\documentclass[dvipsnames,14pt,t]{beamer}
\usepackage{beamerthemeplaincu}
%%\usepackage[T1]{fontenc}
\usepackage[latin1]{inputenc}
\usepackage{mathpartir}
\usepackage[absolute,overlay]{textpos}
\usepackage{ifthen}
\usepackage{tikz}
\usepackage{pgf}
\usepackage{calc} 
\usepackage{ulem}
\usepackage{courier}
\usepackage{listings}
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\usepackage{graphicx} 

\definecolor{javared}{rgb}{0.6,0,0} % for strings
\definecolor{javagreen}{rgb}{0.25,0.5,0.35} % comments
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	basicstyle=\consolas,
	keywordstyle=\color{javapurple}\bfseries,
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  morekeywords={abstract,case,catch,class,def,%
    do,else,extends,false,final,finally,%
    for,if,implicit,import,match,mixin,%
    new,null,object,override,package,%
    private,protected,requires,return,sealed,%
    super,this,throw,trait,true,try,%
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  morecomment=[l]{//},
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  morestring=[b]"""
}

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	basicstyle=\consolas,
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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}

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


\end{frame}}
 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%     

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

They are often used to recognise:\medskip

\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}}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%   


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

Last week I showed you a regular expression matcher 
which works provably in all cases.

\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}}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%   

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\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}}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%   

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\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}}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%   


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\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}

The matching algorithm works similarly, just over regular expression than sets.
\end{frame}}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%   


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\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
\item finally check whether the latter regular expression can match the empty string
\end{enumerate}

\end{frame}}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%   

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

We proved already

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

by induction on the regular expression.

\end{frame}}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%   

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\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}}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%   


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\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}}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%   

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\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}}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%   



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\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

\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}}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%   

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\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}}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%   



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

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


Do you think there are languages that are {\bf not} regular?
\end{frame}}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%   

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\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}}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%   

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\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}}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%   


\end{document}

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