% Chapter Template

% Main chapter title

\chapter{Correctness of Bit-coded Algorithm with Simplification}

\label{Bitcoded2} % Change X to a consecutive number; for referencing this chapter elsewhere, use \ref{ChapterX}

%Then we illustrate how the algorithm without bitcodes falls short for such aggressive 

%simplifications and therefore introduce our version of the bitcoded algorithm and 

%its correctness proof in 

%Chapter 3\ref{Chapter3}. 

\section{Overview}

This chapter

is the point from which novel contributions of this PhD project are introduced

in detail, 

and previous

chapters are essential background work for setting the scene of the formal proof we

are about to describe.

The proof details are necessary materials for this thesis

because it provides necessary context to explain why we need a

new framework for the proof of $\blexersimp$, which involves

simplifications that cause structural changes to the regular expression.

a new formal proof of the correctness of $\blexersimp$, where the 

proof of $\blexer$

is not applicatble in the sense that we cannot straightforwardly extend the

proof of theorem \ref{blexerCorrect} because lemma \ref{retrieveStepwise} does

not hold anymore.

%This is because the structural induction on the stepwise correctness

%of $\inj$ breaks due to each pair of $r_i$ and $v_i$ described

%in chapter \ref{Inj} and \ref{Bitcoded1} no longer correspond to

%each other.

%In this chapter we introduce simplifications

%for annotated regular expressions that can be applied to 

%each intermediate derivative result. This allows

%us to make $\blexer$ much more efficient.

%Sulzmann and Lu already introduced some simplifications for bitcoded regular expressions,

%but their simplification functions could have been more efficient and in some cases needed fixing.

In particular, the correctness theorem 

of the un-optimised bit-coded lexer $\blexer$ in 

chapter \ref{Bitcoded1} formalised by Ausaf et al.

relies on lemma \ref{retrieveStepwise} that says

any value can be retrieved in a stepwise manner:

\begin{center}	

	$\vdash v : (r\backslash c) \implies \retrieve \; (r \backslash c)  \;  v= \retrieve \; r \; (\inj \; r\; c\; v)$

\end{center}

This no longer holds once we introduce simplifications.

Simplifications are necessary to control the size of regular expressions 

during derivatives by eliminating redundant 

sub-expression with some procedure we call $\textit{bsimp}$.

We want to prove the correctness of $\blexersimp$ which integrates

$\textit{bsimp}$ by applying it after each call to the derivative:

\begin{center}

\begin{tabular}{lcl}

	$r \backslash_{bsimps} (c\!::\!s) $ & $\dn$ & $(\textit{bsimp} \; (r \backslash\, c)) \backslash_{bsimps}\, s$ \\

$r \backslash_{bsimps} [\,] $ & $\dn$ & $r$

\end{tabular}

\begin{tabular}{lcl}

  $\textit{blexer\_simp}\;r\,s$ & $\dn$ &

      $\textit{let}\;a = (r^\uparrow)\backslash_{bsimp}\, s\;\textit{in}$\\                

  & & $\;\;\textit{if}\; \textit{bnullable}(a)$\\

  & & $\;\;\textit{then}\;\textit{decode}\,(\textit{bmkeps}\,a)\,r$\\

  & & $\;\;\textit{else}\;\textit{None}$

\end{tabular}

\end{center}

\noindent

Previously without $\textit{bsimp}$ the exact structure of each intermediate 

regular expression is preserved, allowing pairs of inhabitation relations in the form $\vdash v : r_{c} $ and

$\vdash v^{c} : r $ to hold in lemma \ref{retrieveStepwise}(if 

we use the convenient notation $r_{c} \dn r\backslash c$

and $v_{r}^{c} \dn \inj \;r \; c \; v$),

but $\blexersimp$ introduces simplification after the derivative,

getting us trouble in aligning the pairs:

\begin{center}

	$\vdash v: \textit{bsimp} \; r_{c} \implies \retrieve \; \textit{bsimp} \; r_c \; v =\retrieve \; r  \;(\mathord{?} v_{r}^{c}) $

\end{center}

\noindent

It is quite clear that once we made 

$v$ to align with $\textit{bsimp} \; r_{c}$

in the inhabitation relation, something different than $v_{r}^{c}$ needs to be plugged

in for the above statement to hold.

Ausaf et al. \cite{AusafUrbanDyckhoff2016}

made some initial attempts with this idea, see \cite{FahadThesis}

for details.

The other route is to dispose of lemma \ref{retrieveStepwise},

and prove a slightly weakened inductive invariant instead.

We adopt this approach in this thesis.

We first introduce why the inductive invariant in $\blexer$'s proof

is too strong, and suggest a few possible fixes, which leads to

our proof which we believe was the most natural and effective method.

\section{Why Lemma \ref{retrieveStepwise}'s Requirement is too Strong}

%From this chapter we start with the main contribution of this thesis, which

The $\blexer$ proof relies on a lockstep POSIX

correspondence between the lexical value and the

regular expression in each derivative and injection.

%

%

%which is essential for getting an understanding this thesis

%in chapter \ref{Bitcoded1}, which is necessary for understanding why

%the proof 

%

%In this chapter,

%We contrast our simplification function 

%with Sulzmann and Lu's, indicating the simplicity of our algorithm.

%This is another case for the usefulness 

%and reliability of formal proofs on algorithms.

%These ``aggressive'' simplifications would not be possible in the injection-based 

%lexing we introduced in chapter \ref{Inj}.

%We then prove the correctness with the improved version of 

%$\blexer$, called $\blexersimp$, by establishing 

%$\blexer \; r \; s= \blexersimp \; r \; s$ using a term rewriting system.

%

\section{Simplifications by Sulzmann and Lu}

The algorithms $\lexer$ and $\blexer$ work beautifully as functional 

programs, but not as practical code. One main reason for the slowness is due

to the size of intermediate representations--the derivative regular expressions

tend to grow unbounded if the matching involved a large number of possible matches.

Consider the derivatives of the following example $(a^*a^*)^*$:

%and $(a^* + (aa)^*)^*$:

\begin{center}

	\begin{tabular}{lcl}

		$(a^*a^*)^*$ & $ \stackrel{\backslash a}{\longrightarrow}$ & 

		$ (a^*a^* + a^*)\cdot(a^*a^*)^*$\\

			     & 

		$ \stackrel{\backslash a}{\longrightarrow} $ & 

	$((a^*a^* + a^*) + a^*)\cdot(a^*a^*)^* + (a^*a^* + a^*)\cdot(a^*a^*)^*$\\

							     & $\stackrel{\backslash a}{

	\longrightarrow} $ & $\ldots$\\

	\end{tabular}

\end{center}

\noindent

As can be seen, there are several duplications.

A simple-minded simplification function cannot simplify

the third regular expression in the above chain of derivative

regular expressions, namely

\begin{center}

$((a^*a^* + a^*) + a^*)\cdot(a^*a^*)^* + (a^*a^* + a^*)\cdot(a^*a^*)^*$

\end{center}

because the duplicates are

not next to each other, and therefore the rule

$r+ r \rightarrow r$ from $\textit{simp}$ does not fire.

One would expect a better simplification function to work in the 

following way:

\begin{gather*}

	((a^*a^* + \underbrace{a^*}_\text{A})+\underbrace{a^*}_\text{duplicate of A})\cdot(a^*a^*)^* + 

	\underbrace{(a^*a^* + a^*)\cdot(a^*a^*)^*}_\text{further simp removes this}.\\

	\bigg\downarrow (1) \\

	(a^*a^* + a^* 

	\color{gray} + a^* \color{black})\cdot(a^*a^*)^* + 

	\underbrace{(a^*a^* + a^*)\cdot(a^*a^*)^*}_\text{further simp removes this} \\

	\bigg\downarrow (2) \\

	(a^*a^* + a^* 

	)\cdot(a^*a^*)^*  

	\color{gray} + (a^*a^* + a^*) \cdot(a^*a^*)^*\\

	\bigg\downarrow (3) \\

	(a^*a^* + a^* 

	)\cdot(a^*a^*)^*  

\end{gather*}

\noindent

In the first step, the nested alternative regular expression

$(a^*a^* + a^*) + a^*$ is flattened into $a^*a^* + a^* + a^*$.

Now the third term $a^*$ can clearly be identified as a duplicate

and therefore removed in the second step. 

This causes the two

top-level terms to become the same and the second $(a^*a^*+a^*)\cdot(a^*a^*)^*$ 

removed in the final step.

Sulzmann and Lu's simplification function (using our notations) can achieve this

simplification:

\begin{center}

	\begin{tabular}{lcl}

		$\textit{simp}\_{SL} \; _{bs}(_{bs'}\ONE \cdot r)$ & $\dn$ & 

		$\textit{if} \; (\textit{zeroable} \; r)\; \textit{then} \;\; \ZERO$\\

						   & &$\textit{else}\;\; \fuse \; (bs@ bs') \; r$\\

		$\textit{simp}\_{SL} \;(_{bs}r_1\cdot r_2)$ & $\dn$ & $\textit{if} 

		\; (\textit{zeroable} \; r_1 \; \textit{or} \; \textit{zeroable}\; r_2)\;

		\textit{then} \;\; \ZERO$\\

							    & & $\textit{else}\;\;_{bs}((\textit{simp}\_{SL} \;r_1)\cdot

							    (\textit{simp}\_{SL} \; r_2))$\\

		$\textit{simp}\_{SL}  \; _{bs}\sum []$ & $\dn$ & $\ZERO$\\

		$\textit{simp}\_{SL}  \; _{bs}\sum ((_{bs'}\sum rs_1) :: rs_2)$ & $\dn$ &

		$_{bs}\sum ((\map \; (\fuse \; bs')\; rs_1) @ rs_2)$\\

		$\textit{simp}\_{SL}  \; _{bs}\sum[r]$ & $\dn$ & $\fuse \; bs \; (\textit{simp}\_{SL}  \; r)$\\

		$\textit{simp}\_{SL}  \; _{bs}\sum(r::rs)$ & $\dn$ & $_{bs}\sum 

		(\nub \; (\filter \; (\neg\zeroable)\;((\textit{simp}\_{SL}  \; r) :: \map \; \textit{simp}\_{SL}  \; rs)))$\\ 

	\end{tabular}

\end{center}

\noindent

The $\textit{zeroable}$ predicate 

tests whether the regular expression

is equivalent to $\ZERO$, and

can be defined as:

\begin{center}

	\begin{tabular}{lcl}

		$\zeroable \; _{bs}\sum (r::rs)$ & $\dn$ & $\zeroable \; r\;\; \land \;\;

		\zeroable \;_{[]}\sum\;rs $\\

		$\zeroable\;_{bs}(r_1 \cdot r_2)$ & $\dn$ & $\zeroable\; r_1 \;\; \lor \;\; \zeroable \; r_2$\\

		$\zeroable\;_{bs}r^*$ & $\dn$ & $\textit{false}$ \\

		$\zeroable\;_{bs}c$ & $\dn$ & $\textit{false}$\\

		$\zeroable\;_{bs}\ONE$ & $\dn$ & $\textit{false}$\\

		$\zeroable\;_{bs}\ZERO$ & $\dn$ & $\textit{true}$

	\end{tabular}

\end{center}

\noindent

The 

\begin{center}

	\begin{tabular}{lcl}

		$\textit{simp}\_{SL}  \; _{bs}\sum ((_{bs'}\sum rs_1) :: rs_2)$ & $\dn$ &

		$_{bs}\sum ((\map \; (\fuse \; bs')\; rs_1) @ rs_2)$\\

	\end{tabular}

\end{center}

\noindent

clause does flatten the alternative as required in step (1),

but $\textit{simp}\_{SL}$ is insufficient if we want to do steps (2) and (3),

as these ``identical'' terms have different bit-annotations.

They also suggested that the $\textit{simp}\_{SL} $ function should be

applied repeatedly until a fixpoint is reached.

We call this construction $\textit{SLSimp}$:

\begin{center}

	\begin{tabular}{lcl}

		$\textit{SLSimp} \; r$ & $\dn$ & 

		$\textit{while}((\textit{simp}\_{SL}  \; r)\; \cancel{=} \; r)$ \\

					 & & $\quad r := \textit{simp}\_{SL}  \; r$\\

		& & $\textit{return} \; r$

	\end{tabular}

\end{center}

We call the operation of alternatingly 

applying derivatives and simplifications

(until the string is exhausted) Sulz-simp-derivative,

written $\backslash_{SLSimp}$:

\begin{center}

\begin{tabular}{lcl}

	$r \backslash_{SLSimp} (c\!::\!s) $ & $\dn$ & $(\textit{SLSimp} \; (r \backslash c)) \backslash_{SLSimp}\, s$ \\

$r \backslash_{SLSimp} [\,] $ & $\dn$ & $r$

\end{tabular}

\end{center}

\noindent

After the derivatives have been taken, the bitcodes

are extracted and decoded in the same manner

as $\blexer$:

\begin{center}

\begin{tabular}{lcl}

  $\textit{blexer\_SLSimp}\;r\,s$ & $\dn$ &

      $\textit{let}\;a = (r^\uparrow)\backslash_{SLSimp}\, s\;\textit{in}$\\                

  & & $\;\;\textit{if}\; \textit{bnullable}(a)$\\

  & & $\;\;\textit{then}\;\textit{decode}\,(\textit{bmkeps}\,a)\,r$\\

  & & $\;\;\textit{else}\;\textit{None}$

\end{tabular}

\end{center}

\noindent

We implemented this lexing algorithm in Scala, 

and found that the final derivative regular expression

size still grows exponentially (note the logarithmic scale):

\begin{figure}[H]

	\centering

\begin{tikzpicture}

\begin{axis}[

    xlabel={$n$},

    ylabel={size},

    ymode = log,

    legend entries={Final Derivative Size},  

    legend pos=north west,

    legend cell align=left]

\addplot[red,mark=*, mark options={fill=white}] table {SulzmannLuLexer.data};

\end{axis}

\end{tikzpicture} 

\caption{Lexing the regular expression $(a^*a^*)^*$ against strings of the form

$\protect\underbrace{aa\ldots a}_\text{n \textit{a}s}

$ using Sulzmann and Lu's lexer}\label{SulzmannLuLexer}

\end{figure}

\noindent

At $n= 20$ we already get an out-of-memory error with Scala's normal 

JVM heap size settings.

In fact their simplification does not improve much over

the simple-minded simplifications we have shown in \ref{fig:BetterWaterloo}.

The time required also grows exponentially:

\begin{figure}[H]

	\centering

\begin{tikzpicture}

\begin{axis}[

    xlabel={$n$},

    ylabel={time},

    %ymode = log,

    legend entries={time in secs},  

    legend pos=north west,

    legend cell align=left]

\addplot[red,mark=*, mark options={fill=white}] table {SulzmannLuLexerTime.data};

\end{axis}

\end{tikzpicture} 

\caption{Lexing the regular expression $(a^*a^*)^*$ against strings of the form

$\protect\underbrace{aa\ldots a}_\text{n \textit{a}s}

$ using Sulzmann and Lu's lexer}\label{SulzmannLuLexerTime}

\end{figure}

\noindent

which seems like a counterexample for 

Sulzmann and Lu's linear complexity claim

in their paper \cite{Sulzmann2014}:

\begin{quote}\it

``Linear-Time Complexity Claim \\It is easy to see that each call of one of the functions/operations:

simp, fuse, mkEpsBC and isPhi leads to subcalls whose number is bound by the size of the regular expression involved. We claim that thanks to aggressively applying simp this size remains finite. Hence, we can argue that the above mentioned functions/operations have constant time complexity which implies that we can incrementally compute bit-coded parse trees in linear time in the size of the input.'' 

\end{quote}

\noindent

The assumption that the size of the regular expressions

in the algorithm

would stay below a finite constant is not true, at least not in the

examples we considered.

The main reason behind this is that (i) Haskell's $\textit{nub}$

function requires identical annotations between two 

annotated regular expressions to qualify as duplicates,

and therefore cannot simplify cases like $_{SZZ}a^*+_{SZS}a^*$

even if both $a^*$ denote the same language, and

(ii) the ``flattening'' only applies to the head of the list

in the 

\begin{center}

	\begin{tabular}{lcl}

		$\textit{simp}\_{SL}  \; _{bs}\sum ((_{bs'}\sum rs_1) :: rs_2)$ & $\dn$ &

		$_{bs}\sum ((\map \; (\fuse \; bs')\; rs_1) @ rs_2)$\\

	\end{tabular}

\end{center}

\noindent

clause, and therefore is not strong enough to simplify all

needed parts of the regular expression. Moreover,

the $\textit{simp}\_{SL}$ function is applied repeatedly

in each derivative step until a fixed point is reached, 

which makes the algorithm even more

unpredictable and inefficient.

%To not get ``caught off guard'' by

%these counterexamples,

%one needs to be more careful when designing the

%simplification function and making claims about them.

\section{Our $\textit{Simp}$ Function}

We will now introduce our own simplification function.

%by making a contrast with $\textit{simp}\_{SL}$.

We also describe

the ideas behind Sulzmann and Lu's $\textit{simp}\_{SL}$

algorithm 

and why it fails to achieve the desired effect of keeping the sizes of derivatives finitely bounded. 

In addition, our simplification function will come with a formal

correctness proof.

\subsection{Flattening Nested Alternatives}

The idea behind the clause

\begin{center}

	$\textit{simp}\_{SL}  \; _{bs}\sum ((_{bs'}\sum rs_1) :: rs_2) \quad \dn \quad

	       _{bs}\sum ((\map \; (\fuse \; bs')\; rs_1) @ rs_2)$

\end{center}

is that it allows

duplicate removal of regular expressions at different

``levels'' of alternatives.

For example, this would help with the

following simplification:

\begin{center}

$(a+r)+r \longrightarrow a+r$

\end{center}

The problem is that only the head element

is ``spilled out''.

It is more desirable

to flatten

an entire list to open up possibilities for further simplifications

with later regular expressions.

Not flattening the rest of the elements also means that

the later de-duplication process 

does not fully remove further duplicates.

For example,

using $\textit{simp}\_{SL}$ we cannot

simplify

\begin{center}

	$((a^* a^*)+\underline{(a^* + a^*)})\cdot (a^*a^*)^*+

((a^*a^*)+a^*)\cdot (a^*a^*)^*$

\end{center}

due to the underlined part not being the head 

of the alternative.

We define our flatten operation so that it flattens 

the entire list: 

 \begin{center}

  \begin{tabular}{@{}lcl@{}}

  $\textit{flts} \; (_{bs}\sum \textit{as}) :: \textit{as'}$ & $\dn$ & $(\textit{map} \;

     (\textit{fuse}\;bs)\; \textit{as}) \; @ \; \textit{flts} \; as' $ \\

  $\textit{flts} \; \ZERO :: as'$ & $\dn$ & $ \textit{flts} \;  \textit{as'} $ \\

    $\textit{flts} \; a :: as'$ & $\dn$ & $a :: \textit{flts} \; \textit{as'}$ \quad(otherwise) 

\end{tabular}    

\end{center}  

\noindent

Our $\flts$ operation 

also throws away $\ZERO$s

as they do not contribute to a lexing result.

\subsection{Duplicate Removal}

After flattening is done, we can deduplicate.

The de-duplicate function is called $\distinctBy$,

and that is where we make our second improvement over

Sulzmann and Lu's simplification method.

The process goes as follows:

\begin{center}

$rs \stackrel{\textit{flts}}{\longrightarrow} 

rs_{flat} 

\xrightarrow{\distinctBy \; 

rs_{flat} \; \rerases\; \varnothing} rs_{distinct}$

%\stackrel{\distinctBy \; 

%rs_{flat} \; \erase\; \varnothing}{\longrightarrow} \; rs_{distinct}$

\end{center}

where the $\distinctBy$ function is defined as:

\begin{center}

	\begin{tabular}{@{}lcl@{}}

		$\distinctBy \; [] \; f\; acc $ & $ =$ & $ []$\\

		$\distinctBy \; (x :: xs) \; f \; acc$ & $=$ & $\quad \textit{if} (f \; x \in acc)\;\; \textit{then} \;\; \distinctBy \; xs \; f \; acc$\\

						       & & $\quad \textit{else}\;\; x :: (\distinctBy \; xs \; f \; (\{f \; x\} \cup acc))$ 

	\end{tabular}

\end{center}

\noindent

The reason we define a distinct function under a mapping $f$ is because

we want to eliminate regular expressions that are syntactically the same,

but have different bit-codes.

For example, we can remove the second $a^*a^*$ from

$_{ZSZ}a^*a^* + _{SZZ}a^*a^*$, because it

represents a match with shorter initial sub-match 

(and therefore is definitely not POSIX),

and will be discarded by $\bmkeps$ later.

\begin{center}

	$_{ZSZ}\underbrace{a^*}_{ZS:\; match \; 1\; times\quad}\underbrace{a^*}_{Z: \;match\; 1 \;times} + 

	_{SZZ}\underbrace{a^*}_{S: \; match \; 0 \; times\quad}\underbrace{a^*}_{ZZ: \; match \; 2 \; times}

	$

\end{center}

%$_{bs1} r_1 + _{bs2} r_2 \text{where} (r_1)_{\downarrow} = (r_2)_{\downarrow}$

Due to the way our algorithm works,

the matches that conform to the POSIX standard 

will always be placed further to the left. When we 

traverse the list from left to right,

regular expressions we have already seen

will definitely not contribute to a POSIX value,

even if they are attached with different bitcodes.

These duplicates therefore need to be removed.

To achieve this, we call $\rerases$ as the function $f$ during the distinction

operation. The function

$\rerases$ is very similar to $\erase$, except that it preserves the structure

when erasing an alternative regular expression.

The reason why we use $\rerases$ instead of $\erase$ is that

it keeps the structures of alternative 

annotated regular expressions

whereas $\erase$ would turn it back into a binary  tree structure.

Not having to mess with the structure 

greatly simplifies the finiteness proof in chapter 

\ref{Finite}.

We give the definitions of $\rerases$ here together with

the new datatype used by $\rerases$ (as our plain

regular expression datatype does not allow non-binary alternatives).

For now we can think of 

$\rerases$ as the function $(\_)_\downarrow$ defined in chapter \ref{Bitcoded1}

and $\rrexp$ as plain regular expressions, but having a general list constructor