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\chapter{Chapter Title Here} % Main chapter title

\label{ChapterX} % Change X to a consecutive number; for referencing this chapter elsewhere, use \ref{ChapterX}




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\section{Properties of $\backslash c$}

To have a clear idea of what we can and 
need to prove about the algorithms involving
Brzozowski's derivatives, there are a few 
properties we need to be clear about 
it.
\subsection{function $\backslash c$ is not 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, which is
a "meaningful" regex because it contains strings.
We have rediscovered Arden's lemma:\\
\begin{center}
	$A^*B = A\cdot A^* \cdot B + B$
\end{center}

	
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\subsection{Subsection 1}

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\section{Finiteness Property}
In this section let us sketch our argument for why the size of the simplified
derivatives with the aggressive simplification function can be finitely bounded.
For convenience, we use a new datatype which we call $\textit{rrexp}$ to denote
the difference between it and annotated regular expressions. 
For $\rrexp$ we throw away the bitcodes on the annotated regular expressions, but keeps
everything else intact.
It is similar to annotated regular expression being $\erase$ed, but with all its structure being intact
(the $\erase$ function unfortunately does not preserve structure in the case of empty and singleton
$\AALTS$.
We denote the operation of erasing the bits and turning an annotated regular expression 
into an $\rrexp$ as $\rerase$.
%TODO: definition of rerase
That we can bound the size of annotated regular expressions by 
$\rrexp$ is that the bitcodes grow linearly w.r.t the input string, and don't contribute to the structural size of 
an annotated regular expression:
$\rsize (\rerase a) = \asize a$

 Suppose
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
(we omit the precise definition; ditto for lists $\llbracket r\!s\rrbracket$). For this we show that for every $r$
there exists a bound $N$
such that 

\begin{center}
$\forall s. \; \llbracket{\bderssimp r s}\rrbracket \leq N$
\end{center}

\noindent
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 $\ASEQ \textit{bs} r_1 r_2$. In this case our induction
hypotheses state $\exists N_1. \forall s. \; \llbracket \bderssimp(r_1, s) \rrbracket \leq N_1$ 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\, \textit{bs} \, r_1 \,r_2),  s) \rrbracket$\\
& $ = $ & $\llbracket bsimp\,(\textit{ALTs}\;bs\;((ASEQ [] (\bderssimp r_1 s) r_2) ::
    [\bderssimp r_2 s' \;|\; s' \in \textit{Suffix}( r_1, s)]))\rrbracket $ & (1) \\
& $\leq$ &
    $\llbracket\textit{\distinctWith}\,((\ASEQ [] (\bderssimp r_1 s) r_2$) ::
    [$\bderssimp r_2 s' \;|\; s' \in \textit{Suffix}( r_1, s)])\,\approx\;{}\rrbracket + 1 $ & (2) \\
& $\leq$ & $\llbracket\ASEQ [] (\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}

% tell Chengsong about Indian paper of closed forms of derivatives

\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) since it we know that you 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  $\llbracket\ASEQ [] (\bderssimp r_1 s) r_2\rrbracket$ 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

\noindent
Clearly we give in this finiteness argument (Step (5)) a very loose bound that is
far from the actual bound we can expect. We can do better than this, but this does not improve
the finiteness property we are proving. If we are interested in a polynomial bound,
one would hope to obtain a similar tight bound as for partial
derivatives introduced by Antimirov \cite{Antimirov95}. After all the idea with
 $\distinctWith$ is to maintain a ``set'' of alternatives (like the sets in
partial derivatives). Unfortunately to obtain the exact same bound would mean
we need to introduce simplifications, such as
 $(r_1 + r_2) \cdot r_3 \longrightarrow (r_1 \cdot r_3) + (r_2 \cdot r_3)$,
which exist for partial derivatives. However, if we introduce them in our
setting we would lose the POSIX property of our calculated values. We leave better
bounds for future work.\\[-6.5mm]


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\section{"Closed Forms" of regular expressions' derivatives w.r.t strings}
There is a nice property about the compound regular expression 
$r_1 \cdot r_2$ in general,
that the derivatives of it against a string $s$ can be described by
the derivatives w.r.t $r_1$ and $r_2$ over substrings of $s$:
\begin{lemma}
$\textit{sflat}\_{aux} ( (r_1 \cdot r_2) \backslash s) = (r_1 \backslash s) \cdot r_2 :: (\map (r_2 \backslash \_) (\vsuf s r_1))$
\end{lemma}


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\section{interaction between $\distinctWith$ and $\flts$}
Note that it is not immediately obvious that 
$\llbracket \distinctWith (\flts \textit{rs}) = \phi \rrbracket   \leq \llbracket \distinctWith \textit{rs} = \phi \rrbracket  $

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