\documentclass[dvipsnames,14pt,t]{beamer}

\usepackage{../slides}

\usepackage{../graphics}

\usepackage{../langs}

\usepackage{../data}

\hfuzz=220pt 

\pgfplotsset{compat=1.11}

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

\renewcommand{\slidecaption}{AFL 04, King's College London}

\begin{document}

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\begin{frame}[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\\

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

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

  \end{tabular}

  \end{center}

\end{frame}

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\begin{frame}[c]

\frametitle{Regexps and Automata}

\begin{center}

\begin{tikzpicture}

\node (rexp)  {\bl{\bf Regexps}};

\node (nfa) [right=of rexp] {\bl{\bf NFAs}};

\node (dfa) [right=of nfa] {\bl{\bf DFAs}};

\node (mdfa) [right=of dfa] {\bl{\bf \begin{tabular}{c}minimal\\ DFAs\end{tabular}}};

\path[->,red, line width=2mm] (rexp) edge node [above=4mm, black] 

     {\begin{tabular}{c@{\hspace{9mm}}}Thompson's\\[-1mm] construction\end{tabular}} (nfa);

\path[->,red, line width=2mm] (nfa) edge node [above=4mm, black] 

     {\begin{tabular}{c}subset\\[-1mm] construction\end{tabular}}(dfa);

\path[->, red, line width=2mm] (dfa) edge node [below=5mm, black] {minimisation} (mdfa);

\end{tikzpicture}\\

\end{center}

\end{frame}

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\begin{frame}[t]

\begin{tikzpicture}

\begin{axis}[xlabel={\pcode{a}s},ylabel={time in secs},

    enlargelimits=false,

    xtick={0,5,...,30},

    xmax=30,

    ymax=35,

    ytick={0,5,...,30},

    scaled ticks=false,

    axis lines=left,

    width=10cm,

    height=7cm, 

    legend entries={Python,Ruby,my NFA},  

    legend pos=north west,

    legend cell align=left]

\addplot[blue,mark=*, mark options={fill=white}] 

  table {re-python.data};

\addplot[brown,mark=pentagon*, mark options={fill=white}] 

  table {re-ruby.data};  

\addplot[red,mark=triangle*, mark options={fill=white}] 

  table {nfasearch.data};	  

\end{axis}

\end{tikzpicture}

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\begin{frame}[c]

\frametitle{DFA to Rexp}

\begin{center}

\begin{tikzpicture}[scale=2,>=stealth',very thick,

                    every state/.style={minimum size=0pt,

                    draw=blue!50,very thick,fill=blue!20},]

  \node[state, initial]        (q0) at ( 0,1) {$q_0$};

  \node[state]                    (q1) at ( 1,1) {$q_1$};

  \node[state, accepting] (q2) at ( 2,1) {$q_2$};

  \path[->] (q0) edge[bend left] node[above] {\alert{$a$}} (q1)

            (q1) edge[bend left] node[above] {\alert{$b$}} (q0)

            (q2) edge[bend left=50] node[below] {\alert{$b$}} (q0)

            (q1) edge node[above] {\alert{$a$}} (q2)

            (q2) edge [loop right] node {\alert{$a$}} ()

            (q0) edge [loop below] node {\alert{$b$}} ();

\end{tikzpicture}

\end{center}\bigskip

\begin{center}

\begin{tabular}{r@ {\hspace{2mm}}c@ {\hspace{2mm}}l@{\hspace{7mm}}l}

\bl{$q_0$} & \bl{$=$} & \bl{$\epsilon + q_0\,b + q_1\,b +  q_2\,b$} & (start state)\\

\bl{$q_1$} & \bl{$=$} & \bl{$q_0\,a$}\\

\bl{$q_2$} & \bl{$=$} & \bl{$q_1\,a + q_2\,a$}\\

\end{tabular}

\end{center}

Arden's Lemma:

\begin{center}

If \bl{$q = q\,r + s$}\; then\; \bl{$q = s\, r^*$}

\end{center}

\end{frame}

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\begin{frame}[c]

\frametitle{DFA Minimisation}

\begin{enumerate}

\item Take all pairs \bl{$(q, p)$} with \bl{$q \not= p$}

\item Mark all pairs that accepting and non-accepting states

\item For  all unmarked pairs \bl{$(q, p)$} and all characters \bl{$c$} test whether

\begin{center}

\bl{$(\delta(q, c), \delta(p,c))$}

\end{center} 

are marked. If yes, then also mark \bl{$(q, p)$}.

\item Repeat last step until no change.

\item All unmarked pairs can be merged.

\end{enumerate}

\end{frame}

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\begin{frame}[c]

\begin{center}

\begin{tabular}{@{\hspace{-8mm}}cc@{}}

\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}

&

\raisebox{9mm}{\begin{tikzpicture}[scale=0.6,line width=0.8mm]

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

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

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

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

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

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

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

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

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

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

\draw (0.5,-0.5) node {$q_0$}; 

\draw (1.5,-0.5) node {$q_1$}; 

\draw (2.5,-0.5) node {$q_2$}; 

\draw (3.5,-0.5) node {$q_3$};

\draw (-0.5, 3.5) node {$q_1$}; 

\draw (-0.5, 2.5) node {$q_2$}; 

\draw (-0.5, 1.5) node {$q_3$}; 

\draw (-0.5, 0.5) node {$q_4$}; 

\draw (0.5,0.5) node {\large$\star$}; 

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

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

\draw (3.5,0.5) node {\large$\star$};

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

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

\draw (0.5,3.5) node {\large$\star$}; 

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

\end{tikzpicture}}

\end{tabular}

\end{center}

\mbox{}\\[-20mm]\mbox{}

\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_02)  {$q_{0, 2}$};

\node[state] (q_13) [right=of q_02] {$q_{1, 3}$};

\node[state, accepting] (q_4) [right=of q_13] {$q_{4\phantom{,0}}$};

\path[->] (q_02) edge [bend left] node [above]  {\alert{$a$}} (q_13);

\path[->] (q_13) edge [bend left] node [below]  {\alert{$b$}} (q_02);

\path[->] (q_02) edge [loop below] node  {\alert{$b$}} ();

\path[->] (q_13) edge node [above]  {\alert{$a$}} (q_4);

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

\end{tikzpicture}\\

minimal automaton

\end{center}

\end{frame}

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\begin{frame}[c]

\frametitle{Regular Languages}

Two equivalent definitions:\bigskip

\begin{quote}\rm A language is \alert{regular} iff there exists a

regular expression that recognises all its strings.

\end{quote}

\begin{quote}\rm A language is \alert{regular} iff there exists an

automaton that recognises all its strings.

\end{quote}\bigskip\bigskip

\small

for example $a^nb^n$ is not regular

\end{frame}

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\begin{frame}[c]

\frametitle{Negation}

Regular languages are closed under negation:\bigskip

\begin{center}

\begin{tikzpicture}[scale=2,>=stealth',very thick,

                    every state/.style={minimum size=0pt,

                    draw=blue!50,very thick,fill=blue!20}]

  \only<1>{\node[state,initial] (q0) at ( 0,1) {$q_0$};}

  \only<2>{\node[state,initial,accepting] (q0) at ( 0,1) {$q_0$};}

  \only<1>{\node[state] (q1) at ( 1,1) {$q_1$};}

  \only<2>{\node[state,accepting] (q1) at ( 1,1) {$q_1$};}

  \only<1>{\node[state, accepting] (q2) at ( 2,1) {$q_2$};}

  \only<2>{\node[state] (q2) at ( 2,1) {$q_2$};}

  \path[->] (q0) edge[bend left] node[above] {\alert{$a$}} (q1)

            (q1) edge[bend left] node[above] {\alert{$b$}} (q0)

            (q2) edge[bend left=50] node[below] {\alert{$b$}} (q0)

            (q1) edge node[above] {\alert{$a$}} (q2)

            (q2) edge [loop right] node {\alert{$a$}} ()

            (q0) edge [loop below] node {\alert{$b$}} ();

\end{tikzpicture}

\end{center}\bigskip\bigskip

\end{frame}

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\begin{frame}[c]

\mbox{\lstinputlisting[language=While]{../progs/fib.while}}

\end{frame}

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\begin{frame}[c]

\frametitle{??}

\mbox{\lstinputlisting[language=While]{../progs/collatz.while}}

\end{frame}

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\begin{frame}[t]

\tt

\begin{center}\large

\code{"if true then then 42 else +"}

\end{center}

\begin{tabular}{@{}l}

KEYWORD: \\

\hspace{5mm}{if}, {then}, {else},\\ 

WHITESPACE:\\

\hspace{5mm}{" "}, {$\backslash$n},\\ 

IDENT:\\

\hspace{5mm}LETTER $\cdot$ (LETTER + DIGIT + {\_})$^*$\\ 

NUM:\\

\hspace{5mm}(NONZERODIGIT $\cdot$ DIGIT$^*$) + {0}\\

OP:\\

\hspace{5mm}{+}\\

COMMENT:\\

\hspace{5mm}{$\slash$*} $\cdot$ $\sim$(ALL$^*$ $\cdot$ (*$\slash$) $\cdot$ ALL$^*$) $\cdot$ {*$\slash$}

\end{tabular}

\end{frame}

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\begin{frame}[t]

\tt

\begin{center}\large

\code{"if true then then 42 else +"}

\end{center}

\only<1>{

\small\begin{tabular}{l}

KEYWORD(if),\\ 

WHITESPACE,\\ 

IDENT(true),\\ 

WHITESPACE,\\ 

KEYWORD(then),\\ 

WHITESPACE,\\ 

KEYWORD(then),\\ 

WHITESPACE,\\ 

NUM(42),\\ 

WHITESPACE,\\ 

KEYWORD(else),\\ 

WHITESPACE,\\ 

OP(+)

\end{tabular}}

\only<2>{

\small\begin{tabular}{l}

KEYWORD(if),\\ 

IDENT(true),\\ 

KEYWORD(then),\\ 

KEYWORD(then),\\ 

NUM(42),\\ 

KEYWORD(else),\\ 

OP(+)

\end{tabular}}

\end{frame}

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\begin{frame}[c]

There is one small problem with the tokenizer. How should we 

tokenize:

\begin{center}

\large\code{"x-3"}

\end{center}

\tt

\begin{tabular}{@{}l}

ID: \ldots\\

OP:\\

\hspace{5mm}\texttt{"+"}, \texttt{"-"}\\

NUM:\\

\hspace{5mm}(NONZERODIGIT $\cdot$ DIGIT$^*$) + {''0''}\\

NUMBER:\\

\hspace{5mm}NUM +  (\texttt{"-"} $\cdot$ NUM)\\

\end{tabular}

\end{frame}

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\begin{frame}[c]

\frametitle{POSIX: Two Rules}

\begin{itemize}

\item Longest match rule (``maximal munch rule''): The 

longest initial substring matched by any regular expression is taken

as the next token.\bigskip

\item Rule priority:

For a particular longest initial substring, the first regular

expression that can match determines the token.

\end{itemize}\bigskip\bigskip\pause

\small

\hfill most posix matchers are buggy\\

\footnotesize

\hfill \url{http://www.haskell.org/haskellwiki/Regex_Posix}

%\url{http://www.technologyreview.com/tr10/?year=2011}  

%finite deterministic automata/ nondeterministic automaton

%\item problem with infix operations, for example i-12

\end{frame}

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\begin{frame}[c]

\frametitle{Sulzmann Matcher}