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-% Chapter Template
-
-\chapter{Regular Expressions and POSIX Lexing} % Main chapter title
-
-\label{Chapter2} % In chapter 2 \ref{Chapter2} we will introduce the concepts
-%and notations we
-%use for describing the lexing algorithm by Sulzmann and Lu,
-%and then give the algorithm and its variant, and discuss
-%why more aggressive simplifications are needed.
-
-
-\section{Basic Concepts and Notations for Strings, Languages, and Regular Expressions}
-We have a primitive datatype char, denoting characters.
-\[ char ::= a
- \mid b
- \mid c
- \mid \ldots
- \mid z
-\]
-(which one is better?)
-\begin{center}
-\begin{tabular}{lcl}
-$\textit{char}$ & $\dn$ & $a | b | c | \ldots$\\
-\end{tabular}
-\end{center}
-They can form strings by lists:
-\begin{center}
-\begin{tabular}{lcl}
-$\textit{string}$ & $\dn$ & $[] | c :: cs$\\
-& & $(c\; \text{has char type})$
-\end{tabular}
-\end{center}
-And strings can be concatenated to form longer strings:
-\begin{center}
-\begin{tabular}{lcl}
-$[] @ s_2$ & $\dn$ & $s_2$\\
-$(c :: s_1) @ s_2$ & $\dn$ & $c :: (s_1 @ s_2)$
-\end{tabular}
-\end{center}
-
-A set of strings can operate with another set of strings:
-\begin{center}
-\begin{tabular}{lcl}
-$A @ B $ & $\dn$ & $\{s_A @ s_B \mid s_A \in A; s_B \in B \}$\\
-\end{tabular}
-\end{center}
-We also call the above "language concatenation".
-The power of a language is defined recursively, using the
-concatenation operator $@$:
-\begin{center}
-\begin{tabular}{lcl}
-$A^0 $ & $\dn$ & $\{ [] \}$\\
-$A^{n+1}$ & $\dn$ & $A^n @ A$
-\end{tabular}
-\end{center}
-The union of all the natural number powers of a language
-is denoted by the Kleene star operator:
-\begin{center}
-\begin{tabular}{lcl}
-$\bigcup_{i \geq 0} A^i$ & $\denote$ & $A^*$\\
-\end{tabular}
-\end{center}
-
-In Isabelle of course we cannot easily get a counterpart of
-the $\bigcup_{i \geq 0}$ operator, so we instead define the Kleene star
-as an inductive set:
-\begin{center}
-\begin{tabular}{lcl}
-$[] \in A^*$ & &\\
-$s_1 \in A \land \; s_2 \in A^* $ & $\implies$ & $s_1 @ s_2 \in A^*$\\
-\end{tabular}
-\end{center}
-
-We also define an operation of chopping off a character from
-a language:
-\begin{center}
-\begin{tabular}{lcl}
-$\textit{Der} \;c \;A$ & $\dn$ & $\{ s \mid c :: s \in A \}$\\
-\end{tabular}
-\end{center}
-
-This can be generalised to chopping off a string from all strings within set $A$:
-\begin{center}
-\begin{tabular}{lcl}
-$\textit{Ders} \;w \;A$ & $\dn$ & $\{ s \mid w@s \in A \}$\\
-\end{tabular}
-\end{center}
-
-which is essentially the left quotient $A \backslash L'$ of $A$ against
-the singleton language $L' = \{w\}$
-in formal language theory.
-For this dissertation the $\textit{Ders}$ notation would suffice, there is
-no need for a more general derivative definition.
-
-With the sequencing, Kleene star, and $\textit{Der}$ operator on languages,
-we have a few properties of how the language derivative can be defined using
-sub-languages.
-\begin{lemma}
-$\Der \; c \; (A @ B) = \textit{if} \; [] \in A \; \textit{then} ((\Der \; c \; A) @ B ) \cup \Der \; c\; B \quad \textit{else}\; (\Der \; c \; A) @ B$
-\end{lemma}
-\noindent
-This lemma states that if $A$ contains the empty string, $\Der$ can "pierce" through it
-and get to $B$.
-
-The language $A^*$'s derivative can be described using the language derivative
-of $A$:
-\begin{lemma}
-$\textit{Der} \;c \;A^* = (\textit{Der}\; c A) @ (A^*)$\\
-\end{lemma}
-\begin{proof}
-\begin{itemize}
-\item{$\subseteq$}
-The set
-\[ \{s \mid c :: s \in A^*\} \]
-is enclosed in the set
-\[ \{s_1 @ s_2 \mid s_1 \, s_2. s_1 \in \{s \mid c :: s \in A\} \land s_2 \in A^* \} \]
-because whenever you have a string starting with a character
-in the language of a Kleene star $A^*$, then that character together with some sub-string
-immediately after it will form the first iteration, and the rest of the string will
-be still in $A^*$.
-\item{$\supseteq$}
-Note that
-\[ \Der \; c \; A^* = \Der \; c \; (\{ [] \} \cup (A @ A^*) ) \]
-and
-\[ \Der \; c \; (\{ [] \} \cup (A @ A^*) ) = \Der\; c \; (A @ A^*) \]
-where the $\textit{RHS}$ of the above equatioin can be rewritten
-as \[ (\Der \; c\; A) @ A^* \cup A' \], $A'$ being a possibly empty set.
-\end{itemize}
-\end{proof}
-Before we define the $\textit{Der}$ and $\textit{Ders}$ counterpart
-for regular languages, we need to first give definitions for regular expressions.
-
-\section{Regular Expressions and Their Language Interpretation}
-Suppose we have an alphabet $\Sigma$, the strings whose characters
-are from $\Sigma$
-can be expressed as $\Sigma^*$.
-
-We use patterns to define a set of strings concisely. Regular expressions
-are one of such patterns systems:
-The basic regular expressions are defined inductively
- by the following grammar:
-\[ r ::= \ZERO \mid \ONE
- \mid c
- \mid r_1 \cdot r_2
- \mid r_1 + r_2
- \mid r^*
-\]
-
-The language or set of strings defined by regular expressions are defined as
-%TODO: FILL in the other defs
-\begin{center}
-\begin{tabular}{lcl}
-$L \; (r_1 + r_2)$ & $\dn$ & $ L \; (r_1) \cup L \; ( r_2)$\\
-$L \; (r_1 \cdot r_2)$ & $\dn$ & $ L \; (r_1) \cap L \; (r_2)$\\
-\end{tabular}
-\end{center}
-Which are also called the "language interpretation".
-
-
-
-
-% Derivatives of a
-%regular expression, written $r \backslash c$, give a simple solution
-%to the problem of matching a string $s$ with a regular
-%expression $r$: if the derivative of $r$ w.r.t.\ (in
-%succession) all the characters of the string matches the empty string,
-%then $r$ matches $s$ (and {\em vice versa}).
-
-
-\section{Brzozowski Derivatives of Regular Expressions}
-Now with semantic derivatives of a language and regular expressions and
-their language interpretations, we are ready to define derivatives on regexes.
-
-The Brzozowski derivative w.r.t character $c$ is an operation on the regex,
-where the operation transforms the regex to a new one containing
-strings without the head character $c$.
-
-The derivative of regular expression, denoted as
-$r \backslash c$, is a function that takes parameters
-$r$ and $c$, and returns another regular expression $r'$,
-which is computed by the following recursive function:
-
-\begin{center}
-\begin{tabular}{lcl}
- $\ZERO \backslash c$ & $\dn$ & $\ZERO$\\
- $\ONE \backslash c$ & $\dn$ & $\ZERO$\\
- $d \backslash c$ & $\dn$ &
- $\mathit{if} \;c = d\;\mathit{then}\;\ONE\;\mathit{else}\;\ZERO$\\
-$(r_1 + r_2)\backslash c$ & $\dn$ & $r_1 \backslash c \,+\, r_2 \backslash c$\\
-$(r_1 \cdot r_2)\backslash c$ & $\dn$ & $\mathit{if} \, nullable(r_1)$\\
- & & $\mathit{then}\;(r_1\backslash c) \cdot r_2 \,+\, r_2\backslash c$\\
- & & $\mathit{else}\;(r_1\backslash c) \cdot r_2$\\
- $(r^*)\backslash c$ & $\dn$ & $(r\backslash c) \cdot r^*$\\
-\end{tabular}
-\end{center}
-\noindent
-The function derivative, written $r\backslash c$,
-defines how a regular expression evolves into
-a new regular expression after all the string it contains
-is chopped off a certain head character $c$.
-The most involved cases are the sequence
-and star case.
-The sequence case says that if the first regular expression
-contains an empty string then the second component of the sequence
-might be chosen as the target regular expression to be chopped
-off its head character.
-The star regular expression's derivative unwraps the iteration of
-regular expression and attaches the star regular expression
-to the sequence's second element to make sure a copy is retained
-for possible more iterations in later phases of lexing.
-
-
-The $\nullable$ function tests whether the empty string $""$
-is in the language of $r$:
-
-
-\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}
-\noindent
-The empty set does not contain any string and
-therefore not the empty string, the empty string
-regular expression contains the empty string
-by definition, the character regular expression
-is the singleton that contains character only,
-and therefore does not contain the empty string,
-the alternative regular expression (or "or" expression)
-might have one of its children regular expressions
-being nullable and any one of its children being nullable
-would suffice. The sequence regular expression
-would require both children to have the empty string
-to compose an empty string and the Kleene star
-operation naturally introduced the empty string.
-
-We have the following property where the derivative on regular
-expressions coincides with the derivative on a set of strings:
-
-\begin{lemma}
-$\textit{Der} \; c \; L(r) = L (r\backslash c)$
-\end{lemma}
-
-\noindent
-
-
-The main property of the derivative operation
-that enables us to reason about the correctness of
-an algorithm using derivatives is
-
-\begin{center}
-$c\!::\!s \in L(r)$ holds
-if and only if $s \in L(r\backslash c)$.
-\end{center}
-
-\noindent
-We can generalise the derivative operation shown above for single characters
-to strings as follows:
-
-\begin{center}
-\begin{tabular}{lcl}
-$r \backslash (c\!::\!s) $ & $\dn$ & $(r \backslash c) \backslash s$ \\
-$r \backslash [\,] $ & $\dn$ & $r$
-\end{tabular}
-\end{center}
-
-\noindent
-and then define Brzozowski's regular-expression matching algorithm as:
-
-\begin{definition}
-$match\;s\;r \;\dn\; nullable(r\backslash s)$
-\end{definition}
-
-\noindent
-Assuming the string is given as a sequence of characters, say $c_0c_1..c_n$,
-this algorithm presented graphically is as follows:
-
-\begin{equation}\label{graph:successive_ders}
-\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 $s \in L(r_0)$).
-
-Beautiful and simple definition.
-
-If we implement the above algorithm naively, however,
-the algorithm can be excruciatingly slow.
-
-
-\begin{figure}
-\centering
-\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=10000,
- ytick={0,1000,...,10000},
- 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 {EightThousandNodes.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}
-\caption{EightThousandNodes} \label{fig:EightThousandNodes}
-\end{figure}
-
-
-(8000 node data to be added here)
-For example, when starting with the regular
-expression $(a + aa)^*$ and building a few successive derivatives (around 10)
-w.r.t.~the character $a$, one obtains a derivative regular expression
-with more than 8000 nodes (when viewed as a tree)\ref{EightThousandNodes}.
-The reason why $(a + aa) ^*$ explodes so drastically is that without
-pruning, the algorithm will keep records of all possible ways of matching:
-\begin{center}
-$(a + aa) ^* \backslash [aa] = (\ZERO + \ONE \ONE)\cdot(a + aa)^* + (\ONE + \ONE a) \cdot (a + aa)^*$
-\end{center}
-
-\noindent
-Each of the above alternative branches correspond to the match
-$aa $, $a \quad a$ and $a \quad a \cdot (a)$(incomplete).
-These different ways of matching will grow exponentially with the string length,
-and without simplifications that throw away some of these very similar matchings,
-it is no surprise that these expressions grow so quickly.
-Operations like
-$\backslash$ and $\nullable$ need to traverse such trees and
-consequently the bigger the size of the derivative the slower the
-algorithm.
-
-Brzozowski was quick in finding that during this process a lot useless
-$\ONE$s and $\ZERO$s are generated and therefore not optimal.
-He also introduced some "similarity rules", such
-as $P+(Q+R) = (P+Q)+R$ to merge syntactically
-different but language-equivalent sub-regexes to further decrease the size
-of the intermediate regexes.
-
-More simplifications are possible, such as deleting duplicates
-and opening up nested alternatives to trigger even more simplifications.
-And suppose we apply simplification after each derivative step, and compose
-these two operations together as an atomic one: $a \backslash_{simp}\,c \dn
-\textit{simp}(a \backslash c)$. Then we can build
-a matcher with simpler regular expressions.
-
-If we want the size of derivatives in the algorithm to
-stay even lower, we would need more aggressive simplifications.
-Essentially we need to delete useless $\ZERO$s and $\ONE$s, as well as
-delete duplicates whenever possible. For example, the parentheses in
-$(a+b) \cdot c + b\cdot c$ can be opened up to get $a\cdot c + b \cdot c + b
-\cdot c$, and then simplified to just $a \cdot c + b \cdot c$. Another
-example is simplifying $(a^*+a) + (a^*+ \ONE) + (a +\ONE)$ to just
-$a^*+a+\ONE$. These more aggressive simplification rules are for
- a very tight size bound, possibly as low
- as that of the \emph{partial derivatives}\parencite{Antimirov1995}.
-
-Building derivatives and then simplifying them.
-So far, so good. But what if we want to
-do lexing instead of just getting a YES/NO answer?
-This requires us to go back again to the world
-without simplification first for a moment.
-Sulzmann and Lu~\cite{Sulzmann2014} first came up with a nice and
-elegant(arguably as beautiful as the original
-derivatives definition) solution for this.
-
-\subsection*{Values and the Lexing Algorithm by Sulzmann and Lu}
-
-
-They first defined the datatypes for storing the
-lexing information 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
-
-Building on top of Sulzmann and Lu's attempt to formalise the
-notion of POSIX lexing rules \parencite{Sulzmann2014},
-Ausaf and Urban\parencite{AusafDyckhoffUrban2016} modelled
-POSIX matching as a ternary relation recursively defined in a
-natural deduction style.
-With the formally-specified rules for what a POSIX matching is,
-they proved in Isabelle/HOL that the algorithm gives correct results.
-
-But having a correct result is still not enough,
-we want at least some degree of $\mathbf{efficiency}$.
-
-
-
-One regular expression can have multiple lexical values. For example
-for the regular expression $(a+b)^*$, it has a infinite list of
-values corresponding to it: $\Stars\,[]$, $\Stars\,[\Left(Char(a))]$,
-$\Stars\,[\Right(Char(b))]$, $\Stars\,[\Left(Char(a),\,\Right(Char(b))]$,
-$\ldots$, and vice versa.
-Even for the regular expression matching a particular string, there could
-be more than one value corresponding to it.
-Take the example where $r= (a^*\cdot a^*)^*$ and the string
-$s=\underbrace{aa\ldots a}_\text{n \textit{a}s}$.
-If we do not allow any empty iterations in its lexical values,
-there will be $n - 1$ "splitting points" on $s$ we can choose to
-split or not so that each sub-string
-segmented by those chosen splitting points will form different iterations:
-\begin{center}
-\begin{tabular}{lcr}
-$a \mid aaa $ & $\rightarrow$ & $\Stars\, [v_{iteration \,a},\, v_{iteration \,aaa}]$\\
-$aa \mid aa $ & $\rightarrow$ & $\Stars\, [v_{iteration \, aa},\, v_{iteration \, aa}]$\\
-$a \mid aa\mid a $ & $\rightarrow$ & $\Stars\, [v_{iteration \, a},\, v_{iteration \, aa}, \, v_{iteration \, a}]$\\
- & $\textit{etc}.$ &
- \end{tabular}
-\end{center}
-
-And for each iteration, there are still multiple ways to split
-between the two $a^*$s.
-It is not surprising there are exponentially many lexical values
-that are distinct for the regex and string pair $r= (a^*\cdot a^*)^*$ and
-$s=\underbrace{aa\ldots a}_\text{n \textit{a}s}$.
-
-A lexer to keep all the possible values will naturally
-have an exponential runtime on ambiguous regular expressions.
-Somehow one has to decide which lexical value to keep and
-output in a lexing algorithm.
-In practice, we are usually
-interested in POSIX values, which by intuition always
-\begin{itemize}
-\item
-match the leftmost regular expression when multiple options of matching
-are available
-\item
-always match a subpart as much as possible before proceeding
-to the next token.
-\end{itemize}
-The formal definition of a $\POSIX$ value $v$ for a regular expression
-$r$ and string $s$, denoted as $(s, r) \rightarrow v$, can be specified
-in the following set of rules:
-(TODO: write the entire set of inference rules for POSIX )
-\newcommand*{\inference}[3][t]{%
- \begingroup
- \def\and{\\}%
- \begin{tabular}[#1]{@{\enspace}c@{\enspace}}
- #2 \\
- \hline
- #3
- \end{tabular}%
- \endgroup
-}
-\begin{center}
-\inference{$s_1 @ s_2 = s$ \and $(\nexists s_3 s_4 s_5. s_1 @ s_5 = s_3 \land s_5 \neq [] \land s_3 @ s_4 = s \land (s_3, r_1) \rightarrow v_3 \land (s_4, r_2) \rightarrow v_4)$ \and $(s_1, r_1) \rightarrow v_1$ \and $(s_2, r_2) \rightarrow v_2$ }{$(s, r_1 \cdot r_2) \rightarrow \Seq(v_1, v_2)$ }
-\end{center}
-
-The reason why we are interested in $\POSIX$ values is that they can
-be practically used in the lexing phase of a compiler front end.
-For instance, when lexing a code snippet
-$\textit{iffoo} = 3$ with the regular expression $\textit{keyword} + \textit{identifier}$, we want $\textit{iffoo}$ to be recognized
-as an identifier rather than a keyword.
-
- For example, the above $r= (a^*\cdot a^*)^*$ and
-$s=\underbrace{aa\ldots a}_\text{n \textit{a}s}$ example has the POSIX value
-$ \Stars\,[\Seq(Stars\,[\underbrace{\Char(a),\ldots,\Char(a)}_\text{n iterations}], Stars\,[])]$.
-The output of an algorithm we want would be a POSIX matching
-encoded as a value.
-
-
-
-
-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:successive_ders}). In this second phase, a POSIX value
-is generated if the regular expression matches the string.
-How can we construct a value out of regular expressions and character
-sequences only?
-Two functions are involved: $\inj$ and $\mkeps$.
-The function $\mkeps$ constructs a value from the last
-one of all the successive derivatives:
-\begin{ceqn}
-\begin{equation}\label{graph:mkeps}
-\begin{tikzcd}
-r_0 \arrow[r, "\backslash c_0"] & r_1 \arrow[r, "\backslash c_1"] & r_2 \arrow[r, dashed] & r_n \arrow[d, "mkeps" description] \\
- & & & v_n
-\end{tikzcd}
-\end{equation}
-\end{ceqn}
-
-It tells us how can an empty string be matched by a
-regular expression, in a $\POSIX$ way:
-
- \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
-We favour the left to match an empty string if there is a choice.
-When there is a star for us to match the empty string,
-we give the $\Stars$ constructor an empty list, meaning
-no iterations are taken.
-
-
-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 $s$. The POSIX value is maintained throughout the process.
-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.
-\begin{ceqn}
-\begin{equation}\label{graph:inj}
-\begin{tikzcd}
-r_1 \arrow[r, dashed] \arrow[d]& r_i \arrow[r, "\backslash c_i"] \arrow[d] & r_{i+1} \arrow[r, dashed] \arrow[d] & r_n \arrow[d, "mkeps" description] \\
-v_1 \arrow[u] & v_i \arrow[l, dashed] & v_{i+1} \arrow[l,"inj_{r_i} c_i"] & v_n \arrow[l, dashed]
-\end{tikzcd}
-\end{equation}
-\end{ceqn}
-
-
-\noindent
-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.
-The clauses do one thing--identifying the ``hole'' on a
-value to inject the character back into.
-For instance, in the last clause for injecting back to a value
-that would turn into a new star value that corresponds to a star,
-we know it must be a sequence value. And we know that the first
-value of that sequence corresponds to the child regex of the star
-with the first character being chopped off--an iteration of the star
-that had just been unfolded. This value is followed by the already
-matched star iterations we collected before. So we inject the character
-back to the first value and form a new value with this latest iteration
-being added to the previous list of iterations, all under the $\Stars$
-top level.
-
-Putting all the functions $\inj$, $\mkeps$, $\backslash$ together,
-we have a lexer with the following recursive definition:
-\begin{center}
-\begin{tabular}{lcr}
-$\lexer \; r \; [] $ & $=$ & $\mkeps \; r$\\
-$\lexer \; r \;c::s$ & $=$ & $\inj \; r \; c (\lexer (r\backslash c) s)$
-\end{tabular}
-\end{center}
-
-Pictorially, the 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 right to left. For the first value $v_n$, we call the
-function $\textit{mkeps}$, which builds a POSIX lexical value
-for how the empty string has been matched by the (nullable) regular
-expression $r_n$. This function is defined as
-
-
-
-We have mentioned before that derivatives without simplification
-can get clumsy, and this is true for values as well--they reflect
-the size of the regular expression by definition.
-
-One can introduce simplification on the regex and values but have to
-be careful not to break the correctness, as the injection
-function heavily relies on the structure of the regexes and values
-being correct and matching each other.
-It can be achieved by recording some extra rectification functions
-during the derivatives step, and applying these rectifications in
-each run during the injection phase.
-And we can prove that the POSIX value of how
-regular expressions match strings will not be affected---although it is much harder
-to establish.
-Some initial results in this regard have been
-obtained in \cite{AusafDyckhoffUrban2016}.
-
-
-
-%Brzozowski, after giving the derivatives and simplification,
-%did not explore lexing with simplification, or he may well be
-%stuck on an efficient simplification with proof.
-%He went on to examine the use of derivatives together with
-%automaton, and did not try lexing using products.
-
-We want to get rid of the complex and fragile rectification of values.
-Can we not create those intermediate values $v_1,\ldots v_n$,
-and get the lexing information that should be already there while
-doing derivatives in one pass, without a second injection phase?
-In the meantime, can we make sure that simplifications
-are easily handled without breaking the correctness of the algorithm?
-
-Sulzmann and Lu solved this problem by
-introducing additional information to the
-regular expressions called \emph{bitcodes}.
-
-
-
-
-
-
-