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\usepackage{../langs}

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\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}

  \begin{tabular}{lp{8cm}}

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

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

  Slides: & KEATS (also home work and coursework is there)\\

  \end{tabular}

  \end{center}

\end{frame}}

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\mode<presentation>{

\begin{frame}[c]

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

In programming languages 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}}

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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 

which works provably correctly 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}}

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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\\

  \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}

The matching algorithm works similarly, just over regular expression instead of 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{$der\,c\,(der\,b\,(der\,a\,r))$}\bigskip\pause

\item finally check whether the last regular expression can match the empty string

\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}

by induction on the regular expression.\bigskip\pause

\begin{center}

\huge\bf \alert{Any Questions?}

\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

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

\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}

  \end{center}

How about ranges \bl{$[a\mbox{-}z]$}, \bl{$r^+$} and \bl{$\sim{}r$}? Do they increase the

set of languages we can recognise?

\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{$\sim{}r$}  \hspace{6mm} (everything that \bl{r} cannot recognise)\medskip

\item \bl{$L(\sim{}r) \dn UNIV - L(r)$}\medskip

\item \bl{$nullable (\sim{}r) \dn not\, (nullable(r))$}\medskip

\item \bl{$der\,c\,(\sim{}r) \dn \;\sim{}(der\,c\,r)$}

\end{itemize}\bigskip\pause

Used often for recognising comments:

\[

\bl{/ \cdot * \cdot (\sim{}([a\mbox{-}z]^* \cdot * \cdot / \cdot [a\mbox{-}z]^*)) \cdot * \cdot /}

\]

\end{frame}}

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\mode<presentation>{

\begin{frame}[c]

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

Assume you have an alphabet consisting of the letters \bl{a}, \bl{b} and \bl{c} only.

Find a regular expression that matches all strings \emph{except} \bl{ab} and \bl{ac}.

\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 \alert{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 be everywhere defined

\end{itemize}

\begin{center}

\bl{$A(Q, q_0, F, \delta)$}

\end{center}

\end{frame}}

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\mode<presentation>{

\begin{frame}[c]

\begin{center}

\begin{tikzpicture}[>=stealth',very thick,auto,

                             every state/.style={minimum size=0pt,inner sep=2pt,draw=blue!50,very thick,fill=blue!20},]

\node[state,initial]  (q_0)  {$q_0$};

\node[state] (q_1) [right=of q_0] {$q_1$};

\node[state] (q_2) [below right=of q_0] {$q_2$};

\node[state] (q_3) [right=of q_2] {$q_3$};

\node[state, accepting] (q_4) [right=of q_1] {$q_4$};

\path[->] (q_0) edge node [above]  {\alert{$a$}} (q_1);

\path[->] (q_1) edge node [above]  {\alert{$a$}} (q_4);

\path[->] (q_4) edge [loop right] node  {\alert{$a, b$}} ();

\path[->] (q_3) edge node [right]  {\alert{$a$}} (q_4);

\path[->] (q_2) edge node [above]  {\alert{$a$}} (q_3);

\path[->] (q_1) edge node [right]  {\alert{$b$}} (q_2);

\path[->] (q_0) edge node [above]  {\alert{$b$}} (q_2);

\path[->] (q_2) edge [loop left] node  {\alert{$b$}} ();

\path[->] (q_3) edge [bend left=95, looseness=1.3] node [below]  {\alert{$b$}} (q_0);

\end{tikzpicture}

\end{center}\pause

\begin{itemize}

\item the start state can be an accepting state

\item it is possible that there is no accepting state

\item all states might be accepting (but this does not necessarily mean all strings are accepted)

\end{itemize}

\end{frame}}

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\mode<presentation>{

\begin{frame}[c]

\begin{center}

\begin{tikzpicture}[>=stealth',very thick,auto,

                             every state/.style={minimum size=0pt,inner sep=2pt,draw=blue!50,very thick,fill=blue!20},]

\node[state,initial]  (q_0)  {$q_0$};

\node[state] (q_1) [right=of q_0] {$q_1$};

\node[state] (q_2) [below right=of q_0] {$q_2$};

\node[state] (q_3) [right=of q_2] {$q_3$};

\node[state, accepting] (q_4) [right=of q_1] {$q_4$};

\path[->] (q_0) edge node [above]  {\alert{$a$}} (q_1);

\path[->] (q_1) edge node [above]  {\alert{$a$}} (q_4);

\path[->] (q_4) edge [loop right] node  {\alert{$a, b$}} ();

\path[->] (q_3) edge node [right]  {\alert{$a$}} (q_4);

\path[->] (q_2) edge node [above]  {\alert{$a$}} (q_3);

\path[->] (q_1) edge node [right]  {\alert{$b$}} (q_2);

\path[->] (q_0) edge node [above]  {\alert{$b$}} (q_2);

\path[->] (q_2) edge [loop left] node  {\alert{$b$}} ();

\path[->] (q_3) edge [bend left=95, looseness=1.3] node [below]  {\alert{$b$}} (q_0);

\end{tikzpicture}

\end{center}

for this automaton \bl{$\delta$} is the function\\

\begin{center}

\begin{tabular}{lll}

\bl{$(q_0, a) \rightarrow q_1$} & \bl{$(q_1, a) \rightarrow q_4$} & \bl{$(q_4, a) \rightarrow q_4$}\\

\bl{$(q_0, b) \rightarrow q_2$} & \bl{$(q_1, b) \rightarrow q_2$} & \bl{$(q_4, b) \rightarrow q_4$}\\

\end{tabular}\ldots

\end{center}

\end{frame}}

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\mode<presentation>{

\begin{frame}[t]

\frametitle{\begin{tabular}{c}Accepting a String\end{tabular}}

Given

\begin{center}

\bl{$A(Q, q_0, F, \delta)$}

\end{center}

you can define

\begin{center}

\begin{tabular}{l}

\bl{$\hat{\delta}(q, \texttt{""}) \dn q$}\\

\bl{$\hat{\delta}(q, c::s) \dn \hat{\delta}(\delta(q, c), s)$}\\

\end{tabular}

\end{center}\pause

Whether a string \bl{$s$} is accepted by \bl{$A$}?

\begin{center}

\hspace{5mm}\bl{$\hat{\delta}(q_0, s) \in F$}

\end{center}

\end{frame}}

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\mode<presentation>{

\begin{frame}[c]

\frametitle{\begin{tabular}{c}Non-Deterministic\\[-1mm] Finite Automata\end{tabular}}

A non-deterministic finite automaton consists again of:

\begin{itemize}

\item a finite set of states

\item one of these states is the start state

\item some states are accepting states, and

\item there is transition \alert{relation}\medskip 

\end{itemize}

\begin{center}

\begin{tabular}{c}

\bl{$(q_1, a) \rightarrow q_2$}\\

\bl{$(q_1, a) \rightarrow q_3$}\\