\documentclass[a4paper,UKenglish]{lipics}
\usepackage{graphic}
\usepackage{data}
\usepackage{tikz-cd}
\usepackage{tikz}
\usetikzlibrary{graphs}
\usetikzlibrary{graphdrawing}
\usegdlibrary{trees}
%\usepackage{algorithm}
\usepackage{amsmath}
\usepackage{xcolor}
\usepackage[noend]{algpseudocode}
\usepackage{enumitem}
\usepackage{nccmath}
\usepackage{soul}
\definecolor{darkblue}{rgb}{0,0,0.6}
\hypersetup{colorlinks=true,allcolors=darkblue}
\newcommand{\comment}[1]%
{{\color{red}$\Rightarrow$}\marginpar{\raggedright\small{\bf\color{red}#1}}}
% \documentclass{article}
%\usepackage[utf8]{inputenc}
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% \usepackage{amsthm}
%\usepackage{hyperref}
% \usepackage[margin=0.5in]{geometry}
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\title{POSIX Regular Expression Matching and Lexing}
\author{Chengsong Tan}
\affil{King's College London\\
London, UK\\
\texttt{chengsong.tan@kcl.ac.uk}}
\authorrunning{Chengsong Tan}
\Copyright{Chengsong Tan}
\newcommand{\dn}{\stackrel{\mbox{\scriptsize def}}{=}}%
\newcommand{\ZERO}{\mbox{\bf 0}}
\newcommand{\ONE}{\mbox{\bf 1}}
\def\erase{\textit{erase}}
\def\bders{\textit{bders}}
\def\lexer{\mathit{lexer}}
\def\blexer{\textit{blexer}}
\def\fuse{\textit{fuse}}
\def\flatten{\textit{flatten}}
\def\map{\textit{map}}
\def\blexers{\mathit{blexer\_simp}}
\def\simp{\mathit{simp}}
\def\mkeps{\mathit{mkeps}}
\def\bmkeps{\textit{bmkeps}}
\def\inj{\mathit{inj}}
\def\Empty{\mathit{Empty}}
\def\Left{\mathit{Left}}
\def\Right{\mathit{Right}}
\def\Stars{\mathit{Stars}}
\def\Char{\mathit{Char}}
\def\Seq{\mathit{Seq}}
\def\Der{\mathit{Der}}
\def\nullable{\mathit{nullable}}
\def\Z{\mathit{Z}}
\def\S{\mathit{S}}
\def\flex{\textit{flex}}
\def\rup{r^\uparrow}
\def\retrieve{\textit{retrieve}}
\def\AALTS{\textit{AALTS}}
\def\AONE{\textit{AONE}}
%\theoremstyle{theorem}
%\newtheorem{theorem}{Theorem}
%\theoremstyle{lemma}
%\newtheorem{lemma}{Lemma}
%\newcommand{\lemmaautorefname}{Lemma}
%\theoremstyle{definition}
%\newtheorem{definition}{Definition}
\algnewcommand\algorithmicswitch{\textbf{switch}}
\algnewcommand\algorithmiccase{\textbf{case}}
\algnewcommand\algorithmicassert{\texttt{assert}}
\algnewcommand\Assert[1]{\State \algorithmicassert(#1)}%
% New "environments"
\algdef{SE}[SWITCH]{Switch}{EndSwitch}[1]{\algorithmicswitch\ #1\ \algorithmicdo}{\algorithmicend\ \algorithmicswitch}%
\algdef{SE}[CASE]{Case}{EndCase}[1]{\algorithmiccase\ #1}{\algorithmicend\ \algorithmiccase}%
\algtext*{EndSwitch}%
\algtext*{EndCase}%
\begin{document}
\maketitle
\begin{abstract}
Brzozowski introduced in 1964 a beautifully simple algorithm for
regular expression matching based on the notion of derivatives of
regular expressions. In 2014, Sulzmann and Lu extended this
algorithm to not just give a YES/NO answer for whether or not a
regular expression matches a string, but in case it does also
answers with \emph{how} it matches the string. This is important for
applications such as lexing (tokenising a string). The problem is to
make the algorithm by Sulzmann and Lu fast on all inputs without
breaking its correctness. Being fast depends on a complete set of
simplification rules, some of which
have been put forward by Sulzmann and Lu. We have extended their
rules in order to obtain a tight bound on the size of regular expressions.
We have tested these extended rules, but have not
formally established their correctness. We have also not yet looked
at extended regular expressions, such as bounded repetitions,
negation and back-references.
\end{abstract}
\section{Introduction}
%Regular expressions' derivatives, which have received
%renewed interest in the new millenium, is a beautiful....
While we believe derivatives of regular expressions, written
$r\backslash s$, are a beautiful concept (in terms of ease of
implementing them in functional programming languages and in terms of
reasoning about them formally), they have one major drawback: every
derivative step can make regular expressions grow drastically in
size. This in turn has negative effect on the runtime of the
corresponding lexing algorithms. Consider for example the regular
expression $(a+aa)^*$ and the short string $aaaaaaaaaaaa$. The
corresponding derivative contains already 8668 nodes where we assume
the derivative is given as a tree. The reason for the poor runtime of
the derivative-based lexing algorithms is that they need to traverse
such trees over and over again. The solution is to find a complete set
of simplification rules that keep the sizes of derivatives uniformly
small.
This has been partially addressed by the function $\blexer_{simp}$,
which after the simplification the $(a+aa)^*$ example's 8000 nodes will be
reduced to just 6 and stays constant in each derivative step.
The part that still needs more work is the correctness proof of this
function, namely,
\begin{equation}\label{mainthm}
\blexers \; r \; s = \blexer \;r\;s
\end{equation}
\noindent
and this is what this report is mainly about. A condensed
version of the last report will be provided in the next section
to help the reader understand the report, and the attempts
on the problem will follow.
\section{Recapitulation of Concepts From the Last Report}
\subsection*{Regular Expressions and Derivatives}
Suppose (basic) regular expressions are given by the following grammar:
\[ r ::= \ZERO \mid \ONE
\mid c
\mid r_1 \cdot r_2
\mid r_1 + r_2
\mid r^*
\]
\noindent
The ingenious contribution of Brzozowski is the notion of \emph{derivatives} of
regular expressions, written~$\_ \backslash \_$. It uses the auxiliary notion of
$\nullable$ defined below.
\begin{center}
\begin{tabular}{lcl}
$\nullable(\ZERO)$ & $\dn$ & $\mathit{false}$ \\
$\nullable(\ONE)$ & $\dn$ & $\mathit{true}$ \\
$\nullable(c)$ & $\dn$ & $\mathit{false}$ \\
$\nullable(r_1 + r_2)$ & $\dn$ & $\nullable(r_1) \vee \nullable(r_2)$ \\
$\nullable(r_1\cdot r_2)$ & $\dn$ & $\nullable(r_1) \wedge \nullable(r_2)$ \\
$\nullable(r^*)$ & $\dn$ & $\mathit{true}$ \\
\end{tabular}
\end{center}
\begin{center}
\begin{tabular}{lcl}
$\ZERO \backslash c$ & $\dn$ & $\ZERO$\\
$\ONE \backslash c$ & $\dn$ & $\ZERO$\\
$d \backslash c$ & $\dn$ &
$\mathit{if} \;c = d\;\mathit{then}\;\ONE\;\mathit{else}\;\ZERO$\\
$(r_1 + r_2)\backslash c$ & $\dn$ & $r_1 \backslash c \,+\, r_2 \backslash c$\\
$(r_1 \cdot r_2)\backslash c$ & $\dn$ & $\mathit{if} \, nullable(r_1)$\\
& & $\mathit{then}\;(r_1\backslash c) \cdot r_2 \,+\, r_2\backslash c$\\
& & $\mathit{else}\;(r_1\backslash c) \cdot r_2$\\
$(r^*)\backslash c$ & $\dn$ & $(r\backslash c) \cdot r^*$\\
\end{tabular}
\end{center}
\noindent
And defines how a regular expression evolves into
a new regular expression after all the string it contains
is chopped off a certain head character $c$.
The main property of the derivative operation is that
\begin{center}
$c\!::\!s \in L(r)$ holds
if and only if $s \in L(r\backslash c)$.
\end{center}
\noindent
We can generalise the derivative operation shown above for single characters
to strings as follows:
\begin{center}
\begin{tabular}{lcl}
$r \backslash (c\!::\!s) $ & $\dn$ & $(r \backslash c) \backslash s$ \\
$r \backslash [\,] $ & $\dn$ & $r$
\end{tabular}
\end{center}
\noindent
and then define Brzozowski's regular-expression matching algorithm as:
\[
match\;s\;r \;\dn\; nullable(r\backslash s)
\]
\noindent
Assuming the a string is given as a sequence of characters, say $c_0c_1..c_n$,
this algorithm presented graphically is as follows:
\begin{equation}\label{graph:*}
\begin{tikzcd}
r_0 \arrow[r, "\backslash c_0"] & r_1 \arrow[r, "\backslash c_1"] & r_2 \arrow[r, dashed] & r_n \arrow[r,"\textit{nullable}?"] & \;\textrm{YES}/\textrm{NO}
\end{tikzcd}
\end{equation}
\noindent
where we start with a regular expression $r_0$, build successive
derivatives until we exhaust the string and then use \textit{nullable}
to test whether the result can match the empty string. It can be
relatively easily shown that this matcher is correct (that is given
an $s = c_0...c_{n-1}$ and an $r_0$, it generates YES if and only if $s \in L(r_0)$).
\subsection*{Values and the Lexing Algorithm by Sulzmann and Lu}
One limitation of Brzozowski's algorithm is that it only produces a
YES/NO answer for whether a string is being matched by a regular
expression. Sulzmann and Lu~\cite{Sulzmann2014} extended this algorithm
to allow generation of an actual matching, called a \emph{value} or
sometimes also \emph{lexical value}. These values and regular
expressions correspond to each other as illustrated in the following
table:
\begin{center}
\begin{tabular}{c@{\hspace{20mm}}c}
\begin{tabular}{@{}rrl@{}}
\multicolumn{3}{@{}l}{\textbf{Regular Expressions}}\medskip\\
$r$ & $::=$ & $\ZERO$\\
& $\mid$ & $\ONE$ \\
& $\mid$ & $c$ \\
& $\mid$ & $r_1 \cdot r_2$\\
& $\mid$ & $r_1 + r_2$ \\
\\
& $\mid$ & $r^*$ \\
\end{tabular}
&
\begin{tabular}{@{\hspace{0mm}}rrl@{}}
\multicolumn{3}{@{}l}{\textbf{Values}}\medskip\\
$v$ & $::=$ & \\
& & $\Empty$ \\
& $\mid$ & $\Char(c)$ \\
& $\mid$ & $\Seq\,v_1\, v_2$\\
& $\mid$ & $\Left(v)$ \\
& $\mid$ & $\Right(v)$ \\
& $\mid$ & $\Stars\,[v_1,\ldots\,v_n]$ \\
\end{tabular}
\end{tabular}
\end{center}
\noindent
The contribution of Sulzmann and Lu is an extension of Brzozowski's
algorithm by a second phase (the first phase being building successive
derivatives---see \eqref{graph:*}). In this second phase, a POSIX value
is generated in case the regular expression matches the string.
Pictorially, the Sulzmann and Lu algorithm is as follows:
\begin{ceqn}
\begin{equation}\label{graph:2}
\begin{tikzcd}
r_0 \arrow[r, "\backslash c_0"] \arrow[d] & r_1 \arrow[r, "\backslash c_1"] \arrow[d] & r_2 \arrow[r, dashed] \arrow[d] & r_n \arrow[d, "mkeps" description] \\
v_0 & v_1 \arrow[l,"inj_{r_0} c_0"] & v_2 \arrow[l, "inj_{r_1} c_1"] & v_n \arrow[l, dashed]
\end{tikzcd}
\end{equation}
\end{ceqn}
\noindent
For convenience, we shall employ the following notations: the regular
expression we start with is $r_0$, and the given string $s$ is composed
of characters $c_0 c_1 \ldots c_{n-1}$. In the first phase from the
left to right, we build the derivatives $r_1$, $r_2$, \ldots according
to the characters $c_0$, $c_1$ until we exhaust the string and obtain
the derivative $r_n$. We test whether this derivative is
$\textit{nullable}$ or not. If not, we know the string does not match
$r$ and no value needs to be generated. If yes, we start building the
values incrementally by \emph{injecting} back the characters into the
earlier values $v_n, \ldots, v_0$. This is the second phase of the
algorithm from the right to left. For the first value $v_n$, we call the
function $\textit{mkeps}$, which builds the lexical value
for how the empty string has been matched by the (nullable) regular
expression $r_n$. This function is defined as
\begin{center}
\begin{tabular}{lcl}
$\mkeps(\ONE)$ & $\dn$ & $\Empty$ \\
$\mkeps(r_{1}+r_{2})$ & $\dn$
& \textit{if} $\nullable(r_{1})$\\
& & \textit{then} $\Left(\mkeps(r_{1}))$\\
& & \textit{else} $\Right(\mkeps(r_{2}))$\\
$\mkeps(r_1\cdot r_2)$ & $\dn$ & $\Seq\,(\mkeps\,r_1)\,(\mkeps\,r_2)$\\
$mkeps(r^*)$ & $\dn$ & $\Stars\,[]$
\end{tabular}
\end{center}
\noindent
After the $\mkeps$-call, we inject back the characters one by one in order to build
the lexical value $v_i$ for how the regex $r_i$ matches the string $s_i$
($s_i = c_i \ldots c_{n-1}$ ) from the previous lexical value $v_{i+1}$.
After injecting back $n$ characters, we get the lexical value for how $r_0$
matches $s$. For this Sulzmann and Lu defined a function that reverses
the ``chopping off'' of characters during the derivative phase. The
corresponding function is called \emph{injection}, written
$\textit{inj}$; it takes three arguments: the first one is a regular
expression ${r_{i-1}}$, before the character is chopped off, the second
is a character ${c_{i-1}}$, the character we want to inject and the
third argument is the value ${v_i}$, into which one wants to inject the
character (it corresponds to the regular expression after the character
has been chopped off). The result of this function is a new value. The
definition of $\textit{inj}$ is as follows:
\begin{center}
\begin{tabular}{l@{\hspace{1mm}}c@{\hspace{1mm}}l}
$\textit{inj}\,(c)\,c\,Empty$ & $\dn$ & $Char\,c$\\
$\textit{inj}\,(r_1 + r_2)\,c\,\Left(v)$ & $\dn$ & $\Left(\textit{inj}\,r_1\,c\,v)$\\
$\textit{inj}\,(r_1 + r_2)\,c\,Right(v)$ & $\dn$ & $Right(\textit{inj}\,r_2\,c\,v)$\\
$\textit{inj}\,(r_1 \cdot r_2)\,c\,Seq(v_1,v_2)$ & $\dn$ & $Seq(\textit{inj}\,r_1\,c\,v_1,v_2)$\\
$\textit{inj}\,(r_1 \cdot r_2)\,c\,\Left(Seq(v_1,v_2))$ & $\dn$ & $Seq(\textit{inj}\,r_1\,c\,v_1,v_2)$\\
$\textit{inj}\,(r_1 \cdot r_2)\,c\,Right(v)$ & $\dn$ & $Seq(\textit{mkeps}(r_1),\textit{inj}\,r_2\,c\,v)$\\
$\textit{inj}\,(r^*)\,c\,Seq(v,Stars\,vs)$ & $\dn$ & $Stars((\textit{inj}\,r\,c\,v)\,::\,vs)$\\
\end{tabular}
\end{center}
\noindent This definition is by recursion on the ``shape'' of regular
expressions and values.
\subsection*{Simplification of Regular Expressions}
The main drawback of building successive derivatives according
to Brzozowski's definition is that they can grow very quickly in size.
This is mainly due to the fact that the derivative operation generates
often ``useless'' $\ZERO$s and $\ONE$s in derivatives. As a result, if
implemented naively both algorithms by Brzozowski and by Sulzmann and Lu
are excruciatingly slow. For example when starting with the regular
expression $(a + aa)^*$ and building 12 successive derivatives
w.r.t.~the character $a$, one obtains a derivative regular expression
with more than 8000 nodes (when viewed as a tree). Operations like
$\textit{der}$ and $\nullable$ need to traverse such trees and
consequently the bigger the size of the derivative the slower the
algorithm.
Fortunately, one can simplify regular expressions after each derivative
step.
Various simplifications of regular expressions are possible, such
as the simplification of $\ZERO + r$, $r + \ZERO$, $\ONE\cdot r$, $r
\cdot \ONE$, and $r + r$ to just $r$.
Suppose we apply simplification after each derivative step, and compose
these two operations together as an atomic one: $a \backslash_{simp}\,c \dn
\textit{simp}(a \backslash c)$. Then we can build values without having
a cumbersome regular expression, and fortunately if we are careful
enough in making some extra rectifications, the POSIX value of how
regular expressions match strings will not be affected---although is much harder
to establish. Some initial results in this regard have been
obtained in \cite{AusafDyckhoffUrban2016}.
If we want the size of derivatives in Sulzmann and Lu's algorithm to
stay even lower, we would need more aggressive simplifications.
Essentially we need to delete useless $\ZERO$s and $\ONE$s, as well as
deleting duplicates whenever possible. For example, the parentheses in
$(a+b) \cdot c + b\cdot c$ can be opened up to get $a\cdot c + b \cdot c + b
\cdot c$, and then simplified to just $a \cdot c + b \cdot c$. Another
example is simplifying $(a^*+a) + (a^*+ \ONE) + (a +\ONE)$ to just
$a^*+a+\ONE$. Adding these more aggressive simplification rules help us
to achieve a very tight size bound, namely,
the same size bound as that of the \emph{partial derivatives}.
And we want to get rid of complex and fragile rectification of values.
In order to implement the idea of ``spilling out alternatives'' and to
make them compatible with the $\textit{inj}$-mechanism, we use
\emph{bitcodes}. They were first introduced by Sulzmann and Lu.
Here bits and bitcodes (lists of bits) are defined as:
\begin{center}
$b ::= 1 \mid 0 \qquad
bs ::= [] \mid b::bs
$
\end{center}
\noindent
The $1$ and $0$ are not in bold in order to avoid
confusion with the regular expressions $\ZERO$ and $\ONE$. Bitcodes (or
bit-lists) can be used to encode values (or potentially incomplete values) in a
compact form. This can be straightforwardly seen in the following
coding function from values to bitcodes:
\begin{center}
\begin{tabular}{lcl}
$\textit{code}(\Empty)$ & $\dn$ & $[]$\\
$\textit{code}(\Char\,c)$ & $\dn$ & $[]$\\
$\textit{code}(\Left\,v)$ & $\dn$ & $0 :: code(v)$\\
$\textit{code}(\Right\,v)$ & $\dn$ & $1 :: code(v)$\\
$\textit{code}(\Seq\,v_1\,v_2)$ & $\dn$ & $code(v_1) \,@\, code(v_2)$\\
$\textit{code}(\Stars\,[])$ & $\dn$ & $[0]$\\
$\textit{code}(\Stars\,(v\!::\!vs))$ & $\dn$ & $1 :: code(v) \;@\;
code(\Stars\,vs)$
\end{tabular}
\end{center}
\noindent
Here $\textit{code}$ encodes a value into a bitcodes by converting
$\Left$ into $0$, $\Right$ into $1$, and marks the start of a non-empty
star iteration by $1$. The border where a local star terminates
is marked by $0$. This coding is lossy, as it throws away the information about
characters, and also does not encode the ``boundary'' between two
sequence values. Moreover, with only the bitcode we cannot even tell
whether the $1$s and $0$s are for $\Left/\Right$ or $\Stars$. The
reason for choosing this compact way of storing information is that the
relatively small size of bits can be easily manipulated and ``moved
around'' in a regular expression. In order to recover values, we will
need the corresponding regular expression as an extra information. This
means the decoding function is defined as:
%\begin{definition}[Bitdecoding of Values]\mbox{}
\begin{center}
\begin{tabular}{@{}l@{\hspace{1mm}}c@{\hspace{1mm}}l@{}}
$\textit{decode}'\,bs\,(\ONE)$ & $\dn$ & $(\Empty, bs)$\\
$\textit{decode}'\,bs\,(c)$ & $\dn$ & $(\Char\,c, bs)$\\
$\textit{decode}'\,(0\!::\!bs)\;(r_1 + r_2)$ & $\dn$ &
$\textit{let}\,(v, bs_1) = \textit{decode}'\,bs\,r_1\;\textit{in}\;
(\Left\,v, bs_1)$\\
$\textit{decode}'\,(1\!::\!bs)\;(r_1 + r_2)$ & $\dn$ &
$\textit{let}\,(v, bs_1) = \textit{decode}'\,bs\,r_2\;\textit{in}\;
(\Right\,v, bs_1)$\\
$\textit{decode}'\,bs\;(r_1\cdot r_2)$ & $\dn$ &
$\textit{let}\,(v_1, bs_1) = \textit{decode}'\,bs\,r_1\;\textit{in}$\\
& & $\textit{let}\,(v_2, bs_2) = \textit{decode}'\,bs_1\,r_2$\\
& & \hspace{35mm}$\textit{in}\;(\Seq\,v_1\,v_2, bs_2)$\\
$\textit{decode}'\,(0\!::\!bs)\,(r^*)$ & $\dn$ & $(\Stars\,[], bs)$\\
$\textit{decode}'\,(1\!::\!bs)\,(r^*)$ & $\dn$ &
$\textit{let}\,(v, bs_1) = \textit{decode}'\,bs\,r\;\textit{in}$\\
& & $\textit{let}\,(\Stars\,vs, bs_2) = \textit{decode}'\,bs_1\,r^*$\\
& & \hspace{35mm}$\textit{in}\;(\Stars\,v\!::\!vs, bs_2)$\bigskip\\
$\textit{decode}\,bs\,r$ & $\dn$ &
$\textit{let}\,(v, bs') = \textit{decode}'\,bs\,r\;\textit{in}$\\
& & $\textit{if}\;bs' = []\;\textit{then}\;\textit{Some}\,v\;
\textit{else}\;\textit{None}$
\end{tabular}
\end{center}
%\end{definition}
Sulzmann and Lu's integrated the bitcodes into regular expressions to
create annotated regular expressions \cite{Sulzmann2014}.
\emph{Annotated regular expressions} are defined by the following
grammar:%\comment{ALTS should have an $as$ in the definitions, not just $a_1$ and $a_2$}
\begin{center}
\begin{tabular}{lcl}
$\textit{a}$ & $::=$ & $\ZERO$\\
& $\mid$ & $_{bs}\ONE$\\
& $\mid$ & $_{bs}{\bf c}$\\
& $\mid$ & $_{bs}\sum\,as$\\
& $\mid$ & $_{bs}a_1\cdot a_2$\\
& $\mid$ & $_{bs}a^*$
\end{tabular}
\end{center}
%(in \textit{ALTS})
\noindent
where $bs$ stands for bitcodes, $a$ for $\bold{a}$nnotated regular
expressions and $as$ for a list of annotated regular expressions.
The alternative constructor($\sum$) has been generalized to
accept a list of annotated regular expressions rather than just 2.
We will show that these bitcodes encode information about
the (POSIX) value that should be generated by the Sulzmann and Lu
algorithm.
To do lexing using annotated regular expressions, we shall first
transform the usual (un-annotated) regular expressions into annotated
regular expressions. This operation is called \emph{internalisation} and
defined as follows:
%\begin{definition}
\begin{center}
\begin{tabular}{lcl}
$(\ZERO)^\uparrow$ & $\dn$ & $\ZERO$\\
$(\ONE)^\uparrow$ & $\dn$ & $_{[]}\ONE$\\
$(c)^\uparrow$ & $\dn$ & $_{[]}{\bf c}$\\
$(r_1 + r_2)^\uparrow$ & $\dn$ &
$_{[]}\sum[\textit{fuse}\,[0]\,r_1^\uparrow,\,
\textit{fuse}\,[1]\,r_2^\uparrow]$\\
$(r_1\cdot r_2)^\uparrow$ & $\dn$ &
$_{[]}r_1^\uparrow \cdot r_2^\uparrow$\\
$(r^*)^\uparrow$ & $\dn$ &
$_{[]}(r^\uparrow)^*$\\
\end{tabular}
\end{center}
%\end{definition}
\noindent
We use up arrows here to indicate that the basic un-annotated regular
expressions are ``lifted up'' into something slightly more complex. In the
fourth clause, $\textit{fuse}$ is an auxiliary function that helps to
attach bits to the front of an annotated regular expression. Its
definition is as follows:
\begin{center}
\begin{tabular}{lcl}
$\textit{fuse}\;bs \; \ZERO$ & $\dn$ & $\ZERO$\\
$\textit{fuse}\;bs\; _{bs'}\ONE$ & $\dn$ &
$_{bs @ bs'}\ONE$\\
$\textit{fuse}\;bs\;_{bs'}{\bf c}$ & $\dn$ &
$_{bs@bs'}{\bf c}$\\
$\textit{fuse}\;bs\,_{bs'}\sum\textit{as}$ & $\dn$ &
$_{bs@bs'}\sum\textit{as}$\\
$\textit{fuse}\;bs\; _{bs'}a_1\cdot a_2$ & $\dn$ &
$_{bs@bs'}a_1 \cdot a_2$\\
$\textit{fuse}\;bs\,_{bs'}a^*$ & $\dn$ &
$_{bs @ bs'}a^*$
\end{tabular}
\end{center}
\noindent
After internalising the regular expression, we perform successive
derivative operations on the annotated regular expressions. This
derivative operation is the same as what we had previously for the
basic regular expressions, except that we beed to take care of
the bitcodes:
\iffalse
%\begin{definition}{bder}
\begin{center}
\begin{tabular}{@{}lcl@{}}
$(\textit{ZERO})\,\backslash c$ & $\dn$ & $\textit{ZERO}$\\
$(\textit{ONE}\;bs)\,\backslash c$ & $\dn$ & $\textit{ZERO}$\\
$(\textit{CHAR}\;bs\,d)\,\backslash c$ & $\dn$ &
$\textit{if}\;c=d\; \;\textit{then}\;
\textit{ONE}\;bs\;\textit{else}\;\textit{ZERO}$\\
$(\textit{ALTS}\;bs\,as)\,\backslash c$ & $\dn$ &
$\textit{ALTS}\;bs\,(as.map(\backslash c))$\\
$(\textit{SEQ}\;bs\,a_1\,a_2)\,\backslash c$ & $\dn$ &
$\textit{if}\;\textit{bnullable}\,a_1$\\
& &$\textit{then}\;\textit{ALTS}\,bs\,List((\textit{SEQ}\,[]\,(a_1\,\backslash c)\,a_2),$\\
& &$\phantom{\textit{then}\;\textit{ALTS}\,bs\,}(\textit{fuse}\,(\textit{bmkeps}\,a_1)\,(a_2\,\backslash c)))$\\
& &$\textit{else}\;\textit{SEQ}\,bs\,(a_1\,\backslash c)\,a_2$\\
$(\textit{STAR}\,bs\,a)\,\backslash c$ & $\dn$ &
$\textit{SEQ}\;bs\,(\textit{fuse}\, [\Z] (r\,\backslash c))\,
(\textit{STAR}\,[]\,r)$
\end{tabular}
\end{center}
%\end{definition}
\begin{center}
\begin{tabular}{@{}lcl@{}}
$(\textit{ZERO})\,\backslash c$ & $\dn$ & $\textit{ZERO}$\\
$(_{bs}\textit{ONE})\,\backslash c$ & $\dn$ & $\textit{ZERO}$\\
$(_{bs}\textit{CHAR}\;d)\,\backslash c$ & $\dn$ &
$\textit{if}\;c=d\; \;\textit{then}\;
_{bs}\textit{ONE}\;\textit{else}\;\textit{ZERO}$\\
$(_{bs}\textit{ALTS}\;\textit{as})\,\backslash c$ & $\dn$ &
$_{bs}\textit{ALTS}\;(\textit{as}.\textit{map}(\backslash c))$\\
$(_{bs}\textit{SEQ}\;a_1\,a_2)\,\backslash c$ & $\dn$ &
$\textit{if}\;\textit{bnullable}\,a_1$\\
& &$\textit{then}\;_{bs}\textit{ALTS}\,List((_{[]}\textit{SEQ}\,(a_1\,\backslash c)\,a_2),$\\
& &$\phantom{\textit{then}\;_{bs}\textit{ALTS}\,}(\textit{fuse}\,(\textit{bmkeps}\,a_1)\,(a_2\,\backslash c)))$\\
& &$\textit{else}\;_{bs}\textit{SEQ}\,(a_1\,\backslash c)\,a_2$\\
$(_{bs}\textit{STAR}\,a)\,\backslash c$ & $\dn$ &
$_{bs}\textit{SEQ}\;(\textit{fuse}\, [0] \; r\,\backslash c )\,
(_{bs}\textit{STAR}\,[]\,r)$
\end{tabular}
\end{center}
%\end{definition}
\fi
\begin{center}
\begin{tabular}{@{}lcl@{}}
$(\ZERO)\,\backslash c$ & $\dn$ & $\ZERO$\\
$(_{bs}\ONE)\,\backslash c$ & $\dn$ & $\ZERO$\\
$(_{bs}{\bf d})\,\backslash c$ & $\dn$ &
$\textit{if}\;c=d\; \;\textit{then}\;
_{bs}\ONE\;\textit{else}\;\ZERO$\\
$(_{bs}\sum \;\textit{as})\,\backslash c$ & $\dn$ &
$_{bs}\sum\;(\textit{as.map}(\backslash c))$\\
$(_{bs}\;a_1\cdot a_2)\,\backslash c$ & $\dn$ &
$\textit{if}\;\textit{bnullable}\,a_1$\\
& &$\textit{then}\;_{bs}\sum\,[(_{[]}\,(a_1\,\backslash c)\cdot\,a_2),$\\
& &$\phantom{\textit{then},\;_{bs}\sum\,}(\textit{fuse}\,(\textit{bmkeps}\,a_1)\,(a_2\,\backslash c))]$\\
& &$\textit{else}\;_{bs}\,(a_1\,\backslash c)\cdot a_2$\\
$(_{bs}a^*)\,\backslash c$ & $\dn$ &
$_{bs}(\textit{fuse}\, [0] \; r\,\backslash c)\cdot
(_{[]}r^*))$
\end{tabular}
\end{center}
%\end{definition}
\noindent
For instance, when we do derivative of $_{bs}a^*$ with respect to c,
we need to unfold it into a sequence,
and attach an additional bit $0$ to the front of $r \backslash c$
to indicate that there is one more star iteration. Also the sequence clause
is more subtle---when $a_1$ is $\textit{bnullable}$ (here
\textit{bnullable} is exactly the same as $\textit{nullable}$, except
that it is for annotated regular expressions, therefore we omit the
definition). Assume that $\textit{bmkeps}$ correctly extracts the bitcode for how
$a_1$ matches the string prior to character $c$ (more on this later),
then the right branch of alternative, which is $\textit{fuse} \; \bmkeps \; a_1 (a_2
\backslash c)$ will collapse the regular expression $a_1$(as it has
already been fully matched) and store the parsing information at the
head of the regular expression $a_2 \backslash c$ by fusing to it. The
bitsequence $\textit{bs}$, which was initially attached to the
first element of the sequence $a_1 \cdot a_2$, has
now been elevated to the top-level of $\sum$, as this information will be
needed whichever way the sequence is matched---no matter whether $c$ belongs
to $a_1$ or $ a_2$. After building these derivatives and maintaining all
the lexing information, we complete the lexing by collecting the
bitcodes using a generalised version of the $\textit{mkeps}$ function
for annotated regular expressions, called $\textit{bmkeps}$:
%\begin{definition}[\textit{bmkeps}]\mbox{}
\begin{center}
\begin{tabular}{lcl}
$\textit{bmkeps}\,(_{bs}\ONE)$ & $\dn$ & $bs$\\
$\textit{bmkeps}\,(_{bs}\sum a::\textit{as})$ & $\dn$ &
$\textit{if}\;\textit{bnullable}\,a$\\
& &$\textit{then}\;bs\,@\,\textit{bmkeps}\,a$\\
& &$\textit{else}\;bs\,@\,\textit{bmkeps}\,(_{bs}\sum \textit{as})$\\
$\textit{bmkeps}\,(_{bs} a_1 \cdot a_2)$ & $\dn$ &
$bs \,@\,\textit{bmkeps}\,a_1\,@\, \textit{bmkeps}\,a_2$\\
$\textit{bmkeps}\,(_{bs}a^*)$ & $\dn$ &
$bs \,@\, [0]$
\end{tabular}
\end{center}
%\end{definition}
\noindent
This function completes the value information by travelling along the
path of the regular expression that corresponds to a POSIX value and
collecting all the bitcodes, and using $S$ to indicate the end of star
iterations. If we take the bitcodes produced by $\textit{bmkeps}$ and
decode them, we get the value we expect. The corresponding lexing
algorithm looks as follows:
\begin{center}
\begin{tabular}{lcl}
$\textit{blexer}\;r\,s$ & $\dn$ &
$\textit{let}\;a = (r^\uparrow)\backslash s\;\textit{in}$\\
& & $\;\;\textit{if}\; \textit{bnullable}(a)$\\
& & $\;\;\textit{then}\;\textit{decode}\,(\textit{bmkeps}\,a)\,r$\\
& & $\;\;\textit{else}\;\textit{None}$
\end{tabular}
\end{center}
\noindent
In this definition $\_\backslash s$ is the generalisation of the derivative
operation from characters to strings (just like the derivatives for un-annotated
regular expressions).
\subsection*{Our Simplification Rules}
The main point of the bitcodes and annotated regular expressions is that
we can apply rather aggressive (in terms of size) simplification rules
in order to keep derivatives small. We have developed such
``aggressive'' simplification rules and generated test data that show
that the expected bound can be achieved. Obviously we could only
partially cover the search space as there are infinitely many regular
expressions and strings.
One modification we introduced is to allow a list of annotated regular
expressions in the $\sum$ constructor. This allows us to not just
delete unnecessary $\ZERO$s and $\ONE$s from regular expressions, but
also unnecessary ``copies'' of regular expressions (very similar to
simplifying $r + r$ to just $r$, but in a more general setting). Another
modification is that we use simplification rules inspired by Antimirov's
work on partial derivatives. They maintain the idea that only the first
``copy'' of a regular expression in an alternative contributes to the
calculation of a POSIX value. All subsequent copies can be pruned away from
the regular expression. A recursive definition of our simplification function
that looks somewhat similar to our Scala code is given below:
%\comment{Use $\ZERO$, $\ONE$ and so on.
%Is it $ALTS$ or $ALTS$?}\\
\begin{center}
\begin{tabular}{@{}lcl@{}}
$\textit{simp} \; (_{bs}a_1\cdot a_2)$ & $\dn$ & $ (\textit{simp} \; a_1, \textit{simp} \; a_2) \; \textit{match} $ \\
&&$\quad\textit{case} \; (\ZERO, \_) \Rightarrow \ZERO$ \\
&&$\quad\textit{case} \; (\_, \ZERO) \Rightarrow \ZERO$ \\
&&$\quad\textit{case} \; (\ONE, a_2') \Rightarrow \textit{fuse} \; bs \; a_2'$ \\
&&$\quad\textit{case} \; (a_1', \ONE) \Rightarrow \textit{fuse} \; bs \; a_1'$ \\
&&$\quad\textit{case} \; (a_1', a_2') \Rightarrow _{bs}a_1' \cdot a_2'$ \\
$\textit{simp} \; (_{bs}\sum \textit{as})$ & $\dn$ & $\textit{distinct}( \textit{flatten} ( \textit{as.map(simp)})) \; \textit{match} $ \\
&&$\quad\textit{case} \; [] \Rightarrow \ZERO$ \\
&&$\quad\textit{case} \; a :: [] \Rightarrow \textit{fuse bs a}$ \\
&&$\quad\textit{case} \; as' \Rightarrow _{bs}\sum \textit{as'}$\\
$\textit{simp} \; a$ & $\dn$ & $\textit{a} \qquad \textit{otherwise}$
\end{tabular}
\end{center}
\noindent
The simplification does a pattern matching on the regular expression.
When it detected that the regular expression is an alternative or
sequence, it will try to simplify its children regular expressions
recursively and then see if one of the children turn into $\ZERO$ or
$\ONE$, which might trigger further simplification at the current level.
The most involved part is the $\sum$ clause, where we use two
auxiliary functions $\textit{flatten}$ and $\textit{distinct}$ to open up nested
alternatives and reduce as many duplicates as possible. Function
$\textit{distinct}$ keeps the first occurring copy only and remove all later ones
when detected duplicates. Function $\textit{flatten}$ opens up nested $\sum$s.
Its recursive definition is given below:
\begin{center}
\begin{tabular}{@{}lcl@{}}
$\textit{flatten} \; (_{bs}\sum \textit{as}) :: \textit{as'}$ & $\dn$ & $(\textit{map} \;
(\textit{fuse}\;bs)\; \textit{as}) \; @ \; \textit{flatten} \; as' $ \\
$\textit{flatten} \; \ZERO :: as'$ & $\dn$ & $ \textit{flatten} \; \textit{as'} $ \\
$\textit{flatten} \; a :: as'$ & $\dn$ & $a :: \textit{flatten} \; \textit{as'}$ \quad(otherwise)
\end{tabular}
\end{center}
\noindent
Here $\textit{flatten}$ behaves like the traditional functional programming flatten
function, except that it also removes $\ZERO$s. Or in terms of regular expressions, it
removes parentheses, for example changing $a+(b+c)$ into $a+b+c$.
Having defined the $\simp$ function,
we can use the previous notation of natural
extension from derivative w.r.t.~character to derivative
w.r.t.~string:%\comment{simp in the [] case?}
\begin{center}
\begin{tabular}{lcl}
$r \backslash_{simp} (c\!::\!s) $ & $\dn$ & $(r \backslash_{simp}\, c) \backslash_{simp}\, s$ \\
$r \backslash_{simp} [\,] $ & $\dn$ & $r$
\end{tabular}
\end{center}
\noindent
to obtain an optimised version of the algorithm:
\begin{center}
\begin{tabular}{lcl}
$\textit{blexer\_simp}\;r\,s$ & $\dn$ &
$\textit{let}\;a = (r^\uparrow)\backslash_{simp}\, s\;\textit{in}$\\
& & $\;\;\textit{if}\; \textit{bnullable}(a)$\\
& & $\;\;\textit{then}\;\textit{decode}\,(\textit{bmkeps}\,a)\,r$\\
& & $\;\;\textit{else}\;\textit{None}$
\end{tabular}
\end{center}
\noindent
This algorithm keeps the regular expression size small, for example,
with this simplification our previous $(a + aa)^*$ example's 8000 nodes
will be reduced to just 6 and stays constant, no matter how long the
input string is.
\section{Current Work and Progress}
For reasons beyond this report, it turns out that a complete set of
simplification rules depends on values being encoded as
bitsequences.\footnote{Values are the results the lexing algorithms
generate; they encode how a regular expression matched a string.} We
already know that the lexing algorithm using bitsequences but
\emph{without} simplification is correct, albeilt horribly
slow. Therefore in the past 6 months I was trying to prove that the
algorithm using bitsequences plus our simplification rules is
also correct. Formally this amounts to show that
\begin{equation}\label{mainthm}
\blexers \; r \; s = \blexer \;r\;s
\end{equation}
\noindent
whereby $\blexers$ simplifies (makes derivatives smaller) in each
step, whereas with $\blexer$ the size can grow exponentially. This
would be an important milestone for my thesis, because we already
have a very good idea how to establish that our set our simplification
rules keeps the size of derivativs below a relatively tight bound.
In order to prove the main theorem in \eqref{mainthm}, we need to prove that the
two functions produce the same output. The definition of these two functions
is shown below.
\begin{center}
\begin{tabular}{lcl}
$\textit{blexer}\;r\,s$ & $\dn$ &
$\textit{let}\;a = (r^\uparrow)\backslash s\;\textit{in}$\\
& & $\;\;\textit{if}\; \textit{bnullable}(a)$\\
& & $\;\;\textit{then}\;\textit{decode}\,(\textit{bmkeps}\,a)\,r$\\
& & $\;\;\textit{else}\;\textit{None}$
\end{tabular}
\end{center}
\begin{center}
\begin{tabular}{lcl}
$\blexers \; r \, s$ &$\dn$ &
$\textit{let} \; a = (r^\uparrow)\backslash_{simp}\, s\; \textit{in}$\\
& & $\; \; \textit{if} \; \textit{bnullable}(a)$\\
& & $\; \; \textit{then} \; \textit{decode}\,(\textit{bmkeps}\,a)\,r$\\
& & $\;\; \textit{else}\;\textit{None}$
\end{tabular}
\end{center}
\noindent
In these definitions $(r^\uparrow)$ is a kind of coding function that
is the same in each case, similarly the decode and the \textit{bmkeps}
are functions that are the same in each case. Our main
theorem~\eqref{mainthm} therefore boils down to proving the following
two propositions (depending on which branch the if-else clause
takes). They establish how the derivatives \emph{with} simplification
do not change the computed result:
\begin{itemize}
\item{(a)} If a string $s$ is in the language of $L(r)$, then \\
$\textit{bmkeps} (r^\uparrow)\backslash_{simp}\,s = \textit{bmkeps} (r^\uparrow)\backslash s$,\\
\item{(b)} If a string $s$ is in the language $L(r)$, then
$\rup \backslash_{simp} \,s$ is not nullable.
\end{itemize}
\noindent
We have already proved the second part in Isabelle. This is actually
not too difficult because we can show that simplification does not
change the language of simplified regular expressions.
If we can prove the first part, that is the bitsequence algorithm with
simplification produces the same result as the one without
simplification, then we are done. Unfortunately that part requires
more effort, because simplification does not only need to \emph{not}
change the language, but also not change the value (that is the
computed result).
%\bigskip\noindent\rule[1.5ex]{\linewidth}{5pt}
%Do you want to keep this? You essentially want to say that the old
%method used retrieve, which unfortunately cannot be adopted to
%the simplification rules. You could just say that and give an example.
%However you have to think about how you give the example....nobody knows
%about AZERO etc yet. Maybe it might be better to use normal regexes
%like $a + aa$, but annotate bitsequences as subscript like $_1(_0a + _1aa)$.
%\bigskip\noindent\rule[1.5ex]{\linewidth}{5pt}
%REPLY:\\
%Yes, I am essentially saying that the old method
%cannot be adopted without adjustments.
%But this does not mean we should skip
%the proof of the bit-coded algorithm
%as it is still the main direction we are looking into
%to prove things. We are trying to modify
%the old proof to suit our needs, but not give
%up it totally, that is why i believe the old
%proof is fundamental in understanding
%what we are doing in the past 6 months.
%\bigskip\noindent\rule[1.5ex]{\linewidth}{5pt}
\subsubsection*{Existing Proof}
For this we have started with looking at the original proof that
established that the bitsequence algorrithm produces the same result
as the algorithm not using bitsequences. Formally this proof
established
\begin{equation}\label{lexer}
\blexer \; (r^\uparrow) s = \lexer \;r \;s
\end{equation}
%\noindent
%might provide us insight into proving
%\begin{center}
%$\blexer \; r^\uparrow \;s = \blexers \; r^\uparrow \;s$
%\end{center}
\noindent
The proof uses two ``tricks''. One is that it uses a \flex-function
\begin{center}
\begin{tabular}{lcl}
$\textit{flex} \;r\; f\; (c\!::\!s) $ & $\dn$ & $\textit{flex} \; (r\backslash c) \;(\lambda v. f (inj \; r \; c \; v)) \;s$ \\
$\textit{flex} \;r\; f\; [\,] $ & $\dn$ & $f$
\end{tabular}
\end{center}
\noindent
and then proves for the right-hand side in \eqref{lexer}
\begin{center}
$\lexer \;r\; s = \flex \;\textit{id} \; r\;s \;(\mkeps \; (r\backslash s))$
\end{center}.
\noindent
The $\flex$-function essentially does lexing by
stacking up injection functions while doing derivatives.
explicitly showing the order of characters being
injected back in each step.
With $\flex$ we can write $\lexer$ this way:
\begin{center}
$\lexer \;r\; s = \flex \;id \; r\;s \;(\mkeps r\backslash s)$
\end{center}
%\noindent
%$\flex$ focuses on the injections instead of the derivatives ,
%compared to the original definition of $\lexer$, which puts equal
%amount of emphasis on injection and derivative with respect to each
%character:
%\begin{center}
%\begin{tabular}{lcl}
%$\textit{lexer} \; r\; (c\!::\!s) $ & $\dn$ & $\textit{case} \; \lexer \; (r\backslash c) \;s \; %\textit{of}$ \\
% & & $\textit{None} \; \Longrightarrow \; \textit{None}$\\
% & & $\textbar \; v \; \Longrightarrow \; \inj \; r\;c\;v$\\
%$\textit{lexer} \; r\; [\,] $ & $\dn$ & $\textit{if} \; \nullable (r) \; \textit{then} \; \mkeps% (r) \; \textit{else} \;None$
%\end{tabular}
%\end{center}
\noindent
The crux in the existing proof is how $\flex$ relates to injection, namely
\begin{center}
$\flex \; r \; id \; (s@[c]) \; v = \flex \; r \; id \; s \; (inj \; (r\backslash s) \; c\; v)$.
\end{center}
\noindent
This property allows one to rewrite an induction hypothesis like
\begin{center}
$ \flex \; r\; id\; (s@[c])\; v = \textit{decode} \;( \textit{retrieve}\; (\rup \backslash s) \; (\inj \; (r\backslash s) \;c\;v)\;) r$
\end{center}
\subsubsection{Retrieve Function}
The crucial point is to find the
$\textit{POSIX}$ information of a regular expression and how it is modified,
augmented and propagated
during simplification in parallel with the regular expression that
has not been simplified in the subsequent derivative operations. To aid this,
we use the helper function retrieve described by Sulzmann and Lu:
\begin{center}
\begin{tabular}{@{}l@{\hspace{2mm}}c@{\hspace{2mm}}l@{}}
$\textit{retrieve}\,(_{bs}\ONE)\,\Empty$ & $\dn$ & $bs$\\
$\textit{retrieve}\,(_{bs}{\bf c})\,(\Char\,d)$ & $\dn$ & $bs$\\
$\textit{retrieve}\,(_{bs}\sum a::as)\,(\Left\,v)$ & $\dn$ &
$bs \,@\, \textit{retrieve}\,a\,v$\\
$\textit{retrieve}\,(_{bs}\sum a::as)\,(\Right\,v)$ & $\dn$ &
$\textit{bs} \,@\, \textit{retrieve}\,(_{[]}\sum as)\,v$\\
$\textit{retrieve}\,(_{bs}a_1\cdot a_2)\,(\Seq\,v_1\,v_2)$ & $\dn$ &
$bs \,@\,\textit{retrieve}\,a_1\,v_1\,@\, \textit{retrieve}\,a_2\,v_2$\\
$\textit{retrieve}\,(_{bs}a^*)\,(\Stars\,[])$ & $\dn$ &
$bs \,@\, [0]$\\
$\textit{retrieve}\,(_{bs}a^*)\,(\Stars\,(v\!::\!vs))$ & $\dn$ &\\
\multicolumn{3}{l}{
\hspace{3cm}$bs \,@\, [1] \,@\, \textit{retrieve}\,a\,v\,@\,
\textit{retrieve}\,(_{[]}a^*)\,(\Stars\,vs)$}\\
\end{tabular}
\end{center}
%\comment{Did not read further}\\
This function assembles the bitcode
%that corresponds to a lexical value for how
%the current derivative matches the suffix of the string(the characters that
%have not yet appeared, but will appear as the successive derivatives go on.
%How do we get this "future" information? By the value $v$, which is
%computed by a pass of the algorithm that uses
%$inj$ as described in the previous section).
using information from both the derivative regular expression and the
value. Sulzmann and Lu poroposed this function, but did not prove
anything about it. Ausaf and Urban used it to connect the bitcoded
algorithm to the older algorithm by the following equation:
\begin{center} $inj \;a\; c \; v = \textit{decode} \; (\textit{retrieve}\;
(r^\uparrow)\backslash_{simp} \,c)\,v)$
\end{center}
\noindent
whereby $r^\uparrow$ stands for the internalised version of $r$. Ausaf
and Urban also used this fact to prove the correctness of bitcoded
algorithm without simplification. Our purpose of using this, however,
is to establish
\begin{center}
$ \textit{retrieve} \;
a \; v \;=\; \textit{retrieve} \; (\textit{simp}\,a) \; v'.$
\end{center}
The idea is that using $v'$, a simplified version of $v$ that had gone
through the same simplification step as $\textit{simp}(a)$, we are able
to extract the bitcode that gives the same parsing information as the
unsimplified one. However, we noticed that constructing such a $v'$
from $v$ is not so straightforward. The point of this is that we might
be able to finally bridge the gap by proving
\noindent
By using a property of retrieve we have the $\textit{RHS}$ of the above equality is
$decode (retrieve (r^\uparrow \backslash(s @ [c])) v) r$, and this gives the
main lemma result:
\begin{center}
$ \flex \;r\; id \; (s@[c]) \; v =\textit{decode}(\textit{retrieve} (\rup \backslash (s@[c])) \;v) r$
\end{center}
\noindent
To use this lemma result for our
correctness proof, simply replace the $v$ in the
$\textit{RHS}$ of the above equality with
$\mkeps\;(r\backslash (s@[c]))$, and apply the lemma that
\begin{center}
$\textit{decode} \; \bmkeps \; \rup \; r = \textit{decode} \; (\textit{retrieve} \; \rup \; \mkeps(r)) \;r$
\end{center}
\noindent
We get the correctness of our bit-coded algorithm:
\begin{center}
$\flex \;r\; id \; s \; (\mkeps \; r\backslash s) = \textit{decode} \; \bmkeps \; \rup\backslash s \; r$
\end{center}
\noindent
The bridge between the above chain of equalities
is the use of $\retrieve$,
if we want to use a similar technique for the
simplified version of algorithm,
we face the problem that in the above
equalities,
$\retrieve \; a \; v$ is not always defined.
for example,
$\retrieve \; _0(_1a+_0a) \; \Left(\Empty)$
is defined, but not $\retrieve \; (_{01}a) \;\Left(\Empty)$,
though we can extract the same POSIX
bits from the two annotated regular expressions.
The latter might occur when we try to retrieve from
a simplified regular expression using the same value
as the unsimplified one.
This is because $\Left(\Empty)$ corresponds to
the regular expression structure $\ONE+r_2$ instead of
$\ONE$.
That means, if we
want to prove that
\begin{center}
$\textit{decode} \; \bmkeps \; \rup\backslash s \; r = \textit{decode} \; \bmkeps \; \rup\backslash_{simp} s \; r$
\end{center}
\noindent
holds by using $\retrieve$,
we probably need to prove an equality like below:
\begin{center}
%$\retrieve \; \rup\backslash_{simp} s \; \mkeps(r\backslash_{simp} s)=\textit{retrieve} \; \rup\backslash s \; \mkeps(r\backslash s)$
$\retrieve \; \rup\backslash_{simp} s \; \mkeps(f(r\backslash s))=\textit{retrieve} \; \rup\backslash s \; \mkeps(r\backslash s)$
\end{center}
\noindent
$f$ rectifies $r\backslash s$ so the value $\mkeps(f(r\backslash s))$ becomes
something simpler
to make the retrieve function defined.\\
\subsubsection{Ways to Rectify Value}
One way to do this is to prove the following:
\begin{center}
$\retrieve \; \rup\backslash_{simp} s \; \mkeps(\simp(r\backslash s))=\textit{retrieve} \; \rup\backslash s \; \mkeps(r\backslash s)$
\end{center}
\noindent
The reason why we choose $\simp$ as $f$ is because
$\rup\backslash_{simp} \, s$ and $\simp(\rup\backslash \, s)$
have the same shape:
\begin{center}
$\erase (\rup\backslash_{simp} \, s) = \erase(\simp(\rup\backslash s))$
\end{center}
\noindent
$\erase$ in the above equality means to remove the bit-codes
in an annotated regular expression and only keep the original
regular expression(just like "erasing" the bits). Its definition is omitted.
$\rup\backslash_{simp} \, s$ and $\simp(\rup\backslash s)$
are very closely related, but not identical.
\subsubsection{Example for $\rup\backslash_{simp} \, s \neq \simp(\rup\backslash s)$}
For example, let $r$ be the regular expression
$(a+b)(a+a*)$ and $s$ be the string $aa$, then
both $\erase (\rup\backslash_{simp} \, s)$ and $\erase (\simp (\rup\backslash s))$
are $\ONE + a^*$. However, without $\erase$
\begin{center}
$\rup\backslash_{simp} \, s$ is equal to $_0(_0\ONE +_{11}a^*)$
\end{center}
\noindent
whereas
\begin{center}
$\simp(\rup\backslash s)$ is equal to $(_{00}\ONE +_{011}a^*)$
\end{center}
\noindent
(For the sake of visual simplicity, we use numbers to denote the bits
in bitcodes as we have previously defined for annotated
regular expressions. $\S$ is replaced by
subscript $_1$ and $\Z$ by $_0$.)
What makes the difference?
%Two "rules" might be inferred from the above example.
%First, after erasing the bits the two regular expressions
%are exactly the same: both become $1+a^*$. Here the
%function $\simp$ exhibits the "one in the end equals many times
%at the front"
%property: one simplification in the end causes the
%same regular expression structure as
%successive simplifications done alongside derivatives.
%$\rup\backslash_{simp} \, s$ unfolds to
%$\simp((\simp(r\backslash a))\backslash a)$
%and $\simp(\rup\backslash s)$ unfolds to
%$\simp((r\backslash a)\backslash a)$. The one simplification
%in the latter causes the resulting regular expression to
%become $1+a^*$, exactly the same as the former with
%two simplifications.
%Second, the bit-codes are different, but they are essentially
%the same: if we push the outmost bits ${\bf_0}(_0\ONE +_{11}a^*)$ of $\rup\backslash_{simp} \, s$
%inside then we get $(_{00}\ONE +_{011}a^*)$, exactly the
%same as that of $\rup\backslash \, s$. And this difference
%does not matter when we try to apply $\bmkeps$ or $\retrieve$
%to it. This seems a good news if we want to use $\retrieve$
%to prove things.
%If we look into the difference above, we could see that the
%difference is not fundamental: the bits are just being moved
%around in a way that does not hurt the correctness.
During the first derivative operation,
$\rup\backslash a=(_0\ONE + \ZERO)(_0a + _1a^*)$ is
in the form of a sequence regular expression with
two components, the first
one $\ONE + \ZERO$ being nullable.
Recall the simplification function definition:
\begin{center}
\begin{tabular}{@{}lcl@{}}
$\textit{simp} \; (\textit{SEQ}\;bs\,a_1\,a_2)$ & $\dn$ & $ (\textit{simp} \; a_1, \textit{simp} \; a_2) \; \textit{match} $ \\
&&$\quad\textit{case} \; (\ZERO, \_) \Rightarrow \ZERO$ \\
&&$\quad\textit{case} \; (\_, \ZERO) \Rightarrow \ZERO$ \\
&&$\quad\textit{case} \; (\ONE, a_2') \Rightarrow \textit{fuse} \; bs \; a_2'$ \\
&&$\quad\textit{case} \; (a_1', \ONE) \Rightarrow \textit{fuse} \; bs \; a_1'$ \\
&&$\quad\textit{case} \; (a_1', a_2') \Rightarrow \textit{SEQ} \; bs \; a_1' \; a_2'$ \\
$\textit{simp} \; (\textit{ALTS}\;bs\,as)$ & $\dn$ & $\textit{distinct}( \textit{flatten} ( \textit{as.map(simp)})) \; \textit{match} $ \\
&&$\quad\textit{case} \; [] \Rightarrow \ZERO$ \\
&&$\quad\textit{case} \; a :: [] \Rightarrow \textit{fuse bs a}$ \\
&&$\quad\textit{case} \; as' \Rightarrow \textit{ALTS}\;bs\;as'$\\
$\textit{simp} \; a$ & $\dn$ & $\textit{a} \qquad \textit{otherwise}$
\end{tabular}
\end{center}
\noindent
and the definition of $\flatten$:
\begin{center}
\begin{tabular}{c c c}
$\flatten \; []$ & $\dn$ & $[]$\\
$\flatten \; \ZERO::rs$ & $\dn$ & $rs$\\
$\flatten \;(_{\textit{bs}_1}\sum \textit{rs}_1 ::rs)$ & $\dn$ & $(\map \, (\fuse \, \textit{bs}_1) \,\textit{rs}_1) ::: \flatten(rs)$\\
$\flatten \; r :: rs$ & $\dn$ & $r::\flatten(rs)$
\end{tabular}
\end{center}
\noindent
If we call $\simp$ on $\rup\backslash a$, just as $\backslash_{simp}$
requires, then we would go through the third clause of
the sequence case:$\quad\textit{case} \; (\ONE, a_2') \Rightarrow \textit{fuse} \; bs \; a_2'$.
The $\ZERO$ of $(_0\ONE + \ZERO)$ is thrown away
by $\flatten$ and
$_0\ONE$ merged into $(_0a + _1a^*)$ by simply
putting its bits($_0$) to the front of the second component:
${\bf_0}(_0a + _1a^*)$.
After a second derivative operation,
namely, $(_0(_0a + _1a^*))\backslash a$, we get
$
_0(_0 \ONE + _1(_1\ONE \cdot a^*))
$, and this simplifies to $_0(_0 \ONE + _{11} a^*)$
by the third clause of the alternative case:
\begin{center}
$\quad\textit{case} \; as' \Rightarrow _{bs}\sum{as'}$.
\end{center}
\noindent
The outmost bit $_0$ stays with
the outmost regular expression, rather than being fused to
its child regular expressions, as what we will later see happens
to $\simp(\rup\backslash \, s)$.
If we choose to not simplify in the midst of derivative operations,
but only do it at the end after the string has been exhausted,
namely, $\simp(\rup\backslash \, s)=\simp((\rup\backslash a)\backslash a)$,
then at the {\bf second} derivative of
$(\rup\backslash a)\bf{\backslash a}$
we will go throught the clause of $\backslash$:
\begin{center}
\begin{tabular}{lcl}
$(\textit{SEQ}\;bs\,a_1\,a_2)\,\backslash c$ & $\dn$ &
$(when \; \textit{bnullable}\,a_1)$\\
& &$_{bs}\sum\,\;[_{[]}((a_1\,\backslash c) \cdot \,a_2),$\\
& &$(\textit{fuse}\,(\textit{bmkeps}\,a_1)\,(a_2\,\backslash c))]$\\
\end{tabular}
\end{center}
because
$\rup\backslash a = (_0\ONE + \ZERO)(_0a + _1a^*)$
is a sequence
with the first component being nullable
(unsimplified, unlike the first round of running$\backslash_{simp}$).
Therefore $((_0\ONE + \ZERO)(_0a + _1a^*))\backslash a$ splits into
$([(\ZERO + \ZERO)\cdot(_0a + _1a^*)] + _0( _0\ONE + _1[_1\ONE \cdot a^*]))$.
After these two successive derivatives without simplification,
we apply $\simp$ to this regular expression, which goes through
the alternative clause, and each component of
$([(\ZERO + \ZERO)\cdot(_0a + _1a^*)] + _0( _0\ONE + _1[_1\ONE \cdot a^*]))$
will be simplified, giving us the list:$[\ZERO, _0(_0\ONE + _{11}a^*)]$
This list is then "flattened"--$\ZERO$ will be
thrown away by $\textit{flatten}$; $ _0(_0\ONE + _{11}a^*)$
is opened up to make the list consisting of two separate elements
$_{00}\ONE$ and $_{011}a^*$, note that $flatten$
$\fuse$s the bit(s) $_0$ to the front of $_0\ONE $ and $_{11}a^*$.
The order of simplification, which impacts the order that alternatives
are opened up, causes
the bits to be moved differently.
\subsubsection{A Failed Attempt To Remedy the Problem Above}
A simple class of regular expression and string
pairs $(r, s)$ can be deduced from the above example
which trigger the difference between
$\rup\backslash_{simp} \, s$
and $\simp(\rup\backslash s)$:
\begin{center}
\begin{tabular}{lcl}
$D =\{ (r_1 \cdot r_2,\; [c_1c_2]) \mid $ & $\simp(r_2) = r_2, \simp(r_1 \backslash c_1) = \ONE,$\\
$r_1 \; not \; \nullable, c_2 \in L(r_2),$ & $\exists \textit{rs},\textit{bs}.\; r_2 \backslash c_2 = _{bs}{\sum rs}$\\
$\exists \textit{rs}_1. \; \simp(r_2 \backslash c_2) = _{bs}{\sum \textit{rs}_1}$ & $and \;\simp(r_1 \backslash [c_1c_2]) = \ZERO\}$\\
\end{tabular}
\end{center}
We take a pair $(r, \;s)$ from the set $D$.
Now we compute ${\bf \rup \backslash_{simp} s}$, we get:
\begin{center}
\begin{tabular}{lcl}
$(r_1\cdot r_2)\backslash_{simp} \, [c_1c_2]$ & $= \simp\left[ \big(\simp\left[ \left( r_1\cdot r_2 \right) \backslash c_1\right] \big)\backslash c_2\right]$\\
& $= \simp\left[ \big(\simp \left[ \left(r_1 \backslash c_1\right) \cdot r_2 \right] \big) \backslash c_2 \right]$\\
& $= \simp \left[ (\fuse \; \bmkeps(r_1\backslash c_1) \; \simp(r_2) ) \backslash c_2 \right]$,\\
& $= \simp \left[ (\fuse \; \bmkeps(r_1\backslash c_1) \; r_2 ) \backslash c_2 \right]$,
\end{tabular}
\end{center}
\noindent
from the definition of $D$ we know $r_1 \backslash c_1$ is nullable, therefore
$\bmkeps(r_1\backslash c_1)$ returns a bitcode, we shall call it
$\textit{bs}_2$.
The above term can be rewritten as
\begin{center}
$ \simp \left[ \fuse \; \textit{bs}_2\; r_2 \backslash c_2 \right]$,
\end{center}
which is equal to
\begin{center}
$\simp \left[ \fuse \; \textit{bs}_2 \; _{bs}{\sum rs} \right]$\\
$=\simp \left[ \; _{bs_2++bs}{\sum rs} \right]$\\
$= \; _{bs_2++bs}{\sum \textit{rs}_1} $
\end{center}
\noindent
by using the properties from the set $D$ again
and again(The reason why we set so many conditions
that the pair $(r,s)$ need to satisfy is because we can
rewrite them easily to construct the difference.)
Now we compute ${\bf \simp(\rup \backslash s)}$:
\begin{center}
$\simp \big[(r_1\cdot r_2) \backslash [c_1c_2] \big]= \simp \left[ ((r_1 \cdot r_2 )\backslash c_1) \backslash c_2 \right]$
\end{center}
\noindent
Again, using the properties above, we obtain
the following chain of equalities:
\begin{center}
$\simp(\rup \backslash s)= \simp \left[ ((r_1 \cdot r_2 )\backslash c_1) \backslash c_2 \right]= \simp\left[ \left(r_1 \backslash c_1\right) \cdot r_2 \big) \backslash c_2 \right]$\\
$= \simp \left[ \sum[\big( \left(r_1 \backslash c_1\right) \backslash c_2 \big) \cdot r_2 \; , \; \fuse \; (\bmkeps \;r_1\backslash c_1) \; r_2 \backslash c_2 ] \right]$,
\end{center}
\noindent
as before, we call the bitcode returned by
$\bmkeps(r_1\backslash c_1)$ as
$\textit{bs}_2$.
Also, $\simp(r_2 \backslash c_2)$ is
$_{bs}\sum \textit{rs}_1$,
and $( \left(r_1 \backslash c_1\right) \backslash c_2 \cdot r_2)$
simplifies to $\ZERO$,
so the above term can be expanded as
\begin{center}
\begin{tabular}{l}
$\textit{distinct}(\flatten[\ZERO\;, \; _{\textit{bs}_2++\textit{bs}}\sum \textit{rs}_1] ) \; \textit{match} $ \\
$\textit{case} \; [] \Rightarrow \ZERO$ \\
$\textit{case} \; a :: [] \Rightarrow \textit{\fuse \; \textit{bs} a}$ \\
$\textit{case} \; as' \Rightarrow _{[]}\sum as'$\\
\end{tabular}
\end{center}
\noindent
Applying the definition of $\flatten$, we get
\begin{center}
$_{[]}\sum (\textit{map} \; \fuse (\textit{bs}_2 ++ bs) rs_1)$
\end{center}
\noindent
compared to the result
\begin{center}
$ \; _{bs_2++bs}{\sum \textit{rs}_1} $
\end{center}
\noindent
Note how these two regular expressions only
differ in terms of the position of the bits
$\textit{bs}_2++\textit{bs}$. They are the same otherwise.
What caused this difference to happen?
The culprit is the $\flatten$ function, which spills
out the bitcodes in the inner alternatives when
there exists an outer alternative.
Note how the absence of simplification
caused $\simp(\rup \backslash s)$ to
generate the nested alternatives structure:
\begin{center}
$ \sum[\ZERO \;, \; _{bs}\sum \textit{rs} ]$
\end{center}
and this will always trigger the $\flatten$ to
spill out the inner alternative's bitcode $\textit{bs}$,
whereas when
simplification is done along the way,
the structure of nested alternatives is never created(we can
actually prove that simplification function never allows nested
alternatives to happen, more on this later).
How about we do not allow the function $\simp$
to fuse out the bits when it is unnecessary?
Like, for the above regular expression, we might
just delete the outer layer of alternative
\begin{center}
\st{$ {\sum[\ZERO \;,}$} $_{bs}\sum \textit{rs}$ \st{$]$}
\end{center}
and get $_{bs}\sum \textit{rs}$ instead, without
fusing the bits $\textit{bs}$ inside to every element
of $\textit{rs}$.
This idea can be realized by making the following
changes to the $\simp$-function:
\begin{center}
\begin{tabular}{@{}lcl@{}}
$\textit{simp}' \; (_{\textit{bs}}(a_1 \cdot a_2))$ & $\dn$ & $\textit{as} \; \simp \; \textit{was} \; \textit{before} $ \\
$\textit{simp}' \; (_{bs}\sum as)$ & $\dn$ & \st{$\textit{distinct}( \textit{flatten} ( \textit{map simp as})) \; \textit{match} $} \\
&&\st{$\quad\textit{case} \; [] \Rightarrow \ZERO$} \\
&&\st{$\quad\textit{case} \; a :: [] \Rightarrow \textit{fuse bs a}$} \\
&&\st{$\quad\textit{case} \; as' \Rightarrow \textit{ALTS}\;bs\;as'$}\\
&&$\textit{if}(\textit{hollowAlternatives}( \textit{map \; simp \; as}))$\\
&&$\textit{then} \; \fuse \; \textit{bs}\; \textit{extractAlt}(\textit{map} \; \simp \; \textit{as} )$\\
&&$\textit{else} \; \simp(_{bs} \sum \textit{as})$\\
$\textit{simp}' \; a$ & $\dn$ & $\textit{a} \qquad \textit{otherwise}$
\end{tabular}
\end{center}
\noindent
given the definition of $\textit{hollowAlternatives}$ and $\textit{extractAlt}$ :
\begin{center}
$\textit{hollowAlternatives}( \textit{rs}) \dn
\exists r = (_{\textit{bs}_1}\sum \textit{rs}_1) \in \textit{rs}. \forall r' \in \textit{rs}, \;
\textit{either} \; r' = \ZERO \; \textit{or} \; r' = r $
$\textit{extractAlt}( \textit{rs}) \dn \textit{if}\big(
\exists r = (_{\textit{bs}_1}\sum \textit{rs}_1) \in \textit{rs}. \forall r' \in \textit{rs}, \;
\textit{either} \; r' = \ZERO \; \textit{or} \; r' = r \big)\; \textit{then} \; \textit{return} \; r$
\end{center}
\noindent
Basically, $\textit{hollowAlternatives}$ captures the feature of
a list of regular expression of the shape
\begin{center}
$ \sum[\ZERO \;, \; _{bs}\sum \textit{rs} ]$
\end{center}
and this means we can simply elevate the
inner regular expression $_{bs}\sum \textit{rs}$
to the outmost
and throw away the useless $\ZERO$s and
the outer $\sum$ wrapper.
Using this new definition of simp,
under the example where $r$ is the regular expression
$(a+b)(a+a*)$ and $s$ is the string $aa$
the problem of $\rup\backslash_{simp} \, s \neq \simp(\rup\backslash s)$
is resolved.
Unfortunately this causes new problems:
for the counterexample where
\begin{center}
$r$ is the regular expression
$(ab+(a^*+aa))$ and $s$ is the string $aa$
\end{center}
\noindent
$\rup\backslash_{simp} \, s$ is equal to
$ _1(_{011}a^* + _1\ONE) $ whereas
$ \simp(\rup\backslash s) = (_{1011}a^* + _{11}\ONE)$.
This discrepancy does not appear for the old
version of $\simp$.
Why?
During the first derivative operation,
\begin{center}
$\rup\backslash a=( _0[ \ONE\cdot {\bf b}] + _1( _0[ _1\ONE \cdot {\bf a}^*] + [ \ONE \cdot {\bf a}]) )$,
\end{center}
\noindent
the second derivative gives us
\begin{center}
$\rup\backslash a=(_0( [\ZERO\cdot {\bf b}] + 0) + _1( _0( [\ZERO\cdot {\bf a}^*] + _1[ _1\ONE \cdot {\bf a}^*]) + _1( [\ZERO \cdot {\bf a}] + \ONE) ))$,
\end{center}
\noindent
and this simplifies to
\begin{center}
$ _1(_{011}{\bf a}^* + _1\ONE) $
\end{center}
If, after the first derivative we apply simplification we get
$(_0{\bf b} + _{101}{\bf a}^* + _{11}{\bf a} )$,
and we do another derivative, getting
$(\ZERO + (_{101}(\ONE \cdot _1{\bf a}^*)+_{11}\ONE)$,
which simplifies to
\begin{center}
$ (_{1011}a^* + _{11}\ONE) $
\end{center}
We have changed the algorithm to suppress the old
counterexample, but this gives rise to new counterexamples.
This dilemma causes this amendment not a successful
attempt to make $\rup\backslash_{simp} \, s = \simp(\rup\backslash s)$
under every possible regular expression and string.
\subsection{Properties of the Function $\simp$}
We have proved in Isabelle quite a few properties
of the $\simp$-function, which helps the proof to go forward
and we list them here to aid comprehension.
To start, we need a bit of auxiliary notations,
which is quite basic and is only written here
for clarity.
$\textit{sub}(r)$ computes the set of the
sub-regular expression of $r$:
\begin{center}
$\textit{sub}(\ONE) \dn \{\ONE\}$\\
$\textit{sub}(r_1 \cdot r_2) \dn \textit{sub}(r_1) \cup \textit{sub}(r_2) \cup \{r_1 \cdot r_2\}$\\
$\textit{sub}(r_1 + r_2) \dn \textit{sub}(r_1) \cup \textit{sub}(r_2) \cup \{r_1+r_2\}$\\
\end{center}
$\textit{good}(r) \dn r \neq \ZERO \land \\
\forall r' \in \textit{sub}(r), \textit{if} \; r' = _{bs_1}\sum(rs_1), \;
\textit{then} \nexists r'' \in \textit{rs}_1 \; s.t.\;
r'' = _{bs_2}\sum \textit{rs}_2 $
The properties are mainly the ones below:
\begin{itemize}
\item
\begin{center}
$\simp(\simp(r)) = \simp(r)$
\end{center}
\item
\begin{center}
$\textit{if} r = \simp(r') \textit{then} \; \textit{good}(r) $
\end{center}
\end{itemize}
\subsection{the Contains relation}
$\retrieve$ is a too strong relation in that
it only extracts one bitcode instead of a set of them.
Therefore we try to define another relation(predicate)
to capture the fact the regular expression has bits
being moved around but still has all the bits needed.
The contains symbol, written$\gg$, is a relation that
takes two arguments in an infix form
and returns a truth value.
In other words, from the set of regular expression and
bitcode pairs
$\textit{RV} = \{(r, v) \mid r \text{r is a regular expression, v is a value}\}$,
those that satisfy the following requirements are in the set
$\textit{RV}_Contains$.
Unlike the $\retrieve$
function, which takes two arguments $r$ and $v$ and
produces an only answer $\textit{bs}$, it takes only
one argument $r$ and returns a set of bitcodes that
can be generated by $r$.
\begin{center}
\begin{tabular}{llclll}
& & & $_{bs}\ONE$ & $\gg$ & $\textit{bs}$\\
& & & $_{bs}{\bf c}$ & $\gg$ & $\textit{bs}$\\
$\textit{if} \; r_1 \gg \textit{bs}_1$ & $r_2 \; \gg \textit{bs}_2$
& $\textit{then}$ &
$_{bs}{r_1 \cdot r_2}$ &
$\gg$ &
$\textit{bs} @ \textit{bs}_1 @ \textit{bs}_2$\\
$\textit{if} \; r \gg \textit{bs}_1$ & & $\textit{then}$ &
$_{bs}{\sum(r :: \textit{rs}})$ &
$\gg$ &
$\textit{bs} @ \textit{bs}_1 $\\
$\textit{if} \; _{bs}(\sum \textit{rs}) \gg \textit{bs} @ \textit{bs}_1$ & & $\textit{then}$ &
$_{bs}{\sum(r :: \textit{rs}})$ &
$\gg$ &
$\textit{bs} @ \textit{bs}_1 $\\
$\textit{if} \; r \gg \textit{bs}_1\; \textit{and}$ & $_{bs}r^* \gg \textit{bs} @ \textit{bs}_2$ & $\textit{then}$ &
$_{bs}r^* $ &
$\gg$ &
$\textit{bs} @ [0] @ \textit{bs}_1@ \textit{bs}_2 $\\
& & & $_{bs}r^*$ & $\gg$ & $\textit{bs} @ [1]$\\
\end{tabular}
\end{center}
It is a predicate in the sense that given
a regular expression and a bitcode, it
returns true or false, depending on whether
or not the regular expression can actually produce that
value. If the predicates returns a true, then
we say that the regular expression $r$ contains
the bitcode $\textit{bs}$, written
$r \gg \textit{bs}$.
The $\gg$ operator with the
regular expression $r$ may also be seen as a
machine that does a derivative of regular expressions
on all strings simultaneously, taking
the bits by going throught the regular expression tree
structure in a depth first manner, regardless of whether
the part being traversed is nullable or not.
It put all possible bits that can be produced on such a traversal
into a set.
For example, if we are given the regular expression
$((a+b)(c+d))^*$, the tree structure may be written as
\begin{center}
\begin{tikzpicture}
\tikz[tree layout]\graph[nodes={draw, circle}] {
* ->
{@-> {
{+1 ->
{a , b}
},
{+ ->
{c , d }
}
}
}
};
\end{tikzpicture}
\end{center}
\subsection{the $\textit{ders}_2$ Function}
If we want to prove the result
\begin{center}
$ \textit{blexer}\_{simp}(r, \; s) = \textit{blexer}(r, \; s)$
\end{center}
inductively
on the structure of the regular expression,
then we need to induct on the case $r_1 \cdot r_2$,
it can be good if we could express $(r_1 \cdot r_2) \backslash s$
in terms of $r_1 \backslash s$ and $r_2 \backslash s$,
and this naturally induces the $ders2$ function,
which does a "convolution" on $r_1$ and $r_2$ using the string
$s$.
It is based on the observation that the derivative of $r_1 \cdot r_2$
with respect to a string $s$ can actually be written in an "explicit form"
composed of $r_1$'s derivative of $s$ and $r_2$'s derivative of $s$.
This can be illustrated in the following procedure execution
\begin{center}
$ (r_1 \cdot r_2) \backslash [c_1c_2] = (\textit{if} \; \nullable(r_1)\; \textit{then} \; ((r_1 \backslash c_1) \cdot r_2 + r_2 \backslash c_1) \; \textit{else} \; (r_1\backslash c_1) \cdot r_2) \backslash c_2$
\end{center}
which can also be written in a "convoluted sum"
format:
\begin{center}
$ (r_1 \cdot r_2) \backslash [c_1c_2] = \sum{r_1 \backslash s_i \cdot r_2 \backslash s_j}$
\end{center}
In a more serious manner, we should write
\begin{center}
$ (r_1 \cdot r_2) \backslash [c_1c_2] = \sum{r_1 \backslash s_i \cdot r_2 \backslash s_j}$
\end{center}
Note this differentiates from the previous form in the sense that
it calculates the results $r_1\backslash s_i , r_2 \backslash s_j$ first and then glue them together
through nested alternatives whereas the $r_1 \cdot r_2 \backslash s$ procedure,
used by algorithm $\lexer$, can only produce each component of the
resulting alternatives regular expression altogether rather than
generating each of the children nodes one by one
n a single recursive call that is only for generating that
very expression itself.
We have this
\section{Conclusion}
Under the exhaustive tests we believe the main
result holds, yet a proof still seems elusive.
We have tried out different approaches, and
found a lot of properties of the function $\simp$.
The counterexamples where $\rup\backslash_{simp} \, s \neq \simp(\rup\backslash s)$
are also valuable in the sense that
we get to know better why they are not equal and what
are the subtle differences between a
nested simplified regular expression and a
regular expression that is simplified at the final moment.
We are almost there, but a last step is needed to make the proof work.
Hopefully in the next few weeks we will be able to find one.
%CONSTRUCTION SITE HERE
that is to say, despite the bits are being moved around on the regular expression
(difference in bits), the structure of the (unannotated)regular expression
after one simplification is exactly the same after the
same sequence of derivative operations
regardless of whether we did simplification
along the way.
One way would be to give a function that calls
fuse is the culprit: it causes the order in which alternatives
are opened up to be remembered and finally the difference
appear in $\simp(\rup \backslash s)$ and $\rup \backslash{simp} \,s$.
but we have to use them as they are essential in the simplification:
flatten needs them.
However, without erase the above equality does not hold:
for the regular expression
$(a+b)(a+a*)$,
if we do derivative with respect to string $aa$,
we get
\subsection{Another Proof Strategy}
sdddddr does not equal sdsdsdsr sometimes.\\
For example,
This equicalence class method might still have the potential of proving this,
but not yet
i parallelly tried another method of using retrieve\\
The vsimp function, defined as follows
tries to simplify the value in lockstep with
regular expression:\\
The problem here is that
we used retrieve for the key induction:
$decode (retrieve (r\backslash (s @ [c])) v) r $
$decode (retrieve (r\backslash s) (inj (r\backslash s) c v)) r$
Here, decode recovers a value that corresponds to a match(possibly partial)
from bits, and the bits are extracted by retrieve,
and the key value $v$ that guides retrieve is
$mkeps r\backslash s$, $inj r c (mkeps r\backslash s)$, $inj (inj (v))$, ......
if we can
the problem is that
need vsiimp to make a value that is suitable for decoding
$Some(flex rid(s@[c])v) = Some(flex rids(inj (r\backslash s)cv))$
another way that christian came up with that might circumvent the
prblem of finding suitable value is by not stating the visimp
function but include all possible value in a set that a regex is able to produce,
and proving that both r and sr are able to produce the bits that correspond the POSIX value
produced by feeding the same initial regular expression $r$ and string $s$ to the
two functions $ders$ and $ders\_simp$.
The reason why
Namely, if $bmkeps( r_1) = bmkeps(r_2)$, then we
If we define the equivalence relation $\sim_{m\epsilon}$ between two regular expressions
$r_1$ and $r_2$as follows:
$r_1 \sim_{m\epsilon} r_2 \iff bmkeps(r_1)= bmkeps(r_2)$
(in other words, they $r1$ and $r2$ produce the same output under the function $bmkeps$.)
Then the first goal
might be restated as
$(r^\uparrow)\backslash_{simp}\, s \sim_{m\epsilon} (r^\uparrow)\backslash s$.
I tried to establish an equivalence relation between the regular experssions
like dddr dddsr,.....
but right now i am only able to establish dsr and dr, using structural induction on r.
Those involve multiple derivative operations are harder to prove.
Two attempts have been made:
(1)induction on the number of der operations(or in other words, the length of the string s),
the inductive hypothesis was initially specified as
"For an arbitrary regular expression r,
For all string s in the language of r whose length do not exceed
the number n, ders s r me derssimp s r"
and the proof goal may be stated as
"For an arbitrary regular expression r,
For all string s in the language of r whose length do not exceed
the number n+1, ders s r me derssimp s r"
the problem here is that although we can easily break down
a string s of length n+1 into s1@list(c), it is not that easy
to use the i.h. as a stepping stone to prove anything because s1 may well be not
in the language L(r). This inhibits us from obtaining the fact that
ders s1 r me derssimps s1 r.
Further exploration is needed to amend this hypothesis so it includes the
situation when s1 is not nullable.
For example, what information(bits?
values?) can be extracted
from the regular expression ders(s1,r) so that we can compute or predict the possible
result of bmkeps after another derivative operation. What function f can used to
carry out the task? The possible way of exploration can be
more directly perceived throught the graph below:
find a function
f
such that
f(bders s1 r)
= re1
f(bderss s1 r)
= re2
bmkeps(bders s r) = g(re1,c)
bmkeps(bderssimp s r) = g(re2,c)
and g(re1,c) = g(re2,c)
The inductive hypothesis would be
"For all strings s1 of length <= n,
f(bders s1 r)
= re1
f(bderss s1 r)
= re2"
proving this would be a lemma for the main proof:
the main proof would be
"
bmkeps(bders s r) = g(re1,c)
bmkeps(bderssimp s r) = g(re2,c)
for s = s1@c
"
and f need to be a recursive property for the lemma to be proved:
it needs to store not only the "after one char nullable info",
but also the "after two char nullable info",
and so on so that it is able to predict what f will compute after a derivative operation,
in other words, it needs to be "infinitely recursive"\\
To prove the lemma, in other words, to get
"For all strings s1 of length <= n+1,
f(bders s1 r)
= re3
f(bderss s1 r)
= re4"\\
from\\
"For all strings s1 of length <= n,
f(bders s1 r)
= re1
f(bderss s1 r)
= re2"\\
it might be best to construct an auxiliary function h such that\\
h(re1, c) = re3\\
h(re2, c) = re4\\
and re3 = f(bder c (bders s1 r))\\
re4 = f(simp(bder c (bderss s1 r)))
The key point here is that we are not satisfied with what bders s r will produce under
bmkeps, but also how it will perform after a derivative operation and then bmkeps, and two
derivative operations and so on. In essence, we are preserving the regular expression
itself under the function f, in a less compact way than the regluar expression: we are
not just recording but also interpreting what the regular expression matches.
In other words, we need to prove the properties of bderss s r beyond the bmkeps result,
i.e., not just the nullable ones, but also those containing remaining characters.\\
(2)we observed the fact that
erase sdddddr= erase sdsdsdsr
that is to say, despite the bits are being moved around on the regular expression
(difference in bits), the structure of the (unannotated)regular expression
after one simplification is exactly the same after the
same sequence of derivative operations
regardless of whether we did simplification
along the way.
However, without erase the above equality does not hold:
for the regular expression
$(a+b)(a+a*)$,
if we do derivative with respect to string $aa$,
we get
%TODO
sdddddr does not equal sdsdsdsr sometimes.\\
For example,
This equicalence class method might still have the potential of proving this,
but not yet
i parallelly tried another method of using retrieve\\
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