% !TEX program = xelatex
\documentclass[dvipsnames,14pt,t]{beamer}
\usepackage{../slides}
\usepackage{../graphics}
\usepackage{../langs}
\usepackage{../data}
\usepackage{../grammar}
% beamer stuff
\renewcommand{\slidecaption}{CFL 06, King's College London}
\newcommand{\bl}[1]{\textcolor{blue}{#1}}
%\newcommand{\dn}{\stackrel{\mbox{\scriptsize def}}{=}}% for definitions
\newcommand{\qq}{\mbox{\texttt{"}}}
\begin{document}
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\begin{frame}[t]
\frametitle{%
\begin{tabular}{@ {}c@ {}}
\\[-3mm]
\LARGE Compilers and \\[-2mm]
\LARGE Formal Languages (6)\\[3mm]
\end{tabular}}
\normalsize
\begin{center}
\begin{tabular}{ll}
Email: & christian.urban at kcl.ac.uk\\
Office Hours: & Thursdays 12 -- 14\\
Location: & N7.07 (North Wing, Bush House)\\
Slides \& Progs: & KEATS (also homework is there)\\
\end{tabular}
\end{center}
\end{frame}
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% \begin{frame}[c]
% \small
% \mbox{}\\[5mm]
% %\begin{textblock}{10}(3,5)
% \begin{tikzpicture}[scale=1.5,
% node distance=1cm,
% every node/.style={minimum size=7mm}]
% \node (r1) {\bl{$r_1$}};
% \node (r2) [right=of r1] {\bl{$r_2$}};
% \draw[->,line width=1mm] (r1) -- (r2) node[above,midway] {\bl{$der\,a$}};
% \node (r3) [right=of r2] {\bl{$r_3$}};
% \draw[->,line width=1mm] (r2) -- (r3) node[above,midway] {\bl{$der\,b$}};
% \node (r4) [right=of r3] {\bl{$r_4$}};
% \draw[->,line width=1mm] (r3) -- (r4) node[above,midway] {\bl{$der\,c$}};
% \draw (r4) node[anchor=west] {\;\raisebox{3mm}{\bl{$nullable$}}};
% \node (v4) [below=of r4] {\bl{$v_4$}};
% \draw[->,line width=1mm] (r4) -- (v4);
% \node (v3) [left=of v4] {\bl{$v_3$}};
% \draw[->,line width=1mm] (v4) -- (v3) node[below,midway] {\bl{$inj\,c$}};
% \node (v2) [left=of v3] {\bl{$v_2$}};
% \draw[->,line width=1mm] (v3) -- (v2) node[below,midway] {\bl{$inj\,b$}};
% \node (v1) [left=of v2] {\bl{$v_1$}};
% \draw[->,line width=1mm] (v2) -- (v1) node[below,midway] {\bl{$inj\,a$}};
% \draw[->,line width=0.5mm] (r3) -- (v3);
% \draw[->,line width=0.5mm] (r2) -- (v2);
% \draw[->,line width=0.5mm] (r1) -- (v1);
% \draw (r4) node[anchor=north west] {\;\raisebox{-8mm}{\bl{$mkeps$}}};
% \end{tikzpicture}
% %\end{textblock}
% \begin{center}
% \begin{tabular}{l@{\hspace{1mm}}c@{\hspace{1mm}}l}
% \\[-10mm]
% \bl{$inj\,(c)\,c\,Empty$} & \bl{$\dn$} & \bl{$Char\,c$}\\
% \bl{$inj\,(r_1 + r_2)\,c\,Left(v)$} & \bl{$\dn$} & \bl{$Left(inj\,r_1\,c\,v)$}\\
% \bl{$inj\,(r_1 + r_2)\,c\,Right(v)$} & \bl{$\dn$} & \bl{$Right(inj\,r_2\,c\,v)$}\\
% \bl{$inj\,(r_1 \cdot r_2)\,c\,Seq(v_1,v_2)$} & \bl{$\dn$} & \bl{$Seq(inj\,r_1\,c\,v_1,v_2)$}\\
% \bl{$inj\,(r_1 \cdot r_2)\,c\,Left(Seq(v_1,v_2))$} & \bl{$\dn$} & \bl{$Seq(inj\,r_1\,c\,v_1,v_2)$}\\
% \bl{$inj\,(r_1 \cdot r_2)\,c\,Right(v)$} & \bl{$\dn$} & \bl{$Seq(mkeps(r_1),inj\,r_2\,c\,v)$}\\
% \bl{$inj\,(r^*)\,c\,Seq(v,vs)$} & \bl{$\dn$} & \bl{$inj\,r\,c\,v\,::\,vs$}\\
% \end{tabular}
% \end{center}
% \end{frame}
% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\begin{frame}[t]
\frametitle{Starting Symbol}
\bl{\begin{plstx}[margin=1cm]
: \meta{S} ::= \meta{A}\cdot\meta{S}\cdot\meta{B} |
\meta{B}\cdot\meta{S}\cdot\meta{A} | \epsilon\\
: \meta{A} ::= a | \epsilon\\
: \meta{B} ::= b\\
\end{plstx}}
\end{frame}
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\begin{frame}[c]
\frametitle{Hierarchy of Languages}
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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\begin{frame}[c]
\frametitle{Parser Combinators}
Atomic parsers, for example
\begin{center}
\bl{$1::rest \;\Rightarrow\; \{(1, rest)\}$}
\end{center}\bigskip
\begin{itemize}
\item you consume one or more tokens from the\\
input (stream)
\item also works for characters and strings
\end{itemize}
\end{frame}
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\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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\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:\medskip
\begin{center}
((output$_1$, output$_2$), unparsed part)
\end{center}
\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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\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}
\end{frame}
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\begin{frame}[c]
\frametitle{Types of Parsers}
\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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\begin{frame}[c]
\frametitle{Two Grammars}
Which languages are recognised by the following two grammars?
\bl{\begin{plstx}[margin=3cm]
: \meta{S} ::= \liningnums{1}\cdot\meta{S}\cdot \meta{S} | \epsilon\\
\end{plstx}}
\bl{\begin{plstx}[margin=3cm]
: \meta{U} ::= \liningnums{1}\cdot\meta{U} | \epsilon\\
\end{plstx}}
\end{frame}
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\begin{frame}[t]
\frametitle{Ambiguous Grammars}
\begin{center}
\begin{tikzpicture}
\begin{axis}[xlabel={\pcode{1}s},ylabel={time in secs},
enlargelimits=false,
xtick={0,100,...,1000},
xmax=1050,
ymax=33,
ytick={0,5,...,30},
scaled ticks=false,
axis lines=left,
width=11cm,
height=7cm,
legend entries={unambiguous,ambiguous},
legend pos=north east,
legend cell align=left,
x tick label style={font=\small,/pgf/number format/1000 sep={}}]
\addplot[blue,mark=*, mark options={fill=white}]
table {s-grammar1.data};
\only<2>{
\addplot[red,mark=triangle*, mark options={fill=white}]
table {s-grammar2.data};}
\end{axis}
\end{tikzpicture}
\end{center}
\end{frame}
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\begin{frame}[c]
\frametitle{Arithmetic Expressions}
A grammar for arithmetic expressions and numbers:
\bl{\begin{plstx}[margin=1cm]
: \meta{E} ::= \meta{E} \cdot + \cdot \meta{E}
| \meta{E} \cdot * \cdot \meta{E}
| ( \cdot \meta{E} \cdot ) | \meta{N}\\
: \meta{N} ::= \meta{N} \cdot \meta{N} |
0 | 1 | \ldots | 9\\
\end{plstx}}
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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\begin{frame}[t]
\frametitle{Numbers}
\bl{\begin{plstx}[margin=1cm]
: \meta{N} ::= \meta{N} \cdot \meta{N} |
0 | 1 | \ldots | 9\\
\end{plstx}}
A non-left-recursive, non-ambiguous grammar for numbers:
\bl{\begin{plstx}[margin=1cm]
: \meta{N} ::= 0 \cdot \meta{N} | 1 \cdot \meta{N} | \ldots |
0 | 1 | \ldots | 9\\
\end{plstx}}
\end{frame}
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\begin{frame}[c]
\frametitle{Removing Left-Recursion}
The rule for numbers is directly left-recursive:
\begin{center}
\bl{\begin{tabular}{lcl}
$\meta{N}$ & $::=$ & $\meta{N} \cdot \meta{N} \;|\; 0 \;|\; 1\;\;\;\;(\ldots)$
\end{tabular}}
\end{center}
Translate
\begin{center}
\begin{tabular}{ccc}
\bl{\begin{tabular}{lcl}
$\meta{N}$ & $::=$ & $\meta{N} \cdot \alpha$\\
& $\;|\;$ & $\beta$\\
\\
\end{tabular}}
& {\Large$\Rightarrow$} &
\bl{\begin{tabular}{lcl}
$\meta{N}$ & $::=$ & $\beta \cdot \meta{N}'$\\
$\meta{N}'$ & $::=$ & $\alpha \cdot \meta{N}'$\\
& $\;|\;$ & $\epsilon$
\end{tabular}}
\end{tabular}
\end{center}\pause
Which means in this case:
\begin{center}
\bl{\begin{tabular}{lcl}
$\meta{N}$ & $\rightarrow$ & $0 \cdot \meta{N}' \;|\; 1 \cdot \meta{N}'$\\
$\meta{N}'$ & $\rightarrow$ & $\meta{N} \cdot \meta{N}' \;|\; \epsilon$\\
\end{tabular}}
\end{center}
\end{frame}
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\begin{frame}[c]
\frametitle{Operator Precedences}
To disambiguate
\begin{center}
\bl{\begin{tabular}{lcl}
$\meta{E}$ & $::=$ & $\meta{E} \cdot + \cdot \meta{E} \;|\;\meta{E} \cdot * \cdot \meta{E} \;|\;( \cdot \meta{E} \cdot ) \;|\;\meta{N}$ \\
\end{tabular}}
\end{center}
Decide on how many precedence levels, say\medskip\\
highest for \bl{$()$}, medium for \bl{*}, lowest for \bl{+}
\begin{center}
\bl{\begin{tabular}{lcl}
$\meta{E}_{low}$ & $::=$ & $\meta{E}_{med} \cdot + \cdot \meta{E}_{low} \;|\; \meta{E}_{med}$ \\
$\meta{E}_{med}$ & $::=$ & $\meta{E}_{hi} \cdot * \cdot \meta{E}_{med} \;|\; \meta{E}_{hi}$\\
$\meta{E}_{hi}$ & $::=$ & $( \cdot \meta{E}_{low} \cdot ) \;|\;\meta{N}$ \\
\end{tabular}}
\end{center}\pause
\small What happens with \bl{$1 + 3 + 4$}?
\end{frame}
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\begin{frame}[c]
\frametitle{Chomsky Normal Form}
All rules must be of the form
\begin{center}
\bl{$\meta{A} ::= a$}
\end{center}
or
\begin{center}
\bl{$\meta{A} ::= \meta{B}\cdot \meta{C}$}
\end{center}
No rule can contain \bl{$\epsilon$}.
\end{frame}
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\begin{frame}[c]
\frametitle{\begin{tabular}{c}$\epsilon$-Removal\end{tabular}}
\begin{enumerate}
\item If \bl{$A::= \alpha \cdot B \cdot \beta$} and \bl{$B ::= \epsilon$} are in the grammar,
then add \bl{$A::= \alpha \cdot \beta$} (iterate if necessary).
\item Throw out all \bl{$B ::= \epsilon$}.
\end{enumerate}
\small
\begin{center}
\begin{tabular}{ccc}
\bl{\begin{tabular}{l@{\hspace{1mm}}c@{\hspace{1mm}}l}
$N$ & $::=$ & $0 \cdot N' \;|\; 1\cdot N'$\\
$N'$ & $::=$ & $N \cdot N'\;|\;\epsilon$\\
\\
\\
\\
\\
\\
\end{tabular}} &
\bl{\begin{tabular}{l@{\hspace{1mm}}c@{\hspace{1mm}}l}
\\
$N$ & $::=$ & $0 \cdot N' \;|\; 1\cdot N'\;|\;0\;|\;1$\\
$N'$ & $::=$ & $N \cdot N'\;|\;N\;|\;\epsilon$\\
\\
$N$ & $::=$ & $0 \cdot N' \;|\; 1\cdot N'\;|\;0\;|\;1$\\
$N'$ & $::=$ & $N \cdot N'\;|\;N$\\
\end{tabular}}
\end{tabular}
\end{center}
\pause\normalsize
\begin{center}
\bl{\begin{tabular}{l@{\hspace{1mm}}c@{\hspace{1mm}}l}
$N$ & $::=$ & $0 \cdot N\;|\; 1\cdot N\;|\;0\;|\;1$\\
\end{tabular}}
\end{center}
\end{frame}
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\begin{frame}[c]
\frametitle{CYK Algorithm}
If grammar is in Chomsky normalform \ldots
\begin{center}
\bl{\begin{tabular}{@ {}lcl@ {}}
$\meta{S}$ & $::=$ & $\meta{N}\cdot \meta{P}$ \\
$\meta{P}$ & $::=$ & $\meta{V}\cdot \meta{N}$ \\
$\meta{N}$ & $::=$ & $\meta{N}\cdot \meta{N}$ \\
$\meta{N}$ & $::=$ & $\texttt{students} \;|\; \texttt{Jeff} \;|\; \texttt{geometry} \;|\; \texttt{trains} $ \\
$\meta{V}$ & $::=$ & $\texttt{trains}$
\end{tabular}}
\end{center}
\bl{\texttt{Jeff trains geometry students}}
\end{frame}
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\begin{frame}[c]
\frametitle{CYK Algorithm}
\begin{itemize}
\item fastest possible algorithm for recognition problem
\item runtime is \bl{$O(n^3)$}\bigskip
\item grammars need to be transformed into CNF
\end{itemize}
\end{frame}
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\begin{frame}[c]
\frametitle{The Goal of this Course}
\mbox{}\\[-26mm]\mbox{}
\begin{center}
\begin{tikzpicture}[scale=1,
node/.style={
rectangle,rounded corners=3mm,
very thick,draw=black!50,
minimum height=18mm, minimum width=20mm,
top color=white,bottom color=black!20}]
\node at (3.05, 1.8) {\Large\bf Write a Compiler};
\node (0) at (-2.3,0) {};
\node (A) at (0,0) [node] {};
\node [below right] at (A.north west) {lexer};
\node (B) at (3,0) [node] {};
\node [below right=1mm] at (B.north west)
{\mbox{}\hspace{-1mm}parser};
\node (C) at (6,0) [node] {};
\node [below right] at (C.north west)
{\mbox{}\hspace{-1mm}code gen};
\node (1) at (8.4,0) {};
\draw [->,line width=4mm] (0) -- (A);
\draw [->,line width=4mm] (A) -- (B);
\draw [->,line width=4mm] (B) -- (C);
\draw [->,line width=4mm] (C) -- (1);
\end{tikzpicture}
\end{center}
We have a lexer and a parser\ldots
\end{frame}
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\begin{frame}[c]
\begin{center}
\bl{\begin{tabular}{@{}lcl@{}}
\meta{Stmt} & $::=$ & $\texttt{skip}$\\
& $|$ & \textit{Id}\;\texttt{:=}\;\meta{AExp}\\
& $|$ & \texttt{if}\; \meta{BExp} \;\texttt{then}\; \meta{Block} \;\texttt{else}\; \meta{Block}\\
& $|$ & \texttt{while}\; \meta{BExp} \;\texttt{do}\; \meta{Block}\\
& $|$ & \texttt{read}\;\textit{Id}\\
& $|$ & \texttt{write}\;\textit{Id}\\
& $|$ & \texttt{write}\;\textit{String}\medskip\\
\meta{Stmts} & $::=$ & \meta{Stmt} \;\texttt{;}\; \meta{Stmts}\\
& $|$ & \meta{Stmt}\medskip\\
\meta{Block} & $::=$ & \texttt{\{}\,\meta{Stmts}\,\texttt{\}}\\
& $|$ & \meta{Stmt}\medskip\\
\meta{AExp} & $::=$ & \ldots\\
\meta{BExp} & $::=$ & \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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\begin{frame}[c]
\frametitle{An Interpreter}
\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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\begin{frame}[c]
\frametitle{An Interpreter}
\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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\begin{frame}[c]
\frametitle{An Interpreter (2)}
\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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\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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%\begin{frame}[c]
% \frametitle{Coursework: MkEps}
%
%\begin{center}
%\begin{tabular}{@ {}l@ {\hspace{2mm}}c@ {\hspace{2mm}}l@ {}}
% \bl{$mkeps([c_1 c_2 \ldots c_n])$} & \bl{$\dn$} & $\bl{undefined}$\\
% \bl{$mkeps(r^*)$} & \bl{$\dn$} & $\bl{Stars\,[]}$\\
% \bl{$mkeps(r^{\{n\}})$} & \bl{$\dn$} & $\bl{Stars\,(mkeps(r))^n}$\\
% \bl{$mkeps(r^{\{n..\}})$} & \bl{$\dn$} & $\bl{Stars\,(mkeps(r))^n}$\\
% \bl{$mkeps(r^{\{..n\}})$} & \bl{$\dn$} & $\bl{Stars\,[]}$\\
% \bl{$mkeps(r^{\{n..m\}})$} & \bl{$\dn$} & $\bl{Stars\,(mkeps(r))^n}$\medskip\\
%
% \bl{$mkeps(r^+)$} & \bl{$\dn$} & \bl{$mkeps(r^{\{1..\}})$}\\
% \bl{$mkeps(r^?)$} & \bl{$\dn$} & \bl{$mkeps(r^{\{..1\}})$}\\
%\end{tabular}
%\end{center}
%
%\end{frame}
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%
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%\begin{frame}[c]
% \frametitle{Coursework: Inj}
%
%\begin{center}
%\begin{tabular}{@ {}l@ {\hspace{2mm}}c@ {\hspace{2mm}}l@ {}}
% \bl{$inj([c_1 c_2 \ldots c_n])\,c\,Empty$} & \bl{$\dn$} & $\bl{Chr\,c}$\\
% \bl{$inj(r^*)\,c\;(Seq\,v\,(Stars\,vs))$} & \bl{$\dn$} & $\bl{Stars\,(inj\,r\,c\,v::vs)}$\\
% \bl{$inj(r^{\{n\}})\,c\;(Seq\,v\,(Stars\,vs))$} & \bl{$\dn$} & $\bl{Stars\,(inj\,r\,c\,v::vs)}$\\
% \bl{$inj(r^{\{n..\}})\,c\;(Seq\,v\,(Stars\,vs))$} & \bl{$\dn$} & $\bl{Stars\,(inj\,r\,c\,v::vs)}$\\
% \bl{$inj(r^{\{..n\}})\,c\;(Seq\,v\,(Stars\,vs))$} & \bl{$\dn$} & $\bl{Stars\,(inj\,r\,c\,v::vs)}$\\
% \bl{$inj(r^{\{n..m\}})\,c\;(Seq\,v\,(Stars\,vs))$} & \bl{$\dn$} & $\bl{Stars\,(inj\,r\,c\,v::vs)}$\medskip\\
%
% \bl{$inj(r^+)\,c\,v$} & \bl{$\dn$} & \bl{$inj(r^{\{1..\}})\,c\,v$}\\
% \bl{$inj(r^?)\,c\,v$} & \bl{$\dn$} & \bl{$inj(r^{\{..1\}})\,c\,v$}\\
%\end{tabular}
%\end{center}
%
%\end{frame}
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\end{document}
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