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\documentclass[dvipsnames,14pt,t]{beamer}
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\usepackage{beamerthemeplainculight}
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\usepackage[T1]{fontenc}
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\usepackage[latin1]{inputenc}
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\usepackage{mathpartir}
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\usepackage[absolute,overlay]{textpos}
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\usepackage{ifthen}
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\usepackage{tikz}
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\usepackage{pgf}
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\usepackage{calc}
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\usepackage{ulem}
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\usepackage{courier}
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\usepackage{listings}
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\renewcommand{\uline}[1]{#1}
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\usetikzlibrary{arrows}
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\usetikzlibrary{automata}
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\usetikzlibrary{shapes}
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\usetikzlibrary{shadows}
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\usetikzlibrary{positioning}
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\usetikzlibrary{calc}
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\usepackage{graphicx}
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\definecolor{javared}{rgb}{0.6,0,0} % for strings
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\definecolor{javadocblue}{rgb}{0.25,0.35,0.75} % javadoc
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\lstset{language=Java,
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basicstyle=\ttfamily,
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keywordstyle=\color{javapurple}\bfseries,
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morecomment=[s][\color{javadocblue}]{/**}{*/},
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showspaces=false,
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showstringspaces=false}
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\lstdefinelanguage{scala}{
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morekeywords={abstract,case,catch,class,def,%
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do,else,extends,false,final,finally,%
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for,if,implicit,import,match,mixin,%
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new,null,object,override,package,%
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private,protected,requires,return,sealed,%
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super,this,throw,trait,true,try,%
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type,val,var,while,with,yield},
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otherkeywords={=>,<-,<\%,<:,>:,\#,@},
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sensitive=true,
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morecomment=[l]{//},
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morecomment=[n]{/*}{*/},
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morestring=[b]",
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morestring=[b]',
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morestring=[b]"""
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}
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\lstset{language=Scala,
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basicstyle=\ttfamily,
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keywordstyle=\color{javapurple}\bfseries,
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stringstyle=\color{javagreen},
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commentstyle=\color{javagreen},
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morecomment=[s][\color{javadocblue}]{/**}{*/},
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numbers=left,
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tabsize=2,
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showspaces=false,
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showstringspaces=false}
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% beamer stuff
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\renewcommand{\slidecaption}{AFL 05, King's College London, 24.~October 2012}
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\newcommand{\bl}[1]{\textcolor{blue}{#1}}
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\newcommand{\dn}{\stackrel{\mbox{\scriptsize def}}{=}}% for definitions
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\begin{document}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\mode<presentation>{
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\begin{frame}<1>[t]
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\frametitle{%
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\begin{tabular}{@ {}c@ {}}
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\\[-3mm]
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\LARGE Automata and \\[-2mm]
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\LARGE Formal Languages (5)\\[3mm]
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\end{tabular}}
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\normalsize
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\begin{center}
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\begin{tabular}{ll}
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Email: & christian.urban at kcl.ac.uk\\
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Of$\!$fice: & S1.27 (1st floor Strand Building)\\
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Slides: & KEATS (also home work is there)\\
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\end{tabular}
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\end{center}
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\end{frame}}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\mode<presentation>{
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\begin{frame}[t]
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\frametitle{\begin{tabular}{c}Deterministic Finite Automata\end{tabular}}
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A DFA \bl{$A(Q, q_0, F, \delta)$} consists of:
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\begin{itemize}
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\item a finite set of states \bl{$Q$}
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\item one of these states is the start state \bl{$q_0$}
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\item some states are accepting states \bl{$F$}
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\item a transition function \bl{$\delta$}
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\end{itemize}\pause
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\onslide<2->{
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\begin{center}
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\begin{tabular}{l}
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\bl{$\hat{\delta}(q, \texttt{""}) = q$}\\
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\bl{$\hat{\delta}(q, c\!::\!s) = \hat{\delta}(\delta(q, c), s)$}
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\end{tabular}
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\end{center}}
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\only<3,4>{
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\begin{center}
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\begin{tikzpicture}[scale=2, line width=0.5mm]
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\node[state, initial] (q02) at ( 0,1) {$q_{0}$};
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\node[state] (q13) at ( 1,1) {$q_{1}$};
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\node[state, accepting] (q4) at ( 2,1) {$q_2$};
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\path[->] (q02) edge[bend left] node[above] {$a$} (q13)
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(q13) edge[bend left] node[below] {$b$} (q02)
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(q13) edge node[above] {$a$} (q4)
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(q02) edge [loop below] node {$b$} ()
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(q4) edge [loop right] node {$a, b$} ()
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;
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\end{tikzpicture}
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\end{center}}%
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%
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\only<5>{
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\begin{center}
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\bl{$L(A) \dn \{ s \;|\; \hat{\delta}(q_0, s) \in F\}$}
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\end{center}}
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\end{frame}}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\mode<presentation>{
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\begin{frame}[t]
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\frametitle{\begin{tabular}{c}Non-Deterministic\\[-1mm] Finite Automata\end{tabular}}
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An NFA \bl{$A(Q, q_0, F, \delta)$} consists again of:
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\begin{itemize}
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\item a finite set of states
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\item one of these states is the start state
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\item some states are accepting states
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\item a transition \alert{relation}\medskip
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\end{itemize}
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\begin{center}
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\begin{tabular}{c}
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\bl{(q$_1$, a) $\rightarrow$ q$_2$}\\
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\bl{(q$_1$, a) $\rightarrow$ q$_3$}\\
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\end{tabular}
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\hspace{10mm}
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\begin{tabular}{c}
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\bl{(q$_1$, $\epsilon$) $\rightarrow$ q$_2$}\\
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\end{tabular}
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\end{center}\pause\medskip
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A string \bl{s} is accepted by an NFA, if there is a ``lucky'' sequence to an accepting state.
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\end{frame}}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\mode<presentation>{
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\begin{frame}[c]
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\frametitle{\begin{tabular}{c}Last Week\end{tabular}}
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Last week I showed you\bigskip
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\begin{itemize}
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\item an algorithm for automata minimisation
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\item an algorithm for transforming a regular expression into an NFA
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\item an algorithm for transforming an NFA into a DFA (subset construction)
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\end{itemize}
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\end{frame}}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\mode<presentation>{
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\begin{frame}[c]
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\frametitle{\begin{tabular}{c}This Week\end{tabular}}
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Go over the algorithms again, but with two new things and \ldots\medskip
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\begin{itemize}
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\item with the example: what is the regular expression that accepts every string, except those ending
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in \bl{aa}?\medskip
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\item Go over the proof for \bl{$L(rev(r)) = Rev(L(r))$}.\medskip
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\item Anything else so far.
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\end{itemize}
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\end{frame}}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\mode<presentation>{
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\begin{frame}[c]
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\frametitle{\begin{tabular}{c}Proofs By Induction\end{tabular}}
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\begin{itemize}
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\item \bl{$P$} holds for \bl{$\varnothing$}, \bl{$\epsilon$} and \bl{c}\bigskip
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\item \bl{$P$} holds for \bl{r$_1$ + r$_2$} under the assumption that \bl{$P$} already
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holds for \bl{r$_1$} and \bl{r$_2$}.\bigskip
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\item \bl{$P$} holds for \bl{r$_1$ $\cdot$ r$_2$} under the assumption that \bl{$P$} already
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holds for \bl{r$_1$} and \bl{r$_2$}.
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\item \bl{$P$} holds for \bl{r$^*$} under the assumption that \bl{$P$} already
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holds for \bl{r}.
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\end{itemize}
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\begin{center}
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\bl{$P(r):\;\;L(rev(r)) = Rev(L(r))$}
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\end{center}
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\end{frame}}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\mode<presentation>{
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\begin{frame}[t]
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What is the regular expression that accepts every string, except those ending
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in \bl{aa}?\pause\bigskip
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\begin{center}
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\begin{tabular}{l}
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\bl{(a + b)$^*$ba}\\
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\bl{(a + b)$^*$ab}\\
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\bl{(a + b)$^*$bb}\\\pause
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\bl{a}\\
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\bl{\texttt{""}}
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\end{tabular}
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\end{center}\pause
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What are the strings to be avoided?\pause\medskip
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\begin{center}
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\bl{(a + b)$^*$aa}
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\end{center}
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\end{frame}}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\mode<presentation>{
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\begin{frame}[t]
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An NFA for \bl{(a + b)$^*$aa}
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\begin{center}
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\begin{tikzpicture}[scale=2, line width=0.5mm]
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\node[state, initial] (q0) at ( 0,1) {$q_0$};
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\node[state] (q1) at ( 1,1) {$q_1$};
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\node[state, accepting] (q2) at ( 2,1) {$q_2$};
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\path[->] (q0) edge node[above] {$a$} (q1)
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(q1) edge node[above] {$a$} (q2)
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(q0) edge [loop below] node {$a$} ()
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(q0) edge [loop above] node {$b$} ()
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;
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\end{tikzpicture}
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\end{center}\pause
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Minimisation for DFAs\\
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Subset Construction for NFAs
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\end{frame}}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\mode<presentation>{
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\begin{frame}[c]
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\frametitle{\begin{tabular}{c}DFA Minimisation\end{tabular}}
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\begin{enumerate}
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\item Take all pairs \bl{(q, p)} with \bl{q $\not=$ p}
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\item Mark all pairs that accepting and non-accepting states
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\item For all unmarked pairs \bl{(q, p)} and all characters \bl{c} tests wether
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\begin{center}
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\bl{($\delta$(q,c), $\delta$(p,c))}
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\end{center}
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are marked. If yes, then also mark \bl{(q, p)}.
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\item Repeat last step until nothing changed.
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\item All unmarked pairs can be merged.
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\end{enumerate}
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\end{frame}}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\mode<presentation>{
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\begin{frame}[c]
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Minimal DFA \only<1>{\bl{(a + b)$^*$aa}}\only<2->{\alert{not} \bl{(a + b)$^*$aa}}
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\begin{center}
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\begin{tikzpicture}[scale=2, line width=0.5mm]
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\only<1>{\node[state, initial] (q0) at ( 0,1) {$q_0$};}
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\only<2->{\node[state, initial,accepting] (q0) at ( 0,1) {$q_0$};}
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\only<1>{\node[state] (q1) at ( 1,1) {$q_1$};}
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\only<2->{\node[state,accepting] (q1) at ( 1,1) {$q_1$};}
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\only<1>{\node[state, accepting] (q2) at ( 2,1) {$q_2$};}
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\only<2->{\node[state] (q2) at ( 2,1) {$q_2$};}
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\path[->] (q0) edge[bend left] node[above] {$a$} (q1)
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(q1) edge[bend left] node[above] {$b$} (q0)
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(q2) edge[bend left=50] node[below] {$b$} (q0)
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(q1) edge node[above] {$a$} (q2)
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(q2) edge [loop right] node {$a$} ()
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(q0) edge [loop below] node {$b$} ()
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;
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\end{tikzpicture}
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\end{center}
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\onslide<3>{How to get from a DFA to a regular expression?}
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\end{frame}}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\mode<presentation>{
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\begin{frame}[c]
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\begin{center}
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\begin{tikzpicture}[scale=2, line width=0.5mm]
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\only<1->{\node[state, initial] (q0) at ( 0,1) {$q_0$};}
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\only<1->{\node[state] (q1) at ( 1,1) {$q_1$};}
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\only<1->{\node[state] (q2) at ( 2,1) {$q_2$};}
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\path[->] (q0) edge[bend left] node[above] {$a$} (q1)
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(q1) edge[bend left] node[above] {$b$} (q0)
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(q2) edge[bend left=50] node[below] {$b$} (q0)
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(q1) edge node[above] {$a$} (q2)
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(q2) edge [loop right] node {$a$} ()
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(q0) edge [loop below] node {$b$} ()
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;
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\end{tikzpicture}
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\end{center}\pause\bigskip
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\onslide<2->{
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\begin{center}
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\begin{tabular}{r@ {\hspace{2mm}}c@ {\hspace{2mm}}l}
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\bl{$q_0$} & \bl{$=$} & \bl{$2\, q_0 + 3 \,q_1 + 4\, q_2$}\\
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\bl{$q_1$} & \bl{$=$} & \bl{$2 \,q_0 + 3\, q_1 + 1\, q_2$}\\
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\bl{$q_2$} & \bl{$=$} & \bl{$1\, q_0 + 5\, q_1 + 2\, q_2$}\\
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\end{tabular}
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\end{center}
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}
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\end{frame}}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\mode<presentation>{
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\begin{frame}[c]
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\begin{center}
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\begin{tikzpicture}[scale=2, line width=0.5mm]
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\only<1->{\node[state, initial] (q0) at ( 0,1) {$q_0$};}
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\only<1->{\node[state] (q1) at ( 1,1) {$q_1$};}
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\only<1->{\node[state] (q2) at ( 2,1) {$q_2$};}
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\path[->] (q0) edge[bend left] node[above] {$a$} (q1)
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(q1) edge[bend left] node[above] {$b$} (q0)
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(q2) edge[bend left=50] node[below] {$b$} (q0)
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(q1) edge node[above] {$a$} (q2)
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(q2) edge [loop right] node {$a$} ()
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(q0) edge [loop below] node {$b$} ()
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;
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\end{tikzpicture}
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\end{center}\bigskip
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391 |
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\onslide<2->{
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\begin{center}
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|
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\begin{tabular}{r@ {\hspace{2mm}}c@ {\hspace{2mm}}l}
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|
395 |
\bl{$q_0$} & \bl{$=$} & \bl{$\epsilon + q_0\,b + q_1\,b + q_2\,b$}\\
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\bl{$q_1$} & \bl{$=$} & \bl{$q_0\,a$}\\
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397 |
\bl{$q_2$} & \bl{$=$} & \bl{$q_1\,a + q_2\,a$}\\
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398 |
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399 |
\end{tabular}
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400 |
\end{center}
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}
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402 |
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403 |
\onslide<3->{
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|
404 |
Arden's Lemma:
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|
405 |
\begin{center}
|
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|
406 |
If \bl{$q = q\,r + s$}\; then\; \bl{$q = s\, r^*$}
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407 |
\end{center}
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408 |
}
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409 |
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410 |
\end{frame}}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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412 |
|
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46
|
413 |
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
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|
414 |
\mode<presentation>{
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415 |
\begin{frame}[c]
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|
416 |
\frametitle{\begin{tabular}{c}Algorithms on Automata\end{tabular}}
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417 |
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418 |
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419 |
\begin{itemize}
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|
420 |
\item Reg $\rightarrow$ NFA: Thompson-McNaughton-Yamada method\medskip
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|
421 |
\item NFA $\rightarrow$ DFA: Subset Construction\medskip
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|
422 |
\item DFA $\rightarrow$ Reg: Brzozowski's Algebraic Method\medskip
|
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|
423 |
\item DFA minimisation: Hopcrofts Algorithm\medskip
|
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|
424 |
\item complement DFA
|
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|
425 |
\end{itemize}
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|
426 |
|
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|
427 |
\end{frame}}
|
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|
428 |
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
|
47
|
429 |
\newcommand{\qq}{\mbox{\texttt{"}}}
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|
430 |
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
|
|
431 |
\mode<presentation>{
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|
432 |
\begin{frame}[c]
|
|
|
433 |
\frametitle{\begin{tabular}{c}Grammars\end{tabular}}
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|
434 |
|
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|
435 |
\begin{center}
|
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|
436 |
\bl{\begin{tabular}{lcl}
|
|
|
437 |
$E$ & $\rightarrow$ & $F + (F \cdot \qq*\qq \cdot F) + (F \cdot \qq\backslash\qq \cdot F)$\\
|
|
|
438 |
$F$ & $\rightarrow$ & $T + (T \cdot \qq\texttt{+}\qq \cdot T) + (T \cdot \qq\texttt{-}\qq \cdot T)$\\
|
|
|
439 |
$T$ & $\rightarrow$ & $num + (\qq\texttt{(}\qq \cdot E \cdot \qq\texttt{)}\qq)$\\
|
|
|
440 |
\end{tabular}}
|
|
|
441 |
\end{center}
|
|
|
442 |
|
|
|
443 |
\bl{$E$}, \bl{$F$} and \bl{$T$} are non-terminals\\
|
|
|
444 |
\bl{$E$} is start symbol\\
|
|
|
445 |
\bl{$num$}, \bl{(}, \bl{)}, \bl{+} \ldots are terminals\bigskip\\
|
|
|
446 |
|
|
|
447 |
|
|
|
448 |
\bl{\texttt{(2*3)+(3+4)}}
|
|
|
449 |
|
|
|
450 |
\end{frame}}
|
|
|
451 |
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
|
|
452 |
|
|
|
453 |
|
|
|
454 |
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
|
|
455 |
\mode<presentation>{
|
|
|
456 |
\begin{frame}[c]
|
|
|
457 |
|
|
|
458 |
\begin{center}
|
|
|
459 |
\bl{\begin{tabular}{lcl}
|
|
|
460 |
$E$ & $\rightarrow$ & $F + (F \cdot \qq*\qq \cdot F) + (F \cdot \qq\backslash\qq \cdot F)$\\
|
|
|
461 |
$F$ & $\rightarrow$ & $T + (T \cdot \qq\texttt{+}\qq \cdot T) + (T \cdot \qq\texttt{-}\qq \cdot T)$\\
|
|
|
462 |
$T$ & $\rightarrow$ & $num + (\qq\texttt{(}\qq \cdot E \cdot \qq\texttt{)}\qq)$\\
|
|
|
463 |
\end{tabular}}
|
|
|
464 |
\end{center}
|
|
|
465 |
|
|
|
466 |
\begin{center}
|
|
|
467 |
\begin{tikzpicture}[level distance=8mm, blue]
|
|
|
468 |
\node {E}
|
|
|
469 |
child {node {F}
|
|
|
470 |
child {node {T}
|
|
|
471 |
child {node {\qq(\qq\,E\,\qq)\qq}
|
|
|
472 |
child {node{F \qq*\qq{} F}
|
|
|
473 |
child {node {T} child {node {2}}}
|
|
|
474 |
child {node {T} child {node {3}}}
|
|
|
475 |
}
|
|
|
476 |
}
|
|
|
477 |
}
|
|
|
478 |
child {node {\qq+\qq}}
|
|
|
479 |
child {node {T}
|
|
|
480 |
child {node {\qq(\qq\,E\,\qq)\qq}
|
|
|
481 |
child {node {F}
|
|
|
482 |
child {node {T \qq+\qq{} T}
|
|
|
483 |
child {node {3}}
|
|
|
484 |
child {node {4}}
|
|
|
485 |
}
|
|
|
486 |
}}
|
|
|
487 |
}};
|
|
|
488 |
\end{tikzpicture}
|
|
|
489 |
\end{center}
|
|
|
490 |
|
|
|
491 |
\begin{textblock}{5}(1, 5)
|
|
|
492 |
\bl{\texttt{(2*3)+(3+4)}}
|
|
|
493 |
\end{textblock}
|
|
|
494 |
|
|
|
495 |
\end{frame}}
|
|
|
496 |
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
|
44
|
497 |
|
|
|
498 |
\end{document}
|
|
|
499 |
|
|
|
500 |
%%% Local Variables:
|
|
|
501 |
%%% mode: latex
|
|
|
502 |
%%% TeX-master: t
|
|
|
503 |
%%% End:
|
|
|
504 |
|