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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\lexer{\mathit{lexer}}
\def\mkeps{\mathit{mkeps}}
\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}}

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\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. We have already developed some
  simplification rules for this, but have not yet proved that they
  preserve the correctness of the algorithm. We also have not yet
  looked at extended regular expressions, such as bounded repetitions,
  negation and back-references.
\end{abstract}

\section{Introduction}


This PhD-project is about regular expression matching and
lexing. Given the maturity of this topic, the reader might wonder:
Surely, regular expressions must have already been studied to death?
What could possibly be \emph{not} known in this area? And surely all
implemented algorithms for regular expression matching are blindingly
fast?

Unfortunately these preconceptions are not supported by evidence: Take
for example the regular expression $(a^*)^*\,b$ and ask whether
strings of the form $aa..a$ match this regular
expression. Obviously not---the expected $b$ in the last
position is missing. One would expect that modern regular expression
matching engines can find this out very quickly. Alas, if one tries
this example in JavaScript, Python or Java 8 with strings like 28
$a$'s, one discovers that this decision takes around 30 seconds and
takes considerably longer when adding a few more $a$'s, as the graphs
below show:

\begin{center}
\begin{tabular}{@{}c@{\hspace{0mm}}c@{\hspace{0mm}}c@{}}
\begin{tikzpicture}
\begin{axis}[
    xlabel={$n$},
    x label style={at={(1.05,-0.05)}},
    ylabel={time in secs},
    enlargelimits=false,
    xtick={0,5,...,30},
    xmax=33,
    ymax=35,
    ytick={0,5,...,30},
    scaled ticks=false,
    axis lines=left,
    width=5cm,
    height=4cm, 
    legend entries={JavaScript},  
    legend pos=north west,
    legend cell align=left]
\addplot[red,mark=*, mark options={fill=white}] table {re-js.data};
\end{axis}
\end{tikzpicture}
  &
\begin{tikzpicture}
\begin{axis}[
    xlabel={$n$},
    x label style={at={(1.05,-0.05)}},
    %ylabel={time in secs},
    enlargelimits=false,
    xtick={0,5,...,30},
    xmax=33,
    ymax=35,
    ytick={0,5,...,30},
    scaled ticks=false,
    axis lines=left,
    width=5cm,
    height=4cm, 
    legend entries={Python},  
    legend pos=north west,
    legend cell align=left]
\addplot[blue,mark=*, mark options={fill=white}] table {re-python2.data};
\end{axis}
\end{tikzpicture}
  &
\begin{tikzpicture}
\begin{axis}[
    xlabel={$n$},
    x label style={at={(1.05,-0.05)}},
    %ylabel={time in secs},
    enlargelimits=false,
    xtick={0,5,...,30},
    xmax=33,
    ymax=35,
    ytick={0,5,...,30},
    scaled ticks=false,
    axis lines=left,
    width=5cm,
    height=4cm, 
    legend entries={Java 8},  
    legend pos=north west,
    legend cell align=left]
\addplot[cyan,mark=*, mark options={fill=white}] table {re-java.data};
\end{axis}
\end{tikzpicture}\\
\multicolumn{3}{c}{Graphs: Runtime for matching $(a^*)^*\,b$ with strings 
           of the form $\underbrace{aa..a}_{n}$.}
\end{tabular}    
\end{center}  

\noindent These are clearly abysmal and possibly surprising results. One
would expect these systems to do  much better than that---after all,
given a DFA and a string, deciding whether a string is matched by this
DFA should be linear?

Admittedly, the regular expression $(a^*)^*\,b$ is carefully chosen to
exhibit this exponential behaviour.  But unfortunately, such regular
expressions are not just a few outliers. They are actually 
frequent enough to have a separate name created for
them---\emph{evil regular expressions}. In empiric work, Davis et al
report that they have found thousands of such evil regular expressions
in the JavaScript and Python ecosystems \cite{Davis18}.

This exponential blowup in matching algorithms sometimes causes
considerable grief in real life: for example on 20 July 2016 one evil
regular expression brought the webpage
\href{http://stackexchange.com}{Stack Exchange} to its
knees.\footnote{\url{https://stackstatus.net/post/147710624694/outage-postmortem-july-20-2016}}
In this instance, a regular expression intended to just trim white
spaces from the beginning and the end of a line actually consumed
massive amounts of CPU-resources---causing web servers to
grind to a halt. This happened when a post with 20,000 white spaces was
submitted, but importantly the white spaces were neither at the
beginning nor at the end. 
As a result, the regular expression matching
engine needed to backtrack over many choices.
In this example, the time needed to process the string is not 
exactly the classical exponential case, but rather $O(n^2)$
with respect to the string length. But this is enough for the 
home page of Stack Exchange to respond not fast enough to
the load balancer, which thought that there must be some
attack and therefore stopped the servers from responding to 
requests. This made the whole site become unavailable.
Another very recent example is a global outage of all Cloudflare servers
on 2 July 2019. A poorly written regular expression exhibited
exponential behaviour and exhausted CPUs that serve HTTP traffic.
Although the outage had several causes, at the heart was a regular
expression that was used to monitor network
traffic.\footnote{\url{https://blog.cloudflare.com/details-of-the-cloudflare-outage-on-july-2-2019/}}

The underlying problem is that many ``real life'' regular expression
matching engines do not use DFAs for matching. This is because they
support regular expressions that are not covered by the classical
automata theory, and in this more general setting there are quite a few
research questions still unanswered and fast algorithms still need to be
developed (for example how to treat bounded repetitions, negation and
back-references efficiently).
%question: dfa can have exponential states. isn't this the actual reason why they do not use dfas?
%how do they avoid dfas exponential states if they use them for fast matching?

There is also another under-researched problem to do with regular
expressions and lexing, i.e.~the process of breaking up strings into
sequences of tokens according to some regular expressions. In this
setting one is not just interested in whether or not a regular
expression matches a string, but also in \emph{how}.  Consider for example a regular expression
$r_{key}$ for recognising keywords such as \textit{if}, \textit{then}
and so on; and a regular expression $r_{id}$ for recognising
identifiers (say, a single character followed by characters or
numbers). One can then form the compound regular expression
$(r_{key} + r_{id})^*$ and use it to tokenise strings.  But then how
should the string \textit{iffoo} be tokenised?  It could be tokenised
as a keyword followed by an identifier, or the entire string as a
single identifier.  Similarly, how should the string \textit{if} be
tokenised? Both regular expressions, $r_{key}$ and $r_{id}$, would
``fire''---so is it an identifier or a keyword?  While in applications
there is a well-known strategy to decide these questions, called POSIX
matching, only relatively recently precise definitions of what POSIX
matching actually means has been formalised
\cite{AusafDyckhoffUrban2016,OkuiSuzuki2010,Vansummeren2006}. 
Such a definition has also been given by Sulzmann and  Lu \cite{Sulzmann2014}, but the
corresponding correctness proof turned out to be  faulty \cite{AusafDyckhoffUrban2016}.
Roughly, POSIX matching means matching the longest initial substring.
In the case of a tie, the initial sub-match is chosen according to some priorities attached to the
regular expressions (e.g.~keywords have a higher priority than
identifiers). This sounds rather simple, but according to Grathwohl et
al \cite[Page 36]{CrashCourse2014} this is not the case. They wrote:

\begin{quote}
\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{quote}

\noindent
This is also supported by evidence collected by Kuklewicz
\cite{Kuklewicz} who noticed that a number of POSIX regular expression
matchers calculate incorrect results.

Our focus in this project is on an algorithm introduced by Sulzmann and
Lu in 2014 for regular expression matching according to the POSIX
strategy \cite{Sulzmann2014}. Their algorithm is based on an older
algorithm by Brzozowski from 1964 where he introduced the notion of
derivatives of regular expressions~\cite{Brzozowski1964}. We shall
briefly explain this algorithm next.

\section{The Algorithm by Brzozowski based on Derivatives of Regular
Expressions}

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 intended meaning of the constructors is as follows: $\ZERO$
cannot match any string, $\ONE$ can match the empty string, the
character regular expression $c$ can match the character $c$, and so
on.

The ingenious contribution by Brzozowski is the notion of
\emph{derivatives} of regular expressions.  The idea behind this
notion is as follows: suppose a regular expression $r$ can match a
string of the form $c\!::\! s$ (that is a list of characters starting
with $c$), what does the regular expression look like that can match
just $s$? Brzozowski gave a neat answer to this question. He started
with the definition of $nullable$:
\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}
This function simply tests whether the empty string is in $L(r)$.
He then defined
the following operation on regular expressions, written
$r\backslash c$ (the derivative of $r$ w.r.t.~the character $c$):

\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}

%Assuming the classic notion of a
%\emph{language} of a regular expression, written $L(\_)$, t

\noindent
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
For us the main advantage is that derivatives can be
straightforwardly implemented in any functional programming language,
and are easily definable and reasoned about in theorem provers---the
definitions just consist of inductive datatypes and simple recursive
functions. Moreover, the notion of derivatives can be easily
generalised to cover extended regular expression constructors such as
the not-regular expression, written $\neg\,r$, or bounded repetitions
(for example $r^{\{n\}}$ and $r^{\{n..m\}}$), which cannot be so
straightforwardly realised within the classic automata approach.
For the moment however, we focus only on the usual basic regular expressions.


Now if we want to find out whether a string $s$ matches with a regular
expression $r$, build the derivatives of $r$ w.r.t.\ (in succession)
all the characters of the string $s$. Finally, test whether the
resulting regular expression can match the empty string.  If yes, then
$r$ matches $s$, and no in the negative case. To implement this idea
we can generalise the derivative operation to strings like this:

\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 as  regular-expression matching algorithm: 
\[
match\;s\;r \;\dn\; nullable(r\backslash s)
\]

\noindent
This algorithm looks graphically 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 $s \in L(r_0)$).

 
\section{Values and the 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 values}.  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
No value  corresponds to $\ZERO$; $\Empty$ corresponds to
$\ONE$; $\Char$ to the character regular expression; $\Seq$ to the
sequence regular expression and so on. The idea of values is to encode
a kind of lexical value for how the sub-parts of a regular expression match
the sub-parts of a string. To see this, suppose a \emph{flatten} operation, written
$|v|$ for values. We can use this function to extract the underlying string of a value
$v$. For example, $|\mathit{Seq} \, (\textit{Char x}) \, (\textit{Char
y})|$ is the string $xy$.  Using flatten, we can describe how values
encode parse trees: $\Seq\,v_1\, v_2$ encodes a tree with 2 children nodes
that tells how the string $|v_1| @
|v_2|$ matches the regex $r_1 \cdot r_2$ whereby $r_1$ matches the
substring $|v_1|$ and, respectively, $r_2$ matches the substring
$|v_2|$. Exactly how these two are matched is contained in the
children nodes $v_1$ and $v_2$ of parent $\textit{Seq}$ . 

To give a concrete example of how values work, consider the string $xy$
and the regular expression $(x + (y + xy))^*$. We can view this regular
expression as a tree and if the string $xy$ is matched by two Star
``iterations'', then the $x$ is matched by the left-most alternative in
this tree and the $y$ by the right-left alternative. This suggests to
record this matching as

\begin{center}
$\Stars\,[\Left\,(\Char\,x), \Right(\Left(\Char\,y))]$
\end{center}

\noindent
where $\Stars \; [\ldots]$ records all the
iterations; and $\Left$, respectively $\Right$, which
alternative is used. The value for
matching $xy$ in a single ``iteration'', i.e.~the POSIX value,
would look as follows

\begin{center}
$\Stars\,[\Seq\,(\Char\,x)\,(\Char\,y)]$
\end{center}

\noindent
where $\Stars$ has only a single-element list for the single iteration
and $\Seq$ indicates that $xy$ is matched by a sequence regular
expression.

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 parse tree 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 There are no cases for $\ZERO$ and $c$, since
these regular expression cannot match the empty string. Note
also that in case of alternatives we give preference to the
regular expression on the left-hand side. This will become
important later on.

After this, we inject back the characters one by one in order to build
the parse tree $v_i$ for how the regex $r_i$ matches the string $s_i$
($s_i = c_i \ldots c_{n-1}$ ) from the previous parse tree $v_{i+1}$.
After injecting back $n$ characters, we get the parse tree 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. To understands this definition better consider
the situation when we build the derivative on regular expression $r_{i-1}$.
For this we chop off a character from $r_{i-1}$ to form $r_i$. This leaves a
``hole'' in $r_i$ and its corresponding value $v_i$. 
To calculate $v_{i-1}$, we need to
locate where that hole is and fill it. 
We can find this location by
comparing $r_{i-1}$ and $v_i$. For instance, if $r_{i-1}$ is of shape
$r_a \cdot r_b$, and $v_i$ is of shape $\Left(Seq(v_1,v_2))$, we know immediately that 
%
\[ (r_a \cdot r_b)\backslash c = (r_a\backslash c) \cdot r_b \,+\, r_b\backslash c,\]

\noindent
otherwise if $r_a$ is not nullable,
\[ (r_a \cdot r_b)\backslash c = (r_a\backslash c) \cdot r_b,\]

\noindent
the value $v_i$ should be  $\Seq(\ldots)$, contradicting the fact that
$v_i$ is actually of shape $\Left(\ldots)$. Furthermore, since $v_i$ is of shape
$\Left(\ldots)$ instead of $\Right(\ldots)$, we know that the left
branch of \[ (r_a \cdot r_b)\backslash c =
\bold{\underline{ (r_a\backslash c) \cdot r_b} }\,+\, r_b\backslash c,\](underlined)
 is taken instead of the right one. This means $c$ is chopped off 
from $r_a$ rather than $r_b$.
We have therefore found out 
that the hole will be on $r_a$. So we recursively call $\inj\, 
r_a\,c\,v_a$ to fill that hole in $v_a$. After injection, the value 
$v_i$ for $r_i = r_a \cdot r_b$ should be $\Seq\,(\inj\,r_a\,c\,v_a)\,v_b$.
Other clauses can be understood in a similar way.

%\comment{Other word: insight?}
The following example gives an insight of $\textit{inj}$'s effect
and how Sulzmann and Lu's algorithm works as a whole.
 Suppose we have a
regular expression $((((a+b)+ab)+c)+abc)^*$, and want to match it against
the string $abc$ (when $abc$ is written as a regular expression, the most
standard way of expressing it should be $a \cdot (b \cdot c)$. We omit
the parentheses and dots here for readability). 
This algorithm returns a POSIX value, which means it
will go for the longest matching, i.e.~it should match the string
$abc$ in one star iteration, using the longest alternative $abc$ in the
sub-expression $((((a+b)+ab)+c)+abc)$ (we use $r$ to denote this sub-expression
for conciseness). 
Before $\textit{inj}$ comes into play, 
our lexer first builds derivative using string $abc$ (we simplified some regular expressions like
$0 \cdot b$ to $0$ for conciseness; we also omit parentheses if
they are clear from the context):
%Similarly, we allow
%$\textit{ALT}$ to take a list of regular expressions as an argument
%instead of just 2 operands to reduce the nested depth of
%$\textit{ALT}$
\begin{center}
\begin{tabular}{lcl}
$r^*$ & $\xrightarrow{\backslash a}$ & $r_1 = (\ONE+\ZERO+\ONE \cdot b + \ZERO + \ONE \cdot b \cdot c) \cdot r^*$\\
      & $\xrightarrow{\backslash b}$ & $r_2 = (\ZERO+\ZERO+\ONE \cdot \ONE + \ZERO + \ONE \cdot \ONE \cdot c) \cdot r^* +(\ZERO+\ONE+\ZERO  + \ZERO + \ZERO) \cdot r^*$\\
      & $\xrightarrow{\backslash c}$ & $r_3 = ((\ZERO+\ZERO+\ZERO + \ZERO + \ONE \cdot \ONE \cdot \ONE) \cdot r^* + (\ZERO+\ZERO+\ZERO  + \ONE + \ZERO) \cdot r^*) + $\\ 
      &                              & $\phantom{r_3 = (} ((\ZERO+\ONE+\ZERO  + \ZERO + \ZERO) \cdot r^* + (\ZERO+\ZERO+\ZERO  + \ONE + \ZERO) \cdot r^* )$
\end{tabular}
\end{center}

\noindent
In  case $r_3$ is nullable, we can call $\textit{mkeps}$ 
to construct a parse tree for how $r_3$ matched the string $abc$. 
$\textit{mkeps}$ gives the following value $v_3$: 
\begin{center}
$\Left(\Left(\Seq(\Right(\Seq(\Empty, \Seq(\Empty,\Empty))), \Stars [])))$
\end{center}
The outer $\Left(\Left(\ldots))$ tells us the leftmost nullable part of $r_3$(underlined):

\begin{center}
   $( \underline{(\ZERO+\ZERO+\ZERO+ \ZERO+ \ONE \cdot \ONE \cdot \ONE) \cdot r^*} + (\ZERO+\ZERO+\ZERO + \ONE + \ZERO)
  \cdot r^*) +((\ZERO+\ONE+\ZERO  + \ZERO + \ZERO) \cdot r^*+(\ZERO+\ZERO+\ZERO  + \ONE + \ZERO) \cdot r^* ).$
 \end{center}

\noindent
 Note that the leftmost location of term $((\ZERO+\ZERO+\ZERO + \ZERO + \ONE \cdot \ONE \cdot
 \ONE) \cdot r^*$ (which corresponds to the initial sub-match $abc$) allows
 $\textit{mkeps}$ to pick it up because $\textit{mkeps}$ is defined to always choose the
 left one when it is nullable. In the case of this example, $abc$ is
 preferred over $a$ or $ab$. This $\Left(\Left(\ldots))$ location is
 naturally generated by two applications of the splitting clause

\begin{center}
     $(r_1 \cdot r_2)\backslash c  (when \; r_1 \; nullable) \, = (r_1\backslash c) \cdot r_2 \,+\, r_2\backslash c.$
\end{center}
       
\noindent
By this clause, we put $r_1 \backslash c \cdot r_2 $ at the
$\textit{front}$ and $r_2 \backslash c$ at the $\textit{back}$. This
allows $\textit{mkeps}$ to always pick up among two matches the one with a longer
initial sub-match. Removing the outside $\Left(\Left(...))$, the inside
sub-value 
 
\begin{center}
 $\Seq(\Right(\Seq(\Empty, \Seq(\Empty, \Empty))), \Stars [])$
\end{center}

\noindent
tells us how the empty string $[]$ is matched with $(\ZERO+\ZERO+\ZERO + \ZERO + \ONE \cdot
\ONE \cdot \ONE) \cdot r^*$. We match $[]$ by a sequence of 2 nullable regular
expressions. The first one is an alternative, we take the rightmost
alternative---whose language contains the empty string. The second
nullable regular expression is a Kleene star. $\Stars$ tells us how it
generates the nullable regular expression: by 0 iterations to form
$\ONE$. Now $\textit{inj}$ injects characters back and incrementally
builds a parse tree based on $v_3$. Using the value $v_3$, the character
c, and the regular expression $r_2$, we can recover how $r_2$ matched
the string $[c]$ : $\textit{inj} \; r_2 \; c \; v_3$ gives us
 \begin{center}
 $v_2 = \Left(\Seq(\Right(\Seq(\Empty, \Seq(\Empty, c))), \Stars [])),$
 \end{center}
which tells us how $r_2$ matched $[c]$. After this we inject back the character $b$, and get
\begin{center}
$v_1 = \Seq(\Right(\Seq(\Empty, \Seq(b, c))), \Stars [])$
\end{center}
 for how 
 \begin{center}
 $r_1= (\ONE+\ZERO+\ONE \cdot b + \ZERO + \ONE \cdot b \cdot c) \cdot r*$
 \end{center}
  matched  the string $bc$ before it split into 2 pieces. 
  Finally, after injecting character $a$ back to $v_1$, 
  we get  the parse tree 
  \begin{center}
  $v_0= \Stars [\Right(\Seq(a, \Seq(b, c)))]$
  \end{center}
   for how $r$ matched $abc$. This completes the algorithm.
   
%We omit the details of injection function, which is provided by Sulzmann and Lu's paper \cite{Sulzmann2014}. 
Readers might have noticed that the parse tree information 
is actually already available when doing derivatives. 
For example, immediately after the operation $\backslash a$ we know that if we want to match a string that starts with $a$,
 we can either take the initial match to be 
 \begin{center}
\begin{enumerate}
    \item[1)] just $a$ or
    \item[2)] string $ab$ or 
    \item[3)] string $abc$.
\end{enumerate}
\end{center}

\noindent
In order to differentiate between these choices, we just need to
remember their positions--$a$ is on the left, $ab$ is in the middle ,
and $abc$ is on the right. Which one of these alternatives is chosen
later does not affect their relative position because our algorithm does
not change this order. If this parsing information can be determined and
does not change because of later derivatives, there is no point in
traversing this information twice. This leads to an optimisation---if we
store the information for parse trees inside the regular expression,
update it when we do derivative on them, and collect the information
when finished with derivatives and call $\textit{mkeps}$ for deciding which
branch is POSIX, we can generate the parse tree in one pass, instead of
doing the rest $n$ injections. This leads to Sulzmann and Lu's novel
idea of using bitcodes in derivatives.

In the next section, we shall focus on the bitcoded algorithm and the
process of simplification of regular expressions. This is needed in
order to obtain \emph{fast} versions of the Brzozowski's, and Sulzmann
and Lu's algorithms.  This is where the PhD-project aims to advance the
state-of-the-art.


\section{Simplification of Regular Expressions}

Using bitcodes to guide  parsing is not a novel idea. It was applied to
context free grammars and then adapted by Henglein and Nielson for
efficient regular expression parsing using DFAs~\cite{nielson11bcre}.
Sulzmann and Lu took this idea of bitcodes a step further by integrating
bitcodes into derivatives. The reason why we want to use bitcodes in
this project is that we want to introduce more aggressive
simplification rules in order to keep the size of derivatives small
throughout. This is because 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 $r$. These simplifications do not
affect the answer for whether a regular expression matches a string or
not, but fortunately also do not affect the POSIX strategy of how
regular expressions match strings---although the latter is much harder
to establish. Some initial results in this regard have been
obtained in \cite{AusafDyckhoffUrban2016}. 

Unfortunately, the simplification rules outlined above  are not
sufficient to prevent a size explosion in all cases. We
believe a tighter bound can be achieved that prevents an explosion in
\emph{all} cases. Such a tighter bound is suggested by work of Antimirov who
proved that (partial) derivatives can be bound by the number of
characters contained in the initial regular expression
\cite{Antimirov95}. He defined the \emph{partial derivatives} of regular
expressions as follows:

\begin{center}
\begin{tabular}{lcl}
 $\textit{pder} \; c \; 0$ & $\dn$ & $\emptyset$\\
 $\textit{pder} \; c \; 1$ & $\dn$ & $\emptyset$ \\
 $\textit{pder} \; c \; d$ & $\dn$ & $\textit{if} \; c \,=\, d \; \{  1   \}  \; \textit{else} \; \emptyset$ \\ 
  $\textit{pder} \; c \; r_1+r_2$ & $\dn$ & $pder \; c \; r_1 \cup pder \; c \;  r_2$ \\
   $\textit{pder} \; c \; r_1 \cdot r_2$ & $\dn$ & $\textit{if} \; nullable \; r_1 $\\
     & & $\textit{then} \; \{  r \cdot r_2 \mid r \in pder \; c \; r_1   \}  \cup pder \; c \; r_2 \;$\\
     & & $\textit{else} \; \{  r \cdot r_2 \mid r \in pder \; c \; r_1   \} $ \\ 
     $\textit{pder} \; c \; r^*$ & $\dn$ & $ \{  r' \cdot r^* \mid r' \in pder \; c \; r   \}  $ \\  
 \end{tabular}
 \end{center}

\noindent
A partial derivative of a regular expression $r$ is essentially a set of
regular expressions that are either $r$'s children expressions or a
concatenation of them. Antimirov has proved a tight bound of the size of
\emph{all} partial derivatives no matter what the string looks like.
Roughly speaking the size will be quadruple in the size of the regular
expression. If we want the size of derivatives in Sulzmann and Lu's
algorithm to stay equal or below this bound, 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 $(a+b) \cdot c + bc$ 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 helps us to achieve the same size bound as that of the partial
derivatives. In order to implement the idea of ``spilling out alternatives''
and to make them compatible with the $\text{inj}$-mechanism, we use \emph{bitcodes}.
Bits and bitcodes (lists of bits) are just:
%This allows us to prove a tight
%bound on the size of regular expression during the running time of the
%algorithm if we can establish the connection between our simplification
%rules and partial derivatives.

 %We believe, and have generated test
%data, that a similar bound can be obtained for the derivatives in
%Sulzmann and Lu's algorithm. Let us give some details about this next.


\begin{center}
		$b ::=   S \mid  Z \qquad
bs ::= [] \mid b:bs    
$
\end{center}

\noindent
The names $S$ and $Z$  here are quite arbitrary in order to avoid 
confusion with the regular expressions $\ZERO$ and $\ONE$. Bitcodes (or
bit-lists) can be used to encode values (or 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}
  $\textit{code}(\Empty)$ & $\dn$ & $[]$\\
  $\textit{code}(\Char\,c)$ & $\dn$ & $[]$\\
  $\textit{code}(\Left\,v)$ & $\dn$ & $\Z :: code(v)$\\
  $\textit{code}(\Right\,v)$ & $\dn$ & $\S :: code(v)$\\
  $\textit{code}(\Seq\,v_1\,v_2)$ & $\dn$ & $code(v_1) \,@\, code(v_2)$\\
  $\textit{code}(\Stars\,[])$ & $\dn$ & $[\Z]$\\
  $\textit{code}(\Stars\,(v\!::\!vs))$ & $\dn$ & $\S :: code(v) \;@\;
                                                 code(\Stars\,vs)$
\end{tabular}    
\end{center} 

\noindent
Here $\textit{code}$ encodes a value into a bitcodes by converting
$\Left$ into $\Z$, $\Right$ into $\S$, the start point of a non-empty
star iteration into $\S$, and the border where a local star terminates
into $\Z$. 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 $\S$s and $\Z$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@{}}
  $\textit{decode}'\,bs\,(\ONE)$ & $\dn$ & $(\Empty, bs)$\\
  $\textit{decode}'\,bs\,(c)$ & $\dn$ & $(\Char\,c, bs)$\\
  $\textit{decode}'\,(\Z\!::\!bs)\;(r_1 + r_2)$ & $\dn$ &
     $\textit{let}\,(v, bs_1) = \textit{decode}'\,bs\,r_1\;\textit{in}\;
       (\Left\,v, bs_1)$\\
  $\textit{decode}'\,(\S\!::\!bs)\;(r_1 + r_2)$ & $\dn$ &
     $\textit{let}\,(v, bs_1) = \textit{decode}'\,bs\,r_2\;\textit{in}\;
       (\Right\,v, bs_1)$\\                           
  $\textit{decode}'\,bs\;(r_1\cdot r_2)$ & $\dn$ &
        $\textit{let}\,(v_1, bs_1) = \textit{decode}'\,bs\,r_1\;\textit{in}$\\
  & &   $\textit{let}\,(v_2, bs_2) = \textit{decode}'\,bs_1\,r_2$\\
  & &   \hspace{35mm}$\textit{in}\;(\Seq\,v_1\,v_2, bs_2)$\\
  $\textit{decode}'\,(\Z\!::\!bs)\,(r^*)$ & $\dn$ & $(\Stars\,[], bs)$\\
  $\textit{decode}'\,(\S\!::\!bs)\,(r^*)$ & $\dn$ & 
         $\textit{let}\,(v, bs_1) = \textit{decode}'\,bs\,r\;\textit{in}$\\
  & &   $\textit{let}\,(\Stars\,vs, bs_2) = \textit{decode}'\,bs_1\,r^*$\\
  & &   \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:

\begin{center}
\begin{tabular}{lcl}
  $\textit{a}$ & $::=$  & $\textit{ZERO}$\\
                  & $\mid$ & $\textit{ONE}\;\;bs$\\
                  & $\mid$ & $\textit{CHAR}\;\;bs\,c$\\
                  & $\mid$ & $\textit{ALT}\;\;bs\,a_1 \, a_2$\\
                  & $\mid$ & $\textit{SEQ}\;\;bs\,a_1\,a_2$\\
                  & $\mid$ & $\textit{STAR}\;\;bs\,a$
\end{tabular}    
\end{center}  
%(in \textit{ALT})

\noindent
where $bs$ stands for bitcodes, and $a$  for $\bold{a}$nnotated regular
expressions. 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}
  $(\ZERO)^\uparrow$ & $\dn$ & $\textit{ZERO}$\\
  $(\ONE)^\uparrow$ & $\dn$ & $\textit{ONE}\,[]$\\
  $(c)^\uparrow$ & $\dn$ & $\textit{CHAR}\,[]\,c$\\
  $(r_1 + r_2)^\uparrow$ & $\dn$ &
         $\textit{ALT}\;[]\,(\textit{fuse}\,[\Z]\,r_1^\uparrow)\,
                            (\textit{fuse}\,[\S]\,r_2^\uparrow)$\\
  $(r_1\cdot r_2)^\uparrow$ & $\dn$ &
         $\textit{SEQ}\;[]\,r_1^\uparrow\,r_2^\uparrow$\\
  $(r^*)^\uparrow$ & $\dn$ &
         $\textit{STAR}\;[]\,r^\uparrow$\\
\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}
  $\textit{fuse}\;bs\,(\textit{ZERO})$ & $\dn$ & $\textit{ZERO}$\\
  $\textit{fuse}\;bs\,(\textit{ONE}\,bs')$ & $\dn$ &
     $\textit{ONE}\,(bs\,@\,bs')$\\
  $\textit{fuse}\;bs\,(\textit{CHAR}\,bs'\,c)$ & $\dn$ &
     $\textit{CHAR}\,(bs\,@\,bs')\,c$\\
  $\textit{fuse}\;bs\,(\textit{ALT}\,bs'\,a_1\,a_2)$ & $\dn$ &
     $\textit{ALT}\,(bs\,@\,bs')\,a_1\,a_2$\\
  $\textit{fuse}\;bs\,(\textit{SEQ}\,bs'\,a_1\,a_2)$ & $\dn$ &
     $\textit{SEQ}\,(bs\,@\,bs')\,a_1\,a_2$\\
  $\textit{fuse}\;bs\,(\textit{STAR}\,bs'\,a)$ & $\dn$ &
     $\textit{STAR}\,(bs\,@\,bs')\,a$
\end{tabular}    
\end{center}  

\noindent
After internalise we do successive derivative operations on the
annotated regular expressions. This derivative operation is the same as
what we previously have for the simple regular expressions, except that
we beed to take special care of the bitcodes:\comment{You need to be consitent with  ALTS and ALT; ALT is just 
an abbreviation; derivations and so on are defined for ALTS}

 %\begin{definition}{bder}
\begin{center}
  \begin{tabular}{@{}lcl@{}}
  $(\textit{ZERO})\,\backslash c$ & $\dn$ & $\textit{ZERO}$\\  
  $(\textit{ONE}\;bs)\,\backslash c$ & $\dn$ & $\textit{ZERO}$\\  
  $(\textit{CHAR}\;bs\,d)\,\backslash c$ & $\dn$ &
        $\textit{if}\;c=d\; \;\textit{then}\;
         \textit{ONE}\;bs\;\textit{else}\;\textit{ZERO}$\\  
  $(\textit{ALT}\;bs\,a_1\,a_2)\,\backslash c$ & $\dn$ &
        $\textit{ALT}\;bs\,(a_1\,\backslash c)\,(a_2\,\backslash c)$\\
  $(\textit{SEQ}\;bs\,a_1\,a_2)\,\backslash c$ & $\dn$ &
     $\textit{if}\;\textit{bnullable}\,a_1$\\
  & &$\textit{then}\;\textit{ALT}\,bs\,(\textit{SEQ}\,[]\,(a_1\,\backslash c)\,a_2)$\\
  & &$\phantom{\textit{then}\;\textit{ALT}\,bs\,}(\textit{fuse}\,(\textit{bmkeps}\,a_1)\,(a_2\,\backslash c))$\\
  & &$\textit{else}\;\textit{SEQ}\,bs\,(a_1\,\backslash c)\,a_2$\\
  $(\textit{STAR}\,bs\,a)\,\backslash c$ & $\dn$ &
      $\textit{SEQ}\;bs\,(\textit{fuse}\, [\Z] (r\,\backslash c))\,
       (\textit{STAR}\,[]\,r)$
\end{tabular}    
\end{center}    
%\end{definition}

\noindent
For instance, when we unfold $\textit{STAR} \; bs \; a$ into a sequence,
we need to attach an additional bit $Z$ to the front of $r \backslash c$
to indicate that there is one more star iteration. Also the $SEQ$ 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 $bmkeps$ correctly extracts the bitcode for how
$a_1$ matches the string prior to character $c$ (more on this later),
then the right branch of $ALTS$, which is $fuse \; bmkeps \;  a_1 (a_2
\backslash c)$ 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 $bs$, which was initially attached to the head of $SEQ$, has
now been elevated to the top-level of ALT, as this information will be
needed whichever way the $SEQ$ 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}
  $\textit{bmkeps}\,(\textit{ONE}\;bs)$ & $\dn$ & $bs$\\
  $\textit{bmkeps}\,(\textit{ALT}\;bs\,a_1\,a_2)$ & $\dn$ &
     $\textit{if}\;\textit{bnullable}\,a_1$\\
  & &$\textit{then}\;bs\,@\,\textit{bmkeps}\,a_1$\\
  & &$\textit{else}\;bs\,@\,\textit{bmkeps}\,a_2$\\
  $\textit{bmkeps}\,(\textit{SEQ}\;bs\,a_1\,a_2)$ & $\dn$ &
     $bs \,@\,\textit{bmkeps}\,a_1\,@\, \textit{bmkeps}\,a_2$\\
  $\textit{bmkeps}\,(\textit{STAR}\;bs\,a)$ & $\dn$ &
     $bs \,@\, [\S]$
\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).

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 \textit{ALTS} 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 $ALT$ or $ALTS$?}\\

\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} $ \\
   &&$\quad\textit{case} \; (0, \_) \Rightarrow  0$ \\
   &&$\quad\textit{case} \; (\_, 0) \Rightarrow  0$ \\
   &&$\quad\textit{case} \;  (1, a_2') \Rightarrow  \textit{fuse} \; bs \;  a_2'$ \\
   &&$\quad\textit{case} \; (a_1', 1) \Rightarrow  \textit{fuse} \; bs \;  a_1'$ \\
   &&$\quad\textit{case} \; (a_1', a_2') \Rightarrow  \textit{SEQ} \; bs \; a_1' \;  a_2'$ \\

  $\textit{simp} \; (\textit{ALTS}\;bs\,as)$ & $\dn$ & $\textit{distinct}( \textit{flatten} ( \textit{map simp as})) \; \textit{match} $ \\
  &&$\quad\textit{case} \; [] \Rightarrow  0$ \\
   &&$\quad\textit{case} \; a :: [] \Rightarrow  \textit{fuse bs a}$ \\
   &&$\quad\textit{case} \;  as' \Rightarrow  \textit{ALT}\;bs\;as'$\\ 

   $\textit{simp} \; a$ & $\dn$ & $\textit{a} \qquad \textit{otherwise}$   
\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 $\textit{ALTS}$ clause, where we use two
auxiliary functions flatten and distinct to open up nested
$\textit{ALTS}$ and reduce as many duplicates as possible. Function
distinct  keeps the first occurring copy only and remove all later ones
when detected duplicates. Function flatten opens up nested \textit{ALT}.
Its recursive definition is given below:

 \begin{center}
  \begin{tabular}{@{}lcl@{}}
  $\textit{flatten} \; (\textit{ALT}\;bs\,as) :: as'$ & $\dn$ & $(\textit{map} \;
     (\textit{fuse}\;bs)\; \textit{as}) \; @ \; \textit{flatten} \; as' $ \\
  $\textit{flatten} \; \textit{ZERO} :: as'$ & $\dn$ & $ \textit{flatten} \;  as' $ \\
    $\textit{flatten} \; a :: as'$ & $\dn$ & $a :: \textit{flatten} \; as'$ \quad(otherwise) 
\end{tabular}    
\end{center}  

\noindent
Here flatten behaves like the traditional functional programming flatten
function, except that it also removes $\ZERO$s. What it does is
basically removing parentheses like changing $a+(b+c)$ into $a+b+c$.

Suppose we apply simplification after each derivative step, and view
these two operations as an atomic one: $a \backslash_{simp}\,c \dn
\textit{simp}(a \backslash c)$. Then we can use the previous 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 [\,] $ & $\dn$ & $r$
\end{tabular}
\end{center}

\noindent
we 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}

We are currently engaged in two tasks related to this algorithm. The
first task is proving that our simplification rules actually do not
affect the POSIX value that should be generated by the algorithm
according to the specification of a POSIX value and furthermore obtain a
much tighter bound on the sizes of derivatives. The result is that our
algorithm should be correct and faster on all inputs.  The original
blow-up, as observed in JavaScript, Python and Java, would be excluded
from happening in our algorithm. For this proof we use the theorem prover
Isabelle. Once completed, this result will advance the state-of-the-art:
Sulzmann and Lu wrote in their paper~\cite{Sulzmann2014} about the
bitcoded ``incremental parsing method'' (that is the lexing algorithm
outlined in this section):

\begin{quote}\it
  ``Correctness Claim: We further claim that the incremental parsing
  method in Figure~5 in combination with the simplification steps in
  Figure 6 yields POSIX parse trees. We have tested this claim
  extensively by using the method in Figure~3 as a reference but yet
  have to work out all proof details.''
\end{quote}  

\noindent
We would settle this correctness claim. It is relatively straightforward
to establish that after one simplification step, the part of a nullable
derivative that corresponds to a POSIX value remains intact and can
still be collected, in other words, we can show that\comment{Double-check....I think this  is not the case}

\begin{center}
$\textit{bmkeps} \; r = \textit{bmkeps} \; \textit{simp} \; r\;( r\; \textit{nullable})$
\end{center}

\noindent
as this basically comes down to proving actions like removing the
additional $r$ in $r+r$  does not delete important POSIX information in
a regular expression. The hard part of this proof is to establish that

\begin{center}
$\textit{bmkeps} \; \textit{blexer}\_{simp} \; r = \textit{bmkeps} \; \textit{blexer} \; \textit{simp} \; r$
\end{center}

\noindent\comment{OK from here on you still need to work. Did not read.}
That is, if we do derivative on regular expression r and the simplified version, 
they can still provide the same POSIX value if there is one . 
This is not as straightforward as the previous proposition, as the two regular expressions $r$ and $\textit{simp}\; r$
  might become very different regular expressions after repeated application of $\textit{simp}$ and derivative.
The crucial point is to find the indispensable information of 
a regular expression and how it is kept intact during simplification so that it performs 
as good as a regular expression that has not been simplified in the subsequent derivative operations.
To aid this, we use the helping function retrieve described by Sulzmann and Lu:
\\definition of retrieve\\
 
This function assembled the bitcode that corresponds to a parse tree for
how the current derivative matches the suffix of the string(the
characters that have not yet appeared, but is stored in the value).
Sulzmann and Lu used this to connect the bitcoded algorithm to the older
algorithm by the following equation:
 
 \begin{center}
 $inj \;a\; c \; v = \textit{decode} \; (\textit{retrieve}\; ((\textit{internalise}\; r)\backslash_{simp} c) v)$
 \end{center}
 A little fact that needs to be stated to help comprehension:
 \begin{center}
 $r^\uparrow = a$($a$ stands for $\textit{annotated}).$
 \end{center}
 Ausaf and Urban also used this fact to prove  the correctness of bitcoded algorithm without simplification.
 Our purpose of using this, however, is try to establish \\
$ \textit{retrieve} \; a \; v \;=\; \textit{retrieve}  \; \textit{simp}(a) \; v'.$\\
 The idea is that using $v'$,
  a simplified version of $v$ that possibly had gone through the same simplification step as $\textit{simp}(a)$ we are still  able to extract the bit-sequence that gives the same parsing information as the unsimplified one.
 After establishing this, we might be able to finally bridge the gap of proving\\
 $\textit{retrieve} \; r   \backslash  s \; v = \;\textit{retrieve} \; \textit{simp}(r)  \backslash  s \; v'$\\
 and subsequently\\
 $\textit{retrieve} \; r \backslash  s \; v\; = \; \textit{retrieve} \; r  \backslash_{simp}   s \; v'$.\\
 This proves that our simplified version of regular expression still contains all the bitcodes needed.


The second task is to speed up the more aggressive simplification.
Currently it is slower than a naive simplification(the naive version as
implemented in ADU of course can explode in some cases). So it needs to
be explored how to make it faster. Our possibility would be to explore
again the connection to DFAs. This is very much work in progress.

\section{Conclusion}

In this PhD-project we are interested in fast algorithms for regular
expression matching. While this seems to be a ``settled'' area, in
fact interesting research questions are popping up as soon as one steps
outside the classic automata theory (for example in terms of what kind
of regular expressions are supported). The reason why it is
interesting for us to look at the derivative approach introduced by
Brzozowski for regular expression matching, and then much further
developed by Sulzmann and Lu, is that derivatives can elegantly deal
with some of the regular expressions that are of interest in ``real
life''. This includes the not-regular expression, written $\neg\,r$
(that is all strings that are not recognised by $r$), but also bounded
regular expressions such as $r^{\{n\}}$ and $r^{\{n..m\}}$). There is
also hope that the derivatives can provide another angle for how to
deal more efficiently with back-references, which are one of the
reasons why regular expression engines in JavaScript, Python and Java
choose to not implement the classic automata approach of transforming
regular expressions into NFAs and then DFAs---because we simply do not
know how such back-references can be represented by DFAs.


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