ChengsongTanPhdThesis/Chapters/Finite.tex
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% Chapter Template
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\chapter{Finiteness Bound} % Main chapter title
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\label{Finite} 
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%  In Chapter 4 \ref{Chapter4} we give the second guarantee
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%of our bitcoded algorithm, that is a finite bound on the size of any 
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%regex's derivatives. 
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\begin{figure}
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\begin{center}
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	\begin{tabular}{ccc}
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		$\llbracket _{bs}\ONE \rrbracket$ & $\dn$ & $1$\\
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		$\llbracket \ZERO \rrbracket$ & $\dn$ & $1$ \\
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		$\llbracket _{bs} r_1 \cdot r_2 \rrbracket$ & $\dn$ & $\llbracket r_1 \rrbracket + \llbracket r_2 \rrbracket + 1$\\
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		$\llbracket _{bs}\mathbf{c} \rrbracket $ & $\dn$ & $1$\\
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		$\llbracket _{bs}\sum as \rrbracket $ & $\dn$ & $\map \; (\llbracket \_ \rrbracket)\; as   + 1$\\
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		$\llbracket _{bs} a^* \rrbracket $ & $\dn$ & $\llbracket a \rrbracket + 1$.
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	\end{tabular}
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\end{center}
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\caption{The size function of bitcoded regular expressions}\label{brexpSize}
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\end{figure}
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In this chapter we give a guarantee in terms of size: 
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given an annotated regular expression $a$, for any string $s$
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our algorithm $\blexersimp$'s internal annotated regular expression 
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size  is finitely bounded
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by a constant $N_a$ that only depends on $a$:
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\begin{center}
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	$\llbracket \bderssimp{a}{s} \rrbracket \leq N_a$
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\end{center}
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\noindent
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where the size of an annotated regular expression is defined
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in terms of the number of nodes in its tree structure (see figure \ref{brexpSize}).
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We believe this size bound
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is important in the context of POSIX lexing, because 
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\begin{itemize}
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	\item
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		It is a stepping stone towards an ``absence of catastrophic-backtracking''
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		guarantee. 
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		If the internal data structures used by our algorithm cannot grow very large,
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		then our algorithm (which traverses those structures) cannot be too slow.
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		The next step would be to refine the bound $N_a$ so that it
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		is polynomial on $\llbracket a\rrbracket$.
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	\item
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		Having it formalised gives us a higher confidence that
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		our simplification algorithm $\simp$ does not ``mis-behave''
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		like $\simpsulz$ does.
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		The bound is universal, which is an advantage over work which 
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		only gives empirical evidence on some test cases.
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\end{itemize}
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\section{Formalising About Size}
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\noindent
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In our lexer ($\blexersimp$),
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we take an annotated regular expression as input,
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and repeately take derivative of and simplify it:
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\begin{figure}[H]
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	\begin{tikzpicture}[scale=2,
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		every node/.style={minimum size=11mm},
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		->,>=stealth',shorten >=1pt,auto,thick
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		]
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		\node (r0) [rectangle, draw=black, thick, minimum size = 5mm, draw=blue] {$a$};
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		\node (r1) [rectangle, draw=black, thick, right=of r0, minimum size = 7mm]{$a_1$};
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		\draw[->,line width=0.2mm](r0)--(r1) node[above,midway] {$\backslash c_1$};
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		\node (r1s) [rectangle, draw=blue, thick, right=of r1, minimum size=6mm]{$a_{1s}$};
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		\draw[->, line width=0.2mm](r1)--(r1s) node[above, midway] {$\simp$};
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		\node (r2) [rectangle, draw=black, thick,  right=of r1s, minimum size = 12mm]{$a_2$};
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		\draw[->,line width=0.2mm](r1s)--(r2) node[above,midway] {$\backslash c_2$};
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		\node (r2s) [rectangle, draw = blue, thick, right=of r2,minimum size=6mm]{$a_{2s}$};
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		\draw[->,line width=0.2mm](r2)--(r2s) node[above,midway] {$\simp$};
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		\node (rns) [rectangle, draw = blue, thick, right=of r2s,minimum size=6mm]{$a_{ns}$};
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		\draw[->,line width=0.2mm, dashed](r2s)--(rns) node[above,midway] {$\backslash \ldots$};
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		\node (v) [circle, thick, draw, right=of rns, minimum size=6mm, right=1.7cm]{$v$};
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		\draw[->, line width=0.2mm](rns)--(v) node[above, midway] {\bmkeps} node [below, midway] {\decode};
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	\end{tikzpicture}
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	\caption{Regular expression size change during our $\blexersimp$ algorithm}\label{simpShrinks}
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\end{figure}
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\noindent
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Each time
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a derivative is taken, the regular expression might grow.
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However, the simplification that is immediately afterwards will always shrink it so that 
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it stays small.
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This intuition is depicted by the relative size
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change between the black and blue nodes:
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After $\simp$ the node always shrinks.
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Our proof says that all the blue nodes
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stay below a size bound $N_a$ determined by the input $a$.
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\noindent
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Sulzmann and Lu's assumed a similar picture about their algorithm,
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though in fact their algorithm's size might be better depicted by the following graph:
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\begin{figure}[H]
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	\begin{tikzpicture}[scale=2,
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		every node/.style={minimum size=11mm},
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		->,>=stealth',shorten >=1pt,auto,thick
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		]
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		\node (r0) [rectangle, draw=black, thick, minimum size = 5mm, draw=blue] {$a$};
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		\node (r1) [rectangle, draw=black, thick, right=of r0, minimum size = 7mm]{$a_1$};
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		\draw[->,line width=0.2mm](r0)--(r1) node[above,midway] {$\backslash c_1$};
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		\node (r1s) [rectangle, draw=blue, thick, right=of r1, minimum size=7mm]{$a_{1s}$};
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		\draw[->, line width=0.2mm](r1)--(r1s) node[above, midway] {$\simp'$};
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		\node (r2) [rectangle, draw=black, thick,  right=of r1s, minimum size = 17mm]{$a_2$};
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		\draw[->,line width=0.2mm](r1s)--(r2) node[above,midway] {$\backslash c_2$};
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		\node (r2s) [rectangle, draw = blue, thick, right=of r2,minimum size=14mm]{$a_{2s}$};
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		\draw[->,line width=0.2mm](r2)--(r2s) node[above,midway] {$\simp'$};
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		\node (r3) [rectangle, draw = black, thick, right= of r2s, minimum size = 22mm]{$a_3$};
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		\draw[->,line width=0.2mm](r2s)--(r3) node[above,midway] {$\backslash c_3$};
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		\node (rns) [right = of r3, draw=blue, minimum size = 20mm]{$a_{3s}$};
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		\draw[->,line width=0.2mm] (r3)--(rns) node [above, midway] {$\simp'$};
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		\node (rnn) [right = of rns, minimum size = 1mm]{};
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		\draw[->, dashed] (rns)--(rnn) node [above, midway] {$\ldots$};
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	\end{tikzpicture}
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	\caption{Regular expression size change during our $\blexersimp$ algorithm}\label{sulzShrinks}
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\end{figure}
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\noindent
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The picture means that on certain cases their lexer (where they use $\simpsulz$ 
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as the simplification function)
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will have an indefinite size explosion, causing the running time 
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of each derivative step to grow continuously (for example 
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in \ref{SulzmannLuLexerTime}).
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They tested out the run time of their
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lexer on particular examples such as $(a+b+ab)^*$
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and claimed that their algorithm is linear w.r.t to the input.
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With our mecahnised proof, we avoid this type of unintentional
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generalisation.\\
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Before delving in to the details of the formalisation,
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we are going to provide an overview of it.
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In the next subsection, we draw a picture of the bird's eye view
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of the proof.
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\subsection{Overview of the Proof}
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Here is a bird's eye view of the main components of the finiteness proof,
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which involves three steps:
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\begin{figure}[H]
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	\begin{tikzpicture}[scale=1,font=\bf,
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		node/.style={
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			rectangle,rounded corners=3mm,
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			ultra thick,draw=black!50,minimum height=18mm, 
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			minimum width=20mm,
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		top color=white,bottom color=black!20}]
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		\node (0) at (-5,0) 
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			[node, text width=1.8cm, text centered] 
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			{$\llbracket \bderssimp{a}{s} \rrbracket$};
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		\node (A) at (0,0) 
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			[node,text width=1.6cm,  text centered] 
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			{$\llbracket \rderssimp{r}{s} \rrbracket_r$};
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		\node (B) at (3,0) 
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			[node,text width=3.0cm, anchor=west, minimum width = 40mm] 
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			{$\llbracket \textit{ClosedForm}(r, s)\rrbracket_r$};
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		\node (C) at (9.5,0) [node, minimum width=10mm] {$N_r$};
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		\draw [->,line width=0.5mm] (0) -- 
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			node [above,pos=0.45] {=} (A) node [below, pos = 0.45] {$(r = a \downarrow_r)$} (A); 
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		\draw [->,line width=0.5mm] (A) -- 
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			node [above,pos=0.35] {$\quad =\ldots=$} (B); 
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		\draw [->,line width=0.5mm] (B) -- 
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			node [above,pos=0.35] {$\quad \leq \ldots \leq$} (C); 
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	\end{tikzpicture}
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	%\caption{
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\end{figure}
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\noindent
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We explain the steps one by one:
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\begin{itemize}
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	\item
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		We first introduce the operations such as 
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		derivatives, simplification, size calculation, etc.
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		associated with $\rrexp$s, which we have given
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		a very brief introduction to in chapter \ref{Bitcoded2}.
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		The operations on $\rrexp$s are identical to those on
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		annotated regular expressions except that they are unaware
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		of bitcodes. This means that all proofs about size of $\rrexp$s will apply to
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		annotated regular expressions.
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	\item
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		We prove that $\rderssimp{r}{s} = \textit{ClosedForm}(r, s)$,
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		where $\textit{ClosedForm}(r, s)$ is entirely 
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		written in the derivatives of their children regular 
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		expressions.
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		We call the right-hand-side the \emph{Closed Form}
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		of the derivative $\rderssimp{r}{s}$.
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	\item
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		We estimate $\llbracket \textit{ClosedForm}(r, s) \rrbracket_r$.
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		The key observation is that $\distinctBy$'s output is
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		a list with a constant length bound.
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\end{itemize}
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We will expand on these steps in the next sections.\\
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\section{The $\textit{Rrexp}$ Datatype}
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The first step is to define 
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$\textit{rrexp}$s.
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They are without bitcodes,
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allowing a much simpler size bound proof.
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Of course, the bits which encode the lexing information 
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would grow linearly with respect 
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to the input, which should be taken into account when we wish to tackle the runtime comlexity.
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But for the sake of the structural size 
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we can safely ignore them.\\
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To recapitulate, the datatype 
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definition of the $\rrexp$, called
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\emph{r-regular expressions},
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was initially defined in \ref{rrexpDef}.
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The reason for the prefix $r$ is
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to make a distinction  
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with basic regular expressions.
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\[			\rrexp ::=   \RZERO \mid  \RONE
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	\mid  \RCHAR{c}  
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	\mid  \RSEQ{r_1}{r_2}
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	\mid  \RALTS{rs}
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	\mid \RSTAR{r}        
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\]
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The size of an r-regular expression is
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written $\llbracket r\rrbracket_r$, 
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whose definition mirrors that of an annotated regular expression. 
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\begin{center}
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	\begin{tabular}{ccc}
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		$\llbracket _{bs}\ONE \rrbracket_r$ & $\dn$ & $1$\\
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		$\llbracket \ZERO \rrbracket_r$ & $\dn$ & $1$ \\
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		$\llbracket _{bs} r_1 \cdot r_2 \rrbracket_r$ & $\dn$ & $\llbracket r_1 \rrbracket_r + \llbracket r_2 \rrbracket_r + 1$\\
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		$\llbracket _{bs}\mathbf{c} \rrbracket_r $ & $\dn$ & $1$\\
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		$\llbracket _{bs}\sum as \rrbracket_r $ & $\dn$ & $\map \; (\llbracket \_ \rrbracket_r)\; as   + 1$\\
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		$\llbracket _{bs} a^* \rrbracket_r $ & $\dn$ & $\llbracket a \rrbracket_r + 1$.
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	\end{tabular}
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\end{center}
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\noindent
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The $r$ in the subscript of $\llbracket \rrbracket_r$ is to 
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differentiate with the same operation for annotated regular expressions.
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Adding $r$ as subscript will be used in 
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other operations as well.\\
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The transformation from an annotated regular expression
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to an r-regular expression is straightforward.
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\begin{center}
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	\begin{tabular}{lcl}
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		$\rerase{\ZERO}$ & $\dn$ & $\RZERO$\\
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		$\rerase{_{bs}\ONE}$ & $\dn$ & $\RONE$\\
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		$\rerase{_{bs}\mathbf{c}}$ & $\dn$ & $\RCHAR{c}$\\
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		$\rerase{_{bs}r_1\cdot r_2}$ & $\dn$ & $\RSEQ{\rerase{r_1}}{\rerase{r_2}}$\\
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		$\rerase{_{bs}\sum as}$ & $\dn$ & $\RALTS{\map \; \rerase{\_} \; as}$\\
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		$\rerase{_{bs} a ^*}$ & $\dn$ & $\rerase{a} ^*$
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	\end{tabular}
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\end{center}
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613
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\subsection{Why a New Datatype?}
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The reason we take all the trouble 
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defining a new datatype is that $\erase$ makes things harder.
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We initially started by using 
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plain regular expressions and tried to prove
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the lemma \ref{rsizeAsize},
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however the $\erase$ function unavoidbly messes with the structure of the 
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annotated regular expression.
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The $+$ constructor
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of basic regular expressions is binary whereas $\sum$ 
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takes a list, and one has to convert between them:
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\begin{center}
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	\begin{tabular}{ccc}
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		$\erase \; _{bs}\sum [] $ & $\dn$ & $\ZERO$\\
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		$\erase \; _{bs}\sum [a]$ & $\dn$ & $a$\\
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		$\erase \; _{bs}\sum a :: as$ & $\dn$ & $a + (\erase \; _{[]} \sum as)\quad \text{if $as$ length over 1}$
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	\end{tabular}
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\end{center}
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\noindent
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An alternative regular expression with an empty argument list
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will be turned into a $\ZERO$.
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The singleton alternative $\sum [r]$ would have $r$ during the
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$\erase$ function.
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The  annotated regular expression $\sum[a, b, c]$ would turn into
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$(a+(b+c))$.
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All these operations change the size and structure of
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an annotated regular expressions, adding unnecessary 
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complexities to the size bound proof.\\
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For example, if we define the size of a basic plain regular expression 
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in the usual way,
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\begin{center}
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	\begin{tabular}{ccc}
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		$\llbracket \ONE \rrbracket_p$ & $\dn$ & $1$\\
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		$\llbracket \ZERO \rrbracket_p$ & $\dn$ & $1$ \\
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		$\llbracket r_1 \cdot r_2 \rrbracket_p$ & $\dn$ & $\llbracket r_1 \rrbracket_p + \llbracket r_2 \rrbracket_p + 1$\\
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		$\llbracket \mathbf{c} \rrbracket_p $ & $\dn$ & $1$\\
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		$\llbracket r_1 \cdot r_2 \rrbracket_p $ & $\dn$ & $\llbracket r_1 \rrbracket_p \; + \llbracket r_2 \rrbracket_p + 1$\\
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		$\llbracket a^* \rrbracket_p $ & $\dn$ & $\llbracket a \rrbracket_p + 1$
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	\end{tabular}
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\end{center}
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\noindent
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Then the property
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\begin{center}
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	$\llbracket a \rrbracket \stackrel{?}{=} \llbracket a_\downarrow \rrbracket_p$
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\end{center}
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does not hold.
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With $\textit{rerase}$, however, 
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only the bitcodes are thrown away.
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Everything about the structure remains intact.
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Therefore it does not change the size
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of an annotated regular expression:
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\begin{lemma}\label{rsizeAsize}
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	$\rsize{\rerase a} = \asize a$
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\end{lemma}
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\begin{proof}
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	By routine structural induction on $a$.
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\end{proof}
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\noindent
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One might be able to prove an inequality such as
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$\llbracket a \rrbracket  \leq \llbracket  a_\downarrow \rrbracket_p $
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and then estimate $\llbracket  a_\downarrow \rrbracket_p$,
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but we found our approach more straightforward.\\
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613
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\subsection{Functions for R-regular Expressions}
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We shall define the r-regular expression version
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of $\blexer$ and $\blexersimp$ related functions.
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We use $r$ as the prefix or subscript to differentiate
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with the bitcoded version.
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For example,$\backslash_r$, $\rdistincts$, and $\rsimp$
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as opposed to $\backslash$, $\distinctBy$, and $\bsimp$.
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As promised, they are much simpler than their bitcoded counterparts.
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%The operations on r-regular expressions are 
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%almost identical to those of the annotated regular expressions,
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%except that no bitcodes are used. For example,
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The derivative operation becomes simpler:\\
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\begin{center}
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	\begin{tabular}{@{}lcl@{}}
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		$(\ZERO)\,\backslash_r c$ & $\dn$ & $\ZERO$\\  
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		$(\ONE)\,\backslash_r c$ & $\dn$ &
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		$\textit{if}\;c=d\; \;\textit{then}\;
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		\ONE\;\textit{else}\;\ZERO$\\  
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		$(\sum \;\textit{rs})\,\backslash_r c$ & $\dn$ &
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		$\sum\;(\textit{map} \; (\_\backslash_r c) \; rs )$\\
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		$(r_1\cdot r_2)\,\backslash_r c$ & $\dn$ &
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		$\textit{if}\;(\textit{rnullable}\,r_1)$\\
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						 & &$\textit{then}\;\sum\,[(r_1\,\backslash_r c)\cdot\,r_2,$\\
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						 & &$\phantom{\textit{then},\;\sum\,}((r_2\,\backslash_r c))]$\\
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						 & &$\textit{else}\;\,(r_1\,\backslash_r c)\cdot r_2$\\
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		$(r^*)\,\backslash_r c$ & $\dn$ &
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		$( r\,\backslash_r c)\cdot
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		(_{[]}r^*))$
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	\end{tabular}    
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\end{center}  
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\noindent
594
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Similarly, $\distinctBy$ does not need 
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a function checking equivalence because
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there are no bit annotations causing superficial differences
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between syntactically equal terms.
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\begin{center}
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	\begin{tabular}{lcl}
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		$\rdistinct{[]}{rset} $ & $\dn$ & $[]$\\
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		$\rdistinct{r :: rs}{rset}$ & $\dn$ & 
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		$\textit{if}(r \in \textit{rset}) \; \textit{then} \; \rdistinct{rs}{rset}$\\
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					    &        & $\textit{else}\; \;
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					    r::\rdistinct{rs}{(rset \cup \{r\})}$
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	\end{tabular}
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\end{center}
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%TODO: definition of rsimp (maybe only the alternative clause)
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\noindent
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We would like to make clear
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a difference between our $\rdistincts$ and
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the Isabelle $\textit {distinct}$ predicate.
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In Isabelle $\textit{distinct}$ is a function that returns a boolean
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rather than a list.
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It tests if all the elements of a list are unique.\\
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With $\textit{rdistinct}$,
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and the flatten function for $\rrexp$s:
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 \begin{center}
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  \begin{tabular}{@{}lcl@{}}
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  $\textit{rflts} \; (\sum \textit{as}) :: \textit{as'}$ & $\dn$ & $as \; @ \; \textit{rflts} \; as' $ \\
595
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  $\textit{rflts} \; \ZERO :: as'$ & $\dn$ & $ \textit{rflts} \;  \textit{as'} $ \\
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    $\textit{rflts} \; a :: as'$ & $\dn$ & $a :: \textit{rflts} \; \textit{as'}$ \quad(otherwise) 
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\end{tabular}    
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\end{center}  
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\noindent
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one can chain together all the other modules
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such as $\rsimpalts$:
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\begin{center}
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  \begin{tabular}{@{}lcl@{}}
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	  $\rsimpalts \;\; nil$ & $\dn$ & $\RZERO$\\
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	  $\rsimpalts \;\; r::nil$ & $\dn$ & $r$\\
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	  $\rsimpalts \;\; rs$ & $\dn$ & $\sum rs$\\
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\end{tabular}    
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\end{center}  
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\noindent
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and $\rsimpseq$:
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\begin{center}
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  \begin{tabular}{@{}lcl@{}}
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	  $\rsimpseq \;\; \RZERO \; \_ $ &   $=$ &   $\RZERO$\\
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	  $\rsimpseq \;\; \_ \; \RZERO $ &   $=$ &   $\RZERO$\\
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	  $\rsimpseq \;\; \RONE \cdot r_2$ & $\dn$ & $r_2$\\
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	  $\rsimpseq \;\; r_1 r_2$ & $\dn$ & $r_1 \cdot r_2$\\
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\end{tabular}    
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\end{center}  
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and get $\textit{rsimp}$ and $\rderssimp{\_}{\_}$:
595
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\begin{center}
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  \begin{tabular}{@{}lcl@{}}
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596
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	  $\textit{rsimp} \; (r_1\cdot r_2)$ & $\dn$ & $ \textit{rsimp}_{SEQ} \; bs \;(\textit{rsimp} \; r_1) \; (\textit{rsimp}  \; r_2)  $ \\
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	  $\textit{rsimp} \; (_{bs}\sum \textit{rs})$ & $\dn$ & $\textit{rsimp}_{ALTS} \; \textit{bs} \; (\textit{rdistinct} \; ( \textit{rflts} ( \textit{map} \; rsimp \; rs)) \; \rerases \; \varnothing) $ \\
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   $\textit{rsimp} \; r$ & $\dn$ & $\textit{r} \qquad \textit{otherwise}$   
595
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\end{tabular}    
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\end{center} 
596
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\begin{center}
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	\begin{tabular}{@{}lcl@{}}
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		$r\backslash_{rsimp} \, c$ & $\dn$ & $\rsimp \; (r\backslash_r \, c)$
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	\end{tabular}
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\end{center}
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\begin{center}
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	\begin{tabular}{@{}lcl@{}}
601
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$r \backslash_{rsimps} \; \; c\!::\!s $ & $\dn$ & $(r \backslash_{rsimp}\, c) \backslash_{rsimps}\, s$ \\
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$r \backslash_{rsimps} [\,] $ & $\dn$ & $r$
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	\end{tabular}
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\end{center}
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\noindent
601
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We do not define an r-regular expression version of $\blexersimp$,
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as our proof does not involve its use 
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(and there is no bitcode to decode into a lexical value). 
613
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Now we are ready to introduce how r-regular expressions allow
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us to prove the size bound on bitcoded regular expressions.
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\subsection{Using R-regular Expressions to Bound Bit-coded Regular Expressions}
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Everything about the size of annotated regular expressions after the application
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of function $\bsimp$ and $\backslash_{simps}$
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can be calculated via the size of r-regular expressions after the application
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of $\rsimp$ and $\backslash_{rsimps}$:
564
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\begin{lemma}\label{sizeRelations}
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	The following equalities hold:
543
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	\begin{itemize}
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		\item
601
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			$\asize{\bsimps \; a} = \rsize{\rsimp{ \rerase{a}}}$
554
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		\item
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			$\asize{\bderssimp{a}{s}} =  \rsize{\rderssimp{\rerase{a}}{s}}$
554
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diff changeset
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	\end{itemize}
532
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\end{lemma}
601
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\begin{proof}
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	The first part is by induction on the inductive cases
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   444
	of $\textit{bsimp}$.
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	The second part is by induction on the string $s$,
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   446
	where the inductive step follows from part one.
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\end{proof}
543
b2bea5968b89 thesis_thys
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\noindent
596
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   449
With lemma \ref{sizeRelations},
601
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we will be able to focus on 
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   451
estimating only
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$\rsize{\rderssimp{\rerase{a}}{s}}$
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   453
in later parts because
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   454
\begin{center}
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   455
	$\rsize{\rderssimp{\rerase{a}}{s}} \leq N_r \quad$
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   456
	implies
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diff changeset
   457
	$\quad \llbracket a \backslash_{bsimps} s \rrbracket \leq N_r$.
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diff changeset
   458
\end{center}
613
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   459
From now on we 
601
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   460
Unless stated otherwise in the rest of this 
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chapter all regular expressions without
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bitcodes are seen as r-regular expressions ($\rrexp$s).
601
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diff changeset
   463
For the binary alternative r-regular expression $\RALTS{[r_1, r_2]}$,
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   464
we use the notation $r_1 + r_2$
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   465
for brevity.
532
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cc54ce075db5 restructured
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cc54ce075db5 restructured
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%-----------------------------------
596
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%	SUB SECTION ROADMAP RREXP BOUND
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%-----------------------------------
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596
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%\subsection{Roadmap to a Bound for $\textit{Rrexp}$}
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596
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%The way we obtain the bound for $\rrexp$s is by two steps:
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%\begin{itemize}
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   476
%	\item
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   477
%		First, we rewrite $r\backslash s$ into something else that is easier
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   478
%		to bound. This step is especially important for the inductive case 
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   479
%		$r_1 \cdot r_2$ and $r^*$, where the derivative can grow and bloat in a wild way,
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   480
%		but after simplification they will always be equal or smaller to a form consisting of an alternative
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%		list of regular expressions $f \; (g\; (\sum rs))$ with some functions applied to it, where each element will be distinct after the function application.
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   482
%	\item
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   483
%		Then, for such a sum  list of regular expressions $f\; (g\; (\sum rs))$, we can control its size
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   484
%		by estimation, since $\distinctBy$ and $\flts$ are well-behaved and working together would only 
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   485
%		reduce the size of a regular expression, not adding to it.
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   486
%\end{itemize}
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%
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   488
%\section{Step One: Closed Forms}
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   489
%We transform the function application $\rderssimp{r}{s}$
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   490
%into an equivalent 
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diff changeset
   491
%form $f\; (g \; (\sum rs))$.
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diff changeset
   492
%The functions $f$ and $g$ can be anything from $\flts$, $\distinctBy$ and other helper functions from $\bsimp{\_}$.
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   493
%This way we get a different but equivalent way of expressing : $r\backslash s = f \; (g\; (\sum rs))$, we call the
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   494
%right hand side the "closed form" of $r\backslash s$.
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   495
%
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   496
%\begin{quote}\it
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   497
%	Claim: For regular expressions $r_1 \cdot r_2$, we claim that
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diff changeset
   498
%\end{quote}
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   499
%\noindent
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   500
%We explain in detail how we reached those claims.
601
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   501
If we attempt to prove 
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\begin{center}
609
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   503
	$\forall r. \; \exists N_r.\;\; s.t. \llbracket r\backslash_{rsimps} s \rrbracket_r \leq N_r$
601
Chengsong
parents: 596
diff changeset
   504
\end{center}
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parents: 596
diff changeset
   505
using a naive induction on the structure of $r$,
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diff changeset
   506
then we are stuck at the inductive cases such as
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diff changeset
   507
$r_1\cdot r_2$.
Chengsong
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diff changeset
   508
The inductive hypotheses are:
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parents: 596
diff changeset
   509
\begin{center}
Chengsong
parents: 596
diff changeset
   510
	1: $\text{for } r_1, \text{there exists } N_{r_1}.\;\; s.t. 
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parents: 601
diff changeset
   511
	\;\;\forall s.  \llbracket r_1 \backslash_{rsimps} s \rrbracket_r \leq N_{r_1}. $\\
601
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parents: 596
diff changeset
   512
	2: $\text{for } r_2, \text{there exists } N_{r_2}.\;\; s.t. 
609
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parents: 601
diff changeset
   513
	\;\; \forall s. \llbracket r_2 \backslash_{rsimps} s \rrbracket_r \leq N_{r_2}. $
601
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parents: 596
diff changeset
   514
\end{center}
Chengsong
parents: 596
diff changeset
   515
The inductive step to prove would be 
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diff changeset
   516
\begin{center}
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parents: 596
diff changeset
   517
	$\text{there exists } N_{r_1\cdot r_2}. \;\; s.t. \forall s. 
609
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parents: 601
diff changeset
   518
	\llbracket (r_1 \cdot r_2) \backslash_{rsimps} s \rrbracket_r \leq N_{r_1\cdot r_2}.$
601
Chengsong
parents: 596
diff changeset
   519
\end{center}
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parents: 596
diff changeset
   520
The problem is that it is not clear what 
609
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parents: 601
diff changeset
   521
$(r_1\cdot r_2) \backslash_{rsimps} s$ looks like,
601
Chengsong
parents: 596
diff changeset
   522
and therefore $N_{r_1}$ and $N_{r_2}$ in the
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diff changeset
   523
inductive hypotheses cannot be directly used.
609
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diff changeset
   524
We have already seen that $(r_1 \cdot r_2)\backslash s$ 
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diff changeset
   525
and $(r^*)\backslash s$ can grow in a wild way.
613
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diff changeset
   526
609
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diff changeset
   527
The point is that they will be equivalent to a list of
61139fdddae0 chap1 totally done
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diff changeset
   528
terms $\sum rs$, where each term in $rs$ will
61139fdddae0 chap1 totally done
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diff changeset
   529
be made of $r_1 \backslash s' $, $r_2\backslash s'$,
61139fdddae0 chap1 totally done
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diff changeset
   530
and $r \backslash s'$ with $s' \in \textit{SubString} \; s$.
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diff changeset
   531
The list $\sum rs$ will then be de-duplicated by $\textit{rdistinct}$
61139fdddae0 chap1 totally done
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parents: 601
diff changeset
   532
in the simplification which saves $rs$ from growing indefinitely.
61139fdddae0 chap1 totally done
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diff changeset
   533
613
Chengsong
parents: 611
diff changeset
   534
Based on this idea, we develop a proof in two steps.
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parents: 611
diff changeset
   535
First, we show the equality (where
609
61139fdddae0 chap1 totally done
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diff changeset
   536
$f$ and $g$ are functions that do not increase the size of the input)
61139fdddae0 chap1 totally done
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diff changeset
   537
\begin{center}
613
Chengsong
parents: 611
diff changeset
   538
$r\backslash_{rsimps} s = f\; (\textit{rdistinct} \; (g\; \sum rs))$,
609
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Chengsong
parents: 601
diff changeset
   539
\end{center}
613
Chengsong
parents: 611
diff changeset
   540
where $r = r_1 \cdot r_2$ or $r = r_0^*$ and so on.
Chengsong
parents: 611
diff changeset
   541
For example, for $r_1 \cdot r_2$ we have the equality as
Chengsong
parents: 611
diff changeset
   542
	\begin{center}
Chengsong
parents: 611
diff changeset
   543
		$ \rderssimp{r_1 \cdot r_2}{s} = 
Chengsong
parents: 611
diff changeset
   544
		\rsimp{(\sum (r_1 \backslash s \cdot r_2 ) \; :: \;(\map \; \rderssimp{r2}{\_} \;(\vsuf{s}{r_1})))}$
Chengsong
parents: 611
diff changeset
   545
	\end{center}
609
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parents: 601
diff changeset
   546
We call the right-hand-side the 
61139fdddae0 chap1 totally done
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diff changeset
   547
\emph{Closed Form} of $(r_1 \cdot r_2)\backslash_{rsimps} s$.
613
Chengsong
parents: 611
diff changeset
   548
Second, we will bound the closed form of r-regular expressions
Chengsong
parents: 611
diff changeset
   549
using some estimation techniques
Chengsong
parents: 611
diff changeset
   550
and then piece it together
Chengsong
parents: 611
diff changeset
   551
with lemma \ref{sizeRelations} to show the bitcoded regular expressions
Chengsong
parents: 611
diff changeset
   552
in our $\blexersimp$ are finitely bounded.
609
61139fdddae0 chap1 totally done
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parents: 601
diff changeset
   553
613
Chengsong
parents: 611
diff changeset
   554
We will flesh out the first step of the proof we 
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diff changeset
   555
sketched just now in the next section.
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parents: 611
diff changeset
   556
Chengsong
parents: 611
diff changeset
   557
\section{Closed Forms}
609
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diff changeset
   558
In this section we introduce in detail
61139fdddae0 chap1 totally done
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diff changeset
   559
how the closed forms are obtained for regular expressions'
613
Chengsong
parents: 611
diff changeset
   560
derivatives.
Chengsong
parents: 611
diff changeset
   561
We start by proving some basic identities
609
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
   562
involving the simplification functions for r-regular expressions.
613
Chengsong
parents: 611
diff changeset
   563
After that we introduce the rewrite relations
Chengsong
parents: 611
diff changeset
   564
$\rightsquigarrow_h$, $\rightsquigarrow^*_{scf}$
Chengsong
parents: 611
diff changeset
   565
$\rightsquigarrow_f$ and $\rightsquigarrow_g$.
Chengsong
parents: 611
diff changeset
   566
These relations involves similar techniques in chapter \ref{Bitcoded2}.
Chengsong
parents: 611
diff changeset
   567
Finally, we use these identities to establish the
Chengsong
parents: 611
diff changeset
   568
closed forms of the alternative regular expression,
Chengsong
parents: 611
diff changeset
   569
the sequence regular expression, and the star regular expression.
609
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
   570
%$r_1\cdot r_2$, $r^*$ and $\sum rs$.
601
Chengsong
parents: 596
diff changeset
   571
Chengsong
parents: 596
diff changeset
   572
609
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
   573
613
Chengsong
parents: 611
diff changeset
   574
\subsection{Some Basic Identities}
609
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
   575
613
Chengsong
parents: 611
diff changeset
   576
We now introduce lemmas 
611
Chengsong
parents: 610
diff changeset
   577
that are repeatedly used in later proofs.
Chengsong
parents: 610
diff changeset
   578
Note that for the $\textit{rdistinct}$ function there
Chengsong
parents: 610
diff changeset
   579
will be a lot of conversion from lists to sets.
613
Chengsong
parents: 611
diff changeset
   580
We use $set$ to refere to the 
611
Chengsong
parents: 610
diff changeset
   581
function that converts a list $rs$ to the set
Chengsong
parents: 610
diff changeset
   582
containing all the elements in $rs$.
Chengsong
parents: 610
diff changeset
   583
\subsubsection{$\textit{rdistinct}$'s Does the Job of De-duplication}
543
b2bea5968b89 thesis_thys
Chengsong
parents: 532
diff changeset
   584
The $\textit{rdistinct}$ function, as its name suggests, will
613
Chengsong
parents: 611
diff changeset
   585
de-duplicate an r-regular expression list.
Chengsong
parents: 611
diff changeset
   586
It will also remove any elements that 
Chengsong
parents: 611
diff changeset
   587
is already in the accumulator set.
555
Chengsong
parents: 554
diff changeset
   588
\begin{lemma}\label{rdistinctDoesTheJob}
609
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
   589
	%The function $\textit{rdistinct}$ satisfies the following
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
   590
	%properties:
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
   591
	Assume we have the predicate $\textit{isDistinct}$\footnote{We omit its
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
   592
	recursive definition here, its Isabelle counterpart would be $\textit{distinct}$.} 
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
   593
	readily defined
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
   594
	for testing
613
Chengsong
parents: 611
diff changeset
   595
	whether a list's elements are unique. Then the following
609
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
   596
	properties about $\textit{rdistinct}$ hold:
543
b2bea5968b89 thesis_thys
Chengsong
parents: 532
diff changeset
   597
	\begin{itemize}
b2bea5968b89 thesis_thys
Chengsong
parents: 532
diff changeset
   598
		\item
b2bea5968b89 thesis_thys
Chengsong
parents: 532
diff changeset
   599
			If $a \in acc$ then $a \notin (\rdistinct{rs}{acc})$.
b2bea5968b89 thesis_thys
Chengsong
parents: 532
diff changeset
   600
		\item
609
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
   601
			%If list $rs'$ is the result of $\rdistinct{rs}{acc}$,
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
   602
			$\textit{isDistinct} \;\;\; (\rdistinct{rs}{acc})$.
555
Chengsong
parents: 554
diff changeset
   603
		\item
609
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
   604
			$\textit{set} \; (\rdistinct{rs}{acc}) 
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
   605
			= (textit{set} \; rs) - acc$
543
b2bea5968b89 thesis_thys
Chengsong
parents: 532
diff changeset
   606
	\end{itemize}
b2bea5968b89 thesis_thys
Chengsong
parents: 532
diff changeset
   607
\end{lemma}
555
Chengsong
parents: 554
diff changeset
   608
\noindent
543
b2bea5968b89 thesis_thys
Chengsong
parents: 532
diff changeset
   609
\begin{proof}
b2bea5968b89 thesis_thys
Chengsong
parents: 532
diff changeset
   610
	The first part is by an induction on $rs$.
555
Chengsong
parents: 554
diff changeset
   611
	The second and third part can be proven by using the 
609
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
   612
	inductive cases of $\textit{rdistinct}$.
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
   613
543
b2bea5968b89 thesis_thys
Chengsong
parents: 532
diff changeset
   614
\end{proof}
b2bea5968b89 thesis_thys
Chengsong
parents: 532
diff changeset
   615
b2bea5968b89 thesis_thys
Chengsong
parents: 532
diff changeset
   616
\noindent
613
Chengsong
parents: 611
diff changeset
   617
%$\textit{rdistinct}$ will out all regular expression terms
Chengsong
parents: 611
diff changeset
   618
%that are in the accumulator, therefore 
Chengsong
parents: 611
diff changeset
   619
Concatenating a list $rs_a$ at the front of another
Chengsong
parents: 611
diff changeset
   620
list $rs$ whose elements are all from the accumulator, and then calling $\textit{rdistinct}$
Chengsong
parents: 611
diff changeset
   621
on the merged list, the output will be as if we had called $\textit{rdistinct}$
543
b2bea5968b89 thesis_thys
Chengsong
parents: 532
diff changeset
   622
without the prepending of $rs$:
609
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
   623
\begin{lemma}\label{rdistinctConcat}
554
Chengsong
parents: 553
diff changeset
   624
	The elements appearing in the accumulator will always be removed.
Chengsong
parents: 553
diff changeset
   625
	More precisely,
Chengsong
parents: 553
diff changeset
   626
	\begin{itemize}
Chengsong
parents: 553
diff changeset
   627
		\item
Chengsong
parents: 553
diff changeset
   628
			If $rs \subseteq rset$, then 
Chengsong
parents: 553
diff changeset
   629
			$\rdistinct{rs@rsa }{acc} = \rdistinct{rsa }{acc}$.
Chengsong
parents: 553
diff changeset
   630
		\item
609
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
   631
			More generally, if $a \in rset$ and $\rdistinct{rs}{\{a\}} = []$,
554
Chengsong
parents: 553
diff changeset
   632
			then $\rdistinct{(rs @ rs')}{rset} = \rdistinct{rs'}{rset}$
Chengsong
parents: 553
diff changeset
   633
	\end{itemize}
543
b2bea5968b89 thesis_thys
Chengsong
parents: 532
diff changeset
   634
\end{lemma}
554
Chengsong
parents: 553
diff changeset
   635
543
b2bea5968b89 thesis_thys
Chengsong
parents: 532
diff changeset
   636
\begin{proof}
609
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
   637
	By induction on $rs$ and using \ref{rdistinctDoesTheJob}.
543
b2bea5968b89 thesis_thys
Chengsong
parents: 532
diff changeset
   638
\end{proof}
b2bea5968b89 thesis_thys
Chengsong
parents: 532
diff changeset
   639
\noindent
b2bea5968b89 thesis_thys
Chengsong
parents: 532
diff changeset
   640
On the other hand, if an element $r$ does not appear in the input list waiting to be deduplicated,
b2bea5968b89 thesis_thys
Chengsong
parents: 532
diff changeset
   641
then expanding the accumulator to include that element will not cause the output list to change:
611
Chengsong
parents: 610
diff changeset
   642
\begin{lemma}\label{rdistinctOnDistinct}
543
b2bea5968b89 thesis_thys
Chengsong
parents: 532
diff changeset
   643
	The accumulator can be augmented to include elements not appearing in the input list,
b2bea5968b89 thesis_thys
Chengsong
parents: 532
diff changeset
   644
	and the output will not change.	
b2bea5968b89 thesis_thys
Chengsong
parents: 532
diff changeset
   645
	\begin{itemize}
b2bea5968b89 thesis_thys
Chengsong
parents: 532
diff changeset
   646
		\item
611
Chengsong
parents: 610
diff changeset
   647
			If $r \notin rs$, then $\rdistinct{rs}{acc} = \rdistinct{rs}{(\{r\} \cup acc)}$.
543
b2bea5968b89 thesis_thys
Chengsong
parents: 532
diff changeset
   648
		\item
611
Chengsong
parents: 610
diff changeset
   649
			Particularly, if $\;\;\textit{isDistinct} \; rs$, then we have\\
543
b2bea5968b89 thesis_thys
Chengsong
parents: 532
diff changeset
   650
			\[ \rdistinct{rs}{\varnothing} = rs \]
b2bea5968b89 thesis_thys
Chengsong
parents: 532
diff changeset
   651
	\end{itemize}
b2bea5968b89 thesis_thys
Chengsong
parents: 532
diff changeset
   652
\end{lemma}
b2bea5968b89 thesis_thys
Chengsong
parents: 532
diff changeset
   653
\begin{proof}
b2bea5968b89 thesis_thys
Chengsong
parents: 532
diff changeset
   654
	The first half is by induction on $rs$. The second half is a corollary of the first.
b2bea5968b89 thesis_thys
Chengsong
parents: 532
diff changeset
   655
\end{proof}
b2bea5968b89 thesis_thys
Chengsong
parents: 532
diff changeset
   656
\noindent
611
Chengsong
parents: 610
diff changeset
   657
The function $\textit{rdistinct}$ removes duplicates from anywhere in a list.
Chengsong
parents: 610
diff changeset
   658
Despite being seemingly obvious, 
Chengsong
parents: 610
diff changeset
   659
the induction technique is not as straightforward.
554
Chengsong
parents: 553
diff changeset
   660
\begin{lemma}\label{distinctRemovesMiddle}
Chengsong
parents: 553
diff changeset
   661
	The two properties hold if $r \in rs$:
Chengsong
parents: 553
diff changeset
   662
	\begin{itemize}
Chengsong
parents: 553
diff changeset
   663
		\item
555
Chengsong
parents: 554
diff changeset
   664
			$\rdistinct{rs}{rset} = \rdistinct{(rs @ [r])}{rset}$\\
Chengsong
parents: 554
diff changeset
   665
			and\\
554
Chengsong
parents: 553
diff changeset
   666
			$\rdistinct{(ab :: rs @ [ab])}{rset'} = \rdistinct{(ab :: rs)}{rset'}$
Chengsong
parents: 553
diff changeset
   667
		\item
555
Chengsong
parents: 554
diff changeset
   668
			$\rdistinct{ (rs @ rs') }{rset} = \rdistinct{rs @ [r] @ rs'}{rset}$\\
Chengsong
parents: 554
diff changeset
   669
			and\\
554
Chengsong
parents: 553
diff changeset
   670
			$\rdistinct{(ab :: rs @ [ab] @ rs'')}{rset'} = 
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
   671
			\rdistinct{(ab :: rs @ rs'')}{rset'}$
554
Chengsong
parents: 553
diff changeset
   672
	\end{itemize}
Chengsong
parents: 553
diff changeset
   673
\end{lemma}
Chengsong
parents: 553
diff changeset
   674
\noindent
Chengsong
parents: 553
diff changeset
   675
\begin{proof}
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
   676
	By induction on $rs$. All other variables are allowed to be arbitrary.
611
Chengsong
parents: 610
diff changeset
   677
	The second part of the lemma requires the first.
Chengsong
parents: 610
diff changeset
   678
	Note that for each part, the two sub-propositions need to be proven concurrently,
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
   679
	so that the induction goes through.
554
Chengsong
parents: 553
diff changeset
   680
\end{proof}
555
Chengsong
parents: 554
diff changeset
   681
\noindent
611
Chengsong
parents: 610
diff changeset
   682
This allows us to prove a few more equivalence relations involving 
Chengsong
parents: 610
diff changeset
   683
$\textit{rdistinct}$ (it will be useful later):
555
Chengsong
parents: 554
diff changeset
   684
\begin{lemma}\label{rdistinctConcatGeneral}
611
Chengsong
parents: 610
diff changeset
   685
	\mbox{}
555
Chengsong
parents: 554
diff changeset
   686
	\begin{itemize}
Chengsong
parents: 554
diff changeset
   687
		\item
Chengsong
parents: 554
diff changeset
   688
			$\rdistinct{(rs @ rs')}{\varnothing} = \rdistinct{((\rdistinct{rs}{\varnothing})@ rs')}{\varnothing}$
Chengsong
parents: 554
diff changeset
   689
		\item
Chengsong
parents: 554
diff changeset
   690
			$\rdistinct{(rs @ rs')}{\varnothing} = \rdistinct{(\rdistinct{rs}{\varnothing} @ rs')}{\varnothing}$
Chengsong
parents: 554
diff changeset
   691
		\item
Chengsong
parents: 554
diff changeset
   692
			If $rset' \subseteq rset$, then $\rdistinct{rs}{rset} = 
Chengsong
parents: 554
diff changeset
   693
			\rdistinct{(\rdistinct{rs}{rset'})}{rset}$. As a corollary
Chengsong
parents: 554
diff changeset
   694
			of this,
Chengsong
parents: 554
diff changeset
   695
		\item
Chengsong
parents: 554
diff changeset
   696
			$\rdistinct{(rs @ rs')}{rset} = \rdistinct{
Chengsong
parents: 554
diff changeset
   697
			(\rdistinct{rs}{\varnothing}) @ rs')}{rset}$. This
Chengsong
parents: 554
diff changeset
   698
			gives another corollary use later:
Chengsong
parents: 554
diff changeset
   699
		\item
Chengsong
parents: 554
diff changeset
   700
			If $a \in rset$, then $\rdistinct{(rs @ rs')}{rset} = \rdistinct{
Chengsong
parents: 554
diff changeset
   701
			(\rdistinct{(a :: rs)}{\varnothing} @ rs')}{rset} $,
Chengsong
parents: 554
diff changeset
   702
Chengsong
parents: 554
diff changeset
   703
	\end{itemize}
Chengsong
parents: 554
diff changeset
   704
\end{lemma}
Chengsong
parents: 554
diff changeset
   705
\begin{proof}
Chengsong
parents: 554
diff changeset
   706
	By \ref{rdistinctDoesTheJob} and \ref{distinctRemovesMiddle}.
Chengsong
parents: 554
diff changeset
   707
\end{proof}
611
Chengsong
parents: 610
diff changeset
   708
\noindent
613
Chengsong
parents: 611
diff changeset
   709
The next lemma is a more general form of \ref{rdistinctConcat},
Chengsong
parents: 611
diff changeset
   710
it says that
611
Chengsong
parents: 610
diff changeset
   711
$\textit{rdistinct}$ is composable w.r.t list concatenation:
Chengsong
parents: 610
diff changeset
   712
\begin{lemma}\label{distinctRdistinctAppend}
Chengsong
parents: 610
diff changeset
   713
			If $\;\; \textit{isDistinct} \; rs_1$, 
Chengsong
parents: 610
diff changeset
   714
			and $(set \; rs_1) \cap acc = \varnothing$,
Chengsong
parents: 610
diff changeset
   715
			then applying $\textit{rdistinct}$ on $rs_1 @ rs_a$ does not 
Chengsong
parents: 610
diff changeset
   716
			have an effect on $rs_1$:
Chengsong
parents: 610
diff changeset
   717
			\[\textit{rdistinct}\;  (rs_1 @ rsa)\;\, acc
Chengsong
parents: 610
diff changeset
   718
			= rs_1@(\textit{rdistinct} rsa \; (acc \cup rs_1))\]
Chengsong
parents: 610
diff changeset
   719
\end{lemma}
Chengsong
parents: 610
diff changeset
   720
\begin{proof}
Chengsong
parents: 610
diff changeset
   721
	By an induction on 
Chengsong
parents: 610
diff changeset
   722
	$rs_1$, where $rsa$ and $acc$ are allowed to be arbitrary.
Chengsong
parents: 610
diff changeset
   723
\end{proof}
Chengsong
parents: 610
diff changeset
   724
\noindent
Chengsong
parents: 610
diff changeset
   725
$\textit{rdistinct}$ needs to be applied only once, and 
Chengsong
parents: 610
diff changeset
   726
applying it multiple times does not cause any difference:
Chengsong
parents: 610
diff changeset
   727
\begin{corollary}\label{distinctOnceEnough}
Chengsong
parents: 610
diff changeset
   728
	$\textit{rdistinct} \; (rs @ rsa) {} = \textit{rdistinct} \; (rdistinct \; 
Chengsong
parents: 610
diff changeset
   729
	rs \{ \} @ (\textit{rdistinct} \; rs_a \; (set \; rs)))$
Chengsong
parents: 610
diff changeset
   730
\end{corollary}
Chengsong
parents: 610
diff changeset
   731
\begin{proof}
Chengsong
parents: 610
diff changeset
   732
	By lemma \ref{distinctRdistinctAppend}.
Chengsong
parents: 610
diff changeset
   733
\end{proof}
555
Chengsong
parents: 554
diff changeset
   734
611
Chengsong
parents: 610
diff changeset
   735
\subsubsection{The Properties of $\textit{Rflts}$} 
Chengsong
parents: 610
diff changeset
   736
We give in this subsection some properties
Chengsong
parents: 610
diff changeset
   737
involving $\backslash_r$, $\backslash_{rsimp}$, $\textit{rflts}$ and 
Chengsong
parents: 610
diff changeset
   738
$\textit{rsimp}_{ALTS} $, together with any non-trivial lemmas that lead to them.
543
b2bea5968b89 thesis_thys
Chengsong
parents: 532
diff changeset
   739
These will be helpful in later closed form proofs, when
611
Chengsong
parents: 610
diff changeset
   740
we want to transform derivative terms which have
Chengsong
parents: 610
diff changeset
   741
%the ways in which multiple functions involving
Chengsong
parents: 610
diff changeset
   742
%those are composed together
Chengsong
parents: 610
diff changeset
   743
interleaving derivatives and  simplifications applied to them.
543
b2bea5968b89 thesis_thys
Chengsong
parents: 532
diff changeset
   744
611
Chengsong
parents: 610
diff changeset
   745
\noindent
Chengsong
parents: 610
diff changeset
   746
%When the function $\textit{Rflts}$ 
Chengsong
parents: 610
diff changeset
   747
%is applied to the concatenation of two lists, the output can be calculated by first applying the
Chengsong
parents: 610
diff changeset
   748
%functions on two lists separately, and then concatenating them together.
Chengsong
parents: 610
diff changeset
   749
$\textit{Rflts}$ is composable in terms of concatenation:
554
Chengsong
parents: 553
diff changeset
   750
\begin{lemma}\label{rfltsProps}
543
b2bea5968b89 thesis_thys
Chengsong
parents: 532
diff changeset
   751
	The function $\rflts$ has the below properties:\\
b2bea5968b89 thesis_thys
Chengsong
parents: 532
diff changeset
   752
	\begin{itemize}
b2bea5968b89 thesis_thys
Chengsong
parents: 532
diff changeset
   753
		\item
554
Chengsong
parents: 553
diff changeset
   754
			$\rflts \; (rs_1 @ rs_2) = \rflts \; rs_1 @ \rflts \; rs_2$
Chengsong
parents: 553
diff changeset
   755
		\item
Chengsong
parents: 553
diff changeset
   756
			If $r \neq \RZERO$ and $\nexists rs_1. r = \RALTS{rs}_1$, then $\rflts \; (r::rs) = r :: \rflts \; rs$
Chengsong
parents: 553
diff changeset
   757
		\item
Chengsong
parents: 553
diff changeset
   758
			$\rflts \; (rs @ [\RZERO]) = \rflts \; rs$
Chengsong
parents: 553
diff changeset
   759
		\item
Chengsong
parents: 553
diff changeset
   760
			$\rflts \; (rs' @ [\RALTS{rs}]) = \rflts \; rs'@rs$
Chengsong
parents: 553
diff changeset
   761
		\item
Chengsong
parents: 553
diff changeset
   762
			$\rflts \; (rs @ [\RONE]) = \rflts \; rs @ [\RONE]$
Chengsong
parents: 553
diff changeset
   763
		\item
Chengsong
parents: 553
diff changeset
   764
			If $r \neq \RZERO$ and $\nexists rs'. r = \RALTS{rs'}$ then $\rflts \; (rs @ [r])
Chengsong
parents: 553
diff changeset
   765
			= (\rflts \; rs) @ [r]$
555
Chengsong
parents: 554
diff changeset
   766
		\item
Chengsong
parents: 554
diff changeset
   767
			If $r = \RALTS{rs}$ and $r \in rs'$ then for all $r_1 \in rs. 
Chengsong
parents: 554
diff changeset
   768
			r_1 \in \rflts \; rs'$.
Chengsong
parents: 554
diff changeset
   769
		\item
Chengsong
parents: 554
diff changeset
   770
			$\rflts \; (rs_a @ \RZERO :: rs_b) = \rflts \; (rs_a @ rs_b)$
543
b2bea5968b89 thesis_thys
Chengsong
parents: 532
diff changeset
   771
	\end{itemize}
b2bea5968b89 thesis_thys
Chengsong
parents: 532
diff changeset
   772
\end{lemma}
b2bea5968b89 thesis_thys
Chengsong
parents: 532
diff changeset
   773
\noindent
b2bea5968b89 thesis_thys
Chengsong
parents: 532
diff changeset
   774
\begin{proof}
555
Chengsong
parents: 554
diff changeset
   775
	By induction on $rs_1$ in the first sub-lemma, and induction on $r$ in the second part,
Chengsong
parents: 554
diff changeset
   776
	and induction on $rs$, $rs'$, $rs$, $rs'$, $rs_a$ in the third, fourth, fifth, sixth and 
Chengsong
parents: 554
diff changeset
   777
	last sub-lemma.
543
b2bea5968b89 thesis_thys
Chengsong
parents: 532
diff changeset
   778
\end{proof}
611
Chengsong
parents: 610
diff changeset
   779
\noindent
Chengsong
parents: 610
diff changeset
   780
Now we introduce the property that the operations 
Chengsong
parents: 610
diff changeset
   781
derivative and $\rsimpalts$
Chengsong
parents: 610
diff changeset
   782
commute, this will be used later in deriving the closed form for
Chengsong
parents: 610
diff changeset
   783
the alternative regular expression:
Chengsong
parents: 610
diff changeset
   784
\begin{lemma}\label{rderRsimpAltsCommute}
Chengsong
parents: 610
diff changeset
   785
	$\rder{x}{(\rsimpalts \; rs)} = \rsimpalts \; (\map \; (\rder{x}{\_}) \; rs)$
Chengsong
parents: 610
diff changeset
   786
\end{lemma}
Chengsong
parents: 610
diff changeset
   787
\noindent
614
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   788
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   789
\subsubsection{The $RL$ Function: Language Interpretation of $\textit{Rrexp}$s}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   790
Much like the definition of $L$ on plain regular expressions, one could also 
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   791
define the language interpretation of $\rrexp$s.
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   792
\begin{center}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   793
	\begin{tabular}{lcl}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   794
		$RL \; (\ZERO)$ & $\dn$ & $\phi$\\
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   795
		$RL \; (\ONE)$ & $\dn$ & $\{[]\}$\\
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   796
		$RL \; (c)$ & $\dn$ & $\{[c]\}$\\
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   797
		$RL \; \sum rs$ & $\dn$ & $ \bigcup_{r \in rs} (RL \; r)$\\
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   798
		$RL \; (r_1 \cdot r_2)$ & $\dn$ & $ RL \; (r_1) @ RL \; (r_2)$\\
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   799
		$RL \; (r^*)$ & $\dn$ & $ (RL(r))^*$
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   800
	\end{tabular}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   801
\end{center}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   802
\noindent
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   803
The main use of $RL$ is to establish some connections between $\rsimp{}$ 
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   804
and $\rnullable{}$:
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   805
\begin{lemma}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   806
	The following properties hold:
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   807
	\begin{itemize}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   808
		\item
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   809
			If $\rnullable{r}$, then $\rsimp{r} \neq \RZERO$.
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   810
		\item
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   811
			$\rnullable{r \backslash s} \quad $ if and only if $\quad \rnullable{\rderssimp{r}{s}}$.
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   812
	\end{itemize}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   813
\end{lemma}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   814
\begin{proof}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   815
	The first part is by induction on $r$. 
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   816
	The second part is true because property 
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   817
	\[ RL \; r = RL \; (\rsimp{r})\] holds.
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   818
\end{proof}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   819
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   820
\subsubsection{Simplified $\textit{Rrexp}$s are Good}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   821
We formalise the notion of ``good" regular expressions,
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   822
which means regular expressions that
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   823
are not fully simplified. For alternative regular expressions that means they
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   824
do not contain any nested alternatives like 
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   825
\[ r_1 + (r_2 + r_3) \], un-removed $\RZERO$s like \[\RZERO + r\]
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   826
or duplicate elements in a children regular expression list like \[ \sum [r, r, \ldots]\]:
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   827
\begin{center}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   828
	\begin{tabular}{@{}lcl@{}}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   829
		$\good\; \RZERO$ & $\dn$ & $\textit{false}$\\
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   830
		$\good\; \RONE$ & $\dn$ & $\textit{true}$\\
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   831
		$\good\; \RCHAR{c}$ & $\dn$ & $\btrue$\\
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   832
		$\good\; \RALTS{[]}$ & $\dn$ & $\bfalse$\\
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   833
		$\good\; \RALTS{[r]}$ & $\dn$ & $\bfalse$\\
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   834
		$\good\; \RALTS{r_1 :: r_2 :: rs}$ & $\dn$ & 
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   835
		$\textit{isDistinct} \; (r_1 :: r_2 :: rs) \;$\\
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   836
						   & & $\textit{and}\; (\forall r' \in (r_1 :: r_2 :: rs).\; \good \; r'\; \,  \textit{and}\; \, \textit{nonAlt}\; r')$\\
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   837
		$\good \; \RSEQ{\RZERO}{r}$ & $\dn$ & $\bfalse$\\
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   838
		$\good \; \RSEQ{\RONE}{r}$ & $\dn$ & $\bfalse$\\
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   839
		$\good \; \RSEQ{r}{\RZERO}$ & $\dn$ & $\bfalse$\\
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   840
		$\good \; \RSEQ{r_1}{r_2}$ & $\dn$ & $\good \; r_1 \;\, \textit{and} \;\, \good \; r_2$\\
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   841
		$\good \; \RSTAR{r}$ & $\dn$ & $\btrue$\\
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   842
	\end{tabular}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   843
\end{center}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   844
\noindent
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   845
The predicate $\textit{nonAlt}$ evaluates to true when the regular expression is not an
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   846
alternative, and false otherwise.
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   847
The $\good$ property is preserved under $\rsimp_{ALTS}$, provided that
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   848
its non-empty argument list of expressions are all good themsleves, and $\textit{nonAlt}$, 
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   849
and unique:
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   850
\begin{lemma}\label{rsimpaltsGood}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   851
	If $rs \neq []$ and forall $r \in rs. \textit{nonAlt} \; r$ and $\textit{isDistinct} \; rs$,
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   852
	then $\good \; (\rsimpalts \; rs)$ if and only if forall $r \in rs. \; \good \; r$.
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   853
\end{lemma}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   854
\noindent
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   855
We also note that
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   856
if a regular expression $r$ is good, then $\rflts$ on the singleton
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   857
list $[r]$ will not break goodness:
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   858
\begin{lemma}\label{flts2}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   859
	If $\good \; r$, then forall $r' \in \rflts \; [r]. \; \good \; r'$ and $\textit{nonAlt} \; r'$.
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   860
\end{lemma}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   861
\begin{proof}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   862
	By an induction on $r$.
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   863
\end{proof}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   864
\noindent
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   865
The other observation we make about $\rsimp{r}$ is that it never
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   866
comes with nested alternatives, which we describe as the $\nonnested$
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   867
property:
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   868
\begin{center}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   869
	\begin{tabular}{lcl}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   870
		$\nonnested \; \, \sum []$ & $\dn$ & $\btrue$\\
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   871
		$\nonnested \; \, \sum ((\sum rs_1) :: rs_2)$ & $\dn$ & $\bfalse$\\
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   872
		$\nonnested \; \, \sum (r :: rs)$ & $\dn$ & $\nonnested (\sum rs)$\\
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   873
		$\nonnested \; \, r $ & $\dn$ & $\btrue$
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   874
	\end{tabular}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   875
\end{center}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   876
\noindent
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   877
The $\rflts$ function
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   878
always opens up nested alternatives,
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   879
which enables $\rsimp$ to be non-nested:
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   880
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   881
\begin{lemma}\label{nonnestedRsimp}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   882
	$\nonnested \; (\rsimp{r})$
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   883
\end{lemma}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   884
\begin{proof}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   885
	By an induction on $r$.
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   886
\end{proof}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   887
\noindent
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   888
With this we could prove that a regular expressions
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   889
after simplification and flattening and de-duplication,
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   890
will not contain any alternative regular expression directly:
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   891
\begin{lemma}\label{nonaltFltsRd}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   892
	If $x \in \rdistinct{\rflts\; (\map \; \rsimp{} \; rs)}{\varnothing}$ 
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   893
	then $\textit{nonAlt} \; x$.
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   894
\end{lemma}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   895
\begin{proof}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   896
	By \ref{nonnestedRsimp}.
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   897
\end{proof}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   898
\noindent
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   899
The other thing we know is that once $\rsimp{}$ had finished
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   900
processing an alternative regular expression, it will not
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   901
contain any $\RZERO$s, this is because all the recursive 
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   902
calls to the simplification on the children regular expressions
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   903
make the children good, and $\rflts$ will not take out
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   904
any $\RZERO$s out of a good regular expression list,
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   905
and $\rdistinct{}$ will not mess with the result.
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   906
\begin{lemma}\label{flts3Obv}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   907
	The following are true:
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   908
	\begin{itemize}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   909
		\item
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   910
			If for all $r \in rs. \, \good \; r $ or $r = \RZERO$,
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   911
			then for all $r \in \rflts\; rs. \, \good \; r$.
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   912
		\item
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   913
			If $x \in \rdistinct{\rflts\; (\map \; rsimp{}\; rs)}{\varnothing}$
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   914
			and for all $y$ such that $\llbracket y \rrbracket_r$ less than
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   915
			$\llbracket rs \rrbracket_r + 1$, either
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   916
			$\good \; (\rsimp{y})$ or $\rsimp{y} = \RZERO$,
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   917
			then $\good \; x$.
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   918
	\end{itemize}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   919
\end{lemma}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   920
\begin{proof}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   921
	The first part is by induction on $rs$, where the induction
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   922
	rule is the inductive cases for $\rflts$.
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   923
	The second part is a corollary from the first part.
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   924
\end{proof}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   925
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   926
And this leads to good structural property of $\rsimp{}$,
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   927
that after simplification, a regular expression is
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   928
either good or $\RZERO$:
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   929
\begin{lemma}\label{good1}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   930
	For any r-regular expression $r$, $\good \; \rsimp{r}$ or $\rsimp{r} = \RZERO$.
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   931
\end{lemma}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   932
\begin{proof}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   933
	By an induction on $r$. The inductive measure is the size $\llbracket \rrbracket_r$.
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   934
	Lemma \ref{rsimpSize} says that 
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   935
	$\llbracket \rsimp{r}\rrbracket_r$ is smaller than or equal to
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   936
	$\llbracket r \rrbracket_r$.
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   937
	Therefore, in the $r_1 \cdot r_2$ and $\sum rs$ case,
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   938
	Inductive hypothesis applies to the children regular expressions
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   939
	$r_1$, $r_2$, etc. The lemma \ref{flts3Obv}'s precondition is satisfied
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   940
	by that as well.
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   941
	The lemmas \ref{nonnestedRsimp} and  \ref{nonaltFltsRd} are used
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   942
	to ensure that goodness is preserved at the topmost level.
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   943
\end{proof}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   944
We shall prove that any good regular expression is 
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   945
a fixed-point for $\rsimp{}$.
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   946
First we prove an auxiliary lemma:
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   947
\begin{lemma}\label{goodaltsNonalt}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   948
	If $\good \; \sum rs$, then $\rflts\; rs = rs$.
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   949
\end{lemma}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   950
\begin{proof}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   951
	By an induction on $\sum rs$. The inductive rules are the cases
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   952
	for $\good$.
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   953
\end{proof}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   954
\noindent
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   955
Now we are ready to prove that good regular expressions are invariant
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   956
of $\rsimp{}$ application:
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   957
\begin{lemma}\label{test}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   958
	If $\good \;r$ then $\rsimp{r} = r$.
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   959
\end{lemma}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   960
\begin{proof}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   961
	By an induction on the inductive cases of $\good$, using lemmas
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   962
	\ref{goodaltsNonalt} and \ref{rdistinctOnDistinct}.
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   963
	The lemma \ref{goodaltsNonalt} is used in the alternative
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   964
	case where 2 or more elements are present in the list.
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   965
\end{proof}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   966
\noindent
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   967
Given below is a property involving $\rflts$, $\rdistinct{}{}$, $\rsimp{}$ and $\rsimp_{ALTS}$,
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   968
which requires $\ref{good1}$ to go through smoothly.
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   969
It says that an application of $\rsimp_{ALTS}$ can be "absorbed",
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   970
if it its output is concatenated with a list and then applied to $\rflts$.
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   971
\begin{lemma}\label{flattenRsimpalts}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   972
	$\rflts \; ( (\rsimp_{ALTS} \; 
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   973
	(\rdistinct{(\rflts \; (\map \; \rsimp{}\; rs))}{\varnothing})) :: 
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   974
	\map \; \rsimp{} \; rs' ) = 
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   975
	\rflts \; ( (\rdistinct{(\rflts \; (\map \; \rsimp{}\; rs))}{\varnothing}) @ (
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   976
	\map \; \rsimp{rs'}))$
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   977
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   978
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   979
\end{lemma}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   980
\begin{proof}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   981
	By \ref{good1}.
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   982
\end{proof}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   983
\noindent
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   984
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   985
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   986
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   987
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   988
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   989
We are also ready to prove that $\textit{rsimp}$ is idempotent.
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   990
\subsubsection{$\rsimp$ is Idempotent}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   991
The idempotency of $\rsimp$ is very useful in 
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   992
manipulating regular expression terms into desired
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   993
forms so that key steps allowing further rewriting to closed forms
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   994
are possible.
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   995
\begin{lemma}\label{rsimpIdem}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   996
	$\rsimp{r} = \rsimp{\rsimp{r}}$
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   997
\end{lemma}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   998
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
   999
\begin{proof}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1000
	By \ref{test} and \ref{good1}.
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1001
\end{proof}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1002
\noindent
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1003
This property means we do not have to repeatedly
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1004
apply simplification in each step, which justifies
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1005
our definition of $\blexersimp$.
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1006
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1007
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1008
On the other hand, we could repeat the same $\rsimp{}$ applications
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1009
on regular expressions as many times as we want, if we have at least
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1010
one simplification applied to it, and apply it wherever we would like to:
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1011
\begin{corollary}\label{headOneMoreSimp}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1012
	The following properties hold, directly from \ref{rsimpIdem}:
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1013
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1014
	\begin{itemize}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1015
		\item
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1016
			$\map \; \rsimp{(r :: rs)} = \map \; \rsimp{} \; (\rsimp{r} :: rs)$
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1017
		\item
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1018
			$\rsimp{(\RALTS{rs})} = \rsimp{(\RALTS{\map \; \rsimp{} \; rs})}$
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1019
	\end{itemize}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1020
\end{corollary}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1021
\noindent
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1022
This will be useful in later closed form proof's rewriting steps.
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1023
Similarly, we point out the following useful facts below:
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1024
\begin{lemma}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1025
	The following equalities hold if $r = \rsimp{r'}$ for some $r'$:
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1026
	\begin{itemize}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1027
		\item
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1028
			If $r  = \sum rs$ then $\rsimpalts \; rs = \sum rs$.
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1029
		\item
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1030
			If $r = \sum rs$ then $\rdistinct{rs}{\varnothing} = rs$.
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1031
		\item
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1032
			$\rsimpalts \; (\rdistinct{\rflts \; [r]}{\varnothing}) = r$.
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1033
	\end{itemize}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1034
\end{lemma}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1035
\begin{proof}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1036
	By application of lemmas \ref{rsimpIdem} and \ref{good1}.
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1037
\end{proof}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1038
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1039
\noindent
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1040
With the idempotency of $\rsimp{}$ and its corollaries, 
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1041
we can start proving some key equalities leading to the 
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1042
closed forms.
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1043
Now presented are a few equivalent terms under $\rsimp{}$.
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1044
We use $r_1 \sequal r_2 $ here to denote $\rsimp{r_1} = \rsimp{r_2}$.
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1045
\begin{lemma}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1046
	\begin{itemize}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1047
		The following equivalence hold:
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1048
	\item
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1049
		$\rsimpalts \; (\RZERO :: rs) \sequal \rsimpalts\; rs$
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1050
	\item
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1051
		$\rsimpalts \; rs \sequal \rsimpalts (\map \; \rsimp{} \; rs)$
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1052
	\item
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1053
		$\RALTS{\RALTS{rs}} \sequal \RALTS{rs}$
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1054
	\item
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1055
		$\sum ((\sum rs_a) :: rs_b) \sequal \sum rs_a @ rs_b$
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1056
	\item
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1057
		$\RALTS{rs} = \RALTS{\map \; \rsimp{} \; rs}$
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1058
\end{itemize}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1059
\end{lemma}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1060
\begin{proof}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1061
	By induction on the lists involved.
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1062
\end{proof}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1063
\noindent
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1064
The above allows us to prove
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1065
two similar equalities (which are a bit more involved).
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1066
It says that we could flatten out the elements
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1067
before simplification and still get the same result.
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1068
\begin{lemma}\label{simpFlatten3}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1069
	One can flatten the inside $\sum$ of a $\sum$ if it is being 
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1070
	simplified. Concretely,
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1071
	\begin{itemize}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1072
		\item
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1073
			If for all $r \in rs, rs', rs''$, we have $\good \; r $
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1074
			or $r = \RZERO$, then $\sum (rs' @ rs @ rs'') \sequal 
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1075
			\sum (rs' @ [\sum rs] @ rs'')$ holds. As a corollary,
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1076
		\item
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1077
			$\sum (rs' @ [\sum rs] @ rs'') \sequal \sum (rs' @ rs @ rs'')$
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1078
	\end{itemize}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1079
\end{lemma}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1080
\begin{proof}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1081
	By rewriting steps involving the use of \ref{test} and \ref{rdistinctConcatGeneral}.
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1082
	The second sub-lemma is a corollary of the previous.
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1083
\end{proof}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1084
%Rewriting steps not put in--too long and complicated-------------------------------
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1085
\begin{comment}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1086
	\begin{center}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1087
		$\rsimp{\sum (rs' @ rs @ rs'')}  \stackrel{def of bsimp}{=}$  \\
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1088
		$\rsimpalts \; (\rdistinct{\rflts \; ((\map \; \rsimp{}\; rs') @ (\map \; \rsimp{} \; rs ) @ (\map \; \rsimp{} \; rs''))}{\varnothing})$ \\
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1089
		$\stackrel{by \ref{test}}{=} 
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1090
		\rsimpalts \; (\rdistinct{(\rflts \; rs' @ \rflts \; rs @ \rflts \; rs'')}{
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1091
		\varnothing})$\\
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1092
		$\stackrel{by \ref{rdistinctConcatGeneral}}{=}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1093
		\rsimpalts \; (\rdistinct{\rflts \; rs'}{\varnothing} @ \rdistinct{(
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1094
		\rflts\; rs @ \rflts \; rs'')}{\rflts \; rs'})$\\
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1095
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1096
	\end{center}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1097
\end{comment}
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1098
%Rewriting steps not put in--too long and complicated-------------------------------
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1099
\noindent
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1100
d5e9bcb384ec reorder
Chengsong
parents: 613
diff changeset
  1101
613
Chengsong
parents: 611
diff changeset
  1102
We need more equalities like the above to enable a closed form,
Chengsong
parents: 611
diff changeset
  1103
for which we need to introduce a few rewrite relations
Chengsong
parents: 611
diff changeset
  1104
to help
Chengsong
parents: 611
diff changeset
  1105
us obtain them.
554
Chengsong
parents: 553
diff changeset
  1106
610
d028c662a3df data files
Chengsong
parents: 609
diff changeset
  1107
\subsection{The rewrite relation $\hrewrite$ , $\scfrewrites$ , $\frewrite$ and $\grewrite$}
613
Chengsong
parents: 611
diff changeset
  1108
Inspired by the success we had in the correctness proof 
Chengsong
parents: 611
diff changeset
  1109
in \ref{Bitcoded2},
Chengsong
parents: 611
diff changeset
  1110
we follow suit here, defining atomic simplification
Chengsong
parents: 611
diff changeset
  1111
steps as ``small-step'' rewriting steps. This allows capturing 
555
Chengsong
parents: 554
diff changeset
  1112
similarities between terms that would be otherwise
Chengsong
parents: 554
diff changeset
  1113
hard to express.
Chengsong
parents: 554
diff changeset
  1114
557
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1115
We use $\hrewrite$ for one-step atomic rewrite of 
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1116
regular expression simplification, 
555
Chengsong
parents: 554
diff changeset
  1117
$\frewrite$ for rewrite of list of regular expressions that 
Chengsong
parents: 554
diff changeset
  1118
include all operations carried out in $\rflts$, and $\grewrite$ for
613
Chengsong
parents: 611
diff changeset
  1119
rewriting a list of regular expressions possible in both $\rflts$ and $\textit{rdistinct}$.
555
Chengsong
parents: 554
diff changeset
  1120
Their reflexive transitive closures are used to denote zero or many steps,
Chengsong
parents: 554
diff changeset
  1121
as was the case in the previous chapter.
613
Chengsong
parents: 611
diff changeset
  1122
As we have already
Chengsong
parents: 611
diff changeset
  1123
done something similar, the presentation about
Chengsong
parents: 611
diff changeset
  1124
these rewriting rules will be more concise than that in \ref{Bitcoded2}.
554
Chengsong
parents: 553
diff changeset
  1125
To differentiate between the rewriting steps for annotated regular expressions
Chengsong
parents: 553
diff changeset
  1126
and $\rrexp$s, we add characters $h$ and $g$ below the squig arrow symbol
Chengsong
parents: 553
diff changeset
  1127
to mean atomic simplification transitions 
Chengsong
parents: 553
diff changeset
  1128
of $\rrexp$s and $\rrexp$ lists, respectively.
Chengsong
parents: 553
diff changeset
  1129
555
Chengsong
parents: 554
diff changeset
  1130
Chengsong
parents: 554
diff changeset
  1131
Chengsong
parents: 554
diff changeset
  1132
613
Chengsong
parents: 611
diff changeset
  1133
\begin{figure}[H]
554
Chengsong
parents: 553
diff changeset
  1134
\begin{center}
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1135
	\begin{mathpar}
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1136
		\inferrule[RSEQ0L]{}{\RZERO \cdot r_2 \hrewrite \RZERO\\}
555
Chengsong
parents: 554
diff changeset
  1137
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1138
		\inferrule[RSEQ0R]{}{r_1 \cdot \RZERO \hrewrite \RZERO\\}
555
Chengsong
parents: 554
diff changeset
  1139
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1140
		\inferrule[RSEQ1]{}{(\RONE \cdot r) \hrewrite  r\\}\\	
555
Chengsong
parents: 554
diff changeset
  1141
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1142
		\inferrule[RSEQL]{ r_1 \hrewrite r_2}{r_1 \cdot r_3 \hrewrite r_2 \cdot r_3\\}
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1143
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1144
		\inferrule[RSEQR]{ r_3 \hrewrite r_4}{r_1 \cdot r_3 \hrewrite r_1 \cdot r_4\\}\\
555
Chengsong
parents: 554
diff changeset
  1145
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1146
		\inferrule[RALTSChild]{r \hrewrite r'}{\sum (rs_1 @ [r] @ rs_2) \hrewrite \sum (rs_1 @ [r'] @ rs_2)\\}
555
Chengsong
parents: 554
diff changeset
  1147
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1148
		\inferrule[RALTS0]{}{\sum (rs_a @ [\RZERO] @ rs_b) \hrewrite \sum (rs_a @ rs_b)}
555
Chengsong
parents: 554
diff changeset
  1149
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1150
		\inferrule[RALTSNested]{}{\sum (rs_a @ [\sum rs_1] @ rs_b) \hrewrite \sum (rs_a @ rs_1 @ rs_b)}
555
Chengsong
parents: 554
diff changeset
  1151
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1152
		\inferrule[RALTSNil]{}{ \sum [] \hrewrite \RZERO\\}
555
Chengsong
parents: 554
diff changeset
  1153
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1154
		\inferrule[RALTSSingle]{}{ \sum [r] \hrewrite  r\\}	
555
Chengsong
parents: 554
diff changeset
  1155
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1156
		\inferrule[RALTSDelete]{\\ r_1 = r_2}{\sum rs_a @ [r_1] @ rs_b @ [r_2] @ rsc \hrewrite \sum rs_a @ [r_1] @ rs_b @ rs_c}
555
Chengsong
parents: 554
diff changeset
  1157
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1158
	\end{mathpar}
555
Chengsong
parents: 554
diff changeset
  1159
\end{center}
613
Chengsong
parents: 611
diff changeset
  1160
\caption{List of one-step rewrite rules for r-regular expressions ($\hrewrite$)}\label{hRewrite}
Chengsong
parents: 611
diff changeset
  1161
\end{figure}
554
Chengsong
parents: 553
diff changeset
  1162
557
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1163
613
Chengsong
parents: 611
diff changeset
  1164
Like $\rightsquigarrow_s$, it is
Chengsong
parents: 611
diff changeset
  1165
convenient to define rewrite rules for a list of regular expressions,
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1166
where each element can rewrite in many steps to the other (scf stands for
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1167
li\emph{s}t \emph{c}losed \emph{f}orm). This relation is similar to the 
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1168
$\stackrel{s*}{\rightsquigarrow}$ for annotated regular expressions.
557
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1169
613
Chengsong
parents: 611
diff changeset
  1170
\begin{figure}[H]
557
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1171
\begin{center}
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1172
	\begin{mathpar}
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1173
		\inferrule{}{[] \scfrewrites [] }
613
Chengsong
parents: 611
diff changeset
  1174
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1175
		\inferrule{r \hrewrites r' \\ rs \scfrewrites rs'}{r :: rs \scfrewrites r' :: rs'}
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1176
	\end{mathpar}
557
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1177
\end{center}
613
Chengsong
parents: 611
diff changeset
  1178
\caption{List of one-step rewrite rules for a list of r-regular expressions}\label{scfRewrite}
Chengsong
parents: 611
diff changeset
  1179
\end{figure}
555
Chengsong
parents: 554
diff changeset
  1180
%frewrite
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1181
List of one-step rewrite rules for flattening 
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1182
a list of  regular expressions($\frewrite$):
613
Chengsong
parents: 611
diff changeset
  1183
\begin{figure}[H]
555
Chengsong
parents: 554
diff changeset
  1184
\begin{center}
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1185
	\begin{mathpar}
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1186
		\inferrule{}{\RZERO :: rs \frewrite rs \\}
555
Chengsong
parents: 554
diff changeset
  1187
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1188
		\inferrule{}{(\sum rs) :: rs_a \frewrite rs @ rs_a \\}
555
Chengsong
parents: 554
diff changeset
  1189
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1190
		\inferrule{rs_1 \frewrite rs_2}{r :: rs_1 \frewrite r :: rs_2}
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1191
	\end{mathpar}
555
Chengsong
parents: 554
diff changeset
  1192
\end{center}
613
Chengsong
parents: 611
diff changeset
  1193
\caption{List of one-step rewrite rules characterising the $\rflts$ operation on a list}\label{fRewrites}
Chengsong
parents: 611
diff changeset
  1194
\end{figure}
555
Chengsong
parents: 554
diff changeset
  1195
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1196
Lists of one-step rewrite rules for flattening and de-duplicating
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1197
a list of regular expressions ($\grewrite$):
613
Chengsong
parents: 611
diff changeset
  1198
\begin{figure}[H]
555
Chengsong
parents: 554
diff changeset
  1199
\begin{center}
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1200
	\begin{mathpar}
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1201
		\inferrule{}{\RZERO :: rs \grewrite rs \\}
532
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  1202
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1203
		\inferrule{}{(\sum rs) :: rs_a \grewrite rs @ rs_a \\}
555
Chengsong
parents: 554
diff changeset
  1204
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1205
		\inferrule{rs_1 \grewrite rs_2}{r :: rs_1 \grewrite r :: rs_2}
555
Chengsong
parents: 554
diff changeset
  1206
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1207
		\inferrule[dB]{}{rs_a @ [a] @ rs_b @[a] @ rs_c \grewrite rs_a @ [a] @ rsb @ rsc}
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1208
	\end{mathpar}
555
Chengsong
parents: 554
diff changeset
  1209
\end{center}
613
Chengsong
parents: 611
diff changeset
  1210
\caption{List of one-step rewrite rules characterising the $\rflts$ and $\textit{rdistinct}$
Chengsong
parents: 611
diff changeset
  1211
operations}\label{gRewrite}
Chengsong
parents: 611
diff changeset
  1212
\end{figure}
555
Chengsong
parents: 554
diff changeset
  1213
\noindent
611
Chengsong
parents: 610
diff changeset
  1214
We defined
613
Chengsong
parents: 611
diff changeset
  1215
two separate list rewriting relations $\frewrite$ and $\grewrite$.
611
Chengsong
parents: 610
diff changeset
  1216
The rewriting steps that take place during
Chengsong
parents: 610
diff changeset
  1217
flattening is characterised by $\frewrite$.
Chengsong
parents: 610
diff changeset
  1218
$\grewrite$ characterises both flattening and de-duplicating.
557
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1219
Sometimes $\grewrites$ is slightly too powerful
613
Chengsong
parents: 611
diff changeset
  1220
so we would rather use $\frewrites$ to prove
Chengsong
parents: 611
diff changeset
  1221
%because we only
Chengsong
parents: 611
diff changeset
  1222
equalities related to $\rflts$.
Chengsong
parents: 611
diff changeset
  1223
%certain equivalence under the rewriting steps of $\frewrites$.
556
Chengsong
parents: 555
diff changeset
  1224
For example, when proving the closed-form for the alternative regular expression,
613
Chengsong
parents: 611
diff changeset
  1225
one of the equalities needed is:
Chengsong
parents: 611
diff changeset
  1226
\begin{center}
557
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1227
	$\sum (\rDistinct \;\; (\map \; (\_ \backslash x) \; (\rflts \; rs)) \;\; \varnothing) \sequal
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1228
	\sum (\rDistinct \;\;  (\rflts \; (\map \; (\_ \backslash x) \; rs)) \;\; \varnothing)
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1229
	$
613
Chengsong
parents: 611
diff changeset
  1230
\end{center}
556
Chengsong
parents: 555
diff changeset
  1231
\noindent
Chengsong
parents: 555
diff changeset
  1232
Proving this is by first showing 
557
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1233
\begin{lemma}\label{earlyLaterDerFrewrites}
556
Chengsong
parents: 555
diff changeset
  1234
	$\map \; (\_ \backslash x) \;  (\rflts \; rs) \frewrites
557
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1235
	\rflts \; (\map \; (\_ \backslash x) \; rs)$
556
Chengsong
parents: 555
diff changeset
  1236
\end{lemma}
Chengsong
parents: 555
diff changeset
  1237
\noindent
613
Chengsong
parents: 611
diff changeset
  1238
and then the equivalence between two terms
Chengsong
parents: 611
diff changeset
  1239
that can reduce in many steps to each other.
556
Chengsong
parents: 555
diff changeset
  1240
\begin{lemma}\label{frewritesSimpeq}
Chengsong
parents: 555
diff changeset
  1241
	If $rs_1 \frewrites rs_2 $, then $\sum (\rDistinct \; rs_1 \; \varnothing) \sequal 
557
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1242
	\sum (\rDistinct \;  rs_2 \;  \varnothing)$.
556
Chengsong
parents: 555
diff changeset
  1243
\end{lemma}
557
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1244
\noindent
613
Chengsong
parents: 611
diff changeset
  1245
Chengsong
parents: 611
diff changeset
  1246
\begin{corollary}
Chengsong
parents: 611
diff changeset
  1247
	$\sum (\rDistinct \;\; (\map \; (\_ \backslash x) \; (\rflts \; rs)) \;\; \varnothing) \sequal
Chengsong
parents: 611
diff changeset
  1248
	\sum (\rDistinct \;\;  (\rflts \; (\map \; (\_ \backslash x) \; rs)) \;\; \varnothing)
Chengsong
parents: 611
diff changeset
  1249
	$
Chengsong
parents: 611
diff changeset
  1250
\end{corollary}
Chengsong
parents: 611
diff changeset
  1251
557
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1252
But this trick will not work for $\grewrites$.
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1253
For example, a rewriting step in proving
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1254
closed forms is:
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1255
\begin{center}
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1256
	$\rsimp{(\rsimpalts \; (\map \; (\_ \backslash x) \; (\rdistinct{(\rflts \; (\map \; (\rsimp{} \; \circ \; (\lambda r. \rderssimp{r}{xs}))))}{\varnothing})))}$\\
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1257
	$=$ \\
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1258
	$\rsimp{(\rsimpalts \; (\rdistinct{(\map \; (\_ \backslash x) \; (\rflts \; (\map \; (\rsimp{} \; \circ \; (\lambda r. \rderssimp{r}{xs})))) ) }{\varnothing}))} $
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1259
	\noindent
557
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1260
\end{center}
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1261
For this one would hope to have a rewriting relation between the two lists involved,
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1262
similar to \ref{earlyLaterDerFrewrites}. However, it turns out that 
556
Chengsong
parents: 555
diff changeset
  1263
\begin{center}
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1264
	$\map \; (\_ \backslash x) \; (\rDistinct \; rs \; rset) \grewrites \rDistinct \; (\map \;
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1265
	(\_ \backslash x) \; rs) \; ( rset \backslash x)$
556
Chengsong
parents: 555
diff changeset
  1266
\end{center}
Chengsong
parents: 555
diff changeset
  1267
\noindent
557
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1268
does $\mathbf{not}$ hold in general.
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1269
For this rewriting step we will introduce some slightly more cumbersome
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1270
proof technique in later sections.
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1271
The point is that $\frewrite$
613
Chengsong
parents: 611
diff changeset
  1272
allows us to prove equivalence in a straightforward way that is 
Chengsong
parents: 611
diff changeset
  1273
not possible for $\grewrite$. 
555
Chengsong
parents: 554
diff changeset
  1274
556
Chengsong
parents: 555
diff changeset
  1275
557
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1276
\subsubsection{Terms That Can Be Rewritten Using $\hrewrites$, $\grewrites$, and $\frewrites$}
613
Chengsong
parents: 611
diff changeset
  1277
In this part, we present lemmas stating
Chengsong
parents: 611
diff changeset
  1278
pairs of r-regular expressions or r-regular expression lists
Chengsong
parents: 611
diff changeset
  1279
where one could rewrite from one in many steps to the other.
Chengsong
parents: 611
diff changeset
  1280
Most of the proofs to these lemmas are straightforward, using
Chengsong
parents: 611
diff changeset
  1281
an induction on the inductive cases of the corresponding rewriting relations.
Chengsong
parents: 611
diff changeset
  1282
These proofs will therefore be omitted when this is the case.
557
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1283
We present in the below lemma a few pairs of terms that are rewritable via 
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1284
$\grewrites$:
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1285
\begin{lemma}\label{gstarRdistinctGeneral}
613
Chengsong
parents: 611
diff changeset
  1286
	\mbox{}
557
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1287
	\begin{itemize}
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1288
		\item
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1289
			$rs_1 @ rs \grewrites rs_1 @ (\rDistinct \; rs \; rs_1)$
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1290
		\item
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1291
			$rs \grewrites \rDistinct \; rs \; \varnothing$
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1292
		\item
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1293
			$rs_a @ (\rDistinct \; rs \; rs_a) \grewrites rs_a @ (\rDistinct \; 
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1294
			rs \; (\{\RZERO\} \cup rs_a))$
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1295
		\item
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1296
			$rs \;\; @ \;\; \rDistinct \; rs_a \; rset \grewrites rs @  \rDistinct \; rs_a \;
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1297
			(rest \cup rs)$
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1298
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1299
	\end{itemize}
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1300
\end{lemma}
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1301
\noindent
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1302
If a pair of terms $rs_1, rs_2$ are rewritable via $\grewrites$ to each other,
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1303
then they are equivalent under $\rsimp{}$:
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1304
\begin{lemma}\label{grewritesSimpalts}
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1305
	If $rs_1 \grewrites rs_2$, then
613
Chengsong
parents: 611
diff changeset
  1306
	we have the following equivalence:
557
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1307
	\begin{itemize}
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1308
		\item
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1309
			$\sum rs_1 \sequal \sum rs_2$
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1310
		\item
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1311
			$\rsimpalts \; rs_1 \sequal \rsimpalts \; rs_2$
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1312
	\end{itemize}
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1313
\end{lemma}
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1314
\noindent
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1315
Here are a few connecting lemmas showing that
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1316
if a list of regular expressions can be rewritten using $\grewrites$ or $\frewrites $ or
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1317
$\scfrewrites$,
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1318
then an alternative constructor taking the list can also be rewritten using $\hrewrites$:
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1319
\begin{lemma}
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1320
	\begin{itemize}
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1321
		\item
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1322
			If $rs \grewrites rs'$ then $\sum rs \hrewrites \sum rs'$.
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1323
		\item
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1324
			If $rs \grewrites rs'$ then $\sum rs \hrewrites \rsimpalts \; rs'$
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1325
		\item
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1326
			If $rs_1 \scfrewrites rs_2$ then $\sum (rs @ rs_1) \hrewrites \sum (rs @ rs_2)$
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1327
		\item
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1328
			If $rs_1 \scfrewrites rs_2$ then $\sum rs_1 \hrewrites \sum rs_2$
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1329
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1330
	\end{itemize}
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1331
\end{lemma}
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1332
\noindent
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1333
Here comes the meat of the proof, 
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1334
which says that once two lists are rewritable to each other,
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1335
then they are equivalent under $\rsimp{}$:
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1336
\begin{lemma}
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1337
	If $r_1 \hrewrites r_2$ then $r_1 \sequal r_2$.
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1338
\end{lemma}
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1339
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1340
\noindent
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1341
And similar to \ref{Bitcoded2} one can preserve rewritability after taking derivative
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1342
of two regular expressions on both sides:
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1343
\begin{lemma}\label{interleave}
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1344
	If $r \hrewrites r' $ then $\rder{c}{r} \hrewrites \rder{c}{r'}$
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1345
\end{lemma}
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1346
\noindent
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1347
This allows proving more $\mathbf{rsimp}$-equivalent terms, involving $\backslash_r$ now.
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1348
\begin{lemma}\label{insideSimpRemoval}
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1349
	$\rsimp{\rder{c}{\rsimp{r}}} = \rsimp{\rder{c}{r}}  $
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1350
\end{lemma}
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1351
\noindent
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1352
\begin{proof}
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1353
	By \ref{interleave} and \ref{rsimpIdem}.
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1354
\end{proof}
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1355
\noindent
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1356
And this unlocks more equivalent terms:
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1357
\begin{lemma}\label{Simpders}
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1358
	As corollaries of \ref{insideSimpRemoval}, we have
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1359
	\begin{itemize}
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1360
		\item
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1361
			If $s \neq []$ then $\rderssimp{r}{s} = \rsimp{(\rders \; r \; s)}$.
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1362
		\item
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1363
			$\rsimpalts \; (\map \; (\_ \backslash_r x) \;
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1364
			(\rdistinct{rs}{\varnothing})) \sequal
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1365
			\rsimpalts \; (\rDistinct \; 
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1366
			(\map \; (\_ \backslash_r x) rs) \;\varnothing  )$
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1367
	\end{itemize}
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1368
\end{lemma}
611
Chengsong
parents: 610
diff changeset
  1369
\begin{proof}
Chengsong
parents: 610
diff changeset
  1370
	Part 1 is by lemma \ref{insideSimpRemoval},
613
Chengsong
parents: 611
diff changeset
  1371
	part 2 is by lemma \ref{insideSimpRemoval} .%and \ref{distinctDer}.
611
Chengsong
parents: 610
diff changeset
  1372
\end{proof}
557
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1373
\noindent
613
Chengsong
parents: 611
diff changeset
  1374
Chengsong
parents: 611
diff changeset
  1375
\subsection{Closed Forms for $\sum rs$, $r_1\cdot r_2$ and $r^*$}
Chengsong
parents: 611
diff changeset
  1376
\subsubsection{Closed Form for Alternative Regular Expression}
Chengsong
parents: 611
diff changeset
  1377
Lemma \ref{Simpders} leads to the first closed form--
557
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1378
\begin{lemma}\label{altsClosedForm}
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1379
	\begin{center}
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1380
		$\rderssimp{(\sum rs)}{s} \sequal
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1381
		\sum \; (\map \; (\rderssimp{\_}{s}) \; rs)$
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1382
	\end{center}
557
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1383
\end{lemma}
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1384
556
Chengsong
parents: 555
diff changeset
  1385
\noindent
557
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1386
\begin{proof}
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1387
	By a reverse induction on the string $s$.
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1388
	One rewriting step, as we mentioned earlier,
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1389
	involves
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1390
	\begin{center}
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1391
		$\rsimpalts \; (\map \; (\_ \backslash x) \; 
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1392
		(\rdistinct{(\rflts \; (\map \; (\rsimp{} \; \circ \; 
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1393
		(\lambda r. \rderssimp{r}{xs}))))}{\varnothing}))
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1394
		\sequal
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1395
		\rsimpalts \; (\rdistinct{(\map \; (\_ \backslash x) \; 
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1396
			(\rflts \; (\map \; (\rsimp{} \; \circ \; 
557
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1397
		(\lambda r. \rderssimp{r}{xs})))) ) }{\varnothing}) $.
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1398
	\end{center}
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1399
	This can be proven by a combination of 
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1400
	\ref{grewritesSimpalts}, \ref{gstarRdistinctGeneral}, \ref{rderRsimpAltsCommute}, and
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1401
	\ref{insideSimpRemoval}.
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1402
\end{proof}
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1403
\noindent
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1404
This closed form has a variant which can be more convenient in later proofs:
559
Chengsong
parents: 558
diff changeset
  1405
\begin{corollary}{altsClosedForm1}
557
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1406
	If $s \neq []$ then 
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1407
	$\rderssimp \; (\sum \; rs) \; s = 
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1408
	\rsimp{(\sum \; (\map \; \rderssimp{\_}{s} \; rs))}$.
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1409
\end{corollary}
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1410
\noindent
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1411
The harder closed forms are the sequence and star ones.
613
Chengsong
parents: 611
diff changeset
  1412
Before we obtain them, some preliminary definitions
557
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1413
are needed to make proof statements concise.
556
Chengsong
parents: 555
diff changeset
  1414
609
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  1415
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  1416
\subsubsection{Closed Form for Sequence Regular Expressions}
558
Chengsong
parents: 557
diff changeset
  1417
First let's look at a series of derivatives steps on a sequence 
Chengsong
parents: 557
diff changeset
  1418
regular expression, assuming that each time the first
Chengsong
parents: 557
diff changeset
  1419
component of the sequence is always nullable):
557
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1420
\begin{center}
558
Chengsong
parents: 557
diff changeset
  1421
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1422
	$r_1 \cdot r_2 \quad \longrightarrow_{\backslash c}  \quad   r_1  \backslash c \cdot r_2 + r_2 \backslash c \quad \longrightarrow_{\backslash c'} \quad (r_1 \backslash cc' \cdot r_2 + r_2 \backslash c') + r_2 \backslash cc' \longrightarrow_{\backslash c''} \quad$\\
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1423
	$((r_1 \backslash cc'c'' \cdot r_2 + r_2 \backslash c'') + r_2 \backslash c'c'') + r_2 \backslash cc'c''   \longrightarrow_{\backslash c''} \quad
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1424
	\ldots$
558
Chengsong
parents: 557
diff changeset
  1425
557
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1426
\end{center}
558
Chengsong
parents: 557
diff changeset
  1427
Roughly speaking $r_1 \cdot r_2 \backslash s$ can be expresssed as 
Chengsong
parents: 557
diff changeset
  1428
a giant alternative taking a list of terms 
Chengsong
parents: 557
diff changeset
  1429
$[r_1 \backslash_r s \cdot r_2, r_2 \backslash_r s'', r_2 \backslash_r s_1'', \ldots]$,
Chengsong
parents: 557
diff changeset
  1430
where the head of the list is always the term
Chengsong
parents: 557
diff changeset
  1431
representing a match involving only $r_1$, and the tail of the list consisting of
Chengsong
parents: 557
diff changeset
  1432
terms of the shape $r_2 \backslash_r s''$, $s''$ being a suffix of $s$.
557
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1433
This intuition is also echoed by IndianPaper, where they gave
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1434
a pencil-and-paper derivation of $(r_1 \cdot r_2)\backslash s$:
532
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  1435
\begin{center}
558
Chengsong
parents: 557
diff changeset
  1436
	\begin{tabular}{c}
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1437
		$(r_1 \cdot r_2) \backslash_r (c_1 :: c_2 :: \ldots c_n) \myequiv$\\ 
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1438
		\rule{0pt}{3ex} $((r_1 \backslash_r c_1) \cdot r_2 + (\delta\; (\rnullable \; r_1) \; r_2 \backslash_r c_1)) \backslash_r (c_2 :: \ldots c_n) 
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1439
		\myequiv$\\
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1440
		\rule{0pt}{3ex} $((r_1 \backslash_r c_1c_2 \cdot r_2 + (\delta \; (\rnullable \; r_1) \; r_2 \backslash_r c_1c_2))
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1441
		+ (\delta \ (\rnullable \; r_1 \backslash_r c)\; r_2 \backslash_r c_2)) \backslash_r (c_3 \ldots c_n)
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1442
		$
558
Chengsong
parents: 557
diff changeset
  1443
	\end{tabular}
557
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1444
\end{center}
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1445
\noindent
558
Chengsong
parents: 557
diff changeset
  1446
The equality in above should be interpretated
Chengsong
parents: 557
diff changeset
  1447
as language equivalence. 
Chengsong
parents: 557
diff changeset
  1448
The $\delta$ function works similarly to that of
Chengsong
parents: 557
diff changeset
  1449
a Kronecker delta function:
Chengsong
parents: 557
diff changeset
  1450
\[ \delta \; b\; r\]
Chengsong
parents: 557
diff changeset
  1451
will produce $r$
Chengsong
parents: 557
diff changeset
  1452
if $b$ evaluates to true, 
Chengsong
parents: 557
diff changeset
  1453
and $\RZERO$ otherwise.
Chengsong
parents: 557
diff changeset
  1454
Note that their formulation  
Chengsong
parents: 557
diff changeset
  1455
\[
Chengsong
parents: 557
diff changeset
  1456
	((r_1 \backslash_r \, c_1c_2 \cdot r_2 + (\delta \; (\rnullable) \; r_1, r_2 \backslash_r c_1c_2)
Chengsong
parents: 557
diff changeset
  1457
	+ (\delta \; (\rnullable \; r_1 \backslash_r c)\; r_2 \backslash_r c_2)
Chengsong
parents: 557
diff changeset
  1458
\]
Chengsong
parents: 557
diff changeset
  1459
does not faithfully
Chengsong
parents: 557
diff changeset
  1460
represent what the intermediate derivatives would actually look like
Chengsong
parents: 557
diff changeset
  1461
when one or more intermediate results $r_1 \backslash s' \cdot r_2$ are not 
Chengsong
parents: 557
diff changeset
  1462
nullable in the head of the sequence.
Chengsong
parents: 557
diff changeset
  1463
For example, when $r_1$ and $r_1 \backslash_r c_1$ are not nullable,
Chengsong
parents: 557
diff changeset
  1464
the regular expression would not look like 
Chengsong
parents: 557
diff changeset
  1465
\[
Chengsong
parents: 557
diff changeset
  1466
	(r_1 \backslash_r c_1c_2 + \RZERO ) + \RZERO,
Chengsong
parents: 557
diff changeset
  1467
\]
Chengsong
parents: 557
diff changeset
  1468
but actually $r_1 \backslash_r c_1c_2$, the redundant $\RZERO$s will not be created in the
Chengsong
parents: 557
diff changeset
  1469
first place.
Chengsong
parents: 557
diff changeset
  1470
In a closed-form one would want to take into account this 
Chengsong
parents: 557
diff changeset
  1471
and generate the list of
Chengsong
parents: 557
diff changeset
  1472
regular expressions $r_2 \backslash_r s''$ with
Chengsong
parents: 557
diff changeset
  1473
string pairs $(s', s'')$ where $s'@s'' = s$ and
Chengsong
parents: 557
diff changeset
  1474
$r_1 \backslash s'$ nullable.
Chengsong
parents: 557
diff changeset
  1475
We denote the list consisting of such 
Chengsong
parents: 557
diff changeset
  1476
strings $s''$ as $\vsuf{s}{r_1}$.
Chengsong
parents: 557
diff changeset
  1477
Chengsong
parents: 557
diff changeset
  1478
The function $\vsuf{\_}{\_}$ is defined recursively on the structure of the string:
Chengsong
parents: 557
diff changeset
  1479
\begin{center}
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1480
	\begin{tabular}{lcl}
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1481
		$\vsuf{[]}{\_} $ & $=$ &  $[]$\\
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1482
		$\vsuf{c::cs}{r_1}$ & $ =$ & $ \textit{if} (\rnullable{r_1}) \textit{then} \; (\vsuf{cs}{(\rder{c}{r_1})}) @ [c :: cs]$\\
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1483
				    && $\textit{else} \; (\vsuf{cs}{(\rder{c}{r_1}) })  $
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1484
	\end{tabular}
558
Chengsong
parents: 557
diff changeset
  1485
\end{center}
Chengsong
parents: 557
diff changeset
  1486
\noindent
Chengsong
parents: 557
diff changeset
  1487
The list is sorted in the order $r_2\backslash s''$ 
Chengsong
parents: 557
diff changeset
  1488
appears in $(r_1\cdot r_2)\backslash s$.
Chengsong
parents: 557
diff changeset
  1489
In essence, $\vsuf{\_}{\_}$ is doing a 
Chengsong
parents: 557
diff changeset
  1490
"virtual derivative" of $r_1 \cdot r_2$, but instead of producing 
Chengsong
parents: 557
diff changeset
  1491
the entire result $(r_1 \cdot r_2) \backslash s$, 
Chengsong
parents: 557
diff changeset
  1492
it only stores all the strings $s''$ such that $r_2 \backslash s''$
Chengsong
parents: 557
diff changeset
  1493
are occurring terms in $(r_1\cdot r_2)\backslash s$.
Chengsong
parents: 557
diff changeset
  1494
Chengsong
parents: 557
diff changeset
  1495
To make the closed form representation 
Chengsong
parents: 557
diff changeset
  1496
more straightforward,
Chengsong
parents: 557
diff changeset
  1497
the flattetning function $\sflat{\_}$ is used to enable the transformation from 
557
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1498
a left-associative nested sequence of alternatives into 
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1499
a flattened list:
558
Chengsong
parents: 557
diff changeset
  1500
\[
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1501
	\sum [r_1, r_2, r_3, \ldots] \stackrel{\sflat{\_}}{\rightarrow} 
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1502
	(\ldots ((r_1 + r_2) + r_3) + \ldots)
558
Chengsong
parents: 557
diff changeset
  1503
\]
Chengsong
parents: 557
diff changeset
  1504
\noindent
Chengsong
parents: 557
diff changeset
  1505
The definitions $\sflat{\_}$, $\sflataux{\_}$ are given below.
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1506
\begin{center}  
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1507
	\begin{tabular}{ccc}
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1508
		$\sflataux{\AALTS{ }{r :: rs}}$ & $=$ & $\sflataux{r} @ rs$\\
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1509
		$\sflataux{\AALTS{ }{[]}}$ & $ = $ & $ []$\\
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1510
		$\sflataux r$ & $=$ & $ [r]$
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1511
	\end{tabular}
532
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  1512
\end{center}
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  1513
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1514
\begin{center} 
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1515
	\begin{tabular}{ccc}
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1516
		$\sflat{(\sum r :: rs)}$ & $=$ & $\sum (\sflataux{r} @ rs)$\\
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1517
		$\sflat{\sum []}$ & $ = $ & $ \sum []$\\
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1518
		$\sflat r$ & $=$ & $ r$
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1519
	\end{tabular}
557
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1520
\end{center}
558
Chengsong
parents: 557
diff changeset
  1521
\noindent
576
3e1b699696b6 thesis chap5
Chengsong
parents: 564
diff changeset
  1522
$\sflataux{\_}$ breaks up nested alternative regular expressions 
557
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1523
of the $(\ldots((r_1 + r_2) + r_3) + \ldots )$(left-associated) shape
558
Chengsong
parents: 557
diff changeset
  1524
into a "balanced" list: $\AALTS{\_}{[r_1,\, r_2 ,\, r_3, \ldots]}$.
557
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1525
It will return the singleton list $[r]$ otherwise.
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1526
$\sflat{\_}$ works the same  as $\sflataux{\_}$, except that it keeps
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1527
the output type a regular expression, not a list.
558
Chengsong
parents: 557
diff changeset
  1528
$\sflataux{\_}$  and $\sflat{\_}$ are only recursive on the  
Chengsong
parents: 557
diff changeset
  1529
first element of the list.
Chengsong
parents: 557
diff changeset
  1530
Chengsong
parents: 557
diff changeset
  1531
With $\sflataux{}$ a preliminary to the closed form can be stated,
Chengsong
parents: 557
diff changeset
  1532
where the derivative of $r_1 \cdot r_2 \backslash s$ can be
Chengsong
parents: 557
diff changeset
  1533
flattened into a list whose head and tail meet the description
Chengsong
parents: 557
diff changeset
  1534
we gave earlier.
Chengsong
parents: 557
diff changeset
  1535
\begin{lemma}\label{seqSfau0}
Chengsong
parents: 557
diff changeset
  1536
	$\sflataux{\rders{(r_1 \cdot r_2) \backslash s }} = (r_1 \backslash_r s) \cdot r_2 
Chengsong
parents: 557
diff changeset
  1537
	:: (\map \; (r_2 \backslash_r \_) \; (\textit{Suffix} \; s \; r1))$ 
Chengsong
parents: 557
diff changeset
  1538
\end{lemma}
Chengsong
parents: 557
diff changeset
  1539
\begin{proof}
Chengsong
parents: 557
diff changeset
  1540
	By an induction on the string $s$, where the inductive cases 
Chengsong
parents: 557
diff changeset
  1541
	are split as $[]$ and $xs @ [x]$.
Chengsong
parents: 557
diff changeset
  1542
	Note the key identify holds:
Chengsong
parents: 557
diff changeset
  1543
	\[
Chengsong
parents: 557
diff changeset
  1544
		\map \; (r_2 \backslash_r \_) \; (\vsuf{[x]}{(r_1 \backslash_r xs)}) \;\; @ \;\;
Chengsong
parents: 557
diff changeset
  1545
		\map \; (\_ \backslash_r x) \; (\map \; (r_2 \backslash \_) \; (\vsuf{xs}{r_1}))
Chengsong
parents: 557
diff changeset
  1546
	\]
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1547
	=
558
Chengsong
parents: 557
diff changeset
  1548
	\[
Chengsong
parents: 557
diff changeset
  1549
		\map \; (r_2 \backslash_r \_) \; (\vsuf{xs @ [x]}{r_1})
Chengsong
parents: 557
diff changeset
  1550
	\]
Chengsong
parents: 557
diff changeset
  1551
	This enables the inductive case to go through.
Chengsong
parents: 557
diff changeset
  1552
\end{proof}
Chengsong
parents: 557
diff changeset
  1553
\noindent 
Chengsong
parents: 557
diff changeset
  1554
Note that this lemma does $\mathbf{not}$ depend on any
Chengsong
parents: 557
diff changeset
  1555
specific definitions we used,
Chengsong
parents: 557
diff changeset
  1556
allowing people investigating derivatives to get an alternative
Chengsong
parents: 557
diff changeset
  1557
view of what $r_1 \cdot r_2$ is.
532
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  1558
558
Chengsong
parents: 557
diff changeset
  1559
Now we are able to use this for the intuition that 
Chengsong
parents: 557
diff changeset
  1560
the different ways in which regular expressions are 
Chengsong
parents: 557
diff changeset
  1561
nested do not matter under $\rsimp{}$:
557
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1562
\begin{center}
558
Chengsong
parents: 557
diff changeset
  1563
	$\rsimp{r} \stackrel{?}{\sequal} \rsimp{r'}$ if $r = \sum [r_1, r_2, r_3, \ldots]$ 
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1564
	and $r' =(\ldots ((r_1 + r_2) + r_3) + \ldots)$
557
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1565
\end{center}
558
Chengsong
parents: 557
diff changeset
  1566
Simply wrap with $\sum$ constructor and add 
Chengsong
parents: 557
diff changeset
  1567
simplifications to both sides of \ref{seqSfau0}
Chengsong
parents: 557
diff changeset
  1568
and one gets
Chengsong
parents: 557
diff changeset
  1569
\begin{corollary}\label{seqClosedFormGeneral}
Chengsong
parents: 557
diff changeset
  1570
	$\rsimp{\sflat{(r_1 \cdot r_2) \backslash s} }
Chengsong
parents: 557
diff changeset
  1571
	=\rsimp{(\sum (  (r_1 \backslash s) \cdot r_2 :: 
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1572
	\map\; (r_2 \backslash \_) \; (\vsuf{s}{r_1})))}$
558
Chengsong
parents: 557
diff changeset
  1573
\end{corollary}
Chengsong
parents: 557
diff changeset
  1574
Together with the idempotency property of $\rsimp{}$ (lemma \ref{rsimpIdem}),
Chengsong
parents: 557
diff changeset
  1575
it is possible to convert the above lemma to obtain a "closed form"
Chengsong
parents: 557
diff changeset
  1576
for  derivatives nested with simplification:
Chengsong
parents: 557
diff changeset
  1577
\begin{lemma}\label{seqClosedForm}
Chengsong
parents: 557
diff changeset
  1578
	$\rderssimp{(r_1 \cdot r_2)}{s} = \rsimp{(\sum ((r_1 \backslash s) \cdot r_2 ) 
Chengsong
parents: 557
diff changeset
  1579
	:: (\map \; (r_2 \backslash \_) (\vsuf{s}{r_1})))}$
Chengsong
parents: 557
diff changeset
  1580
\end{lemma}
Chengsong
parents: 557
diff changeset
  1581
\begin{proof}
Chengsong
parents: 557
diff changeset
  1582
	By a case analysis of string $s$.
Chengsong
parents: 557
diff changeset
  1583
	When $s$ is empty list, the rewrite is straightforward.
Chengsong
parents: 557
diff changeset
  1584
	When $s$ is a list, one could use the corollary \ref{seqSfau0},
Chengsong
parents: 557
diff changeset
  1585
	and lemma \ref{Simpders} to rewrite the left-hand-side.
Chengsong
parents: 557
diff changeset
  1586
\end{proof}
Chengsong
parents: 557
diff changeset
  1587
As a corollary for this closed form, one can estimate the size 
Chengsong
parents: 557
diff changeset
  1588
of the sequence derivative $r_1 \cdot r_2 \backslash_r s$ using 
Chengsong
parents: 557
diff changeset
  1589
an easier-to-handle expression:
Chengsong
parents: 557
diff changeset
  1590
\begin{corollary}\label{seqEstimate1}
Chengsong
parents: 557
diff changeset
  1591
	\begin{center}
557
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1592
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1593
		$\llbracket \rderssimp{(r_1 \cdot r_2)}{s} \rrbracket_r = \llbracket \rsimp{(\sum ((r_1 \backslash s) \cdot r_2 ) 
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1594
		:: (\map \; (r_2 \backslash \_) (\vsuf{s}{r_1})))} \rrbracket_r$
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1595
558
Chengsong
parents: 557
diff changeset
  1596
	\end{center}
Chengsong
parents: 557
diff changeset
  1597
\end{corollary}
Chengsong
parents: 557
diff changeset
  1598
\noindent
609
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  1599
\subsubsection{Closed Forms for Star Regular Expressions}
564
Chengsong
parents: 562
diff changeset
  1600
We have shown how to control the size of the sequence regular expression $r_1\cdot r_2$ using
Chengsong
parents: 562
diff changeset
  1601
the "closed form" of $(r_1 \cdot r_2) \backslash s$ and then 
Chengsong
parents: 562
diff changeset
  1602
the property of the $\distinct$ function.
Chengsong
parents: 562
diff changeset
  1603
Now we try to get a bound on $r^* \backslash s$ as well.
Chengsong
parents: 562
diff changeset
  1604
Again, we first look at how a star's derivatives evolve, if they grow maximally: 
Chengsong
parents: 562
diff changeset
  1605
\begin{center}
Chengsong
parents: 562
diff changeset
  1606
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1607
	$r^* \quad \longrightarrow_{\backslash c}  \quad   (r\backslash c)  \cdot  r^* \quad \longrightarrow_{\backslash c'}  \quad
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1608
	r \backslash cc'  \cdot r^* + r \backslash c' \cdot r^*  \quad \longrightarrow_{\backslash c''} \quad 
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1609
	(r_1 \backslash cc'c'' \cdot r^* + r \backslash c'') + (r \backslash c'c'' \cdot r^* + r \backslash c'' \cdot r^*)   \quad \longrightarrow_{\backslash c'''}
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1610
	\quad \ldots$
564
Chengsong
parents: 562
diff changeset
  1611
Chengsong
parents: 562
diff changeset
  1612
\end{center}
Chengsong
parents: 562
diff changeset
  1613
When we have a string $s = c :: c' :: c'' \ldots$  such that $r \backslash c$, $r \backslash cc'$, $r \backslash c'$, 
Chengsong
parents: 562
diff changeset
  1614
$r \backslash cc'c''$, $r \backslash c'c''$, $r\backslash c''$ etc. are all nullable,
Chengsong
parents: 562
diff changeset
  1615
the number of terms in $r^* \backslash s$ will grow exponentially, causing the size
Chengsong
parents: 562
diff changeset
  1616
of the derivatives $r^* \backslash s$ to grow exponentially, even if we do not 
Chengsong
parents: 562
diff changeset
  1617
count the possible size explosions of $r \backslash c$ themselves.
Chengsong
parents: 562
diff changeset
  1618
576
3e1b699696b6 thesis chap5
Chengsong
parents: 564
diff changeset
  1619
Thanks to $\rflts$ and $\rDistinct$, we are able to open up regular expressions like
564
Chengsong
parents: 562
diff changeset
  1620
$(r_1 \backslash cc'c'' \cdot r^* + r \backslash c'') + 
Chengsong
parents: 562
diff changeset
  1621
(r \backslash c'c'' \cdot r^* + r \backslash c'' \cdot r^*) $ 
Chengsong
parents: 562
diff changeset
  1622
into $\RALTS{[r_1 \backslash cc'c'' \cdot r^*, r \backslash c'', 
Chengsong
parents: 562
diff changeset
  1623
r \backslash c'c'' \cdot r^*, r \backslash c'' \cdot r^*]}$
Chengsong
parents: 562
diff changeset
  1624
and then de-duplicate terms of the form $r\backslash s' \cdot r^*$ ($s'$ being a substring of $s$).
Chengsong
parents: 562
diff changeset
  1625
This allows us to use a similar technique as $r_1 \cdot r_2$ case,
Chengsong
parents: 562
diff changeset
  1626
where the crux is to get an equivalent form of 
Chengsong
parents: 562
diff changeset
  1627
$\rderssimp{r^*}{s}$ with shape $\rsimp{\sum rs}$.
Chengsong
parents: 562
diff changeset
  1628
This requires generating 
558
Chengsong
parents: 557
diff changeset
  1629
all possible sub-strings $s'$ of $s$
Chengsong
parents: 557
diff changeset
  1630
such that $r\backslash s' \cdot r^*$ will appear 
Chengsong
parents: 557
diff changeset
  1631
as a term in $(r^*) \backslash s$.
Chengsong
parents: 557
diff changeset
  1632
The first function we define is a single-step
Chengsong
parents: 557
diff changeset
  1633
updating function $\starupdate$, which takes three arguments as input:
Chengsong
parents: 557
diff changeset
  1634
the new character $c$ to take derivative with, 
Chengsong
parents: 557
diff changeset
  1635
the regular expression
Chengsong
parents: 557
diff changeset
  1636
$r$ directly under the star $r^*$, and the
Chengsong
parents: 557
diff changeset
  1637
list of strings $sSet$ for the derivative $r^* \backslash s$ 
Chengsong
parents: 557
diff changeset
  1638
up til this point  
Chengsong
parents: 557
diff changeset
  1639
such that $(r^*) \backslash s = \sum_{s' \in sSet} (r\backslash s') \cdot r^*$ 
Chengsong
parents: 557
diff changeset
  1640
(the equality is not exact, more on this later).
Chengsong
parents: 557
diff changeset
  1641
\begin{center}
Chengsong
parents: 557
diff changeset
  1642
	\begin{tabular}{lcl}
Chengsong
parents: 557
diff changeset
  1643
		$\starupdate \; c \; r \; [] $ & $\dn$ & $[]$\\
Chengsong
parents: 557
diff changeset
  1644
		$\starupdate \; c \; r \; (s :: Ss)$ & $\dn$ & \\
Chengsong
parents: 557
diff changeset
  1645
						     & & $\textit{if} \; 
Chengsong
parents: 557
diff changeset
  1646
						     (\rnullable \; (\rders \; r \; s))$ \\
Chengsong
parents: 557
diff changeset
  1647
						     & & $\textit{then} \;\; (s @ [c]) :: [c] :: (
Chengsong
parents: 557
diff changeset
  1648
						     \starupdate \; c \; r \; Ss)$ \\
Chengsong
parents: 557
diff changeset
  1649
						     & & $\textit{else} \;\; (s @ [c]) :: (
Chengsong
parents: 557
diff changeset
  1650
						     \starupdate \; c \; r \; Ss)$
Chengsong
parents: 557
diff changeset
  1651
	\end{tabular}
Chengsong
parents: 557
diff changeset
  1652
\end{center}
Chengsong
parents: 557
diff changeset
  1653
\noindent
Chengsong
parents: 557
diff changeset
  1654
As a generalisation from characters to strings,
Chengsong
parents: 557
diff changeset
  1655
$\starupdates$ takes a string instead of a character
Chengsong
parents: 557
diff changeset
  1656
as the first input argument, and is otherwise the same
Chengsong
parents: 557
diff changeset
  1657
as $\starupdate$.
Chengsong
parents: 557
diff changeset
  1658
\begin{center}
Chengsong
parents: 557
diff changeset
  1659
	\begin{tabular}{lcl}
Chengsong
parents: 557
diff changeset
  1660
		$\starupdates \; [] \; r \; Ss$ & $=$ & $Ss$\\
Chengsong
parents: 557
diff changeset
  1661
		$\starupdates \; (c :: cs) \; r \; Ss$ &  $=$ &  $\starupdates \; cs \; r \; (
Chengsong
parents: 557
diff changeset
  1662
		\starupdate \; c \; r \; Ss)$
Chengsong
parents: 557
diff changeset
  1663
	\end{tabular}
Chengsong
parents: 557
diff changeset
  1664
\end{center}
Chengsong
parents: 557
diff changeset
  1665
\noindent
Chengsong
parents: 557
diff changeset
  1666
For the star regular expression,
Chengsong
parents: 557
diff changeset
  1667
its derivatives can be seen as  a nested gigantic
Chengsong
parents: 557
diff changeset
  1668
alternative similar to that of sequence regular expression's derivatives, 
Chengsong
parents: 557
diff changeset
  1669
and therefore need
Chengsong
parents: 557
diff changeset
  1670
to be ``straightened out" as well.
Chengsong
parents: 557
diff changeset
  1671
The function for this would be $\hflat{}$ and $\hflataux{}$.
Chengsong
parents: 557
diff changeset
  1672
\begin{center}
Chengsong
parents: 557
diff changeset
  1673
	\begin{tabular}{lcl}
Chengsong
parents: 557
diff changeset
  1674
		$\hflataux{r_1 + r_2}$ & $\dn$ & $\hflataux{r_1} @ \hflataux{r_2}$\\
Chengsong
parents: 557
diff changeset
  1675
		$\hflataux{r}$ & $\dn$ & $[r]$
Chengsong
parents: 557
diff changeset
  1676
	\end{tabular}
Chengsong
parents: 557
diff changeset
  1677
\end{center}
557
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1678
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1679
\begin{center}
558
Chengsong
parents: 557
diff changeset
  1680
	\begin{tabular}{lcl}
Chengsong
parents: 557
diff changeset
  1681
		$\hflat{r_1 + r_2}$ & $\dn$ & $\sum (\hflataux {r_1} @ \hflataux {r_2}) $\\
Chengsong
parents: 557
diff changeset
  1682
		$\hflat{r}$ & $\dn$ & $r$
Chengsong
parents: 557
diff changeset
  1683
	\end{tabular}
Chengsong
parents: 557
diff changeset
  1684
\end{center}
Chengsong
parents: 557
diff changeset
  1685
\noindent
Chengsong
parents: 557
diff changeset
  1686
%MAYBE TODO: introduce createdByStar
564
Chengsong
parents: 562
diff changeset
  1687
Again these definitions are tailor-made for dealing with alternatives that have
Chengsong
parents: 562
diff changeset
  1688
originated from a star's derivatives, so we do not attempt to open up all possible 
576
3e1b699696b6 thesis chap5
Chengsong
parents: 564
diff changeset
  1689
regular expressions of the form $\RALTS{rs}$, where $\textit{rs}$ might not contain precisely 2
564
Chengsong
parents: 562
diff changeset
  1690
elements.
Chengsong
parents: 562
diff changeset
  1691
We give a predicate for such "star-created" regular expressions:
Chengsong
parents: 562
diff changeset
  1692
\begin{center}
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1693
	\begin{tabular}{lcr}
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1694
	 &    &       $\createdByStar{(\RSEQ{ra}{\RSTAR{rb}}) }$\\
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1695
		$\createdByStar{r_1} \land \createdByStar{r_2} $ & $ \Longrightarrow$ & $\createdByStar{(r_1 + r_2)}$
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1696
	\end{tabular}
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1697
\end{center}
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1698
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1699
These definitions allows us the flexibility to talk about 
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1700
regular expressions in their most convenient format,
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1701
for example, flattened out $\RALTS{[r_1, r_2, \ldots, r_n]} $
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1702
instead of binary-nested: $((r_1 + r_2) + (r_3 + r_4)) + \ldots$.
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1703
These definitions help express that certain classes of syntatically 
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1704
distinct regular expressions are actually the same under simplification.
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1705
This is not entirely true for annotated regular expressions: 
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1706
%TODO: bsimp bders \neq bderssimp
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1707
\begin{center}
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1708
	$(1+ (c\cdot \ASEQ{bs}{c^*}{c} ))$
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1709
\end{center}
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1710
For bit-codes, the order in which simplification is applied
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1711
might cause a difference in the location they are placed.
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1712
If we want something like
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1713
\begin{center}
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1714
	$\bderssimp{r}{s} \myequiv \bsimp{\bders{r}{s}}$
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1715
\end{center}
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1716
Some "canonicalization" procedure is required,
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1717
which either pushes all the common bitcodes to nodes
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1718
as senior as possible:
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1719
\begin{center}
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1720
	$_{bs}(_{bs_1 @ bs'}r_1 + _{bs_1 @ bs''}r_2) \rightarrow _{bs @ bs_1}(_{bs'}r_1 + _{bs''}r_2) $
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1721
\end{center}
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1722
or does the reverse. However bitcodes are not of interest if we are talking about
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1723
the $\llbracket r \rrbracket$ size of a regex.
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1724
Therefore for the ease and simplicity of producing a
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1725
proof for a size bound, we are happy to restrict ourselves to 
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1726
unannotated regular expressions, and obtain such equalities as
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1727
\begin{lemma}
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1728
	$\rsimp{r_1 + r_2} = \rsimp{\RALTS{\hflataux{r_1} @ \hflataux{r_2}}}$
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1729
\end{lemma}
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1730
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1731
\begin{proof}
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1732
	By using the rewriting relation $\rightsquigarrow$
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1733
\end{proof}
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1734
%TODO: rsimp sflat
564
Chengsong
parents: 562
diff changeset
  1735
And from this we obtain a proof that a star's derivative will be the same
Chengsong
parents: 562
diff changeset
  1736
as if it had all its nested alternatives created during deriving being flattened out:
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1737
For example,
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1738
\begin{lemma}
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1739
	$\createdByStar{r} \implies \rsimp{r} = \rsimp{\RALTS{\hflataux{r}}}$
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1740
\end{lemma}
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1741
\begin{proof}
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1742
	By structural induction on $r$, where the induction rules are these of $\createdByStar{_}$.
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1743
\end{proof}
564
Chengsong
parents: 562
diff changeset
  1744
% The simplification of a flattened out regular expression, provided it comes
Chengsong
parents: 562
diff changeset
  1745
%from the derivative of a star, is the same as the one nested.
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1746
564
Chengsong
parents: 562
diff changeset
  1747
Chengsong
parents: 562
diff changeset
  1748
558
Chengsong
parents: 557
diff changeset
  1749
We first introduce an inductive property
Chengsong
parents: 557
diff changeset
  1750
for $\starupdate$ and $\hflataux{\_}$, 
Chengsong
parents: 557
diff changeset
  1751
it says if we do derivatives of $r^*$
Chengsong
parents: 557
diff changeset
  1752
with a string that starts with $c$,
Chengsong
parents: 557
diff changeset
  1753
then flatten it out,
Chengsong
parents: 557
diff changeset
  1754
we obtain a list
Chengsong
parents: 557
diff changeset
  1755
of the shape $\sum_{s' \in sSet} (r\backslash_r s') \cdot r^*$,
Chengsong
parents: 557
diff changeset
  1756
where $sSet = \starupdates \; s \; r \; [[c]]$.
Chengsong
parents: 557
diff changeset
  1757
\begin{lemma}\label{starHfauInduct}
Chengsong
parents: 557
diff changeset
  1758
	$\hflataux{(\rders{( (\rder{c}{r_0})\cdot(r_0^*))}{s})} = 
Chengsong
parents: 557
diff changeset
  1759
	\map \; (\lambda s_1. (r_0 \backslash_r s_1) \cdot (r_0^*)) \; 
Chengsong
parents: 557
diff changeset
  1760
	(\starupdates \; s \; r_0 \; [[c]])$
Chengsong
parents: 557
diff changeset
  1761
\end{lemma}
Chengsong
parents: 557
diff changeset
  1762
\begin{proof}
Chengsong
parents: 557
diff changeset
  1763
	By an induction on $s$, the inductive cases
Chengsong
parents: 557
diff changeset
  1764
	being $[]$ and $s@[c]$.
Chengsong
parents: 557
diff changeset
  1765
\end{proof}
Chengsong
parents: 557
diff changeset
  1766
\noindent
Chengsong
parents: 557
diff changeset
  1767
Here is a corollary that states the lemma in
Chengsong
parents: 557
diff changeset
  1768
a more intuitive way:
Chengsong
parents: 557
diff changeset
  1769
\begin{corollary}
Chengsong
parents: 557
diff changeset
  1770
	$\hflataux{r^* \backslash_r (c::xs)} = \map \; (\lambda s. (r \backslash_r s) \cdot
Chengsong
parents: 557
diff changeset
  1771
	(r^*))\; (\starupdates \; c\; r\; [[c]])$
Chengsong
parents: 557
diff changeset
  1772
\end{corollary}
Chengsong
parents: 557
diff changeset
  1773
\noindent
Chengsong
parents: 557
diff changeset
  1774
Note that this is also agnostic of the simplification
Chengsong
parents: 557
diff changeset
  1775
function we defined, and is therefore of more general interest.
Chengsong
parents: 557
diff changeset
  1776
Chengsong
parents: 557
diff changeset
  1777
Now adding the $\rsimp{}$ bit for closed forms,
Chengsong
parents: 557
diff changeset
  1778
we have
Chengsong
parents: 557
diff changeset
  1779
\begin{lemma}
Chengsong
parents: 557
diff changeset
  1780
	$a :: rs \grewrites \hflataux{a} @ rs$
Chengsong
parents: 557
diff changeset
  1781
\end{lemma}
Chengsong
parents: 557
diff changeset
  1782
\noindent
Chengsong
parents: 557
diff changeset
  1783
giving us
Chengsong
parents: 557
diff changeset
  1784
\begin{lemma}\label{cbsHfauRsimpeq1}
Chengsong
parents: 557
diff changeset
  1785
	$\rsimp{a+b} = \rsimp{(\sum \hflataux{a} @ \hflataux{b})}$.
Chengsong
parents: 557
diff changeset
  1786
\end{lemma}
Chengsong
parents: 557
diff changeset
  1787
\noindent
Chengsong
parents: 557
diff changeset
  1788
This yields
Chengsong
parents: 557
diff changeset
  1789
\begin{lemma}\label{hfauRsimpeq2}
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1790
	$\rsimp{r} = \rsimp{(\sum \hflataux{r})}$
558
Chengsong
parents: 557
diff changeset
  1791
\end{lemma}
Chengsong
parents: 557
diff changeset
  1792
\noindent
Chengsong
parents: 557
diff changeset
  1793
Together with the rewriting relation
Chengsong
parents: 557
diff changeset
  1794
\begin{lemma}\label{starClosedForm6Hrewrites}
Chengsong
parents: 557
diff changeset
  1795
	$\map \; (\lambda s. (\rsimp{r \backslash_r s}) \cdot (r^*)) \; Ss
Chengsong
parents: 557
diff changeset
  1796
	\scfrewrites
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1797
	\map \; (\lambda s. (\rsimp{r \backslash_r s}) \cdot (r^*)) \; Ss$
558
Chengsong
parents: 557
diff changeset
  1798
\end{lemma}
Chengsong
parents: 557
diff changeset
  1799
\noindent
Chengsong
parents: 557
diff changeset
  1800
We obtain the closed form for star regular expression:
Chengsong
parents: 557
diff changeset
  1801
\begin{lemma}\label{starClosedForm}
Chengsong
parents: 557
diff changeset
  1802
	$\rderssimp{r^*}{c::s} = 
Chengsong
parents: 557
diff changeset
  1803
	\rsimp{
Chengsong
parents: 557
diff changeset
  1804
		(\sum (\map \; (\lambda s. (\rderssimp{r}{s})\cdot r^*) \; 
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1805
		(\starupdates \; s\; r \; [[c]])
558
Chengsong
parents: 557
diff changeset
  1806
		)
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1807
		)
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1808
	}
558
Chengsong
parents: 557
diff changeset
  1809
	$
Chengsong
parents: 557
diff changeset
  1810
\end{lemma}
Chengsong
parents: 557
diff changeset
  1811
\begin{proof}
Chengsong
parents: 557
diff changeset
  1812
	By an induction on $s$.
Chengsong
parents: 557
diff changeset
  1813
	The lemmas \ref{rsimpIdem}, \ref{starHfauInduct}, and \ref{hfauRsimpeq2}
Chengsong
parents: 557
diff changeset
  1814
	are used.	
Chengsong
parents: 557
diff changeset
  1815
\end{proof}
609
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  1816
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  1817
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  1818
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  1819
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  1820
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  1821
613
Chengsong
parents: 611
diff changeset
  1822
%----------------------------------------------------------------------------------------
Chengsong
parents: 611
diff changeset
  1823
%	SECTION ??
Chengsong
parents: 611
diff changeset
  1824
%----------------------------------------------------------------------------------------
Chengsong
parents: 611
diff changeset
  1825
Chengsong
parents: 611
diff changeset
  1826
%-----------------------------------
Chengsong
parents: 611
diff changeset
  1827
%	SECTION syntactic equivalence under simp
Chengsong
parents: 611
diff changeset
  1828
%-----------------------------------
Chengsong
parents: 611
diff changeset
  1829
Chengsong
parents: 611
diff changeset
  1830
Chengsong
parents: 611
diff changeset
  1831
%----------------------------------------------------------------------------------------
Chengsong
parents: 611
diff changeset
  1832
%	SECTION ALTS CLOSED FORM
Chengsong
parents: 611
diff changeset
  1833
%----------------------------------------------------------------------------------------
Chengsong
parents: 611
diff changeset
  1834
%\section{A Closed Form for \textit{ALTS}}
Chengsong
parents: 611
diff changeset
  1835
%Now we prove that  $rsimp (rders\_simp (RALTS rs) s) = rsimp (RALTS (map (\lambda r. rders\_simp r s) rs))$.
Chengsong
parents: 611
diff changeset
  1836
%
Chengsong
parents: 611
diff changeset
  1837
%
Chengsong
parents: 611
diff changeset
  1838
%There are a few key steps, one of these steps is
Chengsong
parents: 611
diff changeset
  1839
%
Chengsong
parents: 611
diff changeset
  1840
%
Chengsong
parents: 611
diff changeset
  1841
%
Chengsong
parents: 611
diff changeset
  1842
%One might want to prove this by something a simple statement like: 
Chengsong
parents: 611
diff changeset
  1843
%
Chengsong
parents: 611
diff changeset
  1844
%For this to hold we want the $\textit{distinct}$ function to pick up
Chengsong
parents: 611
diff changeset
  1845
%the elements before and after derivatives correctly:
Chengsong
parents: 611
diff changeset
  1846
%$r \in rset \equiv (rder x r) \in (rder x rset)$.
Chengsong
parents: 611
diff changeset
  1847
%which essentially requires that the function $\backslash$ is an injective mapping.
Chengsong
parents: 611
diff changeset
  1848
%
Chengsong
parents: 611
diff changeset
  1849
%Unfortunately the function $\backslash c$ is not an injective mapping.
Chengsong
parents: 611
diff changeset
  1850
%
Chengsong
parents: 611
diff changeset
  1851
%\subsection{function $\backslash c$ is not injective (1-to-1)}
Chengsong
parents: 611
diff changeset
  1852
%\begin{center}
Chengsong
parents: 611
diff changeset
  1853
%	The derivative $w.r.t$ character $c$ is not one-to-one.
Chengsong
parents: 611
diff changeset
  1854
%	Formally,
Chengsong
parents: 611
diff changeset
  1855
%	$\exists r_1 \;r_2. r_1 \neq r_2 \mathit{and} r_1 \backslash c = r_2 \backslash c$
Chengsong
parents: 611
diff changeset
  1856
%\end{center}
Chengsong
parents: 611
diff changeset
  1857
%This property is trivially true for the
Chengsong
parents: 611
diff changeset
  1858
%character regex example:
Chengsong
parents: 611
diff changeset
  1859
%\begin{center}
Chengsong
parents: 611
diff changeset
  1860
%	$r_1 = e; \; r_2 = d;\; r_1 \backslash c = \ZERO = r_2 \backslash c$
Chengsong
parents: 611
diff changeset
  1861
%\end{center}
Chengsong
parents: 611
diff changeset
  1862
%But apart from the cases where the derivative
Chengsong
parents: 611
diff changeset
  1863
%output is $\ZERO$, are there non-trivial results
Chengsong
parents: 611
diff changeset
  1864
%of derivatives which contain strings?
Chengsong
parents: 611
diff changeset
  1865
%The answer is yes.
Chengsong
parents: 611
diff changeset
  1866
%For example,
Chengsong
parents: 611
diff changeset
  1867
%\begin{center}
Chengsong
parents: 611
diff changeset
  1868
%	Let $r_1 = a^*b\;\quad r_2 = (a\cdot a^*)\cdot b + b$.\\
Chengsong
parents: 611
diff changeset
  1869
%	where $a$ is not nullable.\\
Chengsong
parents: 611
diff changeset
  1870
%	$r_1 \backslash c = ((a \backslash c)\cdot a^*)\cdot c + b \backslash c$\\
Chengsong
parents: 611
diff changeset
  1871
%	$r_2 \backslash c = ((a \backslash c)\cdot a^*)\cdot c + b \backslash c$
Chengsong
parents: 611
diff changeset
  1872
%\end{center}
Chengsong
parents: 611
diff changeset
  1873
%We start with two syntactically different regular expressions,
Chengsong
parents: 611
diff changeset
  1874
%and end up with the same derivative result.
Chengsong
parents: 611
diff changeset
  1875
%This is not surprising as we have such 
Chengsong
parents: 611
diff changeset
  1876
%equality as below in the style of Arden's lemma:\\
Chengsong
parents: 611
diff changeset
  1877
%\begin{center}
Chengsong
parents: 611
diff changeset
  1878
%	$L(A^*B) = L(A\cdot A^* \cdot B + B)$
Chengsong
parents: 611
diff changeset
  1879
%\end{center}
Chengsong
parents: 611
diff changeset
  1880
\section{Bounding Closed Forms}
Chengsong
parents: 611
diff changeset
  1881
Chengsong
parents: 611
diff changeset
  1882
In this section, we introduce how we formalised the bound
Chengsong
parents: 611
diff changeset
  1883
on closed forms.
Chengsong
parents: 611
diff changeset
  1884
We first prove that functions such as $\rflts$
Chengsong
parents: 611
diff changeset
  1885
will not cause the size of r-regular expressions to grow.
Chengsong
parents: 611
diff changeset
  1886
Putting this together with a general bound 
Chengsong
parents: 611
diff changeset
  1887
on the finiteness of distinct regular expressions
Chengsong
parents: 611
diff changeset
  1888
smaller than a certain size, we obtain a bound on 
Chengsong
parents: 611
diff changeset
  1889
the closed forms.
Chengsong
parents: 611
diff changeset
  1890
Chengsong
parents: 611
diff changeset
  1891
\subsection{$\textit{rsimp}$ Does Not Increment the Size}
Chengsong
parents: 611
diff changeset
  1892
Although it seems evident, we need a series
Chengsong
parents: 611
diff changeset
  1893
of non-trivial lemmas to establish that functions such as $\rflts$
Chengsong
parents: 611
diff changeset
  1894
do not cause the regular expressions to grow.
Chengsong
parents: 611
diff changeset
  1895
\begin{lemma}\label{rsimpMonoLemmas}
Chengsong
parents: 611
diff changeset
  1896
	\mbox{}
Chengsong
parents: 611
diff changeset
  1897
	\begin{itemize}
Chengsong
parents: 611
diff changeset
  1898
		\item
Chengsong
parents: 611
diff changeset
  1899
			\[
Chengsong
parents: 611
diff changeset
  1900
				\llbracket \rsimpalts \; rs \rrbracket_r \leq
Chengsong
parents: 611
diff changeset
  1901
				\llbracket \sum \; rs \rrbracket_r
Chengsong
parents: 611
diff changeset
  1902
			\]
Chengsong
parents: 611
diff changeset
  1903
		\item
Chengsong
parents: 611
diff changeset
  1904
			\[
Chengsong
parents: 611
diff changeset
  1905
				\llbracket \rsimpseq \; r_1 \;  r_2 \rrbracket_r \leq
Chengsong
parents: 611
diff changeset
  1906
				\llbracket r_1 \cdot r_2 \rrbracket_r
Chengsong
parents: 611
diff changeset
  1907
			\]
Chengsong
parents: 611
diff changeset
  1908
		\item
Chengsong
parents: 611
diff changeset
  1909
			\[
Chengsong
parents: 611
diff changeset
  1910
				\llbracket \rflts \; rs \rrbracket_r  \leq
Chengsong
parents: 611
diff changeset
  1911
				\llbracket rs \rrbracket_r 
Chengsong
parents: 611
diff changeset
  1912
			\]
Chengsong
parents: 611
diff changeset
  1913
		\item
Chengsong
parents: 611
diff changeset
  1914
			\[
Chengsong
parents: 611
diff changeset
  1915
				\llbracket \rDistinct \; rs \; ss \rrbracket_r  \leq
Chengsong
parents: 611
diff changeset
  1916
				\llbracket rs \rrbracket_r 
Chengsong
parents: 611
diff changeset
  1917
			\]
Chengsong
parents: 611
diff changeset
  1918
		\item
Chengsong
parents: 611
diff changeset
  1919
			If all elements $a$ in the set $as$ satisfy the property
Chengsong
parents: 611
diff changeset
  1920
			that $\llbracket \textit{rsimp} \; a \rrbracket_r \leq
Chengsong
parents: 611
diff changeset
  1921
			\llbracket a \rrbracket_r$, then we have 
Chengsong
parents: 611
diff changeset
  1922
			\[
Chengsong
parents: 611
diff changeset
  1923
				\llbracket \; \rsimpalts \; (\textit{rdistinct} \;
Chengsong
parents: 611
diff changeset
  1924
				(\textit{rflts} \; (\textit{map}\;\textit{rsimp} as)) \{\})
Chengsong
parents: 611
diff changeset
  1925
				\rrbracket \leq
Chengsong
parents: 611
diff changeset
  1926
				\llbracket \; \sum \; (\rDistinct \; (\rflts \;(\map \;
Chengsong
parents: 611
diff changeset
  1927
				\textit{rsimp} \; x))\; \{ \} ) \rrbracket_r 
Chengsong
parents: 611
diff changeset
  1928
			\]
Chengsong
parents: 611
diff changeset
  1929
	\end{itemize}
Chengsong
parents: 611
diff changeset
  1930
\end{lemma}
Chengsong
parents: 611
diff changeset
  1931
\begin{proof}
Chengsong
parents: 611
diff changeset
  1932
	Point 1, 3, 4 can be proven by an induction on $rs$.
Chengsong
parents: 611
diff changeset
  1933
	Point 2 is by case analysis on $r_1$ and $r_2$.
Chengsong
parents: 611
diff changeset
  1934
	The last part is a corollary of the previous ones.
Chengsong
parents: 611
diff changeset
  1935
\end{proof}
Chengsong
parents: 611
diff changeset
  1936
\noindent
Chengsong
parents: 611
diff changeset
  1937
With the lemmas for each inductive case in place, we are ready to get 
Chengsong
parents: 611
diff changeset
  1938
the non-increasing property as a corollary:
Chengsong
parents: 611
diff changeset
  1939
\begin{corollary}\label{rsimpMono}
Chengsong
parents: 611
diff changeset
  1940
	$\llbracket \textit{rsimp} \; r \rrbracket_r \leq \llbracket r \rrbracket_r$
Chengsong
parents: 611
diff changeset
  1941
\end{corollary}
Chengsong
parents: 611
diff changeset
  1942
\begin{proof}
Chengsong
parents: 611
diff changeset
  1943
	By \ref{rsimpMonoLemmas}.
Chengsong
parents: 611
diff changeset
  1944
\end{proof}
Chengsong
parents: 611
diff changeset
  1945
609
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  1946
\subsection{Estimating the Closed Forms' sizes}
558
Chengsong
parents: 557
diff changeset
  1947
We now summarize the closed forms below:
Chengsong
parents: 557
diff changeset
  1948
\begin{itemize}
Chengsong
parents: 557
diff changeset
  1949
	\item
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1950
		$\rderssimp{(\sum rs)}{s} \sequal
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1951
		\sum \; (\map \; (\rderssimp{\_}{s}) \; rs)$
558
Chengsong
parents: 557
diff changeset
  1952
	\item
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1953
		$\rderssimp{(r_1 \cdot r_2)}{s} \sequal \sum ((r_1 \backslash s) \cdot r_2 ) 
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1954
		:: (\map \; (r_2 \backslash \_) (\vsuf{s}{r_1}))$
558
Chengsong
parents: 557
diff changeset
  1955
	\item
Chengsong
parents: 557
diff changeset
  1956
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1957
		$\rderssimp{r^*}{c::s} = 
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1958
		\rsimp{
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1959
			(\sum (\map \; (\lambda s. (\rderssimp{r}{s})\cdot r^*) \; 
558
Chengsong
parents: 557
diff changeset
  1960
			(\starupdates \; s\; r \; [[c]])
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1961
			)
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1962
			)
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1963
		}
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  1964
		$
558
Chengsong
parents: 557
diff changeset
  1965
\end{itemize}	
Chengsong
parents: 557
diff changeset
  1966
\noindent	
Chengsong
parents: 557
diff changeset
  1967
The closed forms on the left-hand-side
Chengsong
parents: 557
diff changeset
  1968
are all of the same shape: $\rsimp{ (\sum rs)} $.
Chengsong
parents: 557
diff changeset
  1969
Such regular expression will be bounded by the size of $\sum rs'$, 
Chengsong
parents: 557
diff changeset
  1970
where every element in $rs'$ is distinct, and each element 
Chengsong
parents: 557
diff changeset
  1971
can be described by some inductive sub-structures 
Chengsong
parents: 557
diff changeset
  1972
(for example when $r = r_1 \cdot r_2$ then $rs'$ 
Chengsong
parents: 557
diff changeset
  1973
will be solely comprised of $r_1 \backslash s'$ 
Chengsong
parents: 557
diff changeset
  1974
and $r_2 \backslash s''$, $s'$ and $s''$ being 
Chengsong
parents: 557
diff changeset
  1975
sub-strings of $s$).
Chengsong
parents: 557
diff changeset
  1976
which will each have a size uppder bound 
Chengsong
parents: 557
diff changeset
  1977
according to inductive hypothesis, which controls $r \backslash s$.
557
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  1978
558
Chengsong
parents: 557
diff changeset
  1979
We elaborate the above reasoning by a series of lemmas
Chengsong
parents: 557
diff changeset
  1980
below, where straightforward proofs are omitted.
532
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  1981
\begin{lemma}
558
Chengsong
parents: 557
diff changeset
  1982
	If $\forall r \in rs. \rsize{r} $ is less than or equal to $N$,
Chengsong
parents: 557
diff changeset
  1983
	and $\textit{length} \; rs$ is less than or equal to $l$,
Chengsong
parents: 557
diff changeset
  1984
	then $\rsize{\sum rs}$ is less than or equal to $l*N + 1$.
Chengsong
parents: 557
diff changeset
  1985
\end{lemma}
Chengsong
parents: 557
diff changeset
  1986
\noindent
Chengsong
parents: 557
diff changeset
  1987
If we define all regular expressions with size no
Chengsong
parents: 557
diff changeset
  1988
more than $N$ as $\sizeNregex \; N$:
Chengsong
parents: 557
diff changeset
  1989
\[
Chengsong
parents: 557
diff changeset
  1990
	\sizeNregex \; N \dn  \{r \mid \rsize{r} \leq N \}
Chengsong
parents: 557
diff changeset
  1991
\]
Chengsong
parents: 557
diff changeset
  1992
Then such set is finite:
Chengsong
parents: 557
diff changeset
  1993
\begin{lemma}\label{finiteSizeN}
Chengsong
parents: 557
diff changeset
  1994
	$\textit{isFinite}\; (\sizeNregex \; N)$
Chengsong
parents: 557
diff changeset
  1995
\end{lemma}
Chengsong
parents: 557
diff changeset
  1996
\begin{proof}
Chengsong
parents: 557
diff changeset
  1997
	By overestimating the set $\sizeNregex \; N + 1$
Chengsong
parents: 557
diff changeset
  1998
	using union of sets like
Chengsong
parents: 557
diff changeset
  1999
	$\{r_1 \cdot r_2 \mid r_1 \in A
Chengsong
parents: 557
diff changeset
  2000
		\text{and}
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2001
	r_2 \in A\}
558
Chengsong
parents: 557
diff changeset
  2002
	$ where $A = \sizeNregex \; N$.
Chengsong
parents: 557
diff changeset
  2003
\end{proof}
Chengsong
parents: 557
diff changeset
  2004
\noindent
Chengsong
parents: 557
diff changeset
  2005
From this we get a corollary that
Chengsong
parents: 557
diff changeset
  2006
if forall $r \in rs$, $\rsize{r} \leq N$, then the output of 
Chengsong
parents: 557
diff changeset
  2007
$\rdistinct{rs}{\varnothing}$ is a list of regular
Chengsong
parents: 557
diff changeset
  2008
expressions of finite size depending on $N$ only. 
561
486fb297ac7c more done
Chengsong
parents: 559
diff changeset
  2009
\begin{corollary}\label{finiteSizeNCorollary}
558
Chengsong
parents: 557
diff changeset
  2010
	Assumes that for all $r \in rs. \rsize{r} \leq N$,
Chengsong
parents: 557
diff changeset
  2011
	and the cardinality of $\sizeNregex \; N$ is $c_N$
Chengsong
parents: 557
diff changeset
  2012
	then$\rsize{\rdistinct{rs}{\varnothing}} \leq c*N$.
Chengsong
parents: 557
diff changeset
  2013
\end{corollary}
Chengsong
parents: 557
diff changeset
  2014
\noindent
Chengsong
parents: 557
diff changeset
  2015
We have proven that the output of $\rdistinct{rs'}{\varnothing}$
Chengsong
parents: 557
diff changeset
  2016
is bounded by a constant $c_N$ depending only on $N$,
Chengsong
parents: 557
diff changeset
  2017
provided that each of $rs'$'s element
Chengsong
parents: 557
diff changeset
  2018
is bounded by $N$.
Chengsong
parents: 557
diff changeset
  2019
We want to apply it to our setting $\rsize{\rsimp{\sum rs}}$.
Chengsong
parents: 557
diff changeset
  2020
609
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2021
We show that $\rdistinct$ and $\rflts$
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2022
working together is at least as 
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2023
good as $\rdistinct{}{}$ alone, which can be written as
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2024
\begin{center}
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2025
	$\llbracket \rdistinct{(\rflts \; \textit{rs})}{\varnothing} \rrbracket_r 
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2026
	\leq 
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2027
	\llbracket \rdistinct{rs}{\varnothing}  \rrbracket_r  $.
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2028
\end{center}
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2029
We need this so that we know the outcome of our real 
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2030
simplification is better than or equal to a rough estimate,
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2031
and therefore can be bounded by that estimate.
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2032
This is a bit harder to establish compared with proving
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2033
$\textit{flts}$ does not make a list larger (which can
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2034
be proven using routine induction):
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2035
\begin{center}
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2036
	$\llbracket  \textit{rflts}\; rs \rrbracket_r \leq
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2037
	\llbracket  \textit{rs} \rrbracket_r$
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2038
\end{center}
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2039
We cannot simply prove how each helper function
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2040
reduces the size and then put them together:
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2041
From
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2042
\begin{center}
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2043
$\llbracket  \textit{rflts}\; rs \rrbracket_r \leq
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2044
	\llbracket \; \textit{rs} \rrbracket_r$
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2045
\end{center}
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2046
and
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2047
\begin{center}
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2048
     $\llbracket  \textit{rdistinct} \; rs \; \varnothing \leq
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2049
     \llbracket rs \rrbracket_r$
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2050
\end{center}
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2051
one cannot imply
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2052
\begin{center}
558
Chengsong
parents: 557
diff changeset
  2053
	$\llbracket \rdistinct{(\rflts \; \textit{rs})}{\varnothing} \rrbracket_r 
Chengsong
parents: 557
diff changeset
  2054
	\leq 
Chengsong
parents: 557
diff changeset
  2055
	\llbracket \rdistinct{rs}{\varnothing}  \rrbracket_r  $.
609
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2056
\end{center}
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2057
What we can imply is that 
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2058
\begin{center}
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2059
	$\llbracket \rdistinct{(\rflts \; \textit{rs})}{\varnothing} \rrbracket_r 
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2060
	\leq
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2061
	\llbracket rs \rrbracket_r$
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2062
\end{center}
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2063
but this estimate is too rough and $\llbracket rs \rrbracket_r$	is unbounded.
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2064
The way we 
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2065
get through this is by first proving a more general lemma 
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2066
(so that the inductive case goes through):
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2067
\begin{lemma}\label{fltsSizeReductionAlts}
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2068
	If we have three accumulator sets:
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2069
	$noalts\_set$, $alts\_set$ and $corr\_set$,
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2070
	satisfying:
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2071
	\begin{itemize}
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2072
		\item
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2073
			$\forall r \in noalts\_set. \; \nexists xs.\; r = \sum  xs$
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2074
		\item
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2075
			$\forall r \in alts\_set. \; \exists xs. \; r = \sum xs
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2076
			\; \textit{and} \; set \; xs \subseteq corr\_set$
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2077
	\end{itemize}
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2078
	then we have that
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2079
	\begin{center}
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2080
	\begin{tabular}{lcl}
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2081
	$\llbracket  (\textit{rdistinct} \; (\textit{rflts} \; as) \;
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2082
	(noalts\_set \cup corr\_set)) \rrbracket_r$ & $\leq$ &\\
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2083
						    $\llbracket  (\textit{rdistinct} \; as \; (noalts\_set \cup alts\_set \cup
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2084
	\{ \ZERO \} )) \rrbracket_r$ & & \\ 
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2085
	\end{tabular}
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2086
	\end{center}
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2087
		holds.
532
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2088
\end{lemma}
558
Chengsong
parents: 557
diff changeset
  2089
\noindent
609
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2090
We need to split the accumulator into two parts: the part
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2091
which contains alternative regular expressions ($alts\_set$), and 
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2092
the part without any of them($noalts\_set$).
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2093
The set $corr\_set$ is the corresponding set
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2094
of $alts\_set$ with all elements under the $\sum$ constructor
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2095
spilled out.
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2096
\begin{proof}
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2097
	By induction on the list $as$. We make use of lemma \ref{rdistinctConcat}.
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2098
\end{proof}
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2099
By setting all three sets to the empty set, one gets the desired size estimate:
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2100
\begin{corollary}\label{interactionFltsDB}
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2101
	$\llbracket \rdistinct{(\rflts \; \textit{rs})}{\varnothing} \rrbracket_r 
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2102
	\leq 
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2103
	\llbracket \rdistinct{rs}{\varnothing}  \rrbracket_r  $.
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2104
\end{corollary}
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2105
\begin{proof}
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2106
	By using the lemma \ref{fltsSizeReductionAlts}.
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2107
\end{proof}
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2108
\noindent
558
Chengsong
parents: 557
diff changeset
  2109
The intuition is that if we remove duplicates from the $\textit{LHS}$, at least the same amount of 
Chengsong
parents: 557
diff changeset
  2110
duplicates will be removed from the list $\textit{rs}$ in the $\textit{RHS}$. 
Chengsong
parents: 557
diff changeset
  2111
Chengsong
parents: 557
diff changeset
  2112
Now this $\rsimp{\sum rs}$ can be estimated using $\rdistinct{rs}{\varnothing}$:
Chengsong
parents: 557
diff changeset
  2113
\begin{lemma}\label{altsSimpControl}
Chengsong
parents: 557
diff changeset
  2114
	$\rsize{\rsimp{\sum rs}} \leq \rsize{\rdistinct{rs}{\varnothing}}+ 1$
532
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2115
\end{lemma}
558
Chengsong
parents: 557
diff changeset
  2116
\begin{proof}
Chengsong
parents: 557
diff changeset
  2117
	By using \ref{interactionFltsDB}.
Chengsong
parents: 557
diff changeset
  2118
\end{proof}
Chengsong
parents: 557
diff changeset
  2119
\noindent
609
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2120
This is a key lemma in establishing the bounds on all the 
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2121
closed forms.
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2122
With this we are now ready to control the sizes of
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2123
$(r_1 \cdot r_2 )\backslash s$, $r^* \backslash s$.
558
Chengsong
parents: 557
diff changeset
  2124
\begin{theorem}
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2125
	For any regex $r$, $\exists N_r. \forall s. \; \rsize{\rderssimp{r}{s}} \leq N_r$
558
Chengsong
parents: 557
diff changeset
  2126
\end{theorem}
Chengsong
parents: 557
diff changeset
  2127
\noindent
Chengsong
parents: 557
diff changeset
  2128
\begin{proof}
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2129
	We prove this by induction on $r$. The base cases for $\RZERO$,
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2130
	$\RONE $ and $\RCHAR{c}$ are straightforward. 
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2131
	In the sequence $r_1 \cdot r_2$ case,
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2132
	the inductive hypotheses state 
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2133
	$\exists N_1. \forall s. \; \llbracket \rderssimp{r}{s} \rrbracket \leq N_1$ and
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2134
	$\exists N_2. \forall s. \; \llbracket \rderssimp{r_2}{s} \rrbracket \leq N_2$. 
562
Chengsong
parents: 561
diff changeset
  2135
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2136
	When the string $s$ is not empty, we can reason as follows
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2137
	%
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2138
	\begin{center}
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2139
		\begin{tabular}{lcll}
558
Chengsong
parents: 557
diff changeset
  2140
& & $ \llbracket   \rderssimp{r_1\cdot r_2 }{s} \rrbracket_r $\\
561
486fb297ac7c more done
Chengsong
parents: 559
diff changeset
  2141
& $ = $ & $\llbracket \rsimp{(\sum(r_1 \backslash_{rsimp} s \cdot r_2 \; \;  :: \; \; 
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2142
		\map \; (r_2\backslash_{rsimp} \_)\; (\vsuf{s}{r})))} \rrbracket_r $ & (1) \\			
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2143
										     & $\leq$ & $\llbracket \rdistinct{(r_1 \backslash_{rsimp} s \cdot r_2 \; \;  :: \; \; 
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2144
	\map \; (r_2\backslash_{rsimp} \_)\; (\vsuf{s}{r}))}{\varnothing} \rrbracket_r  + 1$ & (2) \\
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2145
											     & $\leq$ & $2 + N_1 + \rsize{r_2} + (N_2 * (card\;(\sizeNregex \; N_2)))$ & (3)\\
558
Chengsong
parents: 557
diff changeset
  2146
\end{tabular}
Chengsong
parents: 557
diff changeset
  2147
\end{center}
561
486fb297ac7c more done
Chengsong
parents: 559
diff changeset
  2148
\noindent
486fb297ac7c more done
Chengsong
parents: 559
diff changeset
  2149
(1) is by the corollary \ref{seqEstimate1}.
486fb297ac7c more done
Chengsong
parents: 559
diff changeset
  2150
(2) is by \ref{altsSimpControl}.
486fb297ac7c more done
Chengsong
parents: 559
diff changeset
  2151
(3) is by \ref{finiteSizeNCorollary}.
562
Chengsong
parents: 561
diff changeset
  2152
Chengsong
parents: 561
diff changeset
  2153
Chengsong
parents: 561
diff changeset
  2154
Combining the cases when $s = []$ and $s \neq []$, we get (4):
Chengsong
parents: 561
diff changeset
  2155
\begin{center}
Chengsong
parents: 561
diff changeset
  2156
	\begin{tabular}{lcll}
Chengsong
parents: 561
diff changeset
  2157
		$\rsize{(r_1 \cdot r_2) \backslash_r s}$ & $\leq$ & 
Chengsong
parents: 561
diff changeset
  2158
		$max \; (2 + N_1 + 
Chengsong
parents: 561
diff changeset
  2159
		\llbracket r_2 \rrbracket_r + 
Chengsong
parents: 561
diff changeset
  2160
		N_2 * (card\; (\sizeNregex \; N_2))) \; \rsize{r_1\cdot r_2}$ & (4)
Chengsong
parents: 561
diff changeset
  2161
	\end{tabular}
Chengsong
parents: 561
diff changeset
  2162
\end{center}
558
Chengsong
parents: 557
diff changeset
  2163
562
Chengsong
parents: 561
diff changeset
  2164
We reason similarly for  $\STAR$.
Chengsong
parents: 561
diff changeset
  2165
The inductive hypothesis is
Chengsong
parents: 561
diff changeset
  2166
$\exists N. \forall s. \; \llbracket \rderssimp{r}{s} \rrbracket \leq N$.
564
Chengsong
parents: 562
diff changeset
  2167
Let $n_r = \llbracket r^* \rrbracket_r$.
562
Chengsong
parents: 561
diff changeset
  2168
When $s = c :: cs$ is not empty,
Chengsong
parents: 561
diff changeset
  2169
\begin{center}
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2170
	\begin{tabular}{lcll}
562
Chengsong
parents: 561
diff changeset
  2171
& & $ \llbracket   \rderssimp{r^* }{c::cs} \rrbracket_r $\\
Chengsong
parents: 561
diff changeset
  2172
& $ = $ & $\llbracket \rsimp{(\sum (\map \; (\lambda s. (r \backslash_{rsimp} s) \cdot r^*) \; (\starupdates\; 
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2173
	cs \; r \; [[c]] )) )} \rrbracket_r $ & (5) \\			
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2174
					      & $\leq$ & $\llbracket 
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2175
					      \rdistinct{
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2176
						      (\map \; 
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2177
						      (\lambda s. (r \backslash_{rsimp} s) \cdot r^*) \; 
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2178
						      (\starupdates\; cs \; r \; [[c]] )
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2179
					      )}
562
Chengsong
parents: 561
diff changeset
  2180
	{\varnothing} \rrbracket_r  + 1$ & (6) \\
Chengsong
parents: 561
diff changeset
  2181
					 & $\leq$ & $1 + (\textit{card} (\sizeNregex \; (N + n_r)))
Chengsong
parents: 561
diff changeset
  2182
	* (1 + (N + n_r)) $ & (7)\\
Chengsong
parents: 561
diff changeset
  2183
\end{tabular}
Chengsong
parents: 561
diff changeset
  2184
\end{center}
Chengsong
parents: 561
diff changeset
  2185
\noindent
Chengsong
parents: 561
diff changeset
  2186
(5) is by the lemma  \ref{starClosedForm}.
Chengsong
parents: 561
diff changeset
  2187
(6) is by \ref{altsSimpControl}.
Chengsong
parents: 561
diff changeset
  2188
(7) is by \ref{finiteSizeNCorollary}.
Chengsong
parents: 561
diff changeset
  2189
Combining with the case when $s = []$, one gets
Chengsong
parents: 561
diff changeset
  2190
\begin{center}
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2191
	\begin{tabular}{lcll}
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2192
		$\rsize{r^* \backslash_r s}$ & $\leq$ & $max \; n_r \; 1 + (\textit{card} (\sizeNregex \; (N + n_r)))
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2193
		* (1 + (N + n_r)) $ & (8)\\
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2194
	\end{tabular}
562
Chengsong
parents: 561
diff changeset
  2195
\end{center}
Chengsong
parents: 561
diff changeset
  2196
\noindent
Chengsong
parents: 561
diff changeset
  2197
Chengsong
parents: 561
diff changeset
  2198
The alternative case is slightly less involved.
Chengsong
parents: 561
diff changeset
  2199
The inductive hypothesis 
Chengsong
parents: 561
diff changeset
  2200
is equivalent to $\exists N. \forall r \in (\map \; (\_ \backslash_r s) \; rs). \rsize{r} \leq N$.
Chengsong
parents: 561
diff changeset
  2201
In the case when $s = c::cs$, we have 
Chengsong
parents: 561
diff changeset
  2202
\begin{center}
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2203
	\begin{tabular}{lcll}
562
Chengsong
parents: 561
diff changeset
  2204
& & $ \llbracket   \rderssimp{\sum rs }{c::cs} \rrbracket_r $\\
Chengsong
parents: 561
diff changeset
  2205
& $ = $ & $\llbracket \rsimp{(\sum (\map \; (\_ \backslash_{rsimp} s)  \; rs) )} \rrbracket_r $ & (9) \\			
Chengsong
parents: 561
diff changeset
  2206
& $\leq$ & $\llbracket (\sum (\map \; (\_ \backslash_{rsimp} s)  \; rs) ) \rrbracket_r $  & (10) \\
Chengsong
parents: 561
diff changeset
  2207
& $\leq$ & $1 + N * (length \; rs) $ & (11)\\
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2208
	\end{tabular}
562
Chengsong
parents: 561
diff changeset
  2209
\end{center}
Chengsong
parents: 561
diff changeset
  2210
\noindent
Chengsong
parents: 561
diff changeset
  2211
(9) is by \ref{altsClosedForm}, (10) by \ref{rsimpSize} and (11) by inductive hypothesis.
Chengsong
parents: 561
diff changeset
  2212
Chengsong
parents: 561
diff changeset
  2213
Combining with the case when $s = []$, one gets
Chengsong
parents: 561
diff changeset
  2214
\begin{center}
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2215
	\begin{tabular}{lcll}
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2216
		$\rsize{\sum rs \backslash_r s}$ & $\leq$ & $max \; \rsize{\sum rs} \; 1+N*(length \; rs)$ 
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2217
						 & (12)\\
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2218
	\end{tabular}
562
Chengsong
parents: 561
diff changeset
  2219
\end{center}
Chengsong
parents: 561
diff changeset
  2220
(4), (8), and (12) are all the inductive cases proven.
558
Chengsong
parents: 557
diff changeset
  2221
\end{proof}
Chengsong
parents: 557
diff changeset
  2222
564
Chengsong
parents: 562
diff changeset
  2223
Chengsong
parents: 562
diff changeset
  2224
\begin{corollary}
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2225
	For any regex $a$, $\exists N_r. \forall s. \; \rsize{\bderssimp{a}{s}} \leq N_r$
564
Chengsong
parents: 562
diff changeset
  2226
\end{corollary}
Chengsong
parents: 562
diff changeset
  2227
\begin{proof}
Chengsong
parents: 562
diff changeset
  2228
	By \ref{sizeRelations}.
Chengsong
parents: 562
diff changeset
  2229
\end{proof}
558
Chengsong
parents: 557
diff changeset
  2230
\noindent
Chengsong
parents: 557
diff changeset
  2231
609
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2232
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2233
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2234
61139fdddae0 chap1 totally done
Chengsong
parents: 601
diff changeset
  2235
558
Chengsong
parents: 557
diff changeset
  2236
%-----------------------------------
Chengsong
parents: 557
diff changeset
  2237
%	SECTION 2
Chengsong
parents: 557
diff changeset
  2238
%-----------------------------------
Chengsong
parents: 557
diff changeset
  2239
532
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2240
557
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  2241
%----------------------------------------------------------------------------------------
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  2242
%	SECTION 3
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  2243
%----------------------------------------------------------------------------------------
812e5d112f49 more changes
Chengsong
parents: 556
diff changeset
  2244
532
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2245
554
Chengsong
parents: 553
diff changeset
  2246
\subsection{A Closed Form for the Sequence Regular Expression}
Chengsong
parents: 553
diff changeset
  2247
\noindent
Chengsong
parents: 553
diff changeset
  2248
Chengsong
parents: 553
diff changeset
  2249
Before we get to the proof that says the intermediate result of our lexer will
Chengsong
parents: 553
diff changeset
  2250
remain finitely bounded, which is an important efficiency/liveness guarantee,
Chengsong
parents: 553
diff changeset
  2251
we shall first develop a few preparatory properties and definitions to 
Chengsong
parents: 553
diff changeset
  2252
make the process of proving that a breeze.
Chengsong
parents: 553
diff changeset
  2253
Chengsong
parents: 553
diff changeset
  2254
We define rewriting relations for $\rrexp$s, which allows us to do the 
Chengsong
parents: 553
diff changeset
  2255
same trick as we did for the correctness proof,
Chengsong
parents: 553
diff changeset
  2256
but this time we will have stronger equalities established.
Chengsong
parents: 553
diff changeset
  2257
532
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2258
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2259
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2260
What guarantee does this bound give us?
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2261
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2262
Whatever the regex is, it will not grow indefinitely.
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2263
Take our previous example $(a + aa)^*$ as an example:
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2264
\begin{center}
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2265
	\begin{tabular}{@{}c@{\hspace{0mm}}c@{\hspace{0mm}}c@{}}
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2266
		\begin{tikzpicture}
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2267
			\begin{axis}[
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2268
				xlabel={number of $a$'s},
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2269
				x label style={at={(1.05,-0.05)}},
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2270
				ylabel={regex size},
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2271
				enlargelimits=false,
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2272
				xtick={0,5,...,30},
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2273
				xmax=33,
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2274
				ymax= 40,
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2275
				ytick={0,10,...,40},
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2276
				scaled ticks=false,
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2277
				axis lines=left,
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2278
				width=5cm,
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2279
				height=4cm, 
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2280
				legend entries={$(a + aa)^*$},  
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2281
				legend pos=north west,
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2282
				legend cell align=left]
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2283
				\addplot[red,mark=*, mark options={fill=white}] table {a_aa_star.data};
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2284
			\end{axis}
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2285
		\end{tikzpicture}
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2286
	\end{tabular}
532
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2287
\end{center}
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2288
We are able to limit the size of the regex $(a + aa)^*$'s derivatives
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2289
with our simplification
532
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2290
rules very effectively.
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2291
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2292
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2293
In our proof for the inductive case $r_1 \cdot r_2$, the dominant term in the bound
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2294
is $l_{N_2} * N_2$, where $N_2$ is the bound we have for $\llbracket \bderssimp{r_2}{s} \rrbracket$.
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2295
Given that $l_{N_2}$ is roughly the size $4^{N_2}$, the size bound $\llbracket \bderssimp{r_1 \cdot r_2}{s} \rrbracket$
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2296
inflates the size bound of $\llbracket \bderssimp{r_2}{s} \rrbracket$ with the function
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2297
$f(x) = x * 2^x$.
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2298
This means the bound we have will surge up at least
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2299
tower-exponentially with a linear increase of the depth.
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2300
For a regex of depth $n$, the bound
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2301
would be approximately $4^n$.
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2302
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2303
Test data in the graphs from randomly generated regular expressions
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2304
shows that the giant bounds are far from being hit.
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2305
%a few sample regular experessions' derivatives
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2306
%size change
576
3e1b699696b6 thesis chap5
Chengsong
parents: 564
diff changeset
  2307
%TODO: giving regex1_size_change.data showing a few regular expressions' size changes 
532
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2308
%w;r;t the input characters number, where the size is usually cubic in terms of original size
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2309
%a*, aa*, aaa*, .....
576
3e1b699696b6 thesis chap5
Chengsong
parents: 564
diff changeset
  2310
%randomly generated regular expressions
611
Chengsong
parents: 610
diff changeset
  2311
\begin{figure}{H}
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2312
	\begin{tabular}{@{}c@{\hspace{0mm}}c@{\hspace{0mm}}c@{}}
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2313
		\begin{tikzpicture}
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2314
			\begin{axis}[
611
Chengsong
parents: 610
diff changeset
  2315
				xlabel={number of characters},
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2316
				x label style={at={(1.05,-0.05)}},
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2317
				ylabel={regex size},
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2318
				enlargelimits=false,
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2319
				xtick={0,5,...,30},
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2320
				xmax=33,
611
Chengsong
parents: 610
diff changeset
  2321
				%ymax=1000,
Chengsong
parents: 610
diff changeset
  2322
				%ytick={0,100,...,1000},
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2323
				scaled ticks=false,
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2324
				axis lines=left,
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2325
				width=5cm,
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2326
				height=4cm, 
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2327
				legend entries={regex1},  
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2328
				legend pos=north west,
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2329
				legend cell align=left]
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2330
				\addplot[red,mark=*, mark options={fill=white}] table {regex1_size_change.data};
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2331
			\end{axis}
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2332
		\end{tikzpicture}
532
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2333
  &
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2334
  \begin{tikzpicture}
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2335
	  \begin{axis}[
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2336
		  xlabel={$n$},
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2337
		  x label style={at={(1.05,-0.05)}},
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2338
		  %ylabel={time in secs},
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2339
		  enlargelimits=false,
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2340
		  xtick={0,5,...,30},
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2341
		  xmax=33,
611
Chengsong
parents: 610
diff changeset
  2342
		  %ymax=1000,
Chengsong
parents: 610
diff changeset
  2343
		  %ytick={0,100,...,1000},
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2344
		  scaled ticks=false,
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2345
		  axis lines=left,
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2346
		  width=5cm,
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2347
		  height=4cm, 
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2348
		  legend entries={regex2},  
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2349
		  legend pos=north west,
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2350
		  legend cell align=left]
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2351
		  \addplot[blue,mark=*, mark options={fill=white}] table {regex2_size_change.data};
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2352
	  \end{axis}
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2353
  \end{tikzpicture}
532
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2354
  &
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2355
  \begin{tikzpicture}
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2356
	  \begin{axis}[
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2357
		  xlabel={$n$},
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2358
		  x label style={at={(1.05,-0.05)}},
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2359
		  %ylabel={time in secs},
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2360
		  enlargelimits=false,
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2361
		  xtick={0,5,...,30},
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2362
		  xmax=33,
611
Chengsong
parents: 610
diff changeset
  2363
		  %ymax=1000,
Chengsong
parents: 610
diff changeset
  2364
		  %ytick={0,100,...,1000},
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2365
		  scaled ticks=false,
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2366
		  axis lines=left,
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2367
		  width=5cm,
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2368
		  height=4cm, 
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2369
		  legend entries={regex3},  
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2370
		  legend pos=north west,
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2371
		  legend cell align=left]
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2372
		  \addplot[cyan,mark=*, mark options={fill=white}] table {regex3_size_change.data};
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2373
	  \end{axis}
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2374
  \end{tikzpicture}\\
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2375
  \multicolumn{3}{c}{Graphs: size change of 3 randomly generated regular expressions $w.r.t.$ input string length.}
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2376
	\end{tabular}    
611
Chengsong
parents: 610
diff changeset
  2377
\end{figure}  
532
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2378
\noindent
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2379
Most of the regex's sizes seem to stay within a polynomial bound $w.r.t$ the 
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2380
original size.
591
b2d0de6aee18 more polishing integrated comments chap2
Chengsong
parents: 590
diff changeset
  2381
We will discuss improvements to this bound in the next chapter.
532
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2382
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2383
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2384
613
Chengsong
parents: 611
diff changeset
  2385
\section{Possible Further Improvements}
590
988e92a70704 more chap5 and chap6 bsimp_idem
Chengsong
parents: 577
diff changeset
  2386
There are two problems with this finiteness result, though.
988e92a70704 more chap5 and chap6 bsimp_idem
Chengsong
parents: 577
diff changeset
  2387
\begin{itemize}
988e92a70704 more chap5 and chap6 bsimp_idem
Chengsong
parents: 577
diff changeset
  2388
	\item
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2389
		First, It is not yet a direct formalisation of our lexer's complexity,
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2390
		as a complexity proof would require looking into 
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2391
		the time it takes to execute {\bf all} the operations
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2392
		involved in the lexer (simp, collect, decode), not just the derivative.
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2393
	\item
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2394
		Second, the bound is not yet tight, and we seek to improve $N_a$ so that
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2395
		it is polynomial on $\llbracket a \rrbracket$.
590
988e92a70704 more chap5 and chap6 bsimp_idem
Chengsong
parents: 577
diff changeset
  2396
\end{itemize}
988e92a70704 more chap5 and chap6 bsimp_idem
Chengsong
parents: 577
diff changeset
  2397
Still, we believe this contribution is fruitful,
988e92a70704 more chap5 and chap6 bsimp_idem
Chengsong
parents: 577
diff changeset
  2398
because
988e92a70704 more chap5 and chap6 bsimp_idem
Chengsong
parents: 577
diff changeset
  2399
\begin{itemize}
988e92a70704 more chap5 and chap6 bsimp_idem
Chengsong
parents: 577
diff changeset
  2400
	\item
988e92a70704 more chap5 and chap6 bsimp_idem
Chengsong
parents: 577
diff changeset
  2401
988e92a70704 more chap5 and chap6 bsimp_idem
Chengsong
parents: 577
diff changeset
  2402
		The size proof can serve as a cornerstone for a complexity
988e92a70704 more chap5 and chap6 bsimp_idem
Chengsong
parents: 577
diff changeset
  2403
		formalisation.
988e92a70704 more chap5 and chap6 bsimp_idem
Chengsong
parents: 577
diff changeset
  2404
		Derivatives are the most important phases of our lexer algorithm.
988e92a70704 more chap5 and chap6 bsimp_idem
Chengsong
parents: 577
diff changeset
  2405
		Size properties about derivatives covers the majority of the algorithm
988e92a70704 more chap5 and chap6 bsimp_idem
Chengsong
parents: 577
diff changeset
  2406
		and is therefore a good indication of complexity of the entire program.
988e92a70704 more chap5 and chap6 bsimp_idem
Chengsong
parents: 577
diff changeset
  2407
	\item
988e92a70704 more chap5 and chap6 bsimp_idem
Chengsong
parents: 577
diff changeset
  2408
		The bound is already a strong indication that catastrophic
988e92a70704 more chap5 and chap6 bsimp_idem
Chengsong
parents: 577
diff changeset
  2409
		backtracking is much less likely to occur in our $\blexersimp$
988e92a70704 more chap5 and chap6 bsimp_idem
Chengsong
parents: 577
diff changeset
  2410
		algorithm.
988e92a70704 more chap5 and chap6 bsimp_idem
Chengsong
parents: 577
diff changeset
  2411
		We refine $\blexersimp$ with $\blexerStrong$ in the next chapter
988e92a70704 more chap5 and chap6 bsimp_idem
Chengsong
parents: 577
diff changeset
  2412
		so that the bound becomes polynomial.
988e92a70704 more chap5 and chap6 bsimp_idem
Chengsong
parents: 577
diff changeset
  2413
\end{itemize}
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2414
532
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2415
%----------------------------------------------------------------------------------------
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2416
%	SECTION 4
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2417
%----------------------------------------------------------------------------------------
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2418
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2419
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2420
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2421
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2422
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2423
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2424
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2425
532
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2426
One might wonder the actual bound rather than the loose bound we gave
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2427
for the convenience of an easier proof.
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2428
How much can the regex $r^* \backslash s$ grow? 
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2429
As  earlier graphs have shown,
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2430
%TODO: reference that graph where size grows quickly
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2431
they can grow at a maximum speed
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2432
exponential $w.r.t$ the number of characters, 
532
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2433
but will eventually level off when the string $s$ is long enough.
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2434
If they grow to a size exponential $w.r.t$ the original regex, our algorithm
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2435
would still be slow.
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2436
And unfortunately, we have concrete examples
576
3e1b699696b6 thesis chap5
Chengsong
parents: 564
diff changeset
  2437
where such regular expressions grew exponentially large before levelling off:
532
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2438
$(a ^ * + (aa) ^ * + (aaa) ^ * + \ldots + 
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2439
(\underbrace{a \ldots a}_{\text{n a's}})^*$ will already have a maximum
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2440
size that is  exponential on the number $n$ 
532
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2441
under our current simplification rules:
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2442
%TODO: graph of a regex whose size increases exponentially.
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2443
\begin{center}
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2444
	\begin{tikzpicture}
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2445
		\begin{axis}[
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2446
			height=0.5\textwidth,
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2447
			width=\textwidth,
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2448
			xlabel=number of a's,
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2449
			xtick={0,...,9},
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2450
			ylabel=maximum size,
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2451
			ymode=log,
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2452
			log basis y={2}
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2453
			]
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2454
			\addplot[mark=*,blue] table {re-chengsong.data};
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2455
		\end{axis}
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2456
	\end{tikzpicture}
532
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2457
\end{center}
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2458
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2459
For convenience we use $(\oplus_{i=1}^{n} (\underbrace{a \ldots a}_{\text{i a's}})^*)^*$
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2460
to express $(a ^ * + (aa) ^ * + (aaa) ^ * + \ldots + 
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2461
(\underbrace{a \ldots a}_{\text{n a's}})^*$ in the below discussion.
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2462
The exponential size is triggered by that the regex
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2463
$\oplus_{i=1}^{n} (\underbrace{a \ldots a}_{\text{i a's}})^*$
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2464
inside the $(\ldots) ^*$ having exponentially many
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2465
different derivatives, despite those difference being minor.
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2466
$(\oplus_{i=1}^{n} (\underbrace{a \ldots a}_{\text{i a's}})^*)^*\backslash \underbrace{a \ldots a}_{\text{m a's}}$
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2467
will therefore contain the following terms (after flattening out all nested 
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2468
alternatives):
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2469
\begin{center}
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2470
	$(\oplus_{i = 1]{n}  (\underbrace{a \ldots a}_{\text{((i - (m' \% i))\%i) a's}})\cdot  (\underbrace{a \ldots a}_{\text{i a's}})^* })\cdot (\oplus_{i=1}^{n} (\underbrace{a \ldots a}_{\text{i a's}})^*)$\\
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2471
	$(1 \leq m' \leq m )$
532
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2472
\end{center}
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2473
These terms are distinct for $m' \leq L.C.M.(1, \ldots, n)$ (will be explained in appendix).
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2474
With each new input character taking the derivative against the intermediate result, more and more such distinct
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2475
terms will accumulate, 
532
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2476
until the length reaches $L.C.M.(1, \ldots, n)$.
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2477
$\textit{distinctBy}$ will not be able to de-duplicate any two of these terms 
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2478
$(\oplus_{i = 1}^{n}  (\underbrace{a \ldots a}_{\text{((i - (m' \% i))\%i) a's}})\cdot  (\underbrace{a \ldots a}_{\text{i a's}})^* )\cdot (\oplus_{i=1}^{n} (\underbrace{a \ldots a}_{\text{i a's}})^*)^*$\\
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2479
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2480
$(\oplus_{i = 1}^{n}  (\underbrace{a \ldots a}_{\text{((i - (m'' \% i))\%i) a's}})\cdot  (\underbrace{a \ldots a}_{\text{i a's}})^* )\cdot (\oplus_{i=1}^{n} (\underbrace{a \ldots a}_{\text{i a's}})^*)^*$\\
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2481
where $m' \neq m''$ \\
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2482
as they are slightly different.
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2483
This means that with our current simplification methods,
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2484
we will not be able to control the derivative so that
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2485
$\llbracket \bderssimp{r}{s} \rrbracket$ stays polynomial %\leq O((\llbracket r\rrbacket)^c)$
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2486
as there are already exponentially many terms.
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2487
These terms are similar in the sense that the head of those terms
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2488
are all consisted of sub-terms of the form: 
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2489
$(\underbrace{a \ldots a}_{\text{j a's}})\cdot  (\underbrace{a \ldots a}_{\text{i a's}})^* $.
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2490
For  $\oplus_{i=1}^{n} (\underbrace{a \ldots a}_{\text{i a's}})^*$, there will be at most
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2491
$n * (n + 1) / 2$ such terms. 
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2492
For example, $(a^* + (aa)^* + (aaa)^*) ^*$'s derivatives
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2493
can be described by 6 terms:
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2494
$a^*$, $a\cdot (aa)^*$, $ (aa)^*$, 
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2495
$aa \cdot (aaa)^*$, $a \cdot (aaa)^*$, and $(aaa)^*$.
532
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2496
The total number of different "head terms",  $n * (n + 1) / 2$,
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2497
is proportional to the number of characters in the regex 
532
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2498
$(\oplus_{i=1}^{n} (\underbrace{a \ldots a}_{\text{i a's}})^*)^*$.
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2499
This suggests a slightly different notion of size, which we call the 
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2500
alphabetic width:
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2501
%TODO:
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2502
(TODO: Alphabetic width def.)
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2503
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2504
532
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2505
Antimirov\parencite{Antimirov95} has proven that 
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2506
$\textit{PDER}_{UNIV}(r) \leq \textit{awidth}(r)$.
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2507
where $\textit{PDER}_{UNIV}(r)$ is a set of all possible subterms
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2508
created by doing derivatives of $r$ against all possible strings.
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2509
If we can make sure that at any moment in our lexing algorithm our 
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2510
intermediate result hold at most one copy of each of the 
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2511
subterms then we can get the same bound as Antimirov's.
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2512
This leads to the algorithm in the next chapter.
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2513
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2514
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2515
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2516
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2517
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2518
%----------------------------------------------------------------------------------------
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2519
%	SECTION 1
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2520
%----------------------------------------------------------------------------------------
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2521
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2522
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2523
%-----------------------------------
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2524
%	SUBSECTION 1
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2525
%-----------------------------------
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2526
\subsection{Syntactic Equivalence Under $\simp$}
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2527
We prove that minor differences can be annhilated
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2528
by $\simp$.
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2529
For example,
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2530
\begin{center}
593
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2531
	$\simp \;(\simpALTs\; (\map \;(\_\backslash \; x)\; (\distinct \; \mathit{rs}\; \phi))) = 
83fab852d72d more chap5
Chengsong
parents: 591
diff changeset
  2532
	\simp \;(\simpALTs \;(\distinct \;(\map \;(\_ \backslash\; x) \; \mathit{rs}) \; \phi))$
532
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2533
\end{center}
cc54ce075db5 restructured
Chengsong
parents:
diff changeset
  2534