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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{javagreen}{rgb}{0.25,0.5,0.35} % comments
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\definecolor{javapurple}{rgb}{0.5,0,0.35} % keywords
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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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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}Negation\end{tabular}}
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Assume you have an alphabet consisting of the letters \bl{a}, \bl{b} and \bl{c} only.
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Find a regular expression that matches all strings \emph{except} \bl{ab}, \bl{ac} and \bl{cba}.
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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 it is possible that there is no accepting state
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\item all states might be accepting (but does not necessarily mean all strings are accepted)
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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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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\mode<presentation>{
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\begin{frame}[c]
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35
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\begin{center}
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36
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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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35
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406 |
\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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33
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35
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\begin{center}
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36
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\begin{tabular}[t]{ll}
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418 |
\bl{r$_1$ + r$_2$} & \includegraphics[scale=0.7]{pics/alt.jpg}\\\\
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419 |
\end{tabular}
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35
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420 |
\end{center}
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36
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35
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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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428 |
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429 |
\begin{center}
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36
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430 |
\begin{tabular}[b]{ll}
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431 |
\bl{r$^*$} & \includegraphics[scale=0.7]{pics/star.jpg}\\
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432 |
\end{tabular}
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38
|
433 |
\end{center}\pause\bigskip
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434 |
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|
435 |
Why can't we just have an epsilon transition from the accepting states to the starting state?
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36
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436 |
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437 |
\end{frame}}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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439 |
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441 |
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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442 |
\mode<presentation>{
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443 |
\begin{frame}[c]
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38
|
444 |
\frametitle{\begin{tabular}{c}Subset Construction\end{tabular}}
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36
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445 |
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38
|
446 |
|
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|
447 |
\begin{textblock}{5}(1,2.5)
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36
|
448 |
\includegraphics[scale=0.5]{pics/ch5.jpg}
|
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37
|
449 |
\end{textblock}
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450 |
|
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38
|
451 |
\begin{textblock}{11}(6.5,4.5)
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37
|
452 |
\begin{tabular}{r|cl}
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|
453 |
& a & b\\
|
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|
454 |
\hline
|
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|
455 |
$\varnothing$ \onslide<2>{\textcolor{white}{*}} & $\varnothing$ & $\varnothing$\\
|
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|
456 |
$\{0\}$ \onslide<2>{\textcolor{white}{*}} & $\{0,1,2\}$ & $\{2\}$\\
|
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|
457 |
$\{1\}$ \onslide<2>{\textcolor{white}{*}} &$\{1\}$ & $\varnothing$\\
|
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|
458 |
$\{2\}$ \onslide<2>{*} & $\varnothing$ &$\{2\}$\\
|
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|
459 |
$\{0,1\}$ \onslide<2>{\textcolor{white}{*}} &$\{0,1,2\}$ &$\{2\}$\\
|
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|
460 |
$\{0,2\}$ \onslide<2>{*}&$\{0,1,2\}$ &$\{2\}$\\
|
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|
461 |
$\{1,2\}$ \onslide<2>{*}& $\{1\}$ & $\{2\}$\\
|
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|
462 |
\onslide<2>{s:} $\{0,1,2\}$ \onslide<2>{*}&$\{0,1,2\}$ &$\{2\}$\\
|
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|
463 |
\end{tabular}
|
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|
464 |
\end{textblock}
|
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|
465 |
|
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36
|
466 |
|
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|
467 |
\end{frame}}
|
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|
468 |
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
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|
469 |
|
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|
470 |
|
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38
|
471 |
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
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|
472 |
\mode<presentation>{
|
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|
473 |
\begin{frame}[c]
|
|
|
474 |
\frametitle{\begin{tabular}{c}Regular Languages\end{tabular}}
|
|
|
475 |
|
|
|
476 |
A language is \alert{regular} iff there exists
|
|
|
477 |
a regular expression that recognises all its strings.\bigskip\medskip
|
|
|
478 |
|
|
|
479 |
or equivalently\bigskip\medskip
|
|
|
480 |
|
|
|
481 |
A language is \alert{regular} iff there exists
|
|
|
482 |
a deterministic finite automaton that recognises all its strings.\bigskip\pause
|
|
|
483 |
|
|
|
484 |
Why is every finite set of strings a regular language?
|
|
|
485 |
\end{frame}}
|
|
|
486 |
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
|
|
487 |
|
|
36
|
488 |
|
|
|
489 |
|
|
|
490 |
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
|
|
491 |
\mode<presentation>{
|
|
|
492 |
\begin{frame}[c]
|
|
|
493 |
|
|
|
494 |
\begin{center}
|
|
38
|
495 |
\includegraphics[scale=0.5]{pics/ch3.jpg}
|
|
|
496 |
\end{center}
|
|
|
497 |
|
|
|
498 |
\begin{center}
|
|
|
499 |
\includegraphics[scale=0.5]{pics/ch4.jpg}\\
|
|
|
500 |
minimal automaton
|
|
35
|
501 |
\end{center}
|
|
|
502 |
|
|
|
503 |
\end{frame}}
|
|
|
504 |
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
|
|
505 |
|
|
39
|
506 |
|
|
|
507 |
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
|
|
508 |
\mode<presentation>{
|
|
|
509 |
\begin{frame}[c]
|
|
|
510 |
|
|
|
511 |
\begin{enumerate}
|
|
|
512 |
\item Take all pairs \bl{(q, p)} with \bl{q $\not=$ p}
|
|
|
513 |
\item Mark all pairs that accepting and non-accepting states
|
|
|
514 |
\item For all unmarked pairs \bl{(q, p)} and all characters \bl{c} tests wether
|
|
|
515 |
\begin{center}
|
|
|
516 |
\bl{($\delta$(q,c), $\delta$(p,c))}
|
|
|
517 |
\end{center}
|
|
41
|
518 |
are marked. If yes, then also mark \bl{(q, p)}
|
|
39
|
519 |
\item Repeat last step until no chance.
|
|
|
520 |
\item All unmarked pairs can be merged.
|
|
|
521 |
\end{enumerate}
|
|
|
522 |
|
|
|
523 |
\end{frame}}
|
|
|
524 |
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
|
|
525 |
|
|
|
526 |
|
|
|
527 |
|
|
38
|
528 |
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
|
|
529 |
\mode<presentation>{
|
|
|
530 |
\begin{frame}[c]
|
|
|
531 |
|
|
|
532 |
Given the function
|
|
|
533 |
|
|
|
534 |
\begin{center}
|
|
|
535 |
\bl{\begin{tabular}{r@{\hspace{1mm}}c@{\hspace{1mm}}l}
|
|
|
536 |
$rev(\varnothing)$ & $\dn$ & $\varnothing$\\
|
|
|
537 |
$rev(\epsilon)$ & $\dn$ & $\epsilon$\\
|
|
|
538 |
$rev(c)$ & $\dn$ & $c$\\
|
|
|
539 |
$rev(r_1 + r_2)$ & $\dn$ & $rev(r_1) + rev(r_2)$\\
|
|
|
540 |
$rev(r_1 \cdot r_2)$ & $\dn$ & $rev(r_2) \cdot rev(r_1)$\\
|
|
|
541 |
$rev(r^*)$ & $\dn$ & $rev(r)^*$\\
|
|
|
542 |
\end{tabular}}
|
|
|
543 |
\end{center}
|
|
|
544 |
|
|
|
545 |
|
|
|
546 |
and the set
|
|
|
547 |
|
|
|
548 |
\begin{center}
|
|
|
549 |
\bl{$Rev\,A \dn \{s^{-1} \;|\; s \in A\}$}
|
|
|
550 |
\end{center}
|
|
|
551 |
|
|
|
552 |
prove whether
|
|
|
553 |
|
|
|
554 |
\begin{center}
|
|
|
555 |
\bl{$L(rev(r)) = Rev (L(r))$}
|
|
|
556 |
\end{center}
|
|
|
557 |
|
|
|
558 |
\end{frame}}
|
|
|
559 |
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
|
35
|
560 |
|
|
|
561 |
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
|
|
562 |
\mode<presentation>{
|
|
|
563 |
\begin{frame}[c]
|
|
33
|
564 |
|
|
38
|
565 |
\begin{itemize}
|
|
|
566 |
\item The star-case in our proof about the matcher needs the following lemma
|
|
|
567 |
\begin{center}
|
|
|
568 |
\bl{Der\,c\,A$^*$ $=$ (Der c A)\,@\, A$^*$}
|
|
|
569 |
\end{center}
|
|
|
570 |
\end{itemize}\bigskip\bigskip
|
|
33
|
571 |
|
|
38
|
572 |
\begin{itemize}
|
|
|
573 |
\item If \bl{\texttt{""} $\in$ A}, then\\ \bl{Der\,c\,(A @ B) $=$ (Der\,c\,A) @ B $\cup$ (Der\,c\,B)}\medskip
|
|
|
574 |
\item If \bl{\texttt{""} $\not\in$ A}, then\\ \bl{Der\,c\,(A @ B) $=$ (Der\,c\,A) @ B}
|
|
|
575 |
|
|
|
576 |
\end{itemize}
|
|
|
577 |
|
|
33
|
578 |
\end{frame}}
|
|
|
579 |
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
|
|
580 |
|
|
|
581 |
|
|
|
582 |
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
|
|
583 |
\mode<presentation>{
|
|
|
584 |
\begin{frame}[c]
|
|
|
585 |
|
|
|
586 |
\begin{itemize}
|
|
35
|
587 |
\item Assuming you have the alphabet \bl{\{a, b, c\}}\bigskip
|
|
|
588 |
\item Give a regular expression that can recognise all strings that have at least one \bl{b}.
|
|
33
|
589 |
\end{itemize}
|
|
|
590 |
|
|
|
591 |
\end{frame}}
|
|
|
592 |
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
|
|
593 |
|
|
|
594 |
|
|
40
|
595 |
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
|
|
596 |
\mode<presentation>{
|
|
|
597 |
\begin{frame}[c]
|
|
|
598 |
|
|
|
599 |
``I hate coding. I do not want to look at code.''
|
|
|
600 |
|
|
|
601 |
\end{frame}}
|
|
|
602 |
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
|
|
603 |
|
|
33
|
604 |
|
|
|
605 |
|
|
|
606 |
\end{document}
|
|
|
607 |
|
|
|
608 |
%%% Local Variables:
|
|
|
609 |
%%% mode: latex
|
|
|
610 |
%%% TeX-master: t
|
|
|
611 |
%%% End:
|
|
|
612 |
|