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\renewcommand{\slidecaption}{AFL 07, King's College London, 13.~November 2013}

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

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\begin{document}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\mode<presentation>{

\begin{frame}<1>[t]

\frametitle{%

  \begin{tabular}{@ {}c@ {}}

  \\[-3mm]

  \LARGE Automata and \\[-2mm] 

  \LARGE Formal Languages (7)\\[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}}

 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%     

\newcommand{\qq}{\mbox{\texttt{"}}}

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\mode<presentation>{

\begin{frame}[c]

\frametitle{\begin{tabular}{c}Two Weeks Ago: CFGs\end{tabular}}

A \alert{context-free} grammar (CFG) \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{$A \rightarrow \text{rhs}_1 | \text{rhs}_2 | \ldots$}

\end{center}

where \bl{rhs} are sequences involving terminals and nonterminals (can also be empty).\medskip\pause

\end{itemize}

\end{frame}}

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\mode<presentation>{

\begin{frame}[c]

\frametitle{\begin{tabular}{c}Hierarchy of Languages\end{tabular}}

Recall that languages are sets of strings.

\begin{center}

\begin{tikzpicture}

[rect/.style={draw=black!50, top color=white,bottom color=black!20, rectangle, very thick, rounded corners}]

\draw (0,0) node [rect, text depth=39mm, text width=68mm] {all languages};

\draw (0,-0.4) node [rect, text depth=28.5mm, text width=64mm] {decidable languages};

\draw (0,-0.85) node [rect, text depth=17mm] {context sensitive languages};

\draw (0,-1.14) node [rect, text depth=9mm, text width=50mm] {context-free languages};

\draw (0,-1.4) node [rect] {regular languages};

\end{tikzpicture}

\end{center}

\end{frame}}

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\mode<presentation>{

\begin{frame}[c]

\frametitle{\begin{tabular}{c}Arithmetic Expressions\end{tabular}}

A grammar for arithmetic expressions and numbers:

\begin{center}

\bl{\begin{tabular}{lcl}

$E$ & $\rightarrow$ &  $E \cdot + \cdot E \;|\;E \cdot * \cdot E \;|\;( \cdot E \cdot ) \;|\;N$ \\

$N$ & $\rightarrow$ & $N \cdot N \;|\; 0 \;|\; 1 \;|\: \ldots \;|\; 9$ 

\end{tabular}}

\end{center}

Unfortunately it is left-recursive (and ambiguous).\medskip\\

A problem for \alert{recursive descent parsers} (e.g.~parser combinators).

\bigskip\pause

\end{frame}}

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\mode<presentation>{

\begin{frame}[t]

\frametitle{\begin{tabular}{c}Numbers\end{tabular}}

\begin{center}

\bl{\begin{tabular}{lcl}

$N$ & $\rightarrow$ &  $N \cdot N \;|\; 0 \;|\; 1 \;|\; \ldots \;|\; 9$\\

\end{tabular}}

\end{center}

A non-left-recursive, non-ambiguous grammar for numbers:

\begin{center}

\bl{\begin{tabular}{lcl}

$N$ & $\rightarrow$ &  $0 \cdot N \;|\;1 \cdot N \;|\;\ldots\;|\; 0 \;|\; 1 \;|\; \ldots \;|\; 9$\\

\end{tabular}}

\end{center}

\end{frame}}

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\mode<presentation>{

\begin{frame}[c]

\frametitle{\begin{tabular}{c}Operator Precedences\end{tabular}}

To disambiguate

\begin{center}

\bl{\begin{tabular}{lcl}

$E$ & $\rightarrow$ &  $E \cdot + \cdot E \;|\;E \cdot * \cdot E \;|\;( \cdot E \cdot ) \;|\;N$ \\

\end{tabular}}

\end{center}

Decide on how many precedence levels, say\medskip\\

\hspace{5mm}highest for \bl{$()$}, medium for \bl{*}, lowest for \bl{+}

\begin{center}

\bl{\begin{tabular}{lcl}

$E_{low}$ & $\rightarrow$ & $E_{med} \cdot + \cdot E_{low} \;|\; E_{med}$ \\

$E_{med}$ & $\rightarrow$ & $E_{hi} \cdot * \cdot E_{med} \;|\; E_{hi}$\\

$E_{hi}$ & $\rightarrow$ &  $( \cdot E_{low} \cdot ) \;|\;N$ \\

\end{tabular}}

\end{center}\pause

\small What happens with \bl{$1 + 3  + 4$}?

\end{frame}}

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\mode<presentation>{

\begin{frame}[c]

\frametitle{\begin{tabular}{c}Removing Left-Recursion\end{tabular}}

The rule for numbers is directly left-recursive:

\begin{center}

\bl{\begin{tabular}{lcl}

$N$ & $\rightarrow$ & $N \cdot N \;|\; 0 \;|\; 1\;\;\;\;(\ldots)$ 

\end{tabular}}

\end{center}

Translate

\begin{center}

\begin{tabular}{ccc}

\bl{\begin{tabular}{lcl}

$N$ & $\rightarrow$ & $N \cdot \alpha$\\

 &  $\;|\;$ & $\beta$\\

 \\ 

\end{tabular}} 

& {\Large$\Rightarrow$} &

\bl{\begin{tabular}{lcl}

$N$ & $\rightarrow$ & $\beta \cdot N'$\\

$N'$ & $\rightarrow$ & $\alpha \cdot N'$\\

 &  $\;|\;$ & $\epsilon$ 

\end{tabular}}

\end{tabular}

\end{center}\pause

Which means

\begin{center}

\bl{\begin{tabular}{lcl}

$N$ & $\rightarrow$ & $0 \cdot N' \;|\; 1 \cdot N'$\\

$N'$ & $\rightarrow$ & $N \cdot N' \;|\; \epsilon$\\

\end{tabular}}

\end{center}

\end{frame}}

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\mode<presentation>{

\begin{frame}[c]

\frametitle{\begin{tabular}{c}Chomsky Normal Form\end{tabular}}

All rules must be of the form

\begin{center}

\bl{$A \rightarrow a$}

\end{center}

or

\begin{center}

\bl{$A \rightarrow B\cdot C$}

\end{center}

No rule can contain \bl{$\epsilon$}.

\end{frame}}

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\mode<presentation>{

\begin{frame}[c]

\frametitle{\begin{tabular}{c}$\epsilon$-Removal\end{tabular}}

\begin{enumerate}

\item If \bl{$A\rightarrow \alpha \cdot B \cdot \beta$} and \bl{$B \rightarrow \epsilon$} are in the grammar,

then add \bl{$A\rightarrow \alpha \cdot \beta$} (iterate if necessary).

\item Throw out all \bl{$B \rightarrow \epsilon$}.

\end{enumerate}

\small

\begin{center}

\begin{tabular}{ccc}

\bl{\begin{tabular}{l@{\hspace{1mm}}c@{\hspace{1mm}}l}

$N$ & $\rightarrow$ & $0 \cdot N' \;|\; 1\cdot N'$\\

$N'$ & $\rightarrow$ & $N \cdot N'\;|\;\epsilon$\\

\\ 

\\

\\

\\

\\

\end{tabular}} &

\bl{\begin{tabular}{l@{\hspace{1mm}}c@{\hspace{1mm}}l}

\\

$N$ & $\rightarrow$ & $0 \cdot N' \;|\; 1\cdot N'\;|\;0\;|\;1$\\

$N'$ & $\rightarrow$ & $N \cdot N'\;|\;N\;|\;\epsilon$\\

\\

$N$ & $\rightarrow$ & $0 \cdot N' \;|\; 1\cdot N'\;|\;0\;|\;1$\\

$N'$ & $\rightarrow$ & $N \cdot N'\;|\;N$\\

\end{tabular}}

\end{tabular}

\end{center}

\pause\normalsize

\begin{center}

\bl{\begin{tabular}{l@{\hspace{1mm}}c@{\hspace{1mm}}l}

$N$ & $\rightarrow$ & $0 \cdot N\;|\; 1\cdot N\;|\;0\;|\;1$\\

\end{tabular}}

\end{center}

\end{frame}}

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\mode<presentation>{

\begin{frame}[c]

\frametitle{\begin{tabular}{c}CYK Algorithm\end{tabular}}

If grammar is in Chomsky normalform \ldots

\begin{center}

\bl{\begin{tabular}{@ {}lcl@ {}}

$S$ & $\rightarrow$ &  $N\cdot P$ \\

$P$ & $\rightarrow$ &  $V\cdot N$ \\

$N$ & $\rightarrow$ &  $N\cdot N$ \\

$N$ & $\rightarrow$ &  $\texttt{students} \;|\; \texttt{Jeff} \;|\; \texttt{geometry} \;|\; \texttt{trains} $ \\

$V$ & $\rightarrow$ &  $\texttt{trains}$ 

\end{tabular}}

\end{center}

\bl{\texttt{Jeff trains geometry students}}

\end{frame}}

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\mode<presentation>{

\begin{frame}[c]

\frametitle{\begin{tabular}{c}CYK Algorithm\end{tabular}}

\begin{itemize}

\item fastest possible algorithm for recognition problem

\item runtime is \bl{$O(n^3)$}\bigskip

\item grammars need to be transferred into CNF

\end{itemize}

\end{frame}}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%  

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\mode<presentation>{

\begin{frame}[c]

\frametitle{\begin{tabular}{c}Hierarchy of Languages\end{tabular}}

Recall that languages are sets of strings.

\begin{center}

\begin{tikzpicture}

[rect/.style={draw=black!50, top color=white,bottom color=black!20, rectangle, very thick, rounded corners}]

\draw (0,0) node [rect, text depth=39mm, text width=68mm] {all languages};

\draw (0,-0.4) node [rect, text depth=28.5mm, text width=64mm] {decidable languages};

\draw (0,-0.85) node [rect, text depth=17mm] {context sensitive languages};

\draw (0,-1.14) node [rect, text depth=9mm, text width=50mm] {context-free languages};

\draw (0,-1.4) node [rect] {regular languages};

\end{tikzpicture}

\end{center}

\end{frame}}

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\mode<presentation>{

\begin{frame}[c]

\frametitle{\begin{tabular}{c}Context Sensitive Grms\end{tabular}}

\begin{center}

\bl{\begin{tabular}{lcl}

$S$ & $\Rightarrow$ &  $bSAA\;|\; \epsilon$\\

$A$ & $\Rightarrow$ & $a$\\

$bA$ & $\Rightarrow$ & $Ab$\\

\end{tabular}}

\end{center}\pause

\begin{center}

\bl{$S \Rightarrow\ldots\Rightarrow^? "ababaa"$}

\end{center}

\end{frame}}

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\mode<presentation>{

\begin{frame}[c]

\begin{center}

\textit{Stmt} & $\rightarrow$ &  $\texttt{skip}$\\

              & $|$ & \textit{Id}\;\texttt{:=}\;\textit{AExp}\\

              & $|$ & \texttt{if}\; \textit{BExp} \;\texttt{then}\; \textit{Block} \;\texttt{else}\; \textit{Block}\\

              & $|$ & \texttt{while}\; \textit{BExp} \;\texttt{do}\; \textit{Block}\\

              & $|$ & \texttt{read}\;\textit{Id}\\

              & $|$ & \texttt{write}\;\textit{Id}\\

              & $|$ & \texttt{write}\;\textit{String}\medskip\\

\textit{Stmts} & $\rightarrow$ &  \textit{Stmt} \;\texttt{;}\; \textit{Stmts}\\

              & $|$ & \textit{Stmt}\medskip\\

\textit{Block} & $\rightarrow$ &  \texttt{\{}\,\textit{Stmts}\,\texttt{\}}\\

                & $|$ & \textit{Stmt}\medskip\\

\textit{AExp} & $\rightarrow$ & \ldots\\

\textit{BExp} & $\rightarrow$ & \ldots\\

\end{center}

\end{frame}}

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\mode<presentation>{

\begin{frame}[c]

\begin{center}

\end{tabular}}

\end{center}