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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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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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numberstyle=\tiny\color{black},
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stepnumber=1,
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tabsize=2,
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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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numberstyle=\tiny\color{black},
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numbersep=10pt,
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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 04, King's College London, 17.~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 (4)\\[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}[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 a tokenizer taking a list of regular expressions\bigskip
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\item tokenization identifies lexeme in an input stream of characters (or string)
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and cathegorizes them into tokens
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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}Two Rules\end{tabular}}
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\begin{itemize}
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\item Longest match rule (maximal munch rule): The
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longest initial substring matched by any regular expression is taken
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as next token.\bigskip
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\item Rule priority:
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For a particular longest initial substring, the first regular
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expression that can match determines the token.
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\end{itemize}
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%\url{http://www.technologyreview.com/tr10/?year=2011}
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%finite deterministic automata/ nondeterministic automaton
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%\item problem with infix operations, for example i-12
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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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\begin{center}
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\texttt{"if true then then 42 else +"}
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\end{center}
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\begin{tabular}{@{}l}
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KEYWORD: \\
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\hspace{5mm}\texttt{"if"}, \texttt{"then"}, \texttt{"else"},\\
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WHITESPACE:\\
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\hspace{5mm}\texttt{" "}, \texttt{"$\backslash$n"},\\
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IDENT:\\
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\hspace{5mm}LETTER $\cdot$ (LETTER + DIGIT + \texttt{"\_"})$^*$\\
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NUM:\\
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\hspace{5mm}(NONZERODIGIT $\cdot$ DIGIT$^*$) + \texttt{"0"}\\
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OP:\\
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\hspace{5mm}\texttt{"+"}\\
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COMMENT:\\
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\hspace{5mm}\texttt{"$\slash$*"} $\cdot$ (ALL$^*$ $\cdot$ \texttt{"*$\slash$"} $\cdot$ ALL$^*$) $\cdot$ \texttt{"*$\slash$"}
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\end{tabular}
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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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\begin{center}
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\texttt{"if true then then 42 else +"}
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\end{center}
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\only<1>{
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\small\begin{tabular}{l}
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KEYWORD(if),\\
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WHITESPACE,\\
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IDENT(true),\\
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WHITESPACE,\\
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KEYWORD(then),\\
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WHITESPACE,\\
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KEYWORD(then),\\
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WHITESPACE,\\
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NUM(42),\\
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WHITESPACE,\\
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KEYWORD(else),\\
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WHITESPACE,\\
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OP(+)
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\end{tabular}}
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\only<2>{
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\small\begin{tabular}{l}
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KEYWORD(if),\\
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IDENT(true),\\
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KEYWORD(then),\\
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KEYWORD(then),\\
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NUM(42),\\
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KEYWORD(else),\\
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OP(+)
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\end{tabular}}
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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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There is one small problem with the tokenizer. How should we
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tokenize:
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\begin{center}
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\texttt{"x - 3"}
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\end{center}
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\begin{tabular}{@{}l}
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OP:\\
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\hspace{5mm}\texttt{"+"}, \texttt{"-"}\\
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NUM:\\
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\hspace{5mm}(NONZERODIGIT $\cdot$ DIGIT$^*$) + \texttt{"0"}\\
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NUMBER:\\
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\hspace{5mm}NUM + (\texttt{"-"} $\cdot$ NUM)\\
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\end{tabular}
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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}Deterministic Finite Automata\end{tabular}}
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A deterministic finite automaton consists 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, and
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\item there is transition function\medskip
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\small
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which takes a state and a character as arguments and produces a new state\smallskip\\
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this function might not always be defined everywhere
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\end{itemize}
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\begin{center}
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\bl{$A(Q, q_0, F, \delta)$}
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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}[c]
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\begin{center}
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\includegraphics[scale=0.7]{pics/ch3.jpg}
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\end{center}\pause
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\begin{itemize}
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\item start can be an accepting state
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\item there is no accepting state
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\item all states are accepting
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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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\begin{center}
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\includegraphics[scale=0.7]{pics/ch3.jpg}
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\end{center}
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for this automaton \bl{$\delta$} is the function\\
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\begin{center}
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\begin{tabular}{lll}
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\bl{(q$_0$, a) $\rightarrow$ q$_1$} & \bl{(q$_1$, a) $\rightarrow$ q$_4$} & \bl{(q$_4$, a) $\rightarrow$ q$_4$}\\
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\bl{(q$_0$, b) $\rightarrow$ q$_2$} & \bl{(q$_1$, b) $\rightarrow$ q$_2$} & \bl{(q$_4$, b) $\rightarrow$ q$_4$}\\
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\end{tabular}\ldots
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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}Accepting a String\end{tabular}}
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Given
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\begin{center}
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\bl{$A(Q, q_0, F, \delta)$}
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\end{center}
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you can define
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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}\pause
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Whether a string \bl{$s$} is accepted by \bl{$A$}?
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\begin{center}
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\hspace{5mm}\bl{$\hat{\delta}(q_0, s) \in F$}
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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}[c]
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\frametitle{\begin{tabular}{c}Non-Deterministic\\[-1mm] Finite Automata\end{tabular}}
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A non-deterministic finite automaton 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, and
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\item there is 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}
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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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\includegraphics[scale=0.7]{pics/ch5.jpg}
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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}[c]
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\begin{center}
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\begin{tabular}[b]{ll}
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\bl{$\varnothing$} & \includegraphics[scale=0.7]{pics/NULL.jpg}\\\\
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\bl{$\epsilon$} & \includegraphics[scale=0.7]{pics/epsilon.jpg}\\\\
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\bl{c} & \includegraphics[scale=0.7]{pics/char.jpg}\\
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\end{tabular}
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\end{center}
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\end{frame}}
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\mode<presentation>{
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\begin{frame}[c]
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\begin{center}
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\begin{tabular}[t]{ll}
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\bl{r$_1$ $\cdot$ r$_2$} & \includegraphics[scale=0.6]{pics/seq.jpg}\\\\
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\end{tabular}
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\end{center}
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\end{frame}}
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\begin{center}
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\begin{tabular}[t]{ll}
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\bl{r$_1$ + r$_2$} & \includegraphics[scale=0.7]{pics/alt.jpg}\\\\
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\end{tabular}
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\end{center}
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\end{frame}}
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\mode<presentation>{
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\begin{frame}[c]
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\begin{center}
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\begin{tabular}[b]{ll}
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\bl{r$^*$} & \includegraphics[scale=0.7]{pics/star.jpg}\\
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\end{tabular}
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\end{center}
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\end{frame}}
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\mode<presentation>{
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\begin{frame}[c]
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\begin{textblock}{5}(1,1)
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\includegraphics[scale=0.5]{pics/ch5.jpg}
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\end{textblock}
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\begin{textblock}{11}(6.5,3)
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\begin{tabular}{r|cl}
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& a & b\\
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\hline
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$\varnothing$ \onslide<2>{\textcolor{white}{*}} & $\varnothing$ & $\varnothing$\\
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$\{0\}$ \onslide<2>{\textcolor{white}{*}} & $\{0,1,2\}$ & $\{2\}$\\
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$\{1\}$ \onslide<2>{\textcolor{white}{*}} &$\{1\}$ & $\varnothing$\\
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$\{2\}$ \onslide<2>{*} & $\varnothing$ &$\{2\}$\\
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$\{0,1\}$ \onslide<2>{\textcolor{white}{*}} &$\{0,1,2\}$ &$\{2\}$\\
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$\{0,2\}$ \onslide<2>{*}&$\{0,1,2\}$ &$\{2\}$\\
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$\{1,2\}$ \onslide<2>{*}& $\{1\}$ & $\{2\}$\\
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\onslide<2>{s:} $\{0,1,2\}$ \onslide<2>{*}&$\{0,1,2\}$ &$\{2\}$\\
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\end{tabular}
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\end{textblock}
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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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\includegraphics[scale=0.7]{pics/ch4.jpg}
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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}[c]
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\frametitle{\begin{tabular}{c}Languages\end{tabular}}
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33
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A language is \alert{regular} iff there exists
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a regular expression that recognises all its strings.\bigskip\bigskip\pause
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\textcolor{gray}{not all languages are regular, e.g.~\bl{a$^n$b$^n$}.}
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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{itemize}
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\item Assuming you have the alphabet \bl{\{a, b, c\}}\bigskip
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\item Give a regular expression that can recognise all strings that have at least one \bl{b}.
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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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33
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\begin{itemize}
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\item The star-case in our proof needs the following lemma
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\begin{center}
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\bl{Der\,c\,A$^*$ $=$ (Der c A)\,@\, A$^*$}
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\end{center}
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\end{itemize}\bigskip\bigskip
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33
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\begin{itemize}
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\item If \bl{\texttt{""} $\in$ A}, then\\ \bl{Der\,c\,(A @ B) $=$ (Der\,c\,A) @ B $\cup$ (Der\,c\,B)}\medskip
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511 |
\item If \bl{\texttt{""} $\not\in$ A}, then\\ \bl{Der\,c\,(A @ B) $=$ (Der\,c\,A) @ B}
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|
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\end{itemize}
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\end{frame}}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\end{document}
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%%% Local Variables:
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%%% mode: latex
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%%% TeX-master: t
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%%% End:
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