\documentclass[dvipsnames,14pt,t]{beamer}
\usepackage{beamerthemeplaincu}
\usepackage[absolute,overlay]{textpos}
\usepackage{ifthen}
\usepackage{tikz}
\usepackage{pgf}
\usepackage{calc}
\usepackage{ulem}
\usepackage{courier}
\usepackage{listings}
\renewcommand{\uline}[1]{#1}
\usetikzlibrary{arrows}
\usetikzlibrary{automata}
\usetikzlibrary{shapes}
\usetikzlibrary{shadows}
\usetikzlibrary{positioning}
\usetikzlibrary{calc}
\usetikzlibrary{plotmarks}
\usepackage{graphicx}
\usepackage{../langs}
\usepackage{../data}
\makeatletter
\lst@CCPutMacro\lst@ProcessOther {"2D}{\lst@ttfamily{-{}}{-{}}}
\@empty\z@\@empty
\makeatother
% beamer stuff
\renewcommand{\slidecaption}{AFL 06, King's College London, 30.~October 2013}
\newcommand{\bl}[1]{\textcolor{blue}{#1}}
\newcommand{\dn}{\stackrel{\mbox{\scriptsize def}}{=}}% for definitions
\begin{document}
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\mode<presentation>{
\begin{frame}<1>[t]
\frametitle{%
\begin{tabular}{@ {}c@ {}}
\\[-3mm]
\LARGE Automata and \\[-2mm]
\LARGE Formal Languages (6)\\[3mm]
\end{tabular}}
\normalsize
\begin{center}
\begin{tabular}{ll}
Email: & christian.urban at kcl.ac.uk\\
Office: & S1.27 (1st floor Strand Building)\\
Slides: & KEATS (also home work is there)\\
\end{tabular}
\end{center}
\end{frame}}
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\mode<presentation>{
\begin{frame}[c]
\frametitle{\begin{tabular}{c}Regular Languages\end{tabular}}
While regular expressions are very useful for lexing,
there is no regular expression that can recognise the language \bl{$a^nb^n$}.\bigskip
\begin{center}
\bl{$(((()()))())$} \;\;vs.\;\; \bl{$(((()()))()))$}
\end{center}
\end{frame}}
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\newcommand{\qq}{\mbox{\texttt{"}}}
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\mode<presentation>{
\begin{frame}[c]
\frametitle{\begin{tabular}{c}Grammars\end{tabular}}
A (context-free) grammar \bl{$G$} consists of
\begin{itemize}
\item a finite set of nonterminal symbols (upper case)
\item a finite terminal symbols or tokens (lower case)
\item a start symbol (which must be a nonterminal)
\item a set of rules
\begin{center}
\bl{$A \rightarrow \text{rhs}$}
\end{center}
where \bl{rhs} are sequences involving terminals and nonterminals,
including the empty sequence \bl{$\epsilon$}.\medskip\pause
We also allow rules
\begin{center}
\bl{$A \rightarrow \text{rhs}_1 | \text{rhs}_2 | \ldots$}
\end{center}
\end{itemize}
\end{frame}}
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\mode<presentation>{
\begin{frame}[c]
\frametitle{\begin{tabular}{c}Palindromes\end{tabular}}
\begin{center}
\bl{\begin{tabular}{lcl}
$S$ & $\rightarrow$ & $\epsilon$ \\
$S$ & $\rightarrow$ & $a\cdot S\cdot a$ \\
$S$ & $\rightarrow$ & $b\cdot S\cdot b$ \\
\end{tabular}}
\end{center}\pause
or
\begin{center}
\bl{\begin{tabular}{lcl}
$S$ & $\rightarrow$ & $\epsilon \;|\; a\cdot S\cdot a \;|\;b\cdot S\cdot b$ \\
\end{tabular}}
\end{center}
\end{frame}}
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\mode<presentation>{
\begin{frame}[c]
\frametitle{\begin{tabular}{c}Arithmetic Expressions\end{tabular}}
\begin{center}
\bl{\begin{tabular}{lcl}
$E$ & $\rightarrow$ & $num\_token$ \\
$E$ & $\rightarrow$ & $E \cdot + \cdot E$ \\
$E$ & $\rightarrow$ & $E \cdot - \cdot E$ \\
$E$ & $\rightarrow$ & $E \cdot * \cdot E$ \\
$E$ & $\rightarrow$ & $( \cdot E \cdot )$
\end{tabular}}
\end{center}\pause
\bl{\texttt{1 + 2 * 3 + 4}}
\end{frame}}
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\mode<presentation>{
\begin{frame}[c]
\frametitle{\begin{tabular}{c}A CFG Derivation\end{tabular}}
\begin{enumerate}
\item Begin with a string containing only the start symbol, say \bl{$S$}\bigskip
\item Replace any nonterminal \bl{$X$} in the string by the
right-hand side of some production \bl{$X \rightarrow \text{rhs}$}\bigskip
\item Repeat 2 until there are no nonterminals
\end{enumerate}
\begin{center}
\bl{$S \rightarrow \ldots \rightarrow \ldots \rightarrow \ldots \rightarrow \ldots $}
\end{center}
\end{frame}}
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\mode<presentation>{
\begin{frame}[c]
\frametitle{\begin{tabular}{c}Example Derivation\end{tabular}}
\begin{center}
\bl{\begin{tabular}{lcl}
$S$ & $\rightarrow$ & $\epsilon \;|\; a\cdot S\cdot a \;|\;b\cdot S\cdot b$ \\
\end{tabular}}
\end{center}\bigskip
\begin{center}
\begin{tabular}{lcl}
\bl{$S$} & \bl{$\rightarrow$} & \bl{$aSa$}\\
& \bl{$\rightarrow$} & \bl{$abSba$}\\
& \bl{$\rightarrow$} & \bl{$abaSaba$}\\
& \bl{$\rightarrow$} & \bl{$abaaba$}\\
\end{tabular}
\end{center}
\end{frame}}
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\mode<presentation>{
\begin{frame}[c]
\frametitle{\begin{tabular}{c}Example Derivation\end{tabular}}
\begin{center}
\bl{\begin{tabular}{lcl}
$E$ & $\rightarrow$ & $num\_token$ \\
$E$ & $\rightarrow$ & $E \cdot + \cdot E$ \\
$E$ & $\rightarrow$ & $E \cdot - \cdot E$ \\
$E$ & $\rightarrow$ & $E \cdot * \cdot E$ \\
$E$ & $\rightarrow$ & $( \cdot E \cdot )$
\end{tabular}}
\end{center}\bigskip
\begin{center}
\begin{tabular}{@{}c@{}c@{}}
\begin{tabular}{l@{\hspace{1mm}}l@{\hspace{1mm}}l}
\bl{$E$} & \bl{$\rightarrow$} & \bl{$E*E$}\\
& \bl{$\rightarrow$} & \bl{$E+E*E$}\\
& \bl{$\rightarrow$} & \bl{$E+E*E+E$}\\
& \bl{$\rightarrow^+$} & \bl{$1+2*3+4$}\\
\end{tabular} &\pause
\begin{tabular}{l@{\hspace{1mm}}l@{\hspace{1mm}}l}
\bl{$E$} & \bl{$\rightarrow$} & \bl{$E+E$}\\
& \bl{$\rightarrow$} & \bl{$E+E+E$}\\
& \bl{$\rightarrow$} & \bl{$E+E*E+E$}\\
& \bl{$\rightarrow^+$} & \bl{$1+2*3+4$}\\
\end{tabular}
\end{tabular}
\end{center}
\end{frame}}
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\mode<presentation>{
\begin{frame}[c]
\frametitle{\begin{tabular}{c}Language of a CFG\end{tabular}}
Let \bl{$G$} be a context-free grammar with start symbol \bl{$S$}.
Then the language \bl{$L(G)$} is:
\begin{center}
\bl{$\{c_1\ldots c_n \;|\; \forall i.\; c_i \in T \wedge S \rightarrow^* c_1\ldots c_n \}$}
\end{center}\pause
\begin{itemize}
\item Terminals, because there are no rules for replacing them.
\item Once generated, terminals are ``permanent''.
\item Terminals ought to be tokens of the language\\
(but can also be strings).
\end{itemize}
\end{frame}}
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\mode<presentation>{
\begin{frame}[c]
\frametitle{\begin{tabular}{c}Parse Trees\end{tabular}}
\begin{center}
\bl{\begin{tabular}{lcl}
$E$ & $\rightarrow$ & $F \;|\; F \cdot * \cdot F$\\
$F$ & $\rightarrow$ & $T \;|\; T \cdot + \cdot T \;|\; T \cdot - \cdot T$\\
$T$ & $\rightarrow$ & $num\_token \;|\; ( \cdot E \cdot )$\\
\end{tabular}}
\end{center}
\begin{center}
\begin{tikzpicture}[level distance=8mm, blue]
\node {$E$}
child {node {$F$}
child {node {$T$}
child {node {(\,$E$\,)}
child {node{$F$ *{} $F$}
child {node {$T$} child {node {2}}}
child {node {$T$} child {node {3}}}
}
}
}
child {node {+}}
child {node {$T$}
child {node {(\,$E$\,)}
child {node {$F$}
child {node {$T$ +{} $T$}
child {node {3}}
child {node {4}}
}
}}
}};
\end{tikzpicture}
\end{center}
\begin{textblock}{5}(1, 6.5)
\bl{\texttt{(2*3)+(3+4)}}
\end{textblock}
\end{frame}}
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\mode<presentation>{
\begin{frame}[c]
\frametitle{\begin{tabular}{c}Arithmetic Expressions\end{tabular}}
\begin{center}
\bl{\begin{tabular}{lcl}
$E$ & $\rightarrow$ & $num\_token$ \\
$E$ & $\rightarrow$ & $E \cdot + \cdot E$ \\
$E$ & $\rightarrow$ & $E \cdot - \cdot E$ \\
$E$ & $\rightarrow$ & $E \cdot * \cdot E$ \\
$E$ & $\rightarrow$ & $( \cdot E \cdot )$
\end{tabular}}
\end{center}\pause\bigskip
A CFG is \alert{left-recursive} if it has a nonterminal \bl{$E$} such
that \bl{$E \rightarrow^+ E\cdot \ldots$}
\end{frame}}
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\mode<presentation>{
\begin{frame}[c]
\frametitle{\begin{tabular}{c}Ambiguous Grammars\end{tabular}}
A grammar is \alert{ambiguous} if there is a string that has at least two different parse trees.
\begin{center}
\bl{\begin{tabular}{lcl}
$E$ & $\rightarrow$ & $num\_token$ \\
$E$ & $\rightarrow$ & $E \cdot + \cdot E$ \\
$E$ & $\rightarrow$ & $E \cdot - \cdot E$ \\
$E$ & $\rightarrow$ & $E \cdot * \cdot E$ \\
$E$ & $\rightarrow$ & $( \cdot E \cdot )$
\end{tabular}}
\end{center}
\bl{\texttt{1 + 2 * 3 + 4}}
\end{frame}}
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\mode<presentation>{
\begin{frame}[c]
\frametitle{\begin{tabular}{c}Dangling Else\end{tabular}}
Another ambiguous grammar:\bigskip
\begin{center}
\bl{\begin{tabular}{lcl}
$E$ & $\rightarrow$ & if $E$ then $E$\\
& $|$ & if $E$ then $E$ else $E$ \\
& $|$ & \ldots
\end{tabular}}
\end{center}\bigskip
\bl{\texttt{if a then if x then y else c}}
\end{frame}}
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\mode<presentation>{
\begin{frame}[c]
\frametitle{\begin{tabular}{c}Parser Combinators\end{tabular}}
Parser combinators: \bigskip
\begin{minipage}{1.1\textwidth}
\begin{center}
\mbox{}\hspace{-12mm}\mbox{}$\underbrace{\text{list of tokens}}_{\text{input}}$ \bl{$\Rightarrow$}
$\underbrace{\text{set of (parsed input, unparsed input)}}_{\text{output}}$
\end{center}
\end{minipage}\bigskip
\begin{itemize}
\item sequencing
\item alternative
\item semantic action
\end{itemize}
\end{frame}}
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\mode<presentation>{
\begin{frame}[c]
Alternative parser (code \bl{$p\;||\;q$})\bigskip
\begin{itemize}
\item apply \bl{$p$} and also \bl{$q$}; then combine the outputs
\end{itemize}
\begin{center}
\large \bl{$p(\text{input}) \cup q(\text{input})$}
\end{center}
\end{frame}}
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\mode<presentation>{
\begin{frame}[c]
Sequence parser (code \bl{$p\sim q$})\bigskip
\begin{itemize}
\item apply first \bl{$p$} producing a set of pairs
\item then apply \bl{$q$} to the unparsed parts
\item then combine the results:\\ \mbox{}\;\;((output$_1$, output$_2$), unparsed part)
\end{itemize}
\begin{center}
\begin{tabular}{l}
\large \bl{$\{((o_1, o_2), u_2) \;|\;$}\\[2mm]
\large\mbox{}\hspace{15mm} \bl{$(o_1, u_1) \in p(\text{input}) \wedge$}\\[2mm]
\large\mbox{}\hspace{15mm} \bl{$(o_2, u_2) \in q(u_1)\}$}
\end{tabular}
\end{center}
\end{frame}}
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\mode<presentation>{
\begin{frame}[c]
Function parser (code \bl{$p \Rightarrow f$})\bigskip
\begin{itemize}
\item apply \bl{$p$} producing a set of pairs
\item then apply the function \bl{$f$} to each first component
\end{itemize}
\begin{center}
\begin{tabular}{l}
\large \bl{$\{(f(o_1), u_1) \;|\; (o_1, u_1) \in p(\text{input})\}$}
\end{tabular}
\end{center}\bigskip\bigskip\pause
\bl{$f$} is the semantic action (``what to do with the parsed input'')
\end{frame}}
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\mode<presentation>{
\begin{frame}[c]
\frametitle{\begin{tabular}{c}Semantic Actions\end{tabular}}
Addition
\begin{center}
\bl{$T \sim + \sim E \Rightarrow \underbrace{f((x,y), z) \Rightarrow x + z}_{\text{semantic action}}$}
\end{center}\pause
Multiplication
\begin{center}
\bl{$F \sim * \sim T \Rightarrow f((x,y), z) \Rightarrow x * z$}
\end{center}\pause
Parenthesis
\begin{center}
\bl{$\text{(} \sim E \sim \text{)} \Rightarrow f((x,y), z) \Rightarrow y$}
\end{center}
\end{frame}}
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\mode<presentation>{
\begin{frame}[c]
\frametitle{\begin{tabular}{c}Types of Parsers\end{tabular}}
\begin{itemize}
\item {\bf Sequencing}: if \bl{$p$} returns results of type \bl{$T$}, and \bl{$q$} results of type \bl{$S$},
then \bl{$p \sim q$} returns results of type
\begin{center}
\bl{$T \times S$}
\end{center}\pause
\item {\bf Alternative}: if \bl{$p$} returns results of type \bl{$T$} then \bl{$q$} \alert{must} also have results of type \bl{$T$},
and \bl{$p \;||\; q$} returns results of type
\begin{center}
\bl{$T$}
\end{center}\pause
\item {\bf Semantic Action}: if \bl{$p$} returns results of type \bl{$T$} and \bl{$f$} is a function from
\bl{$T$} to \bl{$S$}, then
\bl{$p \Rightarrow f$} returns results of type
\begin{center}
\bl{$S$}
\end{center}
\end{itemize}
\end{frame}}
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\mode<presentation>{
\begin{frame}[c]
\frametitle{\begin{tabular}{c}Input Types of Parsers\end{tabular}}
\begin{itemize}
\item input: \alert{string}
\item output: set of (output\_type, \alert{string})
\end{itemize}\bigskip\pause
actually it can be any input type as long as it is a kind of sequence
(for example a string)
\end{frame}}
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\mode<presentation>{
\begin{frame}[c]
\frametitle{\begin{tabular}{c}Scannerless Parsers\end{tabular}}
\begin{itemize}
\item input: \alert{string}
\item output: set of (output\_type, \alert{string})
\end{itemize}\bigskip
but lexers are better when whitespaces or comments need to be filtered out;
then input is a sequence of tokens
\end{frame}}
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\mode<presentation>{
\begin{frame}[c]
\frametitle{\begin{tabular}{c}Successful Parses\end{tabular}}
\begin{itemize}
\item input: string
\item output: \alert{set of} (output\_type, string)
\end{itemize}\bigskip
a parse is successful whenever the input has been
fully ``consumed'' (that is the second component is empty)
\end{frame}}
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\mode<presentation>{
\begin{frame}[c]
\frametitle{Abstract Parser Class}
\mbox{\lstset{language=Scala}\fontsize{10}{12}\selectfont
\texttt{\lstinputlisting{../progs/app7.scala}}}
\end{frame}}
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\mode<presentation>{
\begin{frame}[c]
{\lstset{language=Scala}\fontsize{10}{12}\selectfont
\texttt{\lstinputlisting{../progs/app8.scala}}}
\end{frame}}
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\mode<presentation>{
\begin{frame}[c]
\frametitle{\begin{tabular}{c}Two Grammars\end{tabular}}
Which languages are recognised by the following two grammars?
\begin{center}
\bl{\begin{tabular}{lcl}
$S$ & $\rightarrow$ & $1 \cdot S \cdot S$\\
& $|$ & $\epsilon$
\end{tabular}}
\end{center}\bigskip
\begin{center}
\bl{\begin{tabular}{lcl}
$U$ & $\rightarrow$ & $1 \cdot U$\\
& $|$ & $\epsilon$
\end{tabular}}
\end{center}
\end{frame}}
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\mode<presentation>{
\begin{frame}[t]
\frametitle{\begin{tabular}{c}Ambiguous Grammars\end{tabular}}
\mbox{}\\[-25mm]\mbox{}
\begin{center}
\begin{tikzpicture}[y=.2cm, x=.009cm]
%axis
\draw (0,0) -- coordinate (x axis mid) (1000,0);
\draw (0,0) -- coordinate (y axis mid) (0,30);
%ticks
\foreach \x in {0, 20, 100, 200,...,1000}
\draw (\x,1pt) -- (\x,-3pt)
node[anchor=north] {\small \x};
\foreach \y in {0,5,...,30}
\draw (1pt,\y) -- (-3pt,\y)
node[anchor=east] {\small\y};
%labels
\node[below=0.6cm] at (x axis mid) {\bl{1}s};
\node[rotate=90, left=1.2cm] at (y axis mid) {secs};
%plots
\draw[color=blue] plot[mark=*, mark options={fill=white}]
file {s-grammar1.data};
\only<2->{\draw[color=red] plot[mark=triangle*, mark options={fill=white} ]
file {s-grammar2.data};}
%legend
\begin{scope}[shift={(400,20)}]
\draw[color=blue] (0,0) --
plot[mark=*, mark options={fill=white}] (0.25,0) -- (0.5,0)
node[right]{\small unambiguous};
\only<2->{\draw[yshift=\baselineskip, color=red] (0,0) --
plot[mark=triangle*, mark options={fill=white}] (0.25,0) -- (0.5,0)
node[right]{\small ambiguous};}
\end{scope}
\end{tikzpicture}
\end{center}
\end{frame}}
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\mode<presentation>{
\begin{frame}[c]
\frametitle{\begin{tabular}{c}While-Language\end{tabular}}
\begin{center}
\bl{\begin{tabular}{@{}lcl@{}}
$Stmt$ & $\rightarrow$ & $\text{skip}$\\
& $|$ & $Id := AExp$\\
& $|$ & $\text{if}\; B\!Exp \;\text{then}\; Block \;\text{else}\; Block$\\
& $|$ & $\text{while}\; B\!Exp \;\text{do}\; Block$\medskip\\
$Stmts$ & $\rightarrow$ & $Stmt \;\text{;}\; Stmts$\\
& $|$ & $Stmt$\medskip\\
$Block$ & $\rightarrow$ & $\{ Stmts \}$\\
& $|$ & $Stmt$\medskip\\
$AExp$ & $\rightarrow$ & \ldots\\
$BExp$ & $\rightarrow$ & \ldots\\
\end{tabular}}
\end{center}
\end{frame}}
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\mode<presentation>{
\begin{frame}[c]
\frametitle{\begin{tabular}{c}An Interpreter\end{tabular}}
\begin{center}
\bl{\begin{tabular}{l}
$\{$\\
\;\;$x := 5 \text{;}$\\
\;\;$y := x * 3\text{;}$\\
\;\;$y := x * 4\text{;}$\\
\;\;$x := u * 3$\\
$\}$
\end{tabular}}
\end{center}
\begin{itemize}
\item the interpreter has to record the value of \bl{$x$} before assigning a value to \bl{$y$}\pause
\item \bl{\text{eval}(stmt, env)}
\end{itemize}
\end{frame}}
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\end{document}
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