\documentclass{beamer}

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\usepackage{proof}

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\newcommand{\smath}[1]{\textcolor{darkblue}{\ensuremath{#1}}}

\newcommand{\dn}{\stackrel{\mbox{\scriptsize def}}{=}}

\newcommand{\Zero}{{\bf 0}}

\newcommand{\One}{{\bf 1}}

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       \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}

	\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}

	\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}

	\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] {$v_4$};

		\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);

	\end{tikzpicture}

	\end{center}

	\end{frame}	

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

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

	\begin{frame}

	\frametitle{Problems}

	\begin{itemize}

	\item Sulzmann: \ldots Let's assume $v$ is not a $POSIX$ value, 

	\item Exists ?

	\begin{center}

	$L(r) \not= \varnothing \;\Rightarrow\; \exists v.\;POSIX(v, r)$

	\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}

	$|v_1| @ |v_2| = |v_1'| @ |v_2'|$

	\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 

	\begin{center}

	\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}

	$s \in r \to v$