draw ( 1.5,-1.0) node {\footnotesize{}\texttt{the}}; 

  \draw ( 2.5,-0.5) node {\footnotesize{}\texttt{student}}; 

  \draw ( 3.5,-1.0) node {\footnotesize{}\texttt{team}};

  \draw (-1.5,0.5) node {$A$}; 

  \draw (-0.5,0.5) node {$N$}; 

  \draw ( 0.5,0.5) node {$N,\!V$}; 

  \draw ( 1.5,0.5) node {$A$}; 

  \draw ( 2.5,0.5) node {$N$}; 

  \draw ( 3.5,0.5) node {$N,\!V$};

  \draw (-1.5,1.5) node {$N$}; 

  \draw (-0.5,1.5) node {$N$}; 

  \draw ( 0.5,1.5) node {$$}; 

  \draw ( 1.5,1.5) node {$N$}; 

  \draw ( 2.5,1.5) node {$N$};

  \draw (-1.5,2.5) node {$N$}; 

  \draw (-0.5,2.5) node {$ $}; 

  \draw ( 0.5,2.5) node {$N,\!P$}; 

  \draw ( 1.5,2.5) node {$N$};

  \draw (-1.5,3.5) node {$$}; 

  \draw (-0.5,3.5) node {$N,\!S$}; 

  \draw ( 0.5,3.5) node {$N,\!P$};

  \draw (-1.5,4.5) node {$N,\!S$}; 

  \draw (-0.5,4.5) node {$N,\!S$};

  \draw (-1.5,5.5) node {$N,\!S$}; 

  \draw (-2.4, 5.5) node {$1$}; 

  \draw (-2.4, 4.5) node {$2$}; 

  \draw (-2.4, 3.5) node {$3$}; 

  \draw (-2.4, 2.5) node {$4$}; 

  \draw (-2.4, 1.5) node {$5$}; 

  \draw (-2.4, 0.5) node {$6$}; 

  \end{tikzpicture}

  \end{center}

  }

\item Transform the grammar

\begin{center}

\begin{tabular}{lcl}

$A$ & $::=$ & $0A1 \;|\; BB$\\

$B$ & $::=$ & $\epsilon \;|\; 2B$

\end{tabular}

\end{center}

\noindent

into Chomsky normal form.

\solution{

  First one has to eliminate $\epsilon$. This means we obtain the rules:

  \begin{center}

    \begin{tabular}{lcl}

      $A$ & $::=$ & $0A1 \;|\; 01 \;|\; BB \;|\; B$\\

      $B$ & $::=$ & $2 \;|\; 2B$

    \end{tabular}

  \end{center}

  Now we have to bring the rules into CNF form by adding additional

  non-terminals, like $Z$, $O$, $T$,  and splitting up the rules into ``twos'':

  \begin{center}

    \begin{tabular}{lcl}

      $A$ & $::=$ & $ZC \;|\; ZO \;|\; BB \;|\; 2$\\

      $B$ & $::=$ & $2 \;|\; TB$\\

      $C$ & $::=$ & $AO$\\

      $Z$ & $::=$ & $0$\\

      $O$ & $::=$ & $1$\\

      $T$ & $::=$ & $2$\\                           

    \end{tabular}

  \end{center}   

}

\item Consider the following grammar $G$

\begin{center}

\begin{tabular}{l}

$S ::= \texttt{if0} \cdot E \cdot \texttt{then} \cdot S$\\

$S ::= \texttt{print} \cdot S$\\

$S ::= \texttt{begin} \cdot B\cdot \texttt{end}$\\

$B ::= S\cdot \texttt{;}$\\

$B ::= S\cdot \texttt{;} \cdot B$\\

$S ::= num$\\

$E ::= num$\\

$B ::= num$

\end{tabular}

\end{center}

where $S$ is the start symbol and $S$, $E$ and $B$ are

non-terminals.

Check each rule below and decide whether, when added to $G$,

the combined grammar is ambiguous. If yes, give a string that

has more than one parse tree.

\begin{center}

\begin{tabular}{rl}

(i)   & $S ::= \texttt{if0} \cdot E\cdot \texttt{then} \cdot S\cdot \texttt{else} \cdot S$\\

(ii)  & $B ::= B \cdot B$\\

(iii) & $E ::= ( \cdot E \cdot )$\\

(iv)  & $E ::= E \cdot + \cdot E$

\end{tabular}

\end{center}

\solution{

  (i) this is ambiguous -> it's an instance of the dangling else;

  (ii) rules of the form $B ::= B \cdot B$ are always ambiguous $B \cdot B\cdot B$

  (iii) this is fine

  (iv) same as (ii) $E\cdot + \cdot E \cdot + \cdot E$

  }

\item Suppose the string $``9-5+2''$. Give all ASTs that

  the following two grammars generate for this string.

Grammar 1, where List is the starting symbol:

\begin{center}

\begin{tabular}{lcl}

$List$ & $::=$ &  $List + Digit \mid List - Digit \mid Digit$\\

$Digit$ & $::=$ & $0 \mid 1 \mid 2 \mid 3 \mid 4 \mid 5 \mid 6 \mid 7 \mid 8 \mid 9$

\end{tabular}

\end{center}

Grammar 2, where String is the starting symbol:

\begin{center}

\begin{tabular}{@{}lcl@{}}

  $String$ & $::=$ & $String + String \mid String - String \mid$\\

  & & $0 \mid 1 \mid 2 \mid 3 \mid 4 \mid 5 \mid 6 \mid 7 \mid 8 \mid 9$

\end{tabular}

\end{center}

\solution{

  The point is that Grammar 1 is un-ambiguous, while the second is ambiguous.

  Grammar 1 parses the strings as (9 - 5) + 2. Grammar 2 is ambiguous and

  there are two possibilities (9 - 5) + 2 and 9 - (5 + 2).

}

%\item {\bf (Optional)} The task is to match strings where the letters are in alphabetical order---for example, 

%\texttt{abcfjz} would pass, but \texttt{acb} would not. Whitespace should be ignored---for example

%\texttt{ab c d} should pass. The point is to try to get the regular expression as short as possible!

%See:

%\begin{center}

%\url{http://callumacrae.github.com/regex-tuesday/challenge11.html}

%\end{center}

\end{enumerate}

\end{document}

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The| trainer trains the student A {N,S} => N 

The trainer |trains the student N {N, P} => N S

The trainer trains |the student N N => N 

The trainer trains the |student 

trainer |trains the student team N o {N, P} => N, S

trainer trains| the student team N o N => N

trainer trains the |student team 

trainer trains the student |team {N, P} o {N, V} => N

The| trainer trains the student team A o (N,S) => N

The trainer| trains the student team N o (N,P) => N, S

The trainer trains| the student team N o N => N

The trainer trains the| student team 

The trainer trains the student| team (N,S) o (N,V) => N