% Chapter Template
\chapter{Finiteness Bound} % Main chapter title
\label{Finite}
% In Chapter 4 \ref{Chapter4} we give the second guarantee
%of our bitcoded algorithm, that is a finite bound on the size of any
%regex's derivatives.
In this chapter we give a guarantee in terms of time complexity:
given a regular expression $r$, for any string $s$
our algorithm's internal data structure is finitely bounded.
To obtain such a proof, we need to
\begin{itemize}
\item
Define an new datatype for regular expressions that makes it easy
to reason about the size of an annotated regular expression.
\item
A set of equalities for this new datatype that enables one to
rewrite $\bderssimp{r_1 \cdot r_2}{s}$ and $\bderssimp{r^*}{s}$ etc.
by their children regexes $r_1$, $r_2$, and $r$.
\item
Using those equalities to actually get those rewriting equations, which we call
"closed forms".
\item
Bound the closed forms, thereby bounding the original
$\blexersimp$'s internal data structures.
\end{itemize}
\section{the $\mathbf{r}$-rexp datatype and the size functions}
We have a size function for bitcoded regular expressions, written
$\llbracket r\rrbracket$, which counts the number of nodes if we regard $r$ as a tree
\begin{center}
\begin{tabular}{ccc}
$\llbracket \ACHAR{bs}{c} \rrbracket $ & $\dn$ & $1$\\
\end{tabular}
\end{center}
(TODO: COMPLETE this defn and for $rs$)
The size is based on a recursive function on the structure of the regex,
not the bitcodes.
Therefore we may as well talk about size of an annotated regular expression
in their un-annotated form:
\begin{center}
$\asize(a) = \size(\erase(a))$.
\end{center}
(TODO: turn equals to roughly equals)
But there is a minor nuisance here:
the erase function actually messes with the structure of the regular expression:
\begin{center}
\begin{tabular}{ccc}
$\erase(\AALTS{bs}{[]} )$ & $\dn$ & $\ZERO$\\
\end{tabular}
\end{center}
(TODO: complete this definition with singleton r in alts)
An alternative regular expression with an empty list of children
is turned into an $\ZERO$ during the
$\erase$ function, thereby changing the size and structure of the regex.
These will likely be fixable if we really want to use plain $\rexp$s for dealing
with size, but we choose a more straightforward (or stupid) method by
defining a new datatype that is similar to plain $\rexp$s but can take
non-binary arguments for its alternative constructor,
which we call $\rrexp$ to denote
the difference between it and plain regular expressions.
\[ \rrexp ::= \RZERO \mid \RONE
\mid \RCHAR{c}
\mid \RSEQ{r_1}{r_2}
\mid \RALTS{rs}
\mid \RSTAR{r}
\]
For $\rrexp$ we throw away the bitcodes on the annotated regular expressions,
but keep everything else intact.
It is similar to annotated regular expressions being $\erase$-ed, but with all its structure preserved
(the $\erase$ function unfortunately does not preserve structure in the case of empty and singleton
$\ALTS$).
We denote the operation of erasing the bits and turning an annotated regular expression
into an $\rrexp{}$ as $\rerase{}$.
\begin{center}
\begin{tabular}{lcl}
$\rerase{\AZERO}$ & $=$ & $\RZERO$\\
$\rerase{\ASEQ{bs}{r_1}{r_2}}$ & $=$ & $\RSEQ{\rerase{r_1}}{\rerase{r_2}}$\\
$\rerase{\AALTS{bs}{rs}}$ & $ =$ & $ \RALTS{rs}$
\end{tabular}
\end{center}
%TODO: FINISH definition of rerase
Similarly we could define the derivative and simplification on
$\rrexp$, which would be identical to those we defined for plain $\rexp$s in chapter1,
except that now they can operate on alternatives taking multiple arguments.
\begin{center}
\begin{tabular}{lcr}
$\RALTS{rs} \backslash c$ & $=$ & $\RALTS{map\; (\_ \backslash c) \;rs}$\\
(other clauses omitted)
\end{tabular}
\end{center}
Now that $\rrexp$s do not have bitcodes on them, we can do the
duplicate removal without an equivalence relation:
\begin{center}
\begin{tabular}{lcl}
$\rdistinct{r :: rs}{rset}$ & $=$ & $\textit{if}(r \in \textit{rset}) \; \textit{then} \; \rdistinct{rs}{rset}$\\
& & $\textit{else}\; r::\rdistinct{rs}{(rset \cup \{r\})}$
\end{tabular}
\end{center}
%TODO: definition of rsimp (maybe only the alternative clause)
The reason why these definitions mirror precisely the corresponding operations
on annotated regualar expressions is that we can calculate sizes more easily:
\begin{lemma}
$\rsize{\rerase a} = \asize a$
\end{lemma}
This lemma says that the size of an r-erased regex is the same as the annotated regex.
this does not hold for plain $\rexp$s due to their ways of how alternatives are represented.
\begin{lemma}
$\asize{\bsimp{a}} = \rsize{\rsimp{\rerase{a}}}$
\end{lemma}
Putting these two together we get a property that allows us to estimate the
size of an annotated regular expression derivative using its un-annotated counterpart:
\begin{lemma}
$\asize{\bderssimp{r}{s}} = \rsize{\rderssimp{\rerase{r}}{s}}$
\end{lemma}
Unless stated otherwise in this chapter all $\textit{rexp}$s without
bitcodes are seen as $\rrexp$s.
We also use $r_1 + r_2$ and $\RALTS{[r_1, r_2]}$ interchageably
as the former suits people's intuitive way of stating a binary alternative
regular expression.
%-----------------------------------
% SECTION ?
%-----------------------------------
\section{preparatory properties for getting a finiteness bound}
Before we get to the proof that says the intermediate result of our lexer will
remain finitely bounded, which is an important efficiency/liveness guarantee,
we shall first develop a few preparatory properties and definitions to
make the process of proving that a breeze.
We define rewriting relations for $\rrexp$s, which allows us to do the
same trick as we did for the correctness proof,
but this time we will have stronger equalities established.
\subsection{"hrewrite" relation}
List of 1-step rewrite rules for regular expressions simplification without bitcodes:
\begin{figure}
\caption{the "h-rewrite" rules}
\[
r_1 \cdot \ZERO \rightarrow_h \ZERO \]
\[
\ZERO \cdot r_2 \rightarrow \ZERO
\]
\end{figure}
And we define an "grewrite" relation that works on lists:
\begin{center}
\begin{tabular}{lcl}
$ \ZERO :: rs$ & $\rightarrow_g$ & $rs$
\end{tabular}
\end{center}
With these relations we prove
\begin{lemma}
$rs \rightarrow rs' \implies \RALTS{rs} \rightarrow \RALTS{rs'}$
\end{lemma}
which enables us to prove "closed forms" equalities such as
\begin{lemma}
$\sflat{(r_1 \cdot r_2) \backslash s} = \RALTS{ (r_1 \backslash s) \cdot r_2 :: (\map (r_2 \backslash \_) (\suffix \; s \; r_1 ))}$
\end{lemma}
But the most involved part of the above lemma is proving the following:
\begin{lemma}
$\bsimp{\RALTS{rs @ \RALTS{rs_1} @ rs'}} = \bsimp{\RALTS{rs @rs_1 @ rs'}} $
\end{lemma}
which is needed in proving things like
\begin{lemma}
$r \rightarrow_f r' \implies \rsimp{r} \rightarrow \rsimp{r'}$
\end{lemma}
Fortunately this is provable by induction on the list $rs$
%-----------------------------------
% SECTION 2
%-----------------------------------
\begin{theorem}
For any regex $r$, $\exists N_r. \forall s. \; \llbracket{\bderssimp{r}{s}}\rrbracket \leq N_r$
\end{theorem}
\noindent
\begin{proof}
We prove this by induction on $r$. The base cases for $\AZERO$,
$\AONE \textit{bs}$ and $\ACHAR \textit{bs} c$ are straightforward. The interesting case is
for sequences of the form $\ASEQ{bs}{r_1}{r_2}$. In this case our induction
hypotheses state $\exists N_1. \forall s. \; \llbracket \bderssimp{r}{s} \rrbracket \leq N_1$ and
$\exists N_2. \forall s. \; \llbracket \bderssimp{r_2}{s} \rrbracket \leq N_2$. We can reason as follows
%
\begin{center}
\begin{tabular}{lcll}
& & $ \llbracket \bderssimp{\ASEQ{bs}{r_1}{r_2} }{s} \rrbracket$\\
& $ = $ & $\llbracket bsimp\,(\textit{ALTs}\;bs\;(\ASEQ{nil}{\bderssimp{ r_1}{s}}{ r_2} ::
[\bderssimp{ r_2}{s'} \;|\; s' \in \textit{Suffix}( r_1, s)]))\rrbracket $ & (1) \\
& $\leq$ &
$\llbracket\textit{\distinctWith}\,((\ASEQ{nil}{\bderssimp{r_1}{s}}{r_2}$) ::
[$\bderssimp{ r_2}{s'} \;|\; s' \in \textit{Suffix}( r_1, s)])\,\approx\;{}\rrbracket + 1 $ & (2) \\
& $\leq$ & $\llbracket\ASEQ{bs}{\bderssimp{ r_1}{s}}{r_2}\rrbracket +
\llbracket\textit{distinctWith}\,[\bderssimp{r_2}{s'} \;|\; s' \in \textit{Suffix}( r_1, s)]\,\approx\;{}\rrbracket + 1 $ & (3) \\
& $\leq$ & $N_1 + \llbracket r_2\rrbracket + 2 +
\llbracket \distinctWith\,[ \bderssimp{r_2}{s'} \;|\; s' \in \textit{Suffix}( r_1, s)] \,\approx\;\rrbracket$ & (4)\\
& $\leq$ & $N_1 + \llbracket r_2\rrbracket + 2 + l_{N_{2}} * N_{2}$ & (5)
\end{tabular}
\end{center}
\noindent
where in (1) the $\textit{Suffix}( r_1, s)$ are the all the suffixes of $s$ where $\bderssimp{ r_1}{s'}$ is nullable ($s'$ being a suffix of $s$).
The reason why we could write the derivatives of a sequence this way is described in section 2.
The term (2) is used to control (1).
That is because one can obtain an overall
smaller regex list
by flattening it and removing $\ZERO$s first before applying $\distinctWith$ on it.
Section 3 is dedicated to its proof.
In (3) we know that $\llbracket \ASEQ{bs}{(\bderssimp{ r_1}{s}}{r_2}\rrbracket$ is
bounded by $N_1 + \llbracket{}r_2\rrbracket + 1$. In (5) we know the list comprehension contains only regular expressions of size smaller
than $N_2$. The list length after $\distinctWith$ is bounded by a number, which we call $l_{N_2}$. It stands
for the number of distinct regular expressions smaller than $N_2$ (there can only be finitely many of them).
We reason similarly for $\STAR$.\medskip
\end{proof}
What guarantee does this bound give us?
Whatever the regex is, it will not grow indefinitely.
Take our previous example $(a + aa)^*$ as an example:
\begin{center}
\begin{tabular}{@{}c@{\hspace{0mm}}c@{\hspace{0mm}}c@{}}
\begin{tikzpicture}
\begin{axis}[
xlabel={number of $a$'s},
x label style={at={(1.05,-0.05)}},
ylabel={regex size},
enlargelimits=false,
xtick={0,5,...,30},
xmax=33,
ymax= 40,
ytick={0,10,...,40},
scaled ticks=false,
axis lines=left,
width=5cm,
height=4cm,
legend entries={$(a + aa)^*$},
legend pos=north west,
legend cell align=left]
\addplot[red,mark=*, mark options={fill=white}] table {a_aa_star.data};
\end{axis}
\end{tikzpicture}
\end{tabular}
\end{center}
We are able to limit the size of the regex $(a + aa)^*$'s derivatives
with our simplification
rules very effectively.
In our proof for the inductive case $r_1 \cdot r_2$, the dominant term in the bound
is $l_{N_2} * N_2$, where $N_2$ is the bound we have for $\llbracket \bderssimp{r_2}{s} \rrbracket$.
Given that $l_{N_2}$ is roughly the size $4^{N_2}$, the size bound $\llbracket \bderssimp{r_1 \cdot r_2}{s} \rrbracket$
inflates the size bound of $\llbracket \bderssimp{r_2}{s} \rrbracket$ with the function
$f(x) = x * 2^x$.
This means the bound we have will surge up at least
tower-exponentially with a linear increase of the depth.
For a regex of depth $n$, the bound
would be approximately $4^n$.
Test data in the graphs from randomly generated regular expressions
shows that the giant bounds are far from being hit.
%a few sample regular experessions' derivatives
%size change
%TODO: giving regex1_size_change.data showing a few regexes' size changes
%w;r;t the input characters number, where the size is usually cubic in terms of original size
%a*, aa*, aaa*, .....
%randomly generated regexes
\begin{center}
\begin{tabular}{@{}c@{\hspace{0mm}}c@{\hspace{0mm}}c@{}}
\begin{tikzpicture}
\begin{axis}[
xlabel={number of $a$'s},
x label style={at={(1.05,-0.05)}},
ylabel={regex size},
enlargelimits=false,
xtick={0,5,...,30},
xmax=33,
ymax=1000,
ytick={0,100,...,1000},
scaled ticks=false,
axis lines=left,
width=5cm,
height=4cm,
legend entries={regex1},
legend pos=north west,
legend cell align=left]
\addplot[red,mark=*, mark options={fill=white}] table {regex1_size_change.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=1000,
ytick={0,100,...,1000},
scaled ticks=false,
axis lines=left,
width=5cm,
height=4cm,
legend entries={regex2},
legend pos=north west,
legend cell align=left]
\addplot[blue,mark=*, mark options={fill=white}] table {regex2_size_change.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=1000,
ytick={0,100,...,1000},
scaled ticks=false,
axis lines=left,
width=5cm,
height=4cm,
legend entries={regex3},
legend pos=north west,
legend cell align=left]
\addplot[cyan,mark=*, mark options={fill=white}] table {regex3_size_change.data};
\end{axis}
\end{tikzpicture}\\
\multicolumn{3}{c}{Graphs: size change of 3 randomly generated regexes $w.r.t.$ input string length.}
\end{tabular}
\end{center}
\noindent
Most of the regex's sizes seem to stay within a polynomial bound $w.r.t$ the
original size.
This suggests a link towrads "partial derivatives"
introduced by Antimirov \cite{Antimirov95}.
\section{Antimirov's partial derivatives}
The idea behind Antimirov's partial derivatives
is to do derivatives in a similar way as suggested by Brzozowski,
but maintain a set of regular expressions instead of a single one:
%TODO: antimirov proposition 3.1, needs completion
\begin{center}
\begin{tabular}{ccc}
$\partial_x(a+b)$ & $=$ & $\partial_x(a) \cup \partial_x(b)$\\
$\partial_x(\ONE)$ & $=$ & $\phi$
\end{tabular}
\end{center}
Rather than joining the calculated derivatives $\partial_x a$ and $\partial_x b$ together
using the alternatives constructor, Antimirov cleverly chose to put them into
a set instead. This breaks the terms in a derivative regular expression up,
allowing us to understand what are the "atomic" components of it.
For example, To compute what regular expression $x^*(xx + y)^*$'s
derivative against $x$ is made of, one can do a partial derivative
of it and get two singleton sets $\{x^* \cdot (xx + y)^*\}$ and $\{x \cdot (xx + y) ^* \}$
from $\partial_x(x^*) \cdot (xx + y) ^*$ and $\partial_x((xx + y)^*)$.
To get all the "atomic" components of a regular expression's possible derivatives,
there is a procedure Antimirov called $\textit{lf}$, short for "linear forms", that takes
whatever character is available at the head of the string inside the language of a
regular expression, and gives back the character and the derivative regular expression
as a pair (which he called "monomial"):
\begin{center}
\begin{tabular}{ccc}
$\lf(\ONE)$ & $=$ & $\phi$\\
$\lf(c)$ & $=$ & $\{(c, \ONE) \}$\\
$\lf(a+b)$ & $=$ & $\lf(a) \cup \lf(b)$\\
$\lf(r^*)$ & $=$ & $\lf(r) \bigodot \lf(r^*)$\\
\end{tabular}
\end{center}
%TODO: completion
There is a slight difference in the last three clauses compared
with $\partial$: instead of a dot operator $ \textit{rset} \cdot r$ that attaches the regular
expression $r$ with every element inside $\textit{rset}$ to create a set of
sequence derivatives, it uses the "circle dot" operator $\bigodot$ which operates
on a set of monomials (which Antimirov called "linear form") and a regular
expression, and returns a linear form:
\begin{center}
\begin{tabular}{ccc}
$l \bigodot (\ZERO)$ & $=$ & $\phi$\\
$l \bigodot (\ONE)$ & $=$ & $l$\\
$\phi \bigodot t$ & $=$ & $\phi$\\
$\{ (x, \ZERO) \} \bigodot t$ & $=$ & $\{(x,\ZERO) \}$\\
$\{ (x, \ONE) \} \bigodot t$ & $=$ & $\{(x,t) \}$\\
$\{ (x, p) \} \bigodot t$ & $=$ & $\{(x,p\cdot t) \}$\\
$\lf(a+b)$ & $=$ & $\lf(a) \cup \lf(b)$\\
$\lf(r^*)$ & $=$ & $\lf(r) \cdot \lf(r^*)$\\
\end{tabular}
\end{center}
%TODO: completion
Some degree of simplification is applied when doing $\bigodot$, for example,
$l \bigodot (\ZERO) = \phi$ corresponds to $r \cdot \ZERO \rightsquigarrow \ZERO$,
and $l \bigodot (\ONE) = l$ to $l \cdot \ONE \rightsquigarrow l$, and
$\{ (x, \ZERO) \} \bigodot t = \{(x,\ZERO) \}$ to $\ZERO \cdot x \rightsquigarrow \ZERO$,
and so on.
With the function $\lf$ one can compute all possible partial derivatives $\partial_{UNIV}(r)$ of a regex $r$ with
an iterative procedure:
\begin{center}
\begin{tabular}{llll}
$\textit{while}$ & $(\Delta_i \neq \phi)$ & & \\
& $\Delta_{i+1}$ & $ =$ & $\lf(\Delta_i) - \PD_i$ \\
& $\PD_{i+1}$ & $ =$ & $\Delta_{i+1} \cup \PD_i$ \\
$\partial_{UNIV}(r)$ & $=$ & $\PD$ &
\end{tabular}
\end{center}
$(r_1 + r_2) \cdot r_3 \longrightarrow (r_1 \cdot r_3) + (r_2 \cdot r_3)$,
However, if we introduce them in our
setting we would lose the POSIX property of our calculated values.
A simple example for this would be the regex $(a + a\cdot b)\cdot(b\cdot c + c)$.
If we split this regex up into $a\cdot(b\cdot c + c) + a\cdot b \cdot (b\cdot c + c)$, the lexer
would give back $\Left(\Seq(\Char(a), \Left(\Char(b \cdot c))))$ instead of
what we want: $\Seq(\Right(ab), \Right(c))$. Unless we can store the structural information
in all the places where a transformation of the form $(r_1 + r_2)\cdot r \rightarrow r_1 \cdot r + r_2 \cdot r$
occurs, and apply them in the right order once we get
a result of the "aggressively simplified" regex, it would be impossible to still get a $\POSIX$ value.
This is unlike the simplification we had before, where the rewriting rules
such as $\ONE \cdot r \rightsquigarrow r$, under which our lexer will give the same value.
We will discuss better
bounds in the last section of this chapter.\\[-6.5mm]
%----------------------------------------------------------------------------------------
% SECTION ??
%----------------------------------------------------------------------------------------
\section{"Closed Forms" of regular expressions' derivatives w.r.t strings}
To embark on getting the "closed forms" of regexes, we first
define a few auxiliary definitions to make expressing them smoothly.
\begin{center}
\begin{tabular}{ccc}
$\sflataux{\AALTS{ }{r :: rs}}$ & $=$ & $\sflataux{r} @ rs$\\
$\sflataux{\AALTS{ }{[]}}$ & $ = $ & $ []$\\
$\sflataux r$ & $=$ & $ [r]$
\end{tabular}
\end{center}
The intuition behind $\sflataux{\_}$ is to break up nested regexes
of the $(\ldots((r_1 + r_2) + r_3) + \ldots )$(left-associated) shape
into a more "balanced" list: $\AALTS{\_}{[r_1,\, r_2 ,\, r_3, \ldots]}$.
It will return the singleton list $[r]$ otherwise.
$\sflat{\_}$ works the same as $\sflataux{\_}$, except that it keeps
the output type a regular expression, not a list:
\begin{center}
\begin{tabular}{ccc}
$\sflat{\AALTS{ }{r :: rs}}$ & $=$ & $\sflataux{r} @ rs$\\
$\sflat{\AALTS{ }{[]}}$ & $ = $ & $ \AALTS{ }{[]}$\\
$\sflat r$ & $=$ & $ [r]$
\end{tabular}
\end{center}
$\sflataux{\_}$ and $\sflat{\_}$ is only recursive in terms of the
first element of the list of children of
an alternative regex ($\AALTS{ }{rs}$), and is purposefully built for dealing with the regular expression
$r_1 \cdot r_2 \backslash s$.
With $\sflat{\_}$ and $\sflataux{\_}$ we make
precise what "closed forms" we have for the sequence derivatives and their simplifications,
in other words, how can we construct $(r_1 \cdot r_2) \backslash s$
and $\bderssimp{(r_1\cdot r_2)}{s}$,
if we are only allowed to use a combination of $r_1 \backslash s'$ ($\bderssimp{r_1}{s'}$)
and $r_2 \backslash s'$ ($\bderssimp{r_2}{s'}$), where $s'$ ranges over
the substring of $s$?
First let's look at a series of derivatives steps on a sequence
regular expression, (assuming) that each time the first
component of the sequence is always nullable):
\begin{center}
$r_1 \cdot r_2 \quad \longrightarrow_{\backslash c} \quad r_1 \backslash c \cdot r_2 + r_2 \backslash c \quad \longrightarrow_{\backslash c'} \quad (r_1 \backslash cc' \cdot r_2 + r_2 \backslash c') + r_2 \backslash cc' \longrightarrow_{\backslash c''} \quad$\\
$((r_1 \backslash cc'c'' \cdot r_2 + r_2 \backslash c'') + r_2 \backslash c'c'') + r_2 \backslash cc'c'' \longrightarrow_{\backslash c''} \quad
\ldots$
\end{center}
%TODO: cite indian paper
Indianpaper have come up with a slightly more formal way of putting the above process:
\begin{center}
$r_1 \cdot r_2 \backslash (c_1 :: c_2 ::\ldots c_n) \myequiv r_1 \backslash (c_1 :: c_2:: \ldots c_n) +
\delta(\nullable(r_1 \backslash (c_1 :: c_2 \ldots c_{n-1}) ), r_2 \backslash (c_n)) + \ldots
+ \delta (\nullable(r_1), r_2\backslash (c_1 :: c_2 ::\ldots c_n))$
\end{center}
where $\delta(b, r)$ will produce $r$
if $b$ evaluates to true,
and $\ZERO$ otherwise.
But the $\myequiv$ sign in the above equation means language equivalence instead of syntactical
equivalence. To make this intuition useful
for a formal proof, we need something
stronger than language equivalence.
With the help of $\sflat{\_}$ we can state the equation in Indianpaper
more rigorously:
\begin{lemma}
$\sflat{(r_1 \cdot r_2) \backslash s} = \RALTS{ (r_1 \backslash s) \cdot r_2 :: (\map (r_2 \backslash \_) (\vsuf{s}{r_1}))}$
\end{lemma}
The function $\vsuf{\_}{\_}$ is defined recursively on the structure of the string:
\begin{center}
\begin{tabular}{lcl}
$\vsuf{[]}{\_} $ & $=$ & $[]$\\
$\vsuf{c::cs}{r_1}$ & $ =$ & $ \textit{if} (\rnullable{r_1}) \textit{then} \; (\vsuf{cs}{(\rder{c}{r_1})}) @ [c :: cs]$\\
&& $\textit{else} \; (\vsuf{cs}{(\rder{c}{r_1}) }) $
\end{tabular}
\end{center}
It takes into account which prefixes $s'$ of $s$ would make $r \backslash s'$ nullable,
and keep a list of suffixes $s''$ such that $s' @ s'' = s$. The list is sorted in
the order $r_2\backslash s''$ appears in $(r_1\cdot r_2)\backslash s$.
In essence, $\vsuf{\_}{\_}$ is doing a "virtual derivative" of $r_1 \cdot r_2$, but instead of producing
the entire result of $(r_1 \cdot r_2) \backslash s$,
it only stores all the $s''$ such that $r_2 \backslash s''$ are occurring terms in $(r_1\cdot r_2)\backslash s$.
With this we can also add simplifications to both sides and get
\begin{lemma}
$\rsimp{\sflat{(r_1 \cdot r_2) \backslash s} }= \rsimp{\AALTS{[[]}{ (r_1 \backslash s) \cdot r_2 :: (\map (r_2 \backslash \_) (\vsuf{s}{r_1}))}}$
\end{lemma}
Together with the idempotency property of $\rsimp$,
%TODO: state the idempotency property of rsimp
\begin{lemma}
$\rsimp{r} = \rsimp{\rsimp{r}}$
\end{lemma}
it is possible to convert the above lemma to obtain a "closed form"
for derivatives nested with simplification:
\begin{lemma}
$\rderssimp{(r_1 \cdot r_2)}{s} = \rsimp{\AALTS{[[]}{ (r_1 \backslash s) \cdot r_2 :: (\map (r_2 \backslash \_) (\vsuf{s}{r_1}))}}$
\end{lemma}
And now the reason we have (1) in section 1 is clear:
$\rsize{\rsimp{\RALTS{ (r_1 \backslash s) \cdot r_2 :: (\map \;(r_2 \backslash \_) \; (\vsuf{s}{r_1}))}}}$,
is equal to $\rsize{\rsimp{\RALTS{ ((r_1 \backslash s) \cdot r_2 :: (\map \; (r_2 \backslash \_) \; (\textit{Suffix}(r1, s))))}}}$
as $\vsuf{s}{r_1}$ is one way of expressing the list $\textit{Suffix}( r_1, s)$.
%----------------------------------------------------------------------------------------
% SECTION 3
%----------------------------------------------------------------------------------------
\section{interaction between $\distinctWith$ and $\flts$}
Note that it is not immediately obvious that
\begin{center}
$\llbracket \distinctWith (\flts \textit{rs}) = \phi \rrbracket \leq \llbracket \distinctWith \textit{rs} = \phi \rrbracket $.
\end{center}
The intuition is that if we remove duplicates from the $\textit{LHS}$, at least the same amount of
duplicates will be removed from the list $\textit{rs}$ in the $\textit{RHS}$.
%-----------------------------------
% SECTION syntactic equivalence under simp
%-----------------------------------
\section{Syntactic Equivalence Under $\simp$}
We prove that minor differences can be annhilated
by $\simp$.
For example,
\begin{center}
$\simp \;(\simpALTs\; (\map \;(\_\backslash \; x)\; (\distinct \; \mathit{rs}\; \phi))) =
\simp \;(\simpALTs \;(\distinct \;(\map \;(\_ \backslash\; x) \; \mathit{rs}) \; \phi))$
\end{center}
%----------------------------------------------------------------------------------------
% SECTION ALTS CLOSED FORM
%----------------------------------------------------------------------------------------
\section{A Closed Form for \textit{ALTS}}
Now we prove that $rsimp (rders\_simp (RALTS rs) s) = rsimp (RALTS (map (\lambda r. rders\_simp r s) rs))$.
There are a few key steps, one of these steps is
\[
rsimp (rsimp\_ALTs (map (rder x) (rdistinct (rflts (map (rsimp \circ (\lambda r. rders\_simp r xs)) rs)) {})))
= rsimp (rsimp\_ALTs (rdistinct (map (rder x) (rflts (map (rsimp \circ (\lambda r. rders\_simp r xs)) rs))) {}))
\]
One might want to prove this by something a simple statement like:
$map (rder x) (rdistinct rs rset) = rdistinct (map (rder x) rs) (rder x) ` rset)$.
For this to hold we want the $\textit{distinct}$ function to pick up
the elements before and after derivatives correctly:
$r \in rset \equiv (rder x r) \in (rder x rset)$.
which essentially requires that the function $\backslash$ is an injective mapping.
Unfortunately the function $\backslash c$ is not an injective mapping.
\subsection{function $\backslash c$ is not injective (1-to-1)}
\begin{center}
The derivative $w.r.t$ character $c$ is not one-to-one.
Formally,
$\exists r_1 \;r_2. r_1 \neq r_2 \mathit{and} r_1 \backslash c = r_2 \backslash c$
\end{center}
This property is trivially true for the
character regex example:
\begin{center}
$r_1 = e; \; r_2 = d;\; r_1 \backslash c = \ZERO = r_2 \backslash c$
\end{center}
But apart from the cases where the derivative
output is $\ZERO$, are there non-trivial results
of derivatives which contain strings?
The answer is yes.
For example,
\begin{center}
Let $r_1 = a^*b\;\quad r_2 = (a\cdot a^*)\cdot b + b$.\\
where $a$ is not nullable.\\
$r_1 \backslash c = ((a \backslash c)\cdot a^*)\cdot c + b \backslash c$\\
$r_2 \backslash c = ((a \backslash c)\cdot a^*)\cdot c + b \backslash c$
\end{center}
We start with two syntactically different regexes,
and end up with the same derivative result.
This is not surprising as we have such
equality as below in the style of Arden's lemma:\\
\begin{center}
$L(A^*B) = L(A\cdot A^* \cdot B + B)$
\end{center}
%----------------------------------------------------------------------------------------
% SECTION 4
%----------------------------------------------------------------------------------------
\section{A Bound for the Star Regular Expression}
We have shown how to control the size of the sequence regular expression $r_1\cdot r_2$ using
the "closed form" of $(r_1 \cdot r_2) \backslash s$ and then
the property of the $\distinct$ function.
Now we try to get a bound on $r^* \backslash s$ as well.
Again, we first look at how a star's derivatives evolve, if they grow maximally:
\begin{center}
$r^* \quad \longrightarrow_{\backslash c} \quad (r\backslash c) \cdot r^* \quad \longrightarrow_{\backslash c'} \quad
r \backslash cc' \cdot r^* + r \backslash c' \cdot r^* \quad \longrightarrow_{\backslash c''} \quad
(r_1 \backslash cc'c'' \cdot r^* + r \backslash c'') + (r \backslash c'c'' \cdot r^* + r \backslash c'' \cdot r^*) \quad \longrightarrow_{\backslash c'''}
\quad \ldots$
\end{center}
When we have a string $s = c :: c' :: c'' \ldots$ such that $r \backslash c$, $r \backslash cc'$, $r \backslash c'$,
$r \backslash cc'c''$, $r \backslash c'c''$, $r\backslash c''$ etc. are all nullable,
the number of terms in $r^* \backslash s$ will grow exponentially, causing the size
of the derivatives $r^* \backslash s$ to grow exponentially, even if we do not
count the possible size explosions of $r \backslash c$ themselves.
Thanks to $\flts$ and $\distinctWith$, we are able to open up regexes like
$(r_1 \backslash cc'c'' \cdot r^* + r \backslash c'') + (r \backslash c'c'' \cdot r^* + r \backslash c'' \cdot r^*) $
into $\RALTS{[r_1 \backslash cc'c'' \cdot r^*, r \backslash c'', r \backslash c'c'' \cdot r^*, r \backslash c'' \cdot r^*]}$
and then de-duplicate terms of the form $r\backslash s' \cdot r^*$ ($s'$ being a substring of $s$).
For this we define $\hflataux{\_}$ and $\hflat{\_}$, similar to $\sflataux{\_}$ and $\sflat{\_}$:
%TODO: definitions of and \hflataux \hflat
\begin{center}
\begin{tabular}{ccc}
$\hflataux{r_1 + r_2}$ & $=$ & $\hflataux{r_1} @ \hflataux{r_2}$\\
$\hflataux r$ & $=$ & $ [r]$
\end{tabular}
\end{center}
\begin{center}
\begin{tabular}{ccc}
$\hflat{r_1 + r_2}$ & $=$ & $\RALTS{\hflataux{r_1} @ \hflataux{r_2}}$\\
$\hflat r$ & $=$ & $ r$
\end{tabular}
\end{center}s
Again these definitions are tailor-made for dealing with alternatives that have
originated from a star's derivatives, so we don't attempt to open up all possible
regexes of the form $\RALTS{rs}$, where $\textit{rs}$ might not contain precisely 2
elements.
We give a predicate for such "star-created" regular expressions:
\begin{center}
\begin{tabular}{lcr}
& & $\createdByStar{(\RSEQ{ra}{\RSTAR{rb}}) }$\\
$\createdByStar{r_1} \land \createdByStar{r_2} $ & $ \Longrightarrow$ & $\createdByStar{(r_1 + r_2)}$
\end{tabular}
\end{center}
These definitions allows us the flexibility to talk about
regular expressions in their most convenient format,
for example, flattened out $\RALTS{[r_1, r_2, \ldots, r_n]} $
instead of binary-nested: $((r_1 + r_2) + (r_3 + r_4)) + \ldots$.
These definitions help express that certain classes of syntatically
distinct regular expressions are actually the same under simplification.
This is not entirely true for annotated regular expressions:
%TODO: bsimp bders \neq bderssimp
\begin{center}
$(1+ (c\cdot \ASEQ{bs}{c^*}{c} ))$
\end{center}
For bit-codes, the order in which simplification is applied
might cause a difference in the location they are placed.
If we want something like
\begin{center}
$\bderssimp{r}{s} \myequiv \bsimp{\bders{r}{s}}$
\end{center}
Some "canonicalization" procedure is required,
which either pushes all the common bitcodes to nodes
as senior as possible:
\begin{center}
$_{bs}(_{bs_1 @ bs'}r_1 + _{bs_1 @ bs''}r_2) \rightarrow _{bs @ bs_1}(_{bs'}r_1 + _{bs''}r_2) $
\end{center}
or does the reverse. However bitcodes are not of interest if we are talking about
the $\llbracket r \rrbracket$ size of a regex.
Therefore for the ease and simplicity of producing a
proof for a size bound, we are happy to restrict ourselves to
unannotated regular expressions, and obtain such equalities as
\begin{lemma}
$\rsimp{r_1 + r_2} = \rsimp{\RALTS{\hflataux{r_1} @ \hflataux{r_2}}}$
\end{lemma}
\begin{proof}
By using the rewriting relation $\rightsquigarrow$
\end{proof}
%TODO: rsimp sflat
And from this we obtain a proof that a star's derivative will be the same
as if it had all its nested alternatives created during deriving being flattened out:
For example,
\begin{lemma}
$\createdByStar{r} \implies \rsimp{r} = \rsimp{\RALTS{\hflataux{r}}}$
\end{lemma}
\begin{proof}
By structural induction on $r$, where the induction rules are these of $\createdByStar{_}$.
\end{proof}
% The simplification of a flattened out regular expression, provided it comes
%from the derivative of a star, is the same as the one nested.
One might wonder the actual bound rather than the loose bound we gave
for the convenience of an easier proof.
How much can the regex $r^* \backslash s$ grow?
As earlier graphs have shown,
%TODO: reference that graph where size grows quickly
they can grow at a maximum speed
exponential $w.r.t$ the number of characters,
but will eventually level off when the string $s$ is long enough.
If they grow to a size exponential $w.r.t$ the original regex, our algorithm
would still be slow.
And unfortunately, we have concrete examples
where such regexes grew exponentially large before levelling off:
$(a ^ * + (aa) ^ * + (aaa) ^ * + \ldots +
(\underbrace{a \ldots a}_{\text{n a's}})^*$ will already have a maximum
size that is exponential on the number $n$
under our current simplification rules:
%TODO: graph of a regex whose size increases exponentially.
\begin{center}
\begin{tikzpicture}
\begin{axis}[
height=0.5\textwidth,
width=\textwidth,
xlabel=number of a's,
xtick={0,...,9},
ylabel=maximum size,
ymode=log,
log basis y={2}
]
\addplot[mark=*,blue] table {re-chengsong.data};
\end{axis}
\end{tikzpicture}
\end{center}
For convenience we use $(\oplus_{i=1}^{n} (\underbrace{a \ldots a}_{\text{i a's}})^*)^*$
to express $(a ^ * + (aa) ^ * + (aaa) ^ * + \ldots +
(\underbrace{a \ldots a}_{\text{n a's}})^*$ in the below discussion.
The exponential size is triggered by that the regex
$\oplus_{i=1}^{n} (\underbrace{a \ldots a}_{\text{i a's}})^*$
inside the $(\ldots) ^*$ having exponentially many
different derivatives, despite those difference being minor.
$(\oplus_{i=1}^{n} (\underbrace{a \ldots a}_{\text{i a's}})^*)^*\backslash \underbrace{a \ldots a}_{\text{m a's}}$
will therefore contain the following terms (after flattening out all nested
alternatives):
\begin{center}
$(\oplus_{i = 1]{n} (\underbrace{a \ldots a}_{\text{((i - (m' \% i))\%i) a's}})\cdot (\underbrace{a \ldots a}_{\text{i a's}})^* })\cdot (\oplus_{i=1}^{n} (\underbrace{a \ldots a}_{\text{i a's}})^*)$\\
$(1 \leq m' \leq m )$
\end{center}
These terms are distinct for $m' \leq L.C.M.(1, \ldots, n)$ (will be explained in appendix).
With each new input character taking the derivative against the intermediate result, more and more such distinct
terms will accumulate,
until the length reaches $L.C.M.(1, \ldots, n)$.
$\textit{distinctBy}$ will not be able to de-duplicate any two of these terms
$(\oplus_{i = 1}^{n} (\underbrace{a \ldots a}_{\text{((i - (m' \% i))\%i) a's}})\cdot (\underbrace{a \ldots a}_{\text{i a's}})^* )\cdot (\oplus_{i=1}^{n} (\underbrace{a \ldots a}_{\text{i a's}})^*)^*$\\
$(\oplus_{i = 1}^{n} (\underbrace{a \ldots a}_{\text{((i - (m'' \% i))\%i) a's}})\cdot (\underbrace{a \ldots a}_{\text{i a's}})^* )\cdot (\oplus_{i=1}^{n} (\underbrace{a \ldots a}_{\text{i a's}})^*)^*$\\
where $m' \neq m''$ \\
as they are slightly different.
This means that with our current simplification methods,
we will not be able to control the derivative so that
$\llbracket \bderssimp{r}{s} \rrbracket$ stays polynomial %\leq O((\llbracket r\rrbacket)^c)$
as there are already exponentially many terms.
These terms are similar in the sense that the head of those terms
are all consisted of sub-terms of the form:
$(\underbrace{a \ldots a}_{\text{j a's}})\cdot (\underbrace{a \ldots a}_{\text{i a's}})^* $.
For $\oplus_{i=1}^{n} (\underbrace{a \ldots a}_{\text{i a's}})^*$, there will be at most
$n * (n + 1) / 2$ such terms.
For example, $(a^* + (aa)^* + (aaa)^*) ^*$'s derivatives
can be described by 6 terms:
$a^*$, $a\cdot (aa)^*$, $ (aa)^*$,
$aa \cdot (aaa)^*$, $a \cdot (aaa)^*$, and $(aaa)^*$.
The total number of different "head terms", $n * (n + 1) / 2$,
is proportional to the number of characters in the regex
$(\oplus_{i=1}^{n} (\underbrace{a \ldots a}_{\text{i a's}})^*)^*$.
This suggests a slightly different notion of size, which we call the
alphabetic width:
%TODO:
(TODO: Alphabetic width def.)
Antimirov\parencite{Antimirov95} has proven that
$\textit{PDER}_{UNIV}(r) \leq \textit{awidth}(r)$.
where $\textit{PDER}_{UNIV}(r)$ is a set of all possible subterms
created by doing derivatives of $r$ against all possible strings.
If we can make sure that at any moment in our lexing algorithm our
intermediate result hold at most one copy of each of the
subterms then we can get the same bound as Antimirov's.
This leads to the algorithm in the next chapter.
%----------------------------------------------------------------------------------------
% SECTION 1
%----------------------------------------------------------------------------------------
\section{Idempotency of $\simp$}
\begin{equation}
\simp \;r = \simp\; \simp \; r
\end{equation}
This property means we do not have to repeatedly
apply simplification in each step, which justifies
our definition of $\blexersimp$.
It will also be useful in future proofs where properties such as
closed forms are needed.
The proof is by structural induction on $r$.
%-----------------------------------
% SUBSECTION 1
%-----------------------------------
\subsection{Syntactic Equivalence Under $\simp$}
We prove that minor differences can be annhilated
by $\simp$.
For example,
\begin{center}
$\simp \;(\simpALTs\; (\map \;(\_\backslash \; x)\; (\distinct \; \mathit{rs}\; \phi))) =
\simp \;(\simpALTs \;(\distinct \;(\map \;(\_ \backslash\; x) \; \mathit{rs}) \; \phi))$
\end{center}