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+\documentclass{beamer}
+\usepackage{tikz}
+\usepackage[english]{babel}
+\usepackage{proof}
+\usetheme{Luebeck}
+\usetikzlibrary{positioning}
+\usetikzlibrary{decorations.pathreplacing}
+\definecolor{darkblue}{rgb}{0,0,.803}
+\definecolor{cream}{rgb}{1,1,.8}
+
+\newcommand{\smath}[1]{\textcolor{darkblue}{\ensuremath{#1}}}
+\newcommand{\dn}{\stackrel{\mbox{\scriptsize def}}{=}}
+\newcommand{\Zero}{{\bf 0}}
+\newcommand{\One}{{\bf 1}}
+
+\newenvironment{bubble}[1][]{%
+\addtolength{\leftmargini}{4mm}%
+\begin{tikzpicture}[baseline=(current bounding box.north)]%
+\draw (0,0) node[inner sep=2mm,fill=cream,ultra thick,draw=red,rounded corners=2mm]%
+\bgroup\begin{minipage}{#1}\raggedright{}}
+{\end{minipage}\egroup;%
+\end{tikzpicture}\bigskip}
+
+\newcommand\grid[1]{%
+\begin{tikzpicture}[baseline=(char.base)]
+ \path[use as bounding box]
+ (0,0) rectangle (1em,1em);
+ \draw[red!50, fill=red!20]
+ (0,0) rectangle (1em,1em);
+ \node[inner sep=1pt,anchor=base west]
+ (char) at (0em,\gridraiseamount) {#1};
+\end{tikzpicture}}
+\newcommand\gridraiseamount{0.12em}
+
+\makeatletter
+\newcommand\Grid[1]{%
+ \@tfor\z:=#1\do{\grid{\z}}}
+\makeatother
+
+\newcommand\Vspace[1][.3em]{%
+ \mbox{\kern.06em\vrule height.3ex}%
+ \vbox{\hrule width#1}%
+ \hbox{\vrule height.3ex}}
+
+\def\VS{\Vspace[0.6em]}
+
+
+\title[POSIX Lexing with Derivatives of Regexes]
+ {\bf POSIX Lexing with\\
+ \bf Derivatives of Regular Expressions\\
+ \bf (Proof Pearl)}
+\author{Fahad Ausaf, Roy Dyckhoff and Christian Urban}
+\date{King's College London, University of St Andrews}
+
+\begin{document}
+\maketitle
+
+
+ %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+ \begin{frame}
+ \frametitle{Brzozowski's Derivatives of Regular Expressions}
+
+ Idea: If \smath{r} matches the string \smath{c\!::\!s},
+ what is a regular expression that matches just \smath{s}? \\
+
+ \begin{center}
+ \begin{tabular}{l@{\hspace{5mm}}lcl}
+ chars:
+ &\smath{\Zero \backslash c} & \smath{\dn} & \smath{\Zero}\\
+ &\smath{\One \backslash c} & \smath{\dn} & \smath{\Zero}\\
+ &\smath{d \backslash c} & \smath{\dn} &
+ \smath{\textit{if}\;d = c\;\textit{then}\;\One\;\textit{else}\;\Zero}\\
+ &\smath{r_1 + r_2 \backslash c} & \smath{\dn} &
+ \smath{r_1 \backslash c \,+\, r_2 \backslash c}\\
+ &\smath{r_1 \cdot r_2 \backslash c} & \smath{\dn} &
+ \smath{\textit{if}\;\textit{nullable}\;r_1}\\
+ && & \smath{\textit{then}\;r_1\backslash c \cdot r_2 \,+\, r_2\backslash c
+ \;\textit{else}\;r_1\backslash c \cdot r_2}\\
+ &\smath{r^* \backslash c} & \smath{\dn} &
+ \smath{r\backslash c \,\cdot\, r^*}\bigskip\\
+
+ strings:
+ &\smath{r\backslash []} & \smath{\dn} & \smath{r}\\
+ &\smath{r\backslash c\!::\!s} & \smath{\dn} &
+ \smath{(r\backslash c)\backslash s}\\
+ \end{tabular}
+ \end{center}
+ \end{frame}
+
+ %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+
+ \begin{frame}
+ \frametitle{Brzozowski's Matcher}
+
+ Does \smath{r_1} match string \smath{abc}?
+
+ \begin{center}
+ \begin{tikzpicture}[scale=2,node distance=1.3cm,every node/.style={minimum size=8mm}]
+ \node (r1) {\smath{r_1}};
+ \node (r2) [right=of r1] {\smath{r_2}};
+ \draw[->,line width=1mm] (r1) -- (r2) node[above,midway] {\smath{\_\backslash a}};
+ \node (r3) [right=of r2] {\smath{r_3}};
+ \draw[->,line width=1mm] (r2) -- (r3) node[above,midway] {\smath{\_\backslash b}};
+ \node (r4) [right=of r3] {\smath{r_4}};
+ \draw[->,line width=1mm] (r3) -- (r4) node[above,midway] {\smath{\_\backslash c}};
+ \draw (r4) node[anchor=west] {\;\raisebox{3mm}{\smath{\;\;\textit{nullable}?}}};
+ \end{tikzpicture}
+ \end{center}\pause
+
+ It leads to an elegant functional program:
+
+ \begin{center}
+ \smath{\textit{matches}\,(r, s) \dn \textit{nullable}\,(r\backslash s)}
+ \end{center}\pause
+
+ It is an easy exercise to formally prove (e.g.~Coq, HOL, Isabelle):
+
+ \begin{center}
+ \smath{\textit{matches}\,(r, s)} if and only if
+ \smath{s \in L(r)}
+ \end{center}\pause
+
+ {\bf But Brzozowski's matcher gives only a yes/no-answer.}
+ \end{frame}
+
+ %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+
+ \begin{frame}
+ \frametitle{Sulzmann and Lu's Matcher}
+
+ Sulzmann and Lu added a second phase in order to answer
+ \alert{\textbf{how}} the regular expression matched the string.
+
+ \begin{center}\small
+ \begin{tikzpicture}[scale=1,node distance=0.8cm,every node/.style={minimum size=7mm}]
+ \node (r1) {\smath{r_1}};
+ \node (r2) [right=of r1] {\smath{r_2}};
+ \draw[->,line width=1mm] (r1) -- (r2) node[above,midway] {\smath{\_\backslash a}};
+ \node (r3) [right=of r2] {\smath{r_3}};
+ \draw[->,line width=1mm] (r2) -- (r3) node[above,midway] {\smath{\_\backslash b}};
+ \node (r4) [right=of r3] {\smath{r_4}};
+ \draw[->,line width=1mm] (r3) -- (r4) node[above,midway] {\smath{\_\backslash c}};
+ \draw (r4) node[anchor=west] {\;\raisebox{3mm}{\smath{\;\;nullable?}}};
+ \node (v4) [below=of r4] {\smath{v_4}};
+ \draw[->,line width=1mm] (r4) -- (v4);
+ \node (v3) [left=of v4] {\smath{v_3}};
+ \draw[->,line width=1mm] (v4) -- (v3) node[below,midway] {\smath{inj\,c}};
+ \node (v2) [left=of v3] {\smath{v_2}};
+ \draw[->,line width=1mm] (v3) -- (v2) node[below,midway] {\smath{inj\,b}};
+ \node (v1) [left=of v2] {\smath{v_1}};
+ \draw[->,line width=1mm] (v2) -- (v1) node[below,midway] {\smath{inj\,a}};
+ \draw (r4) node[anchor=north west] {\;\raisebox{-8mm}{\smath{mkeps}}};
+
+ \draw [decorate,decoration={brace,amplitude=10pt,raise=8mm},
+ line width=1.5mm]
+ (r1) -- (r4) node [black,midway,above, yshift=12mm]
+ {\large\bf first phase};
+ \draw [decorate,decoration={brace,amplitude=10pt,raise=8mm},
+ line width=1.5mm]
+ (v4) -- (v1) node [black,midway,below, yshift=-12mm]
+ {\large\bf second phase};
+ %% first phase
+ \draw[line width=14mm, rounded corners, opacity=0.1,
+ cap=round,join=round,color=yellow!30]
+ (r1.center) -- (r4.center);
+ %% second phase
+ \draw[line width=14.1mm, rounded corners, opacity=0.2,
+ cap=round,join=round,draw=black, fill=white]
+ (r4) -- (v4.center) -- (v1.center);
+ \draw[line width=14mm, rounded corners, opacity=0.2,
+ cap=round,join=round,color=yellow!30]
+ (r4) -- (v4.center) -- (v1.center);
+ \end{tikzpicture}
+ \end{center}
+
+ There are several possible answers for
+ \alert{\textbf{how}}: POSIX, GREEDY, \ldots
+
+ \end{frame}
+
+ %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+
+ \begin{frame}{Regular Expressions and Values}
+ Regular expressions and their corresponding values (for how a
+ regular expression matched a string):\bigskip
+
+ \begin{columns}[c] % the "c" option specifies center vertical alignment
+ \column{.4\textwidth} % column designated by a command
+ \begin{tabular}{ l l l }
+ \smath{r} & ::= & \smath{\Zero} \\
+ & $\mid$ & \smath{\One} \\
+ & $\mid$ & \smath{c} \\
+ & $\mid$ & \smath{r_1 \cdot r_2} \\
+ & $\mid$ & \smath{r_1 + r_2} \\ \\
+ & $\mid$ & \smath{r^*} \\
+ \end{tabular}
+ \column{.4\textwidth}
+ \begin{tabular}{ l l l }
+
+ \smath{v} & ::= & \\
+ & $\mid$ & \smath{Empty} \\
+ & $\mid$ & \smath{Char(c)} \\
+ & $\mid$ & \smath{Seq(v_1\cdot v_2)} \\
+ & $\mid$ & \smath{Left(v)} \\
+ & $\mid$ & \smath{Right(v)} \\
+ & $\mid$ & \smath{[v_1,...,v_n]}
+ \end{tabular}
+ \end{columns}
+ \end{frame}
+
+ %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+
+
+
+ %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+
+ \begin{frame}
+ \frametitle{POSIX Matching (needed for Lexing)}
+
+
+ \begin{bubble}[10cm]
+ {\bf Longest Match Rule:} The longest
+ initial substring matched by any regular expression is taken
+ as the next token.
+ \end{bubble}
+
+ \begin{bubble}[10cm]
+ {\bf Rule Priority:} For a particular longest initial substring,
+ the first regular expression that can match determines the
+ token.
+ \end{bubble}
+
+ For example: \smath{r_{keywords} + r_{identifiers}}:\bigskip
+
+ \begin{itemize}
+ \item \smath{\texttt{\Grid{iffoo\VS bla}}}
+
+ \item \smath{\texttt{\Grid{if\VS bla}}}
+ \end{itemize}
+
+ \end{frame}
+
+
+
+ %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+
+ \begin{frame}
+ \frametitle{Problems with POSIX}
+
+ Grathwohl, Henglein and Rasmussen wrote:
+
+ \begin{bubble}[10cm]
+ \it ``The POSIX strategy is more complicated than the greedy because
+ of the dependence on information
+ about the length of matched strings in the various subexpressions.''
+ \end{bubble}\bigskip
+
+ Also Kuklewicz maintains a unit-test repository for POSIX
+ matching, which indicates that most POSIX mathcers are buggy.
+
+ \begin{center}
+ \url{http://www.haskell.org/haskellwiki/Regex_Posix}
+ \end{center}
+
+ \end{frame}
+
+ %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+
+ \begin{frame}
+ \frametitle{``Correctness'' by Sulzmann and Lu}
+ \begin{itemize}
+ \item Sulzmann \& Lu's idea is to order all possible
+ answer such that they can prove the correct answer is
+ the maximum
+ \item The idea is taken from a GREEDY algorithm (and it
+ works there)\bigskip\pause
+
+ \item {\bf But} we made no progress in formalising
+ Sulzmann \& Lu's idea, because
+
+ \begin{itemize}
+ \item transitivity, existence of maxima etc all fail to
+ turn into real proofs
+ \item the reason: the ordering works only if ....
+ \item though we did find mistakes:
+\begin{center}
+ \small
+ ``How could I miss this? Well, I was rather careless when
+ stating this Lemma :)\smallskip
+
+ Great example how formal machine checked proofs (and
+ proof assistants) can help to spot flawed reasoning steps.''
+
+ \end{center}\pause
+
+ \begin{center}
+ \small
+ ``Well, I don't think there's any flaw. The issue is how to
+ come up with a mechanical proof. In my world mathematical
+ proof $=$ mechanical proof doesn't necessarily hold.''
+
+ \end{center}\pause
+ \end{itemize}
+
+ \end{itemize}
+ \end{frame}
+
+
+ %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+
+ \begin{frame}
+ \frametitle{Sulzmann and Lu Matcher}
+
+ We want to match the string $abc$ using $r_1$\\
+
+ \begin{center}
+ \begin{tikzpicture}[scale=2,node distance=1.3cm,every node/.style={minimum size=8mm}]
+ \node (r1) {$r_1$};
+ \node (r2) [right=of r1] {$r_2$};
+ \draw[->,line width=1mm] (r1) -- (r2) node[above,midway] {$der\,a$};\pause
+ \node (r3) [right=of r2] {$r_3$};
+ \draw[->,line width=1mm] (r2) -- (r3) node[above,midway] {$der\,b$};\pause
+ \node (r4) [right=of r3] {$r_4$};
+ \draw[->,line width=1mm] (r3) -- (r4) node[above,midway] {$der\,c$};\pause
+ \draw (r4) node[anchor=west] {\;\raisebox{3mm}{$\;\;nullable?$}};\pause
+ \node (v4) [below=of r4] {$v_4$};
+ \draw[->,line width=1mm] (r4) -- (v4);\pause
+ \node (v3) [left=of v4] {$v_3$};
+ \draw[->,line width=1mm] (v4) -- (v3) node[below,midway] {$inj\,c$};\pause
+ \node (v2) [left=of v3] {$v_2$};
+ \draw[->,line width=1mm] (v3) -- (v2) node[below,midway] {$inj\,b$};\pause
+ \node (v1) [left=of v2] {$v_1$};
+ \draw[->,line width=1mm] (v2) -- (v1) node[below,midway] {$inj\,a$};\pause
+ \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}{$mkeps$}};
+ \end{tikzpicture}
+ \end{center}
+
+ \end{frame}
+
+ %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+
+
+ %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+
+ \begin{frame}
+ \frametitle{Problems}
+ \begin{itemize}
+
+ \item Sulzmann: \ldots Let's assume $v$ is not a $POSIX$ value,
+ then there must be another one \ldots contradiction.\pause
+
+ \item Exists ?
+ \begin{center}
+ $L(r) \not= \varnothing \;\Rightarrow\; \exists v.\;POSIX(v, r)$
+ \end{center}\pause
+
+ \item In the sequence case
+ $Seq(v_1,v_2)\succ_{r_1\cdot r_2} Seq(v_1', v_2')$,
+ the induction hypotheses require
+ $|v_1| = |v_1'|$ and $|v_2| = |v_2'|$, but you only know
+
+ \begin{center}
+ $|v_1| @ |v_2| = |v_1'| @ |v_2'|$
+ \end{center}\pause
+
+ \item Although one begins with the assumption that the two
+ values have the same flattening, this cannot be maintained
+ as one descends into the induction (alternative, sequence)
+
+ \end{itemize}
+ \end{frame}
+
+ %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+
+ \begin{frame}
+ \frametitle{Our Solution}
+
+ \begin{itemize}
+ \item A direct definition of what a POSIX value is, using the
+ relation \smath{s \in r \to v} (our specification)\bigskip
+
+ \begin{center}
+ \smath{\infer{[] \in \epsilon \to Empty}{}}\hspace{15mm}
+ \smath{\infer{[c] \in c \to Char(c)}{}}\bigskip\medskip
+
+ \smath{\infer{s \in r_1 + r_2 \to Left(v)}
+ {s \in r_1 \to v}}\hspace{10mm}
+ \smath{\infer{s \in r_1 + r_2 \to Right(v)}
+ {s \in r_2 \to v & s \not\in L(r_1)}}\bigskip\medskip
+
+ \smath{\infer{s_1 @ s_2 \in r_1 \cdot r_2 \to Seq(v_1, v_2)}
+ {\small\begin{array}{l}
+ s_1 \in r_1 \to v_1 \\
+ s_2 \in r_2 \to v_2 \\
+ \neg(\exists s_3\,s_4.\; s_3 \not= []
+ \wedge s_3 @ s_4 = s_2 \wedge
+ s_1 @ s_3 \in L(r_1) \wedge
+ s_4 \in L(r_2))
+ \end{array}}}
+
+ {\ldots}
+ \end{center}
+
+ \end{itemize}
+ \end{frame}
+
+ %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+ \begin{frame}[c]
+ \frametitle{Properties}
+
+ It is almost trival to prove:
+
+ \begin{itemize}
+ \item Uniqueness
+ \begin{center}
+ If $s \in r \to v_1$ and $s \in r \to v_2$ then
+ $v_1 = v_2$
+ \end{center}\bigskip
+
+ \item Correctness
+ \begin{center}
+ $lexer(r, s) = v$ if and only if $s \in r \to v$
+ \end{center}
+ \end{itemize}\bigskip\bigskip\pause
+
+
+ You can now start to implement optimisations and derive
+ correctness proofs for them. But we still do not know whether
+
+ \begin{center}
+ $s \in r \to v$
+ \end{center}
+
+ is a POSIX value according to Sulzmann \& Lu's definition
+ (biggest value for $s$ and $r$)
+ \end{frame}
+
+
+\end{document}