\documentclass{article}

\usepackage{../style}

\usepackage{../langs}

\usepackage{../graphics}

\begin{document}

\section*{Handout 4 (Sulzmann \& Lu Algorithm)}

So far our algorithm based on derivatives was only able to say

yes or no depending on whether a string was matched by regular

expression or not. Often a more interesting question is to

find out \emph{how} a regular expression matched a string?

Answering this question will also help us with the problem we

\begin{center}

\begin{tabular}{cc}

\begin{tabular}{@{}rrl@{}}

  $r$ & $::=$  & $\varnothing$\\

      & $\mid$ & $\epsilon$   \\

      & $\mid$ & $c$          \\

      & $\mid$ & $r_1 \cdot r_2$\\

      & $\mid$ & $r_1 + r_2$   \\

  \\

      & $\mid$ & $r^*$         \\

\end{tabular}

&

\begin{tabular}{@{\hspace{0mm}}rrl@{}}

   $v$ & $::=$  & \\

      &        & $Empty$   \\

      & $\mid$ & $Char(c)$          \\

      & $\mid$ & $Seq(v_1,v_2)$\\

      & $\mid$ & $Left(v)$   \\

      & $\mid$ & $Right(v)$  \\

      & $\mid$ & $[v_1,\ldots\,v_n]$ \\

\end{tabular}

\end{tabular}

\end{center}

\noindent The point is that there is a very strong

correspondence between them. There is no value for the

$\varnothing$ regular expression, since it does not match any

string. Otherwise there is exactly one value corresponding to

each regular expression with the exception of $r_1 + r_2$

where there are two values, namely $Left(v)$ and $Right(v)$

corresponding to the two alternatives. Note that $r^*$ is

associated with a list of values, one for each copy of $r$

that was needed to match the string. This means we might also

return the empty list $[]$, if no copy was needed.

To emphasise the connection between regular expressions and

values, I have in my implementation the convention that

regular expressions are written entirely with upper-case

letters, while values just start with a single upper-case

character. My definition of values in Scala is below. I use 

this in the REPL of Scala; when I use the Scala compiler

I need to rename some constructors, because Scala on Macs

does not like classes that are called \pcode{EMPTY} and

\pcode{Empty}.

{\small\lstinputlisting[language=Scala,numbers=none]

{../progs/app02.scala}}

Graphically the algorithm by Sulzmann \& Lu can be illustrated

by the picture in Figure~\ref{Sulz} where the path from the

left to the right involving $der/nullable$ is the first phase

of the algorithm and $mkeps/inj$, the path from right to left,

the second phase. This picture shows the steps required when a

regular expression, say $r_1$, matches the string $abc$. We

first build the three derivatives (according to $a$, $b$ and

$c$). We then use $nullable$ to find out whether the resulting

regular expression can match the empty string. If yes, we call

the function $mkeps$.

\begin{figure}[t]

\begin{center}

\begin{tikzpicture}[scale=2,node distance=1.2cm,

                    every node/.style={minimum size=7mm}]

\node (r1)  {$r_1$};

\node (r2) [right=of r1]{$r_2$};

\draw[->,line width=1mm](r1)--(r2) node[above,midway] {$der\,a$};

\node (r3) [right=of r2]{$r_3$};

\draw[->,line width=1mm](r2)--(r3) node[above,midway] {$der\,b$};

\node (r4) [right=of r3]{$r_4$};

\draw[->,line width=1mm](r3)--(r4) node[above,midway] {$der\,c$};

\draw (r4) node[anchor=west] {\;\raisebox{3mm}{$nullable$}};

\node (v4) [below=of r4]{$v_4$};

\draw[->,line width=1mm](r4) -- (v4);

\node (v3) [left=of v4] {$v_3$};

\draw[->,line width=1mm](v4)--(v3) node[below,midway] {$inj\,c$};

\node (v2) [left=of v3]{$v_2$};

\draw[->,line width=1mm](v3)--(v2) node[below,midway] {$inj\,b$};

\node (v1) [left=of v2] {$v_1$};

\draw[->,line width=1mm](v2)--(v1) node[below,midway] {$inj\,a$};

\draw (r4) node[anchor=north west] {\;\raisebox{-8mm}{$mkeps$}};

\end{tikzpicture}

\end{center}

\caption{The two phases of the algorithm by Sulzmann \& Lu.

\label{Sulz}}

\end{figure}

The $mkeps$ function calculates a value for how a regular

expression has matched the empty string. Its definition

is as follows:

\begin{center}

\begin{tabular}{lcl}

  $mkeps(\epsilon)$       & $\dn$ & $Empty$\\

  $mkeps(r_1 + r_2)$      & $\dn$ & if $nullable(r_1)$  \\

                          &       & then $Left(mkeps(r_1))$\\

                          &       & else $Right(mkeps(r_2))$\\

  $mkeps(r_1 \cdot r_2)$  & $\dn$ & $Seq(mkeps(r_1),mkeps(r_2))$\\

  $mkeps(r^*)$            & $\dn$ & $[]$  \\

\end{tabular}

\end{center}

\noindent There are no cases for $\varnothing$ and $c$, since

these regular expression cannot match the empty string. Note

also that in case of alternatives we give preference to the

regular expression on the left-hand side. This will become

important later on.

The second phase of the algorithm is organised so that it will

calculate a value for how the derivative regular expression

has matched a string whose first character has been chopped

off. Now we need a function that reverses this ``chopping

off'' for values. The corresponding function is called $inj$

for injection. This function takes three arguments: the first

one is a regular expression for which we want to calculate the

value, the second is the character we want to inject and the

third argument is the value where we will inject the

character. The result of this function is a new value. The

definition of $inj$ is as follows: 

\begin{center}

\begin{tabular}{l@{\hspace{1mm}}c@{\hspace{1mm}}l}

  $inj\,(c)\,c\,Empty$            & $\dn$  & $Char\,c$\\

  $inj\,(r_1 + r_2)\,c\,Left(v)$  & $\dn$  & $Left(inj\,r_1\,c\,v)$\\

  $inj\,(r_1 + r_2)\,c\,Right(v)$ & $\dn$  & $Right(inj\,r_2\,c\,v)$\\

  $inj\,(r_1 \cdot r_2)\,c\,Seq(v_1,v_2)$ & $\dn$  & $Seq(inj\,r_1\,c\,v_1,v_2)$\\

  $inj\,(r_1 \cdot r_2)\,c\,Left(Seq(v_1,v_2))$ & $\dn$  & $Seq(inj\,r_1\,c\,v_1,v_2)$\\

  $inj\,(r_1 \cdot r_2)\,c\,Right(v)$ & $\dn$  & $Seq(mkeps(r_1),inj\,r_2\,c\,v)$\\

  $inj\,(r^*)\,c\,Seq(v,vs)$         & $\dn$  & $inj\,r\,c\,v\,::\,vs$\\

\end{tabular}

\end{center}

\noindent This definition is by recursion on the regular

expression and by analysing the shape of the values. Therefore

there are, for example, three cases for sequence regular

expressions. The last clause for the star regular expression

returns a list where the first element is $inj\,r\,c\,v$ and

the other elements are $vs$. That means $\_\,::\,\_$ should be 

read as list cons.

To understand what is going on, it might be best to do some

example calculations and compare them with Figure~\ref{Sulz}.

For this note that we have not yet dealt with the need of

simplifying regular expressions (this will be a topic on its

own later). Suppose the regular expression is $a \cdot (b

\cdot c)$ and the input string is $abc$. The derivatives from

the first phase are as follows:

\begin{center}

\begin{tabular}{ll}

$r_1$: & $a \cdot (b \cdot c)$\\

$r_2$: & $\epsilon \cdot (b \cdot c)$\\

$r_3$: & $(\varnothing \cdot (b \cdot c)) + (\epsilon \cdot c)$\\

$r_4$: & $(\varnothing \cdot (b \cdot c)) + ((\varnothing \cdot c) + \epsilon)$\\

\end{tabular}

\end{center}

\noindent According to the simple algorithm, we would test

whether $r_4$ is nullable, which in this case it is. This

means we can use the function $mkeps$ to calculate a value for

how $r_4$ was able to match the empty string. Remember that

this function gives preference for alternatives on the

left-hand side. However there is only $\epsilon$ on the very

right-hand side of $r_4$ that matches the empty string.

Therefore $mkeps$ returns the value

\begin{center}

$v_4:\;Right(Right(Empty))$

\end{center}

\noindent The point is that from this value we can directly

read off which part of $r_4$ matched the empty string. Next we

have to ``inject'' the last character, that is $c$ in the

running example, into this value $v_4$ in order to calculate

how $r_3$ could have matched the string $c$. According to the

definition of $inj$ we obtain

\begin{center}

$v_3:\;Right(Seq(Empty, Char(c)))$

\end{center}

\noindent This is the correct result, because $r_3$ needs

to use the right-hand alternative, and then $\epsilon$ needs

to match the empty string and $c$ needs to match $c$.

Next we need to inject back the letter $b$ into $v_3$. This

gives

\begin{center}

$v_2:\;Seq(Empty, Seq(Char(b), Char(c)))$

\end{center}

\noindent which is again the correct result for how $r_2$

matched the string $bc$. Finally we need to inject back the

letter $a$ into $v_2$ giving the final result

\begin{center}

$v_1:\;Seq(Char(a), Seq(Char(b), Char(c)))$

\end{center}

\noindent This now corresponds to how the regular

expression $a \cdot (b \cdot c)$ matched the string $abc$.

There are a few auxiliary functions that are of interest

when analysing this algorithm. One is called \emph{flatten},

written $|\_|$, which extracts the string ``underlying'' a 

value. It is defined recursively as

\begin{center}

\begin{tabular}{lcl}

  $|Empty|$     & $\dn$ & $[]$\\

  $|Char(c)|$   & $\dn$ & $[c]$\\

  $|Left(v)|$   & $\dn$ & $|v|$\\

  $|Right(v)|$  & $\dn$ & $|v|$\\

  $|Seq(v_1,v_2)|$& $\dn$ & $|v_1| \,@\, |v_2|$\\

  $|[v_1,\ldots ,v_n]|$ & $\dn$ & $|v_1| \,@\ldots @\, |v_n|$\\

\end{tabular}

\end{center}

\noindent Using flatten we can see what is the string behind 

the values calculated by $mkeps$ and $inj$ in our running 

example are:

\begin{center}

\begin{tabular}{ll}

$|v_4|$: & $[]$\\

$|v_3|$: & $c$\\

$|v_2|$: & $bc$\\

$|v_1|$: & $abc$

\end{tabular}

\end{center}

\noindent This indicates that $inj$ indeed is injecting, or

adding, back a character into the value. Next we look at how

simplification can be included into this algorithm.

\subsubsection*{Simplification}

Generally the matching algorithms based on derivatives do

poorly unless the regular expressions are simplified after

each derivative step. But this is a bit more involved in the

algorithm of Sulzmann \& Lu. So what follows might require you

to read several times before it makes sense and also might

require that you do some example calculations. As a first

example consider the last derivation step in our earlier

example:

\begin{center}

$r_4 = der\,c\,r_3 = 

(\varnothing \cdot (b \cdot c)) + ((\varnothing \cdot c) + \epsilon)$

\end{center}

\noindent Simplifying this regular expression would just give

us $\epsilon$. Running $mkeps$ with this regular expression as

input, however, would then provide us with $Empty$ instead of

$Right(Right(Empty))$ that was obtained without the

simplification. The problem is we need to recreate this more

complicated value, rather than just $Empty$.

This will require what I call \emph{rectification functions}.

They need to be calculated whenever a regular expression gets

simplified. Rectification functions take a value as argument

and return a (rectified) value. Let us first take a look again

at our simplification rules:

\begin{center}

\begin{tabular}{l@{\hspace{2mm}}c@{\hspace{2mm}}l}

$r \cdot \varnothing$ & $\mapsto$ & $\varnothing$\\ 

$\varnothing \cdot r$ & $\mapsto$ & $\varnothing$\\ 

$r \cdot \epsilon$ & $\mapsto$ & $r$\\ 

$\epsilon \cdot r$ & $\mapsto$ & $r$\\ 

$r + \varnothing$ & $\mapsto$ & $r$\\ 

$\varnothing + r$ & $\mapsto$ & $r$\\ 

$r + r$ & $\mapsto$ & $r$\\ 

\end{tabular}

\end{center}

\noindent Applying them to $r_4$ will require several nested

simplifications in order end up with just $\epsilon$. However,

it is possible to apply them in a depth-first, or inside-out,

manner in order to calculate this simplified regular

expression.

The rectification we can implement this by letting simp return

not just a (simplified) regular expression, but also a

rectification function. Let us consider the alternative case,

$r_1 + r_2$, first. By going depth-first, we first simplify

the component regular expressions $r_1$ and $r_2.$ This will

return simplified versions (if they can be simplified), say

$r_{1s}$ and $r_{2s}$, but also two rectification functions

$f_{1s}$ and $f_{2s}$. We need to assemble them in order to

obtain a rectified value for $r_1 + r_2$. In case $r_{1s}$

simplified to $\varnothing$, we continue the derivative

calculation with $r_{2s}$. The Sulzmann \& Lu algorithm would

return a corresponding value, say $v_{2s}$. But now this value

needs to be ``rectified'' to the value 

\begin{center}

$Right(v_{2s})$

\end{center}

\noindent The reason is that we look for the value that tells

us how $r_1 + r_2$ could have matched the string, not just

$r_2$ or $r_{2s}$. Unfortunately, this is still not the right

value in general because there might be some simplifications

that happened inside $r_2$ and for which the simplification

function retuned also a rectification function $f_{2s}$. So in

fact we need to apply this one too which gives

\begin{center}

$Right(f_{2s}(v_{2s}))$

\end{center}

\noindent This is now the correct, or rectified, value. Since

the simplification will be done in the first phase of the

algorithm, but the rectification needs to be done to the

values in the second phase, it is advantageous to calculate

the rectification as a function, remember this function and

then apply the value to this function during the second phase.

So if we want to implement the rectification as function, we 

would need to return

\begin{center}

$\lambda v.\,Right(f_{2s}(v))$

\end{center}

\noindent which is the lambda-calculus notation for

a function that expects a value $v$ and returns everything

after the dot where $v$ is replaced by whatever value is 

given.

Let us package this idea with rectification functions

into a single function (still only considering the alternative

case):

\begin{center}

\begin{tabular}{l}

$simp(r)$:\\

\quad case $r = r_1 + r_2$\\

\qquad let $(r_{1s}, f_{1s}) = simp(r_1)$\\

\qquad \phantom{let} $(r_{2s}, f_{2s}) = simp(r_2)$\smallskip\\

\qquad case $r_{1s} = \varnothing$: 

       return $(r_{2s}, \lambda v. \,Right(f_{2s}(v)))$\\

\qquad case $r_{2s} = \varnothing$: 

       return $(r_{1s}, \lambda v. \,Left(f_{1s}(v)))$\\

\qquad case $r_{1s} = r_{2s}$:

       return $(r_{1s}, \lambda v. \,Left(f_{1s}(v)))$\\

\qquad otherwise: 

       return $(r_{1s} + r_{2s}, f_{alt}(f_{1s}, f_{2s}))$\\

\end{tabular}

\end{center}

\noindent We first recursively call the simplification with

$r_1$ and $r_2$. This gives simplified regular expressions,

$r_{1s}$ and $r_{2s}$, as well as two rectification functions

$f_{1s}$ and $f_{2s}$. We next need to test whether the

simplified regular expressions are $\varnothing$ so as to make

further simplifications. In case $r_{1s}$ is $\varnothing$,

then we can return $r_{2s}$ (the other alternative). However

for this we need to build a corresponding rectification 

function, which as said above is

\begin{center}

$\lambda v.\,Right(f_{2s}(v))$

\end{center}

\noindent The case where $r_{2s} = \varnothing$ is similar:

We return $r_{1s}$ and rectify with $Left(\_)$ and the

other calculated rectification function $f_{1s}$. This gives

\begin{center}

$\lambda v.\,Left(f_{1s}(v))$

\end{center}

\noindent The next case where $r_{1s} = r_{2s}$ can be treated

like the one where $r_{2s} = \varnothing$. We return $r_{1s}$

and rectify with $Left(\_)$ and so on.

The otherwise-case is slightly more complicated. In this case

neither $r_{1s}$ nor $r_{2s}$ are $\varnothing$ and also

$r_{1s} \not= r_{2s}$, which means no further simplification

can be applied. Accordingly, we return $r_{1s} + r_{2s}$ as

the simplified regular expression. In principle we also do not

have to do any rectification, because no simplification was

done in this case. But this is actually not true: There might

have been simplifications inside $r_{1s}$ and $r_{2s}$. We

therefore need to take into account the calculated

rectification functions $f_{1s}$ and $f_{2s}$. We can do this

by defining a rectification function $f_{alt}$ which takes two

rectification functions as arguments and applies them

according to whether the value is of the form $Left(\_)$ or

$Right(\_)$:

\begin{center}

\begin{tabular}{l@{\hspace{1mm}}l}

$f_{alt}(f_1, f_2) \dn$\\

\qquad $\lambda v.\,$ case $v = Left(v')$: 

      & return $Left(f_1(v'))$\\

\qquad \phantom{$\lambda v.\,$} case $v = Right(v')$: 

      & return $Right(f_2(v'))$\\      

\end{tabular}

\end{center}

The other interesting case with simplification is the sequence

case. In this case the main simplification function is as

follows

\begin{center}

\begin{tabular}{l}

$simp(r)$:\qquad\qquad (continued)\\

\quad case $r = r_1 \cdot r_2$\\

\qquad let $(r_{1s}, f_{1s}) = simp(r_1)$\\

\qquad \phantom{let} $(r_{2s}, f_{2s}) = simp(r_2)$\smallskip\\

\qquad case $r_{1s} = \varnothing$: 

       return $(\varnothing, f_{error})$\\

\qquad case $r_{2s} = \varnothing$: 

       return $(\varnothing, f_{error})$\\

\qquad case $r_{1s} = \epsilon$: 

return $(r_{2s}, \lambda v. \,Seq(f_{1s}(Empty), f_{2s}(v)))$\\

\qquad case $r_{2s} = \epsilon$: 

return $(r_{1s}, \lambda v. \,Seq(f_{1s}(v), f_{2s}(Empty)))$\\

\qquad otherwise: 

       return $(r_{1s} \cdot r_{2s}, f_{seq}(f_{1s}, f_{2s}))$\\

\end{tabular}

\end{center}

\noindent whereby in the last line $f_{seq}$ is again pushing

the two rectification functions into the two components of the

Seq-value:

\begin{center}

\begin{tabular}{l@{\hspace{1mm}}l}

$f_{seq}(f_1, f_2) \dn$\\

\qquad $\lambda v.\,$ case $v = Seq(v_1, v_2)$: 

      & return $Seq(f_1(v_1), f_2(v_2))$\\

\end{tabular}

\end{center}

\noindent Note that in the case of $r_{1s} = \varnothing$

(similarly $r_{2s}$) we use the function $f_{error}$ for

rectification. If you think carefully, then you will realise

that this function will actually never been called. This is

because a sequence with $\varnothing$ will never recognise any

string and therefore the second phase of the algorithm would

never been called. The simplification function still expects

us to give a function. So in my own implementation I just

returned a function which raises an error. In the case

where $r_{1s} = \epsilon$ (similarly $r_{2s}$) we have

to create a sequence where the first component is a rectified

version of $Empty$. Therefore we call $f_{1s}$ with $Empty$.