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% beamer stuff 
\renewcommand{\slidecaption}{AFL 04, King's College London, 17.~October 2012}
\newcommand{\bl}[1]{\textcolor{blue}{#1}}       
\newcommand{\dn}{\stackrel{\mbox{\scriptsize def}}{=}}% for definitions

\begin{document}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\mode<presentation>{
\begin{frame}<1>[t]
\frametitle{%
  \begin{tabular}{@ {}c@ {}}
  \\[-3mm]
  \LARGE Automata and \\[-2mm] 
  \LARGE Formal Languages (4)\\[3mm] 
  \end{tabular}}

  \normalsize
  \begin{center}
  \begin{tabular}{ll}
  Email:  & christian.urban at kcl.ac.uk\\
  Of$\!$fice: & S1.27 (1st floor Strand Building)\\
  Slides: & KEATS (also home work is there)\\
  \end{tabular}
  \end{center}


\end{frame}}
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\mode<presentation>{
\begin{frame}[c]
\frametitle{\begin{tabular}{c}Last Week\end{tabular}}

Last week I showed you

\begin{itemize}
\item tokenizer

\item tokenization identifies lexeme in an input stream of characters (or string)
and categorizes them into tokens

\item maximal munch rule
\end{itemize}

\url{http://www.technologyreview.com/tr10/?year=2011}
  
\end{frame}}
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\mode<presentation>{
\begin{frame}[c]
\frametitle{\begin{tabular}{c}The Derivative of a Rexp\end{tabular}}

\begin{center}
\begin{tabular}{@ {}l@ {\hspace{2mm}}c@ {\hspace{2mm}}l@ {\hspace{-10mm}}l@ {}}
  \bl{der c ($\varnothing$)}            & \bl{$\dn$} & \bl{$\varnothing$} & \\
  \bl{der c ($\epsilon$)}           & \bl{$\dn$} & \bl{$\varnothing$} & \\
  \bl{der c (d)}           & \bl{$\dn$} & \bl{if c $=$ d then $\epsilon$ else $\varnothing$} & \\
  \bl{der c (r$_1$ + r$_2$)} & \bl{$\dn$} & \bl{(der c r$_1$) + (der c r$_2$)} & \\
  \bl{der c (r$_1$ $\cdot$ r$_2$)} & \bl{$\dn$}  & \bl{if nullable r$_1$}\\
  & & \bl{then ((der c r$_1$) $\cdot$ r$_2$) + (der c r$_2$)}\\ 
  & & \bl{else (der c r$_1$) $\cdot$ r$_2$}\\
  \bl{der c (r$^*$)}          & \bl{$\dn$} & \bl{(der c r) $\cdot$ (r$^*$)}\\
  \end{tabular}
\end{center}

``the regular expression after \bl{c} has been recognised'' 

\end{frame}}
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\mode<presentation>{
\begin{frame}[c]

For this we defined the set \bl{Der c A} as

\begin{center}
\bl{Der c A $\dn$ $\{$ s $|$  c::s $\in$ A$\}$ } 
\end{center}

which is called the semantic derivative of a set
and proved 

\begin{center}
\bl{$L$(der c r) $=$ Der c ($L$(r))}
\end{center}


\end{frame}}
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\mode<presentation>{
\begin{frame}[c]
\frametitle{\begin{tabular}{c}The Idea of the Algorithm\end{tabular}}

If we want to recognise the string \bl{abc} with regular expression \bl{r}
then\medskip

\begin{enumerate}
\item \bl{Der a ($L$(r))}\pause
\item \bl{Der b (Der a ($L$(r)))}
\item \bl{Der c (Der b (Der a ($L$(r))))}\pause
\item finally we test whether the empty string is in set\pause\medskip
\end{enumerate}

The matching algorithm works similarly, just over regular expression than sets.
\end{frame}}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\mode<presentation>{
\begin{frame}[c]

Input: string \bl{abc} and regular expression \bl{r} 

\begin{enumerate}
\item \bl{der a r}
\item \bl{der b (der a r)}
\item \bl{der c (der b (der a r))}\pause
\item finally check whether the latter regular expression can match the empty string
\end{enumerate}

\end{frame}}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\mode<presentation>{
\begin{frame}[c]

We need to prove

\begin{center}
\bl{$L$(der c r) $=$ Der c ($L$(r))}
\end{center}

by induction on the regular expression.

\end{frame}}
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\mode<presentation>{
\begin{frame}[c]
\frametitle{\begin{tabular}{c}Proofs about Rexp\end{tabular}}

\begin{itemize}
\item \bl{$P$} holds for \bl{$\varnothing$}, \bl{$\epsilon$} and \bl{c}\bigskip
\item \bl{$P$} holds for \bl{r$_1$ + r$_2$} under the assumption that \bl{$P$} already
holds for \bl{r$_1$} and \bl{r$_2$}.\bigskip
\item \bl{$P$} holds for \bl{r$_1$ $\cdot$ r$_2$} under the assumption that \bl{$P$} already
holds for \bl{r$_1$} and \bl{r$_2$}.
\item \bl{$P$} holds for \bl{r$^*$} under the assumption that \bl{$P$} already
holds for \bl{r}.
\end{itemize}

\end{frame}}
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\mode<presentation>{
\begin{frame}[c]
\frametitle{\begin{tabular}{c}Proofs about Natural Numbers\\ and Strings\end{tabular}}

\begin{itemize}
\item \bl{$P$} holds for \bl{$0$} and
\item \bl{$P$} holds for \bl{$n + 1$} under the assumption that \bl{$P$} already
holds for \bl{$n$}
\end{itemize}\bigskip

\begin{itemize}
\item \bl{$P$} holds for \bl{\texttt{""}} and
\item \bl{$P$} holds for \bl{$c\!::\!s$} under the assumption that \bl{$P$} already
holds for \bl{$s$}
\end{itemize}

\end{frame}}
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\mode<presentation>{
\begin{frame}[t]
\frametitle{\begin{tabular}{c}Regular Expressions\end{tabular}}

\begin{center}
  \begin{tabular}{@ {}rrl@ {\hspace{13mm}}l}
  \bl{r} & \bl{$::=$}  & \bl{$\varnothing$}  & null\\
         & \bl{$\mid$} & \bl{$\epsilon$}        & empty string / "" / []\\
         & \bl{$\mid$} & \bl{c}                         & character\\
         & \bl{$\mid$} & \bl{r$_1$ $\cdot$ r$_2$} & sequence\\
         & \bl{$\mid$} & \bl{r$_1$ + r$_2$}  & alternative / choice\\
         & \bl{$\mid$} & \bl{r$^*$}                   & star (zero or more)\\
  \end{tabular}\bigskip\pause
  \end{center}

\end{frame}}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\mode<presentation>{
\begin{frame}[c]
\frametitle{\begin{tabular}{c}Languages\end{tabular}}

A \alert{language} is a set of strings.\bigskip

A \alert{regular expression} specifies a set of strings or language.\bigskip

A language is \alert{regular} iff there exists
a regular expression that recognises all its strings.\bigskip\bigskip\pause

\textcolor{gray}{not all languages are regular, e.g.~\bl{a$^n$b$^n$}.}
\end{frame}}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%   

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\mode<presentation>{
\begin{frame}[t]
\frametitle{\begin{tabular}{c}Regular Expressions\end{tabular}}

\begin{center}
  \begin{tabular}{@ {}rrl@ {\hspace{13mm}}l}
  \bl{r} & \bl{$::=$}  & \bl{$\varnothing$}  & null\\
         & \bl{$\mid$} & \bl{$\epsilon$}        & empty string / "" / []\\
         & \bl{$\mid$} & \bl{c}                         & character\\
         & \bl{$\mid$} & \bl{r$_1$ $\cdot$ r$_2$} & sequence\\
         & \bl{$\mid$} & \bl{r$_1$ + r$_2$}  & alternative / choice\\
         & \bl{$\mid$} & \bl{r$^*$}                   & star (zero or more)\\
  \end{tabular}\bigskip
  \end{center}

How about ranges \bl{[a-z]}, \bl{r$^\text{+}$} and \bl{!r}?
  
\end{frame}}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\mode<presentation>{
\begin{frame}[c]
\frametitle{\begin{tabular}{c}Negation of Regular Expr's\end{tabular}}

\begin{itemize}
\item \bl{!r}  \hspace{6mm} (everything that \bl{r} cannot recognise)\medskip
\item \bl{$L$(!r) $\dn$ UNIV - $L$(r)}\medskip
\item \bl{nullable (!r) $\dn$ not (nullable(r))}\medskip
\item \bl{der\,c\,(!r) $\dn$ !(der\,c\,r)}
\end{itemize}

\end{frame}}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\mode<presentation>{
\begin{frame}[c]
\frametitle{\begin{tabular}{c}Regular Exp's for Lexing\end{tabular}}

Lexing separates strings into ``words'' / components.

\begin{itemize}
\item Identifiers (non-empty strings of letters or digits, starting with a letter)
\item Numbers (non-empty sequences of digits omitting leading zeros)
\item Keywords (else, if, while, \ldots)
\item White space (a non-empty sequence of blanks, newlines and tabs)
\item Comments
\end{itemize}

\end{frame}}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\mode<presentation>{
\begin{frame}[c]
\frametitle{\begin{tabular}{c}Automata\end{tabular}}

A deterministic finite automaton consists of:

\begin{itemize}
\item a set of states
\item one of these states is the start state
\item some states are accepting states, and
\item there is transition function\medskip 

\small
which takes a state as argument and a character and produces a new state\smallskip\\
this function might not always be defined
\end{itemize}

\end{frame}}
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\end{document}

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