\documentclass[a4paper,UKenglish]{lipics}
\usepackage{graphic}
\usepackage{data}
\usepackage{tikz-cd}
%\usepackage{algorithm}
\usepackage{amsmath}
\usepackage[noend]{algpseudocode}
\usepackage{enumitem}
\usepackage{nccmath}
\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}
%\usepackage[english]{babel}
%\usepackage{listings}
% \usepackage{amsthm}
%\usepackage{hyperref}
% \usepackage[margin=0.5in]{geometry}
%\usepackage{pmboxdraw}
\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\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 size of regular expressions.
We have tested the correctness of these extended rules, but have not
formally established their correctness. We also have not yet looked
at extended regular expressions, such as bounded repetitions,
negation and back-references.
\end{abstract}
\section{Introduction}
While we believe derivatives of regular expressions is a beautiful
concept (interms of ease to implementing them in functional programming
language and ease to reason about them formally), they have one major
drawback: every derivative step can make regular expressions grow
drastically in size. This in turn has negative effects on the runtime of
the corresponding lexing algorithms. Consider for example the regular
expression $(a+aa)^*$ and the short string $aaaaaaaaaaaa$. The size of
the corresponding derivative is already 8668 node assuming the derivatives
is seen as a tree. The reason for the poor runtime of the 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.
For reasons beyond this report, it turns out that a complete set of
simplification rules depend on values being encoded as bitsequences.
(Vlue are the results of the lexing algorithms generate; they encode how
a regular expression matched a string.) We already know that the lexing
algorithm \emph{without} simplification is correct. Therefore in the
past 6 months we were trying to prove that the algorithm using bitsequences plus
our simplification rules is 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, because we already have a very good idea how to
establish that our set our simplification rules keeps the size below a
relatively tight bound.
In order to prove the main theorem \eqref{mainthm}, we need to prove the
two functions produce the same output. The definition of these 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}
functions. Our main theorem \eqref{mainthm} therefore boils down to
proving the following two propositions (depending on which branch the
if-else clause takes). It establishes how the derivatives \emph{with}
simplification do not change the computed result:
\begin{itemize}
\item{} 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{} 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 in Isabelle the second part. This is actually not
too difficult because we can show that simplification does not change
the language of regular expressions. If we can prove the first case,
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 (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}
The correctness proof of
\begin{center}
$\blexer \; r^\uparrow s = \lexer \;r \;s$
\end{center}
\noindent
might provide us insight into proving
\begin{center}
$\blexer \; r^\uparrow \;s = \blexers \; r^\uparrow \;s$
\end{center}
\noindent
(that is also
why we say the new proof builds on the older one).
The proof defined the function $\flex$ as another way of
expressing the $\lexer$ function:
\begin{center}
$\lexer \;r\; s = \flex \;\textit{id} \; r\;s \;(\mkeps \; r\backslash s)$
\end{center}.
\noindent
(proof for the above equality will be explained later)
The definition of $flex$ is as follows:
\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
here $\flex$ 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
Using this feature of $\flex$ we can rewrite the lexing
$w.r.t \; s @ [c]$ in term of lexing
$w.r.t \; s$:
\begin{center}
$\flex \; r \; id \; (s@[c]) \; v = \flex \; r \; id \; s \; (inj \; (r\backslash s) \; c\; v)$.
\end{center}
\noindent
this allows us to use
the inductive hypothesis to get
\begin{center}
$ \flex \; r\; id\; (s@[c])\; v = \textit{decode} \;( \textit{retrieve}\; (\rup \backslash s) \; (\inj \; (r\backslash s) \;c\;v)\;) r$
\end{center}
\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.\\
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.
For example, let $r$ be the regular expression
$(a+b)(a+a*)$ and $s$ be the string $aa$, then
\begin{center}
$\rup\backslash_{simp} \, s$ is equal to $_0(_0\ONE +_{11}a^*)$
\end{center}
\noindent
whereas
\begin{center}
$\rup\backslash \, s$ is equal to $(_{00}\ONE +_{011}a^*)$
\end{center}
\noindent
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 "once equals many times"
property: one simplification in the end causes the
same regular expression structure as
successive simplifications done alongside derivatives.
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.\\
If we look into the difference above, we could see why:
during the first derivative operation,
$\rup\backslash a=(_0\ONE + \ZERO)(_0a + _1a^*)$ gets into a sequence
with the first part being nullable, but not the second part.
and we can use this to construct a set of examples based
on this type of behaviour of two operations.
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\\
%HERE CONSTRUCTION SITE
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\\
\noindent\rule[0.5ex]{\linewidth}{1pt}
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 this is not the case---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 in terms of the size of the regular expression and
the string?
Admittedly, the regular expression $(a^*)^*\,b$ is carefully chosen to
exhibit this super-linear 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}. Static analysis
approach that is both sound and complete exists\cite{17Bir}, but the running
time on certain examples in the RegExLib and Snort regular expressions
libraries is unacceptable. Therefore the problem of efficiency still remains.
This superlinear 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 was $O(n^2)$ with respect to the string length. This
quadratic overhead was enough for the homepage of Stack Exchange to
respond so slowly that the load balancer assumed there must be some
attack and therefore stopped the servers from responding to any
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 efficiently bounded repetitions, negation and
back-references).
%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 have 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 $r$, we can 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 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
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 lexical values: $\Seq\,v_1\,
v_2$ encodes a tree with two 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 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 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 about what value is calculated.
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. 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 standard way of expressing it is $a \cdot (b \cdot c)$. But we
usually omit the parentheses and dots here for better readability. This
algorithm returns a POSIX value, which means it will produce the longest
matching. Consequently, it matches the string $abc$ in one star
iteration, using the longest alternative $abc$ in the sub-expression (we shall use $r$ to denote this
sub-expression for conciseness):
\[((((a+b)+ab)+c)+\underbrace{abc}_r)\]
\noindent
Before $\textit{inj}$ is called, our lexer first builds derivative using
string $abc$ (we simplified some regular expressions like $\ZERO \cdot
b$ to $\ZERO$ 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 lexical value for how $r_3$ matched the string $abc$.
This function 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}
\begin{tabular}{l@{\hspace{2mm}}l}
& $\big(\underline{(\ZERO+\ZERO+\ZERO+ \ZERO+ \ONE \cdot \ONE \cdot \ONE) \cdot r^*}
\;+\; (\ZERO+\ZERO+\ZERO + \ONE + \ZERO) \cdot r^*\big)$ \smallskip\\
$+$ & $\big((\ZERO+\ONE+\ZERO + \ZERO + \ZERO) \cdot r^*
\;+\; (\ZERO+\ZERO+\ZERO + \ONE + \ZERO) \cdot r^* \big)$
\end{tabular}
\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
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 two 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 lexical value 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 two substrings.
Finally, after injecting character $a$ back to $v_1$,
we get the lexical value 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 lexical value 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 of these alternatives is chosen
later does not affect their relative position because the 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 lexical values 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 lexical value 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 lexing 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}
$\textit{pder} \; c \; \ZERO$ & $\dn$ & $\emptyset$\\
$\textit{pder} \; c \; \ONE$ & $\dn$ & $\emptyset$ \\
$\textit{pder} \; c \; d$ & $\dn$ & $\textit{if} \; c \,=\, d \; \{ \ONE \} \; \textit{else} \; \emptyset$ \\
$\textit{pder} \; c \; r_1+r_2$ & $\dn$ & $pder \; c \; r_1 \cup pder \; c \; r_2$ \\
$\textit{pder} \; c \; r_1 \cdot r_2$ & $\dn$ & $\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 sum of
the size of \emph{all} partial derivatives no matter what the string
looks like. Roughly speaking the size sum will be at most cubic in the
size of the regular expression.
If we want the size of derivatives in Sulzmann and Lu's algorithm to
stay 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 $S$ and $Z$ are arbitrary names for the bits 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:%\comment{ALTS should have an $as$ in the definitions, not just $a_1$ and $a_2$}
\begin{center}
\begin{tabular}{lcl}
$\textit{a}$ & $::=$ & $\textit{ZERO}$\\
& $\mid$ & $\textit{ONE}\;\;bs$\\
& $\mid$ & $\textit{CHAR}\;\;bs\,c$\\
& $\mid$ & $\textit{ALTS}\;\;bs\,as$\\
& $\mid$ & $\textit{SEQ}\;\;bs\,a_1\,a_2$\\
& $\mid$ & $\textit{STAR}\;\;bs\,a$
\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($\textit{ALTS}$) 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}
$(\ZERO)^\uparrow$ & $\dn$ & $\textit{ZERO}$\\
$(\ONE)^\uparrow$ & $\dn$ & $\textit{ONE}\,[]$\\
$(c)^\uparrow$ & $\dn$ & $\textit{CHAR}\,[]\,c$\\
$(r_1 + r_2)^\uparrow$ & $\dn$ &
$\textit{ALTS}\;[]\,List((\textit{fuse}\,[\Z]\,r_1^\uparrow),\,
(\textit{fuse}\,[\S]\,r_2^\uparrow))$\\
$(r_1\cdot r_2)^\uparrow$ & $\dn$ &
$\textit{SEQ}\;[]\,r_1^\uparrow\,r_2^\uparrow$\\
$(r^*)^\uparrow$ & $\dn$ &
$\textit{STAR}\;[]\,r^\uparrow$\\
\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}
$\textit{fuse}\;bs\,(\textit{ZERO})$ & $\dn$ & $\textit{ZERO}$\\
$\textit{fuse}\;bs\,(\textit{ONE}\,bs')$ & $\dn$ &
$\textit{ONE}\,(bs\,@\,bs')$\\
$\textit{fuse}\;bs\,(\textit{CHAR}\,bs'\,c)$ & $\dn$ &
$\textit{CHAR}\,(bs\,@\,bs')\,c$\\
$\textit{fuse}\;bs\,(\textit{ALTS}\,bs'\,as)$ & $\dn$ &
$\textit{ALTS}\,(bs\,@\,bs')\,as$\\
$\textit{fuse}\;bs\,(\textit{SEQ}\,bs'\,a_1\,a_2)$ & $\dn$ &
$\textit{SEQ}\,(bs\,@\,bs')\,a_1\,a_2$\\
$\textit{fuse}\;bs\,(\textit{STAR}\,bs'\,a)$ & $\dn$ &
$\textit{STAR}\,(bs\,@\,bs')\,a$
\end{tabular}
\end{center}
\noindent
After 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:
%\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}
\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 $ALTS$, 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}
$\textit{bmkeps}\,(\textit{ONE}\;bs)$ & $\dn$ & $bs$\\
$\textit{bmkeps}\,(\textit{ALTS}\;bs\,a::as)$ & $\dn$ &
$\textit{if}\;\textit{bnullable}\,a$\\
& &$\textit{then}\;bs\,@\,\textit{bmkeps}\,a$\\
& &$\textit{else}\;bs\,@\,\textit{bmkeps}\,(\textit{ALTS}\;bs\,as)$\\
$\textit{bmkeps}\,(\textit{SEQ}\;bs\,a_1\,a_2)$ & $\dn$ &
$bs \,@\,\textit{bmkeps}\,a_1\,@\, \textit{bmkeps}\,a_2$\\
$\textit{bmkeps}\,(\textit{STAR}\;bs\,a)$ & $\dn$ &
$bs \,@\, [\S]$
\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 $ALTS$ 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} $ \\
&&$\quad\textit{case} \; (\ZERO, \_) \Rightarrow \ZERO$ \\
&&$\quad\textit{case} \; (\_, \ZERO) \Rightarrow \ZERO$ \\
&&$\quad\textit{case} \; (\ONE, a_2') \Rightarrow \textit{fuse} \; bs \; a_2'$ \\
&&$\quad\textit{case} \; (a_1', \ONE) \Rightarrow \textit{fuse} \; bs \; a_1'$ \\
&&$\quad\textit{case} \; (a_1', a_2') \Rightarrow \textit{SEQ} \; bs \; a_1' \; a_2'$ \\
$\textit{simp} \; (\textit{ALTS}\;bs\,as)$ & $\dn$ & $\textit{distinct}( \textit{flatten} ( \textit{map simp as})) \; \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
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 $\textit{flatten}$ and $\textit{distinct}$ to open up nested
$\textit{ALTS}$ 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 \textit{ALTS}.
Its recursive definition is given below:
\begin{center}
\begin{tabular}{@{}lcl@{}}
$\textit{flatten} \; (\textit{ALTS}\;bs\,as) :: as'$ & $\dn$ & $(\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 $\textit{flatten}$ 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 $a+b+c$.
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_{simp} [\,] $ & $\dn$ & $r$
\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 tree [our lexical values]. 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 like to 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}
%\comment{If i remember correctly, you have proved this lemma.
%I feel this is indeed not true because you might place arbitrary
%bits on the regex r, however if this is the case, did i remember wrongly that
%you proved something like simplification does not affect $\textit{bmkeps}$ results?
%Anyway, i have amended this a little bit so it does not allow arbitrary bits attached
%to a regex. Maybe it works now.}
\begin{center}
$\textit{bmkeps} \; a = \textit{bmkeps} \; \textit{bsimp} \; a\;($\textit{provided}$ \; a\; is \; \textit{bnullable} )$
\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{blexer}\_{simp}(r, \; s) = \textit{blexer}(r, \; s)$
\end{center}
%comment{This is not true either...look at the definion blexer/blexer-simp}
\noindent That is, if we do derivative on regular expression $r$ and then
simplify it, and repeat this process until we exhaust the string, we get a
regular expression $r''$($\textit{LHS}$) that provides the POSIX matching
information, which is exactly the same as the result $r'$($\textit{RHS}$ of the
normal derivative algorithm that only does derivative repeatedly and has no
simplification at all. This might seem at first glance very unintuitive, as
the $r'$ could be exponentially larger than $r''$, but can be explained in the
following way: we are pruning away the possible matches that are not POSIX.
Since there could be exponentially many
non-POSIX matchings and only 1 POSIX matching, it
is understandable that our $r''$ can be a lot smaller. we 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 $r''$ might have
become very different. 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@{}}
$\textit{retrieve}\,(\textit{ONE}\,bs)\,\Empty$ & $\dn$ & $bs$\\
$\textit{retrieve}\,(\textit{CHAR}\,bs\,c)\,(\Char\,d)$ & $\dn$ & $bs$\\
$\textit{retrieve}\,(\textit{ALTS}\,bs\,a::as)\,(\Left\,v)$ & $\dn$ &
$bs \,@\, \textit{retrieve}\,a\,v$\\
$\textit{retrieve}\,(\textit{ALTS}\,bs\,a::as)\,(\Right\,v)$ & $\dn$ &
$bs \,@\, \textit{retrieve}\,(\textit{ALTS}\,bs\,as)\,v$\\
$\textit{retrieve}\,(\textit{SEQ}\,bs\,a_1\,a_2)\,(\Seq\,v_1\,v_2)$ & $\dn$ &
$bs \,@\,\textit{retrieve}\,a_1\,v_1\,@\, \textit{retrieve}\,a_2\,v_2$\\
$\textit{retrieve}\,(\textit{STAR}\,bs\,a)\,(\Stars\,[])$ & $\dn$ &
$bs \,@\, [\S]$\\
$\textit{retrieve}\,(\textit{STAR}\,bs\,a)\,(\Stars\,(v\!::\!vs))$ & $\dn$ &\\
\multicolumn{3}{l}{
\hspace{3cm}$bs \,@\, [\Z] \,@\, \textit{retrieve}\,a\,v\,@\,
\textit{retrieve}\,(\textit{STAR}\,[]\,a)\,(\Stars\,vs)$}\\
\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
\begin{center}
$\textit{retrieve} \; (r^\uparrow \backslash s) \; v = \;\textit{retrieve} \;
(\textit{simp}(r^\uparrow) \backslash s) \; v'$
\end{center}
\noindent
and subsequently
\begin{center}
$\textit{retrieve} \; (r^\uparrow \backslash s) \; v\; = \; \textit{retrieve} \;
(r^\uparrow \backslash_{simp} \, s) \; v'$.
\end{center}
\noindent
The $\textit{LHS}$ of the above equation is the bitcode we want. This
would prove that our simplified version of regular expression still
contains all the bitcodes needed. The task here is to find a way to
compute the correct $v'$.
The second task is to speed up the more aggressive simplification. Currently
it is slower than the original naive simplification by Ausaf and Urban (the
naive version as implemented by Ausaf and Urban of course can ``explode'' in
some cases). It is therefore not surprising that the speed is also much slower
than regular expression engines in popular programming languages such as Java
and Python on most inputs that are linear. For example, just by rewriting the
example regular expression in the beginning of this report $(a^*)^*\,b$ into
$a^*\,b$ would eliminate the ambiguity in the matching and make the time
for matching linear with respect to the input string size. This allows the
DFA approach to become blindingly fast, and dwarf the speed of our current
implementation. For example, here is a comparison of Java regex engine
and our implementation on this example.
\begin{center}
\begin{tabular}{@{}c@{\hspace{0mm}}c@{\hspace{0mm}}c@{}}
\begin{tikzpicture}
\begin{axis}[
xlabel={$n*1000$},
x label style={at={(1.05,-0.05)}},
ylabel={time in secs},
enlargelimits=false,
xtick={0,5,...,30},
xmax=33,
ymax=9,
scaled ticks=true,
axis lines=left,
width=5cm,
height=4cm,
legend entries={Bitcoded Algorithm},
legend pos=north west,
legend cell align=left]
\addplot[red,mark=*, mark options={fill=white}] table {bad-scala.data};
\end{axis}
\end{tikzpicture}
&
\begin{tikzpicture}
\begin{axis}[
xlabel={$n*1000$},
x label style={at={(1.05,-0.05)}},
%ylabel={time in secs},
enlargelimits=false,
xtick={0,5,...,30},
xmax=33,
ymax=9,
scaled ticks=false,
axis lines=left,
width=5cm,
height=4cm,
legend entries={Java},
legend pos=north west,
legend cell align=left]
\addplot[cyan,mark=*, mark options={fill=white}] table {good-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}
Java regex engine can match string of thousands of characters in a few milliseconds,
whereas our current algorithm gets excruciatingly slow on input of this size.
The running time in theory is linear, however it does not appear to be the
case in an actual implementation. So it needs to be explored how to
make our algorithm faster on all inputs. It could be the recursive calls that are
needed to manipulate bits that are causing the slow down. A possible solution
is to write recursive functions into tail-recusive form.
Another possibility would be to explore
again the connection to DFAs to speed up the algorithm on
subcalls that are small enough. 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.
We also plan to implement the bitcoded algorithm
in some imperative language like C to see if the inefficiency of the
Scala implementation
is language specific. To make this research more comprehensive we also plan
to contrast our (faster) version of bitcoded algorithm with the
Symbolic Regex Matcher, the RE2, the Rust Regex Engine, and the static
analysis approach by implementing them in the same language and then compare
their performance.
\bibliographystyle{plain}
\bibliography{root}
\end{document}