\documentclass[dvipsnames,14pt,t]{beamer}
\usepackage{../slides}
\usepackage{../graphics}
\usepackage{../langs}
\usepackage{../data}


% beamer stuff 
\renewcommand{\slidecaption}{CFL 07, King's College London}
\newcommand{\bl}[1]{\textcolor{blue}{#1}}       
%\newcommand{\dn}{\stackrel{\mbox{\scriptsize def}}{=}}% for definitions

\begin{document}

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\begin{frame}[t]
\frametitle{%
  \begin{tabular}{@ {}c@ {}}
  \\[-3mm]
  \LARGE Compilers and \\[-2mm] 
  \LARGE Formal Languages (7)\\[3mm] 
  \end{tabular}}

  \normalsize
  \begin{center}
  \begin{tabular}{ll}
  Email:  & christian.urban at kcl.ac.uk\\
  Office: & N\liningnums{7.07} (North Wing, Bush House)\\
  Slides: & KEATS (also home work is there)\\
  \end{tabular}
  \end{center}

\end{frame}
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\newcommand{\qq}{\mbox{\texttt{"}}}
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\begin{frame}[c]
\frametitle{CFGs}

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}}
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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}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}
\bl{\begin{tabular}{@{}lcl@{}}
\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{tabular}}
\end{center}
\end{frame}}
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\mode<presentation>{
\begin{frame}[c]

\mbox{\lstinputlisting[language=while]{../progs/fib.while}}

\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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\mode<presentation>{
\begin{frame}[c]
\frametitle{\begin{tabular}{c}Interpreter\end{tabular}}

\begin{center}
\bl{\begin{tabular}{@{}lcl@{}}
$\text{eval}(n, E)$ & $\dn$ & $n$\\
$\text{eval}(x, E)$ & $\dn$ & $E(x)$ \;\;\;\textcolor{black}{lookup \bl{$x$} in \bl{$E$}}\\
$\text{eval}(a_1 + a_2, E)$ & $\dn$ & $\text{eval}(a_1, E) + \text{eval}(a_2, E)$\\
$\text{eval}(a_1 - a_2, E)$ & $\dn$ & $\text{eval}(a_1, E) - \text{eval}(a_2, E)$\\
$\text{eval}(a_1 * a_2, E)$ & $\dn$ & $\text{eval}(a_1, E) * \text{eval}(a_2, E)$\bigskip\\
$\text{eval}(a_1 = a_2, E)$ & $\dn$ & $\text{eval}(a_1, E) = \text{eval}(a_2, E)$\\
$\text{eval}(a_1\,!\!= a_2, E)$ & $\dn$ & $\neg(\text{eval}(a_1, E) = \text{eval}(a_2, E))$\\
$\text{eval}(a_1 < a_2, E)$ & $\dn$ & $\text{eval}(a_1, E) < \text{eval}(a_2, E)$\
\end{tabular}}
\end{center}

\end{frame}}
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\mode<presentation>{
\begin{frame}[c]
\frametitle{\begin{tabular}{c}Interpreter (2)\end{tabular}}

\begin{center}
\bl{\begin{tabular}{@{}lcl@{}}
$\text{eval}(\text{skip}, E)$ & $\dn$ & $E$\\
$\text{eval}(x:=a, E)$ & $\dn$ & \bl{$E(x \mapsto \text{eval}(a, E))$}\\
\multicolumn{3}{@{}l@{}}{$\text{eval}(\text{if}\;b\;\text{then}\;cs_1\;\text{else}\;cs_2 , E) \dn$}\\
\multicolumn{3}{@{}l@{}}{\hspace{2cm}$\text{if}\;\text{eval}(b,E)\;\text{then}\;
\text{eval}(cs_1,E)$}\\
\multicolumn{3}{@{}l@{}}{\hspace{2cm}$\phantom{\text{if}\;\text{eval}(b,E)\;}\text{else}\;\text{eval}(cs_2,E)$}\\
\multicolumn{3}{@{}l@{}}{$\text{eval}(\text{while}\;b\;\text{do}\;cs, E) \dn$}\\
\multicolumn{3}{@{}l@{}}{\hspace{2cm}$\text{if}\;\text{eval}(b,E)$}\\
\multicolumn{3}{@{}l@{}}{\hspace{2cm}$\text{then}\;
\text{eval}(\text{while}\;b\;\text{do}\;cs, \text{eval}(cs,E))$}\\
\multicolumn{3}{@{}l@{}}{\hspace{2cm}$\text{else}\; E$}\\
$\text{eval}(\text{write}\; x, E)$ & $\dn$ & $\{\;\text{println}(E(x))\; ;\;E\;\}$\\
\end{tabular}}
\end{center}

\end{frame}}
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\mode<presentation>{
\begin{frame}[c]
\frametitle{\begin{tabular}{c}Test Program\end{tabular}}

\mbox{}\\[-18mm]\mbox{}

{\lstset{language=While}%%\fontsize{10}{12}\selectfont
\texttt{\lstinputlisting{../progs/loops.while}}}

\end{frame}}
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\mode<presentation>{
\begin{frame}[t]
\frametitle{\begin{tabular}{c}Interpreted Code\end{tabular}}

\begin{center}
\begin{tikzpicture}
\begin{axis}[axis x line=bottom, axis y line=left, xlabel=n, ylabel=secs, legend style=small]
\addplot+[smooth] file {interpreted.data};
\end{axis}
\end{tikzpicture}
\end{center}

\end{frame}}
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\mode<presentation>{
\begin{frame}[c]
\frametitle{\begin{tabular}{c}Java Virtual Machine\end{tabular}}

\begin{itemize}
\item introduced in 1995
\item is a stack-based VM (like Postscript, CLR of .Net)
\item contains a JIT compiler
\item many languages take advantage of JVM's infrastructure (JRE)
\item is garbage collected $\Rightarrow$ no buffer overflows
\item some languages compile to the JVM: Scala, Clojure\ldots
\end{itemize}

\end{frame}}
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\end{document}

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