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%%\usepackage[T1]{fontenc}

\usepackage[latin1]{inputenc}

\usepackage{mathpartir}

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\usepackage{ifthen}

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\usepackage{calc} 

\usepackage{ulem}

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\definecolor{javared}{rgb}{0.6,0,0} % for strings

\definecolor{javagreen}{rgb}{0.25,0.5,0.35} % comments

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\lstset{language=Java,

	basicstyle=\consolas,

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    private,protected,requires,return,sealed,%

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  morestring=[b]"""

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% beamer stuff 

\renewcommand{\slidecaption}{AFL 03, King's College London, 9.~October 2013}

\newcommand{\bl}[1]{\textcolor{blue}{#1}}       

\newcommand{\dn}{\stackrel{\mbox{\scriptsize def}}{=}}% for definitions

\begin{document}

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\mode<presentation>{

\begin{frame}<1>[t]

\frametitle{%

  \begin{tabular}{@ {}c@ {}}

  \\[-3mm]

  \LARGE Automata and \\[-2mm] 

  \LARGE Formal Languages (3)\\[3mm] 

  \end{tabular}}

  %\begin{center}

  %\includegraphics[scale=0.3]{pics/ante1.jpg}\hspace{5mm}

  %\includegraphics[scale=0.31]{pics/ante2.jpg}\\

  %\footnotesize\textcolor{gray}{Antikythera automaton, 100 BC (Archimedes?)}

  %\end{center}

\normalsize

  \begin{center}

  Email:  & christian.urban at kcl.ac.uk\\

  Office: & S1.27 (1st floor Strand Building)\\

  \end{tabular}

  \end{center}

\end{frame}}

 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%     

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\mode<presentation>{

\begin{frame}[c]

\frametitle{\begin{tabular}{c}Regular Expressions\end{tabular}}

\begin{itemize}

\item symbols, digits

\item identifiers

\item numbers (non-leading zeros)

\item keywords

\item comments

\end{itemize}\bigskip

\mbox{}\hfill\bl{\url{http://www.regexper.com}}

\end{frame}}

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\mode<presentation>{

\begin{frame}[c]

\frametitle{\begin{tabular}{c}Last Week\end{tabular}}

Last week I showed you a regular expression matcher 

\begin{center}

\bl{$matcher\,r\,s$} \;\;if and only if \;\; \bl{$s \in L(r)$}

\end{center}\bigskip\bigskip 

\small

\textcolor{gray}{\mbox{}\hfill{}by Janusz Brzozowski (1964)}

\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{$(der\,c\,r) \cdot (r^*)$} &\smallskip\\\pause

  \bl{$der\!s\, []\, r$}     & \bl{$\dn$} & \bl{$r$} & \\

  \bl{$der\!s\, (c\!::\!s)\, r$} & \bl{$\dn$} & \bl{$der\!s\,s\,(der\,c\,r)$} & \\

  \end{tabular}

\end{center}

\end{frame}}

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\mode<presentation>{

\begin{frame}[c]

To see what is going on, define

\begin{center}

\bl{$Der\,c\,A \dn \{ s \;|\;  c\!::\!s \in A\}$ } 

\end{center}\bigskip\bigskip

\small

For \bl{$A = \{"f\!oo", "bar", "f\!rak"\}$} then

\begin{center}

\bl{\begin{tabular}{l@{\hspace{2mm}}c@{\hspace{2mm}}l}

$Der\,f\,A$ & $=$ & $\{"oo", "rak"\}$\\

$Der\,b\,A$ & $=$ &  $\{"ar"\}$\\  

$Der\,a\,A$ & $=$ & $\varnothing$\\

\end{tabular}}

\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{$Der\,b\,(Der\,a\,(L(r)))$}\pause

\item \bl{$Der\,c\,(Der\,b\,(Der\,a\,(L(r))))$}\pause

\item finally we test whether the empty string is in this set\pause\medskip

\end{enumerate}

\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{$der\,c\,(der\,b\,(der\,a\,r))$}\bigskip\pause

\end{enumerate}

\end{frame}}

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\mode<presentation>{

\begin{frame}[c]

We proved already

\begin{center}

\bl{$nullable(r)$} \;if and only if\;  \bl{$"" \in L(r)$}

\end{center}

\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 Rexps\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{$r_2$}.\bigskip

\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\\[-1mm] 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{$""$} 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}[c]

\frametitle{\begin{tabular}{c}Languages\end{tabular}}

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

A \alert{regular expression} specifies a language.\bigskip

A language is \alert{regular} iff there exists

a regular expression that recognises all its strings.\bigskip\bigskip\pause

\end{frame}}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%   

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\mode<presentation>{

\begin{frame}[t]

\frametitle{\begin{tabular}{c}Regular Expressions\end{tabular}}

\begin{center}

  \end{center}

\end{frame}}

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\mode<presentation>{

\begin{frame}[c]

\frametitle{\begin{tabular}{c}Negation of Regular Expr's\end{tabular}}

\begin{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}}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%   

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\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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