Solving of Regular Equations Revisited (extended version)

TL;DR

This paper revisits the solving of regular equations using Arden's Lemma, providing a detailed algorithm, a computational interpretation, and applications to convert automata to unambiguous expressions and express operations algebraically.

Contribution

It offers a concise algorithmic specification, a novel interpretation of solving via coercions, and new algebraic methods for automata operations without conversion.

Findings

Identifies conditions for unambiguous solutions of regular equations.

Provides a method to convert DFA to unambiguous regular expressions.

Expresses subtraction and shuffling operations algebraically via regular equations.

Abstract

Solving of regular equations via Arden's Lemma is folklore knowledge. We first give a concise algorithmic specification of all elementary solving steps. We then discuss a computational interpretation of solving in terms of coercions that transform parse trees of regular equations into parse trees of solutions. Thus, we can identify some conditions on the shape of regular equations under which resulting solutions are unambiguous. We apply our result to convert a DFA to an unambiguous regular expression. In addition, we show that operations such as subtraction and shuffling can be expressed via some appropriate set of regular equations. Thus, we obtain direct (algebraic) methods without having to convert to and from finite automaton.

Equations

$$\begin{array}[]{lcl}R,S&::=&\emptyset\mid\{\varepsilon\}\mid\{x\}\mid(R+S)\mid(R\cdot S)\mid(R^{*})\ \ \ \ \ \mbox{where $x\in\Sigma$.}\end{array}$$

$$\begin{array}[]{lcl}r,s&::=&\phi\mid\varepsilon\mid x\mid(r+s)\mid(r\cdot s)\mid(r^{*})\ \ \ \ \ \mbox{where $x\in\Sigma$.}\end{array}$$

$$\begin{array}[]{lcll}\alpha&&::=&r\cdot R\mid r\mid\alpha+\alpha\end{array}$$

$$\begin{array}[]{c}\mbox{(E1)}\ \ \gamma_{1}\cdot(\gamma_{2}+\gamma_{3})\simeq\gamma_{1}\cdot\gamma_{2}+\gamma_{1}\cdot\gamma_{3}\ \ \ \ \mbox{(E2)}\ \ \gamma_{1}\cdot(\gamma_{2}\cdot\gamma_{3})\simeq(\gamma_{1}\cdot\gamma_{2})\cdot\gamma_{3}\\ \\ \mbox{(E3)}\ \ \gamma_{1}+(\gamma_{2}+\gamma_{3})\simeq(\gamma_{1}+\gamma_{2})+\gamma_{3}\ \ \ \ \mbox{(E4)}\ \ \gamma_{2}\cdot\gamma_{1}+\gamma_{3}\cdot\gamma_{1}\simeq(\gamma_{2}+\gamma_{3})\cdot\gamma_{1}\\ \mbox{(E5)}\ \ \gamma_{1}+\gamma_{2}\simeq\gamma_{2}+\gamma_{1}\ \ \ \ \mbox{(E6)}\ \ {\begin{array}[]{c}\gamma_{1}\simeq\gamma_{2}\\ \hline\cr\beta[\gamma_{1}]\simeq\beta[\gamma_{2}]\end{array}}\ \ \ \ \mbox{(E7)}\ \ {\begin{array}[]{c}\gamma_{1}\simeq\gamma_{2}\ \ \gamma_{2}\simeq\gamma_{3}\\ \hline\cr\gamma_{1}\simeq\gamma_{3}\end{array}}\end{array}$$

$$\begin{array}[]{c}\beta[]::=[]\mid\beta[]+\beta\mid\beta+\beta[]\end{array}$$

$$\begin{array}[]{c}\mbox{(Arden)}\ \ \ \ {\begin{array}[]{c}R\not\in\alpha\\ \hline\cr\langle\psi,R\approx s\cdot R+\alpha\uplus{\cal E}\rangle\Rightarrow\langle\psi,R\approx s^{*}\cdot\alpha\uplus{\cal E}\rangle\end{array}}\\ \\ \mbox{(Subst)}\ \ \ \ {\begin{array}[]{c}R\not\in\alpha\\ \psi^{\prime}=\{R\mapsto\alpha\}\cup\{S\mapsto\{R\mapsto\alpha\}(\gamma)\mid S\mapsto\gamma\in\psi\}\\ {\cal E}^{\prime}=\{R^{\prime}\approx\alpha^{\prime\prime}\mid R^{\prime}\approx\alpha^{\prime}\in{\cal E}\wedge\{R\mapsto\alpha\}(\alpha^{\prime})\simeq\alpha^{\prime\prime}\}\\ \hline\cr\langle\psi,R\approx\alpha\uplus{\cal E}\rangle\Rightarrow\langle\psi^{\prime},{\cal E}^{\prime}\rangle\end{array}}\end{array}$$

$$vars(\mathcal{E})=(\{R_1,\dots ,R_n\},\{\{S_1,\dots ,S_m\}\})$$

$$\begin{array}[]{ll}&\langle\{\},\{R_{1}\approx x\cdot R_{1}+y\cdot R_{2}+\varepsilon,R_{2}\approx y\cdot R_{1}+x\cdot R_{2}+\varepsilon\}\rangle\\ \stackrel{{\scriptstyle\mbox{(Arden)}}}{{\Rightarrow}}&\langle\{\},\{R_{1}\approx x^{*}\cdot(y\cdot R_{2}+\varepsilon),R_{2}\approx y\cdot R_{1}+x\cdot R_{2}+\varepsilon\}\rangle\\ \stackrel{{\scriptstyle\mbox{(Subst)}}}{{\Rightarrow}}&(y\cdot(x^{*}\cdot(y\cdot R_{2}+\varepsilon))+x\cdot R_{2}+\varepsilon\simeq(y\cdot x^{*}\cdot y+x)\cdot R_{2}+y\cdot x^{*}\cdot\varepsilon+\varepsilon)\\ &\langle\{R_{1}\mapsto x^{*}\cdot(y\cdot R_{2}+\varepsilon)\},\{R_{2}\approx(y\cdot x^{*}\cdot y+x)\cdot R_{2}+y\cdot x^{*}\cdot\varepsilon+\varepsilon\}\rangle\\ \stackrel{{\scriptstyle\mbox{(Arden)}}}{{\Rightarrow}}&\langle\{R_{1}\mapsto x^{*}\cdot(y\cdot R_{2}+\varepsilon)\},\{R_{2}\approx(y\cdot x^{*}\cdot y+x)^{*}\cdot(y\cdot x^{*}\cdot\varepsilon+\varepsilon)\}\rangle\\ \stackrel{{\scriptstyle\mbox{(Subst)}}}{{\Rightarrow}}&\langle\{R_{1}\mapsto x^{*}\cdot(y\cdot(y\cdot x^{*}\cdot y+x)^{*}\cdot(y\cdot x^{*}\cdot\varepsilon+\varepsilon)+\varepsilon),\\ &R_{2}\mapsto(y\cdot x^{*}\cdot y+x)^{*}\cdot(y\cdot x^{*}\cdot\varepsilon+\varepsilon)\},\{\}\rangle\end{array}$$

$$\begin{array}[]{ll}&\langle\{\},\{R_{1}\approx x\cdot R_{1}+y\cdot R_{2}+\varepsilon,R_{2}\approx y\cdot R_{1}+x\cdot R_{2}+\varepsilon\}\rangle\\ \Rightarrow^{*}&\langle\{R_{1}\mapsto(x+y\cdot x^{*}\cdot y)^{*}+y\cdot x^{*}+\varepsilon,\\ &R_{2}\mapsto x^{*}\cdot(y\cdot((x+y\cdot x^{*}\cdot y)^{*}+y\cdot x^{*}+\varepsilon)+\varepsilon)\},\{\}\rangle\end{array}$$

$$\begin{array}[]{c}u,v\ ::=\ \textsc{Eps}\mid\textsc{Sym}\ x\mid\textsc{Seq}\ v\ v\mid\textsc{Inl}~{}v\mid\textsc{Inr}~{}v\mid vs\mid\textsc{Fold}\ v\ \ \ \ vs\ ::=\ []\mid v:vs\end{array}$$

$$\begin{array}[]{c}{\cal E}\,\vdash\,[]:r^{*}\ \ \ \ {\cal E}\,\vdash\,\textsc{Eps}:\epsilon\ \ \ \ {\begin{array}[]{c}x\in\Sigma\\ \hline\cr{\cal E}\,\vdash\,\textsc{Sym}\ x:x\end{array}}\\ {\begin{array}[]{c}{\cal E}\,\vdash\,v:r\quad{\cal E}\,\vdash\,vs:r^{*}\\ \hline\cr{\cal E}\,\vdash\,(v:vs):r^{*}\end{array}}\ \ \ \ {\begin{array}[]{c}{\cal E}\,\vdash\,v_{1}:r_{1}\quad{\cal E}\,\vdash\,v_{2}:r_{2}\\ \hline\cr{\cal E}\,\vdash\,\textsc{Seq}\ v_{1}\ v_{2}:r_{1}\cdot r_{2}\end{array}}\\ {\begin{array}[]{c}{\cal E}\,\vdash\,v_{1}:r_{1}\\ \hline\cr{\cal E}\,\vdash\,\textsc{Inl}~{}v_{1}:r_{1}+r_{2}\end{array}}\ \ \ \ {\begin{array}[]{c}{\cal E}\,\vdash\,v_{2}:r_{2}\\ \hline\cr{\cal E}\,\vdash\,\textsc{Inr}~{}v_{2}:r_{1}+r_{2}\end{array}}\ \ \ \ {\begin{array}[]{c}{\cal E}\,\vdash\,v:\alpha\ \ \ \ R\approx\alpha\in{\cal E}\\ \hline\cr{\cal E}\,\vdash\,\textsc{Fold}\ v:R\end{array}}\end{array}$$

$$\mathcal{E}⊢\text{textsc}Fold(\text{textsc}Inl(\text{textsc}Seq(\text{textsc}Symx)(\text{textsc}Fold(\text{textsc}Inr(\text{textsc}Symy))))):R$$

$$\begin{array}[]{lllllllllllllllll}|\textsc{Eps}|&=&\epsilon&|\textsc{Sym}\ x|&=&x&|\textsc{Inl}~{}v|&=&|v|{}{}{}{}&|v:vs|&=&|v|\cdot|vs|\\ |[]|&=&\epsilon{}{}{}&|\textsc{Seq}\ v_{1}\ v_{2}|&=&|v_{1}|\cdot|v_{2}|{}{}{}&|\textsc{Inr}~{}v|&=&|v|&|\textsc{Fold}\ v|&=&|v|\end{array}$$

$$\text{textsc}Fold(\text{textsc}Inl(\text{textsc}Seq(\text{textsc}Symx)(\text{textsc}Fold(\text{textsc}Inr(\text{textsc}Symy)))))$$

$$\begin{array}[]{lrl}c&::=&v\mid k\mid\lambda v.c\mid c\ c\mid\mbox{\sf rec}\ x.c\mid\mbox{\sf case}\ c\ \mbox{\sf of}\ [pat_{1}\Rightarrow c_{1},\ldots,pat_{n}\Rightarrow c_{n}]\\ pat&::=&y\mid k\ pat_{1}\ ...pat_{arity(k)}\end{array}$$

$$\begin{array}[]{llll}f_{A}=&\mbox{\sf rec}\ f.\lambda x.&\mbox{\sf case}\ x\ \mbox{\sf of}\\ &&[\textsc{Inr}\ u\Rightarrow\textsc{Seq}\ []\ u,\\ &&\textsc{Inl}\ (\textsc{Seq}\ u\ (\textsc{Fold}\ v)\Rightarrow&\mbox{\sf case}\ f(v)\ \mbox{\sf of}\\ &&&[\textsc{Seq}\ us\ u_{2}\Rightarrow\textsc{Seq}\ (u:us)\ u_{2}]]\end{array}$$

$$\begin{array}[]{llll}f_{A}^{-1}=&\mbox{\sf rec}\ g.\lambda x.&\mbox{\sf case}\ x\ \mbox{\sf of}\\ &&[\textsc{Seq}\ []\ u\Rightarrow\textsc{Fold}\ (\textsc{Inr}\ u),\\ &&\textsc{Seq}\ (v:vs)\ u\Rightarrow\textsc{Fold}\ (\textsc{Inl}\ (\textsc{Seq}\ v\ (g\ (\textsc{Seq}\ vs\ u))))]\end{array}$$

$$\begin{array}[]{ll}f_{E_{1}}=\lambda v.&\mbox{\sf case}\ v\ \mbox{\sf of}\\ &[\textsc{Seq}\ u\ (\textsc{Inl}\ v)\Rightarrow\textsc{Inl}\ (\textsc{Seq}\ u\ v),\\ &\textsc{Seq}\ u\ (\textsc{Inr}\ v)\Rightarrow\textsc{Inr}\ (\textsc{Seq}\ u\ v)]\end{array}$$

$$\begin{array}[]{ll}f_{E_{1}}^{-1}=\lambda v.&\mbox{\sf case}\ v\ \mbox{\sf of}\\ &[\textsc{Inl}\ (\textsc{Seq}\ u\ v)\Rightarrow\textsc{Seq}\ (\textsc{Inl}\ u)\ v,\\ &\textsc{Inr}\ (\textsc{Seq}\ u\ v)\Rightarrow\textsc{Seq}\ (\textsc{Inr}\ u)\ v]\end{array}$$

$$\begin{array}[]{lcl}\delta\langle\rangle::=r\cdot\langle\rangle\mid\delta\langle\rangle+\delta\langle\rangle\mid\delta\langle\rangle+\alpha\mid\alpha+\delta\langle\rangle\end{array}$$

$$\begin{array}[]{ll}f_{r\cdot\langle\rangle}&=\begin{array}[]{ll}\lambda u.&\mbox{\sf case}\ u\ \mbox{\sf of}\\ &[\textsc{Seq}\ u\ (\textsc{Fold}\ v)\Rightarrow\textsc{Seq}\ u\ v]\end{array}\\ \\ f_{\delta\langle\rangle+\delta\langle\rangle}&=\begin{array}[]{ll}\lambda u.&\mbox{\sf case}\ u\ \mbox{\sf of}\\ &[\textsc{Inl}\ v\Rightarrow\textsc{Inl}\ (f_{\delta\langle\rangle}(v)),\\ &\textsc{Inr}\ v\Rightarrow\textsc{Inr}\ (f_{\delta\langle\rangle}(v))]\end{array}\end{array}$$

$$\begin{array}[]{ccc}\begin{array}[]{ll}f_{\delta\langle\rangle+\alpha}&=\begin{array}[]{ll}\lambda u.&\mbox{\sf case}\ u\ \mbox{\sf of}\\ &[\textsc{Inl}\ v\Rightarrow\textsc{Inl}\ (f_{\delta\langle\rangle}(v)),\\ &\textsc{Inr}\ v\Rightarrow\textsc{Inr}\ v]\end{array}\end{array}&&\begin{array}[]{ll}f_{\alpha+\delta\langle\rangle}&=\begin{array}[]{ll}\lambda u.&\mbox{\sf case}\ u\ \mbox{\sf of}\\ &[\textsc{Inl}\ v\Rightarrow\textsc{Inl}\ v,\\ &\textsc{Inr}\ v\Rightarrow\textsc{Inr}\ (f_{\delta\langle\rangle}(v))]\end{array}\end{array}\end{array}$$

$$\begin{array}[]{c}\mbox{(C-Arden)}\ \ \ \ {\begin{array}[]{c}R\not\in\alpha\\ \hline\cr\langle\psi,\textsc{Fold}\ v:R\approx s\cdot R+\alpha\uplus{\cal E}\rangle\Rightarrow\langle\psi,\textsc{Fold}\ f_{A}(v):R\approx s^{*}\cdot\alpha\uplus{\cal E}\rangle\end{array}}\\ \\ \mbox{(C-Subst)}\ \ \ \ {\begin{array}[]{c}R\not\in\alpha\\ \begin{array}[]{lcl}\psi^{\prime}&=&\{v:R\mapsto\alpha\}\\ &\cup&\{v:R^{\prime}\mapsto\alpha^{\prime}\mid v:R^{\prime}\mapsto\alpha^{\prime}\in\psi\wedge R\not\in\alpha^{\prime}\}\\ &\cup&\{\textsc{Fold}\ f(f_{\delta\langle\rangle}(v)):R^{\prime}\mapsto\alpha^{\prime\prime}\mid\textsc{Fold}\ v:R^{\prime}\mapsto\delta^{\prime}\langle R\rangle\in\psi\wedge\\ &&\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ R\not\in\delta^{\prime}\langle\alpha\rangle\wedge\\ &&\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \delta^{\prime}\langle\alpha\rangle\stackrel{{\scriptstyle f}}{{\simeq}}\alpha^{\prime\prime}\}\end{array}\\ \begin{array}[]{lcl}{\cal E}^{\prime}&=&\{v:R^{\prime}\approx\alpha^{\prime}\mid v:R^{\prime}\approx\alpha^{\prime}\in{\cal E}\wedge R\not\in\alpha^{\prime}\}\\ &\cup&\{\textsc{Fold}\ f(f_{\delta\langle\rangle}(v)):R^{\prime}\approx\alpha^{\prime\prime}\mid\textsc{Fold}\ v:R^{\prime}\approx\delta^{\prime}\langle R\rangle\in{\cal E}\wedge\\ &&\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ R\not\in\delta^{\prime}\langle\alpha\rangle\wedge\\ &&\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \delta^{\prime}\langle\alpha\rangle\stackrel{{\scriptstyle f}}{{\simeq}}\alpha^{\prime\prime}\}\end{array}\\ \hline\cr\langle\psi,\textsc{Fold}\ v:R\approx\alpha\uplus{\cal E}\rangle\Rightarrow\langle\psi^{\prime},{\cal E}^{\prime}\rangle\end{array}}\end{array}$$

$$\begin{array}[]{ll}d_{x}(\phi)=\phi&d_{x}(\varepsilon)=\phi\\ \\ d_{x}(y)=\left\{\begin{array}[]{ll}\varepsilon&\mbox{if $x=y$}\\ \phi&\mbox{otherwise}\end{array}\right.&d_{x}(r+s)=d_{x}(r)+d_{x}(s)\\ \\ d_{x}(r\cdot s)=\left\{\begin{array}[]{ll}d_{x}(r)\cdot s&\mbox{if $\varepsilon\not\in{\cal L}(r)$ \ \ \mbox{}}\\ d_{x}(r)\cdot s+d_{x}(s)&\mbox{otherwise}\end{array}\right.&d_{x}(r^{*})=d_{x}(r)\cdot r^{*}\end{array}$$

$$d_x((x+y{)}^{\ast })=d_x(x+y)\cdot (x+y{)}^{\ast }=(d_x(x)+d_x(y))\cdot (x+y{)}^{\ast }=(\epsilon +\varphi )\cdot (x+y{)}^{\ast }$$

$$\begin{array}[]{c}r\equiv x_{1}\cdot d_{x_{1}}(r)+\cdots+x_{n}\cdot d_{x_{n}}(r)\ \mbox{($+\varepsilon$ if $r$ nullable)}\end{array}$$

$$\begin{array}[]{c}{\cal L}(r)-{\cal L}(s)\equiv\sum_{x\in\Sigma}x\cdot({\cal L}(\mathit{cnf}(d_{x}(r)))-{\cal L}(\mathit{cnf}(d_{x}(s))))+T\end{array}$$

$$\begin{array}[]{c}(x\cdot R)-(x\cdot S)\equiv x\cdot(R-S)\\ (x\cdot R)-(y\cdot S)\equiv x\cdot R\ \ \ \ \ \ \mbox{where $x\not=y$}\\ R-\phi\equiv R\\ (R+S)-T\equiv(R-T)+(S-T)\\ R-(S+T)\equiv(R-S)-T\end{array}$$

$$\begin{array}[]{cr}R_{r^{\prime},s^{\prime}}\approx\sum_{x\in\Sigma}x\cdot R_{\mathit{cnf}(d_{x}(r^{\prime})),\mathit{cnf}(d_{x}(s^{\prime}))}+t&\ \ \ \ (E1)\end{array}$$

$$\begin{array}[]{cr}{\cal L}(r)-{\cal L}(s)\equiv\sum_{x\in\Sigma}x\cdot({\cal L}(\mathit{cnf}(d_{x}(r)))-{\cal L}(\mathit{cnf}(d_{x}(s))))+T&\ \ \ \ (E2)\end{array}$$

Full text

Keywords: regular equations and expressions, parse trees, ambiguity, subtraction, shuffling

1 Introduction

The conversion of a regular expression (RE) into a deterministic finite automaton (DFA) is a well-studied topic. Various methods and optimized implementations exist. The opposite direction has received less attention. In the literature, there are two well-known methods to translate DFAs to REs, namely, state elimination [6] and solving of equations via Arden’s Lemma [3].

The solving method works by algebraic manipulation of equations. Identity laws are applied to change the syntactic form of an equation’s right-hand side such that Arden’s Lemma is applicable. Thus, the set of equations is reduced and in a finite number of steps a solution can be obtained. State elimination has a more operational flavor and reduces states by introducing transitions labeled with regular expressions. The state elimination method appears to be better studied in the literature. For example, see the works [12, 15, 1] that discuss heuristics to obtain short regular expressions.

In this paper, we revisit solving of regular equations via Arden’s Lemma. Specifically, we make the following contributions:

• •

We give a concise algorithmic description of solving of regular equations where we give a precise specification of all algebraic laws applied (Section 3).

• •

We give a computational interpretation of solving by means of coercions that transform parses tree of regular equations into parse trees of solutions. We can identify simple criteria on the shape of regular equations under which resulting solutions are unambiguous (Section 4).

• •

We apply our results to the following scenarios:

• –

We show that regular expressions obtained from DFAs via Brzozowski’s algebraic method are always unambiguous (Section 5).

• –

We provide direct, algebraic methods to obtain the subtraction and shuffle among two regular expressions (Section 6). Correctness follows via some simple coalgebraic reasoning.

We conclude in Section 7 where we also discuss related works.

The appendix contains further details such as proofs and a parser for regular equations. We also report on an implementation for solving of regular equations in Haskell [14] including benchmark results.

2 Preliminaries

Let $\Sigma$ be a finite set of symbols (literals) with $x$, $y$, and $z$ ranging over $\Sigma$. We write $\Sigma^{*}$ for the set of finite words over $\Sigma$, $\varepsilon$ for the empty word, and $v\cdot w$ for the concatenation of words $v$ and $w$. A language is a subset of $\Sigma^{*}$.

Definition 1** (Regular Languages)**

The set ${\cal R}$ of regular languages is defined inductively over some alphabet $\Sigma$ by

[TABLE]

Each regular language is a subset of $\Sigma^{*}$ where we assume that $R\cdot S$ denotes $\{v\cdot w\mid v\in R\wedge w\in S\}$, $R+S$ denotes $R\cup S$ and $R^{*}$ denotes $\{w_{1}\cdot\cdots\cdot w_{n}\mid n\geq 0\wedge\forall i\in\{1,\ldots,n\}.~{}w_{i}\in R\}$.

We write $R\equiv S$ if $R$ and $S$ denote the same set of words.

We often omit parentheses by assuming that ∗ binds tighter than $\cdot$ and $\cdot$ binds tighter than $+$. As it is common, we assume that $+$ and $\cdot$ are right-associative. That is, $R+S+T$ stands for $(R+(S+T))$ and $R+S+R\cdot S\cdot T$ stands for $R+(S+(R\cdot(S\cdot T)))$.

Definition 2** (Regular Expressions)**

The set RE of regular expressions is defined inductively over some alphabet $\Sigma$ by

[TABLE]

Definition 3** (From Regular Expressions to Languages)**

*The meaning function ${\cal L}$ maps a regular expression to a language. It is defined inductively as follows:

${\cal L}(\phi)=\{\}$. ${\cal L}(\varepsilon)=\{\varepsilon\}$. ${\cal L}(x)=\{x\}$. ${\cal L}(r+s)=({\cal L}(r)+{\cal L}(s))$. ${\cal L}(r\cdot s)=({\cal L}(r)\cdot{\cal L}(s))$. ${\cal L}(r^{*})=({\cal L}(r)^{*})$.*

We say that regular expressions $r$ and $s$ are equivalent, $r\equiv s$, if ${\cal L}(r)={\cal L}(s)$.

Definition 4** (Nullability)**

A regular expression $r$ is nullable if $\varepsilon\in{\cal L}(r)$.

Lemma 5** (Arden’s Lemma [3])**

Let $R$, $S$, $T$ be regular languages where $\varepsilon\not\in S$. Then, we have that $R\equiv S\cdot R+T$ iff $R\equiv S^{*}\cdot T$.

The direction from right to left holds in general. For the direction from left to right, pre-condition $\varepsilon\not\in S$ is required. For our purposes, we only require the direction from right to left.

3 Solving Regular Equations

Definition 6** (Regular Equations)**

We write $E$ to denote a regular equation of the form $R\approx\alpha$ where the form of the right-hand side $\alpha$ is as follows.

[TABLE]

In addition to $\alpha$, we will sometimes use $\beta$ to denote right-hand sides.

We will treat regular language symbols $R$ like variables. We write $r,s,t$ to denote expressions that do not refer to symbols $R$.

We write $R\in\alpha$ to denote that $R$ appears in $\alpha$. Otherwise, we write $R\not\in\alpha$.

We write ${\cal E}$ to denote a set $\{R_{1}\approx\alpha_{1},\dots,R_{n}\approx\alpha_{n}\}$ of regular equations. We assume that (1) left-hand sides are distinct by requiring that $R_{i}\not=R_{j}$ for $i\not=j$, and (2) regular language symbols on right-hand sides appear on some left-hand side by requiring that for any $R\in\alpha_{j}$ for some $j$ there exists $i$ such that $R=R_{i}$. We define ${\it dom}({\cal E})=\{R_{1},\dots,R_{n}\}$.

Regular languages are closed under union and concatenation, hence, we can guarantee the existence of solutions of these variables in terms of regular expressions.

Definition 7** (Solutions)**

We write $\{R_{1}\mapsto\gamma_{1},\dots,R_{n}\mapsto\gamma_{n}\}$ to denote an idempotent substitution mapping $R_{i}$ to $\gamma_{i}$ where $\gamma_{i}$ denote expressions that may consist of a mix of regular expressions and regular language symbols $R$.

Let $\psi=\{R_{1}\mapsto\gamma_{1},\dots,R_{n}\mapsto\gamma_{n}\}$ be a substitution and $\gamma$ some expression. Then, $\psi(\gamma)$ is derived from $\gamma$ by replacing each occurrence of $R_{i}$ by $\gamma_{i}$.

Let ${\cal E}=\{R_{1}\approx\alpha_{1},\dots,R_{n}\approx\alpha_{n}\}$. Then, we say that $\psi$ is a solution for ${\cal E}$ if $\psi(R_{i})$, $\psi(\alpha_{i})$ are regular expressions where $\psi(R_{i})\equiv\psi(\alpha_{i})$ for $i=1,\dots,n$.

We solve equations as follows. We apply Arden’s Lemma on equations that are of a certain (normal) form $R\approx s\cdot R+\alpha$ where $R\not\in\alpha$. Thus, we can eliminate this equation by substituting $R$ with $s^{*}\cdot\alpha$ on all right-hand sides. In case of $R\approx\alpha$ where $R\not\in\alpha$ we can substitute directly. We repeat this process until all equations are solved. Below, we formalize the technical details.

Definition 8** (Normal Form)**

We say that $R\approx\alpha$ is in normal form iff either (1) $R\not\in\alpha$, or (2) $\alpha=s_{1}\cdot R_{1}+\dots+s_{n}\cdot R_{n}+t$ such that $R=R_{1}$ and $R_{i}\not=R_{j}$ for $i\not=j$.

Recall that $t$ does not refer to symbols $R$. Every equation can be brought into normal form by applying the following algebraic equivalence laws.

Definition 9** (Equivalence)**

We say two expressions $\gamma_{1}$ and $\gamma_{2}$ are equivalent, written $\gamma_{1}\simeq\gamma_{2}$, if one can be transformed into the other by application of the following rules.

[TABLE]

Rule (E6) assumes expressions with a hole.

[TABLE]

We write $\beta[\gamma]$ to denote the expression where the hole $[]$ is replaced by $\gamma$.

We formulate solving of equations in terms of a rewrite system among a configuration $\langle\psi,{\cal E}\rangle$ where substitution $\psi$ represents the so far accumulated solution and ${\cal E}$ the yet to be solved set of equations.

Definition 10** (Solving)**

Let ${\cal E}=\{R_{1}\approx\alpha_{1},\dots,R_{n}\approx\alpha_{n}\}$. Then, we write $R\approx\alpha\uplus{\cal E}^{\prime}$ to denote the set that equals to ${\cal E}$ where $R\approx\alpha$ refers to some equation in ${\cal E}$ and ${\cal E}^{\prime}$ refers to the set of remaining equations.

[TABLE]

We write $\Rightarrow^{*}$ to denote the transitive and reflexive closure of solving steps $\Rightarrow$.

Initially, all equations are in normal form. Rule (Arden) applies Arden’s Lemma on some equation in normal form. Rule (Subst) removes an equation $R\approx\alpha$ where $R\not\in\alpha$. The substitution $\{R\mapsto\alpha\}$ implied by the equation is applied on all remaining right-hand sides. To retain the normal form property of equations, we normalize right-hand sides by applying rules (E1-7). The details of normalization are described in the proof of the upcoming statement. We then extend the solution accumulated so far by adding $\{R\mapsto\alpha\}$. As we assume substitutions are idempotent, $\{R\mapsto\alpha\}$ is applied on all expressions in the codomain of $\psi$.

Theorem 11** (Regular Equation Solutions)**

Let ${\cal E}$ be a set of regular equations in normal form. Then, $\langle\{\},{\cal E}\rangle\Rightarrow^{*}\langle\psi,\{\}\rangle$ for some substitution $\psi$ where $\psi$ is a solution for ${\cal E}$.

Proof

We first observe that rule (Arden) and (Subst) maintain the normal form property for equations. This immediately applies to rule (Arden).

Consider rule (Subst). Consider $R^{\prime}\approx\alpha^{\prime}$. We need to show that $\{R\mapsto\alpha\}(\alpha^{\prime})$ can be transformed to some form $\alpha^{\prime\prime}$ such that $R^{\prime}\approx\alpha^{\prime\prime}$ is in normal form.

If $R\not\in\alpha^{\prime}$ nothing needs to be done as we assume that equations are initially in normal form.

Otherwise, we consider the possible shapes of $\alpha$ and $\alpha^{\prime}$. W.l.o.g. $\alpha^{\prime}$ is of the form $t_{1}\cdot R_{1}+\dots+r\cdot R+\dots+t_{n}\cdot R_{n}+t^{\prime}$ and $\alpha$ is of the form $s_{1}\cdot T_{1}+\dots+s_{k}\cdot T_{k}+t^{\prime\prime}$. We rely here on rule (E3) that allows us to drop parentheses among summands.

$R$ is replaced by $\alpha$ in $\alpha^{\prime}$. This generates the subterm $r\cdot(s_{1}\cdot T_{1}+\dots+s_{k}\cdot T_{k}+t^{\prime\prime})$. On this subterm, we exhaustively apply rules (E1-2). This yields the subterm $(r\cdot s_{1})\cdot T_{1}+\dots+(r\cdot s_{k})\cdot T_{k}+t^{\prime\prime}$.

This subterm is one of the sums in the term obtained from $\{R\mapsto\alpha\}(\alpha^{\prime})$. Via rules (E6-7) the above transformation steps can be applied on the entire term $\{R\mapsto\alpha\}(\alpha^{\prime})$. Hence, this term can be brought into the form $r_{1}\cdot S_{1}+\dots+r_{m}\cdot S_{m}+t$. Subterm $t$ equals $t^{\prime}+t^{\prime\prime}$ and subterms $r_{i}\cdot S_{i}$ refer to one of the subterms $t_{j}\cdot R_{j}$ or $(r\cdot s_{l})\cdot T_{l}$.

We are not done yet because subterms $r_{i}\cdot S_{i}$ may contain duplicate symbols. That is, $S_{i}=S_{j}$ for $i\not=j$. We apply rule (E4) in combination with rule (E3) and (E4) to combine subterms with the same symbol. Thus, we reach the form $r_{1}^{\prime}\cdot R_{1}^{\prime}+\dots+r_{o}^{\prime}\cdot R_{o}^{\prime}+t$ such that $R_{i}\not=R_{j}$ for $i\not=j$.

If $R^{\prime}\not=R^{\prime}_{i}$ for $i=1,\dots,o$ we are done. Otherwise, $R=R^{\prime}_{i}$ for some $i$. We apply again (E3) and (E5) to ensure that the component $s_{i}\cdot R_{i}$ appears first in the sum.

Next, we show that within a finite number of (Arden) and (Subst) rule applications we reach the configuration $\langle\psi,\{\}\rangle$. For this purpose, we define an ordering relation among configurations $\langle\psi,{\cal E}\rangle$.

For ${\cal E}=\{R_{1}\approx\alpha_{1},\dots,R_{n}\approx\alpha_{n}\}$ we define

[TABLE]

where $\{\!\{\dots\}\!\}$ denotes a multi-set and $S_{j}$ are the distinct occurrences of symbols appearing on some right-hand side $\alpha_{i}$. Recall that by construction $\{S_{1},\dots,S_{m}\}\subseteq\{R_{1},\dots,R_{n}\}$. See (2) in Definition 6. We define $\langle\psi,{\cal E}\rangle<\langle\psi^{\prime},{\cal E}^{\prime}\rangle$ iff either (a) $M\subsetneq M^{\prime}$ or (b) $M=M^{\prime}$ and the number of symbols in $N$ is strictly smaller than the number of symbols in $N^{\prime}$ where $\mathit{vars}({\cal E})=(M,N)$ and $\mathit{vars}({\cal E}^{\prime})=(M^{\prime},N^{\prime})$.

For sets ${\cal E}$ of regular equations as defined in Definition 6 this is a well-founded order. Each of the rules (Subst) and (Arden) yield a smaller configuration w.r.t this order. For rule (Subst) case (a) applies whereas for rule (Arden) case (b) applies. Configuration $\langle\psi,\{\}\rangle$ for some $\psi$ is the minimal element. Hence, in a finite number of rule applications we reach $\langle\psi,\{\}\rangle$.

Substitution $\psi$ must be a solution because (1) normalization steps are equivalence preserving and (2) based on Arden’s Lemma we have that every solution for $R\approx s^{*}\cdot\alpha$ is also a solution for $R\approx s\cdot R+\alpha$. ∎

Example 1

Consider ${\cal E}=\{R_{1}\approx x\cdot R_{1}+y\cdot R_{2}+\varepsilon,R_{2}\approx y\cdot R_{1}+x\cdot R_{2}+\varepsilon\}$. For convenience, we additionally make use of associativity of concatenation ($\cdot$).

[TABLE]

The formulation in Definition 10 leaves the exact order in which equations are solved unspecified. Semantically, this form of non-determinism has no impact on the solution obtained. However, the syntactic shape of solutions is sensitive to the order in which equations are solved.

Suppose we favor the second equation which then yields the following.

[TABLE]

where for convenience, we exploit the law $r\cdot\varepsilon\equiv r$.

4 Computational Interpretation

We characterize under which conditions solutions to regular equations are unambiguous. By unambiguous solutions we mean that the resulting expressions are unambiguous. An expression is ambiguous if there exists a word which can be matched in more than one way. That is, there must be two distinct parse trees which share the same underlying word [4].

We proceed by establishing the notion of a parse tree. Parse trees capture the word that has been matched and also record which parts of the regular expression have been matched. We follow [9] and view expressions as types and parse trees as values.

Definition 12** (Parse Trees)**

[TABLE]

The valid relations among parse trees and regular expressions are defined via a natural deduction style proof system.

[TABLE]

For expressions not referring to variables we write $\,\vdash\,v:r$ as a shorthand for $\{\}\,\vdash\,v:r$.

Parse tree values are built using data constructors. The constant constructor Eps represents the value belonging to the empty word regular expression. For letters, we use the unary constructor Sym to record the symbol. In case of choice ($+$), we use unary constructors Inl and Inr to indicate if either the left or right expression is part of the match. For repetition (Kleene star) we use Haskell style lists where we write $[v_{1},...,v_{n}]$ as a short-hand for the list $v_{1}:...:v_{n}:[]$. In addition to the earlier work [9], we introduce a Fold constructor and a proof rule to (un)fold a regular equation.

Example 2

Consider ${\cal E}=\{R\approx x\cdot R+y\}$. Then, we find that

[TABLE]

The equation is unfolded twice where we first match against the left part $x\cdot R$ and then against the right part $y$.

The relation established in Definition 12 among parse trees, expressions and equations is correct in the sense that (1) flattening of the parse tree yields a word in the language and (2) for each word there exists a parse tree.

Definition 13** (Flattening)**

We can flatten a parse tree to a word as follows:

[TABLE]

Proposition 14

Let ${\cal E}$ be a set of regular equations and $\psi$ a solution. Let $R\in{\it dom}({\cal E})$. (1) If $w\in{\cal L}(\psi(R))$ then ${\cal E}\,\vdash\,v:R$ for some parse tree $v$ such that $|v|=w$. (2) If ${\cal E}\,\vdash\,v:R$ then $|v|\in{\cal L}(\psi(R))$.

The above result follows by providing a parser for regular equations. For (1) it suffices to compute a parse tree if one exists. For (2) we need to enumerate all possible parse trees. This is possible by extending our prior work [20, 21] to the regular equation setting. Details are given in Appendix 0.B.

Parse trees may not be unique because some equations/expressions may be ambiguous in the sense that a word can be matched in more than one way. This means that there are two distinct parse trees representing the same word. We extend the notion of ambiguous expressions [4] to the setting of regular equations.

Definition 15** (Ambiguity)**

Let ${\cal E}$ be a set of regular equations and $r$ be an expression. We say $r$ is ambiguous w.r.t. ${\cal E}$ iff there exist two distinct parse trees $v_{1}$ and $v_{2}$ such that ${\cal E}\,\vdash\,v_{1}:r$ and ${\cal E}\,\vdash\,v_{2}:r$ where $|v_{1}|=|v_{2}|$.

Example 3

$[\textsc{Inl}\ (\textsc{Seq}\ (\textsc{Sym}\ x)\ (\textsc{Sym}\ y))]$ and $[\textsc{Inr}\ (\textsc{Inl}\ (\textsc{Sym}\ x)),\textsc{Inr}\ (\textsc{Inr}\ (\textsc{Sym}\ y))]$ are two distinct parse trees for expression $(x\cdot y+x+y)^{*}$ (where ${\cal E}=\{\}$) and word $x\cdot y$.

On the other hand, the equation from Example 2 is unambiguous due to the following result.

Definition 16** (Non-Overlapping Equations)**

We say an equation $E$ is non-overlapping if $E$ is of the following form $R\approx x_{1}\cdot R_{1}+\dots+x_{n}\cdot R_{n}+t$ where $x_{i}\not=x_{j}$ for $i\not=j$ and either $t=\varepsilon$ or $t=\phi$.

Equation $R\approx x\cdot R+y$ does not exactly match the above definition. However, we can transform ${\cal E}=\{R\approx x\cdot R+y\}$ into the equivalent set ${\cal E}^{\prime}=\{R\approx x\cdot R+y\cdot S,S\approx\varepsilon\}$ that satisfies the non-overlapping condition.

Proposition 17** (Unambiguous Regular Equations)**

Let ${\cal E}$ be a set of non-overlapping equations where $R\in{\it dom}({\cal E})$. Then, we have that $R$ is unambiguous.

Ultimately, we are interested in obtaining a parse tree for the resulting solutions rather than the original set of equations. For instance, the solution for Example 2 is $x^{*}\cdot y$. Hence, we wish to transform the parse tree

[TABLE]

into a parse tree for $x^{*}\cdot y$. Furthermore, we wish to guarantee that if equations are unambiguous so are solutions. We achieve both results by explaining each solving step among regular equations in terms of a (bijective) transformation among the associated parse trees.

We refer to these transformations as coercions as they operate on parse trees. We assume the following term language to represent coercions.

Definition 18** (Coercion Terms)**

Coercion terms $c$ and patterns $pat$ are inductively defined by

[TABLE]

where pattern variables $y$ range overs a denumerable set of variables disjoint from $\Sigma$ and constructors $k$ are taken from the set ${\cal K}=\{\textsc{Eps},\textsc{Seq},\textsc{Inl},\textsc{Inr},\textsc{Fold}\}$. The function $arity(k)$ defines the arity of constructor $k$. Patterns are linear (i.e., all pattern variables are distinct) and we write $\lambda pat.c$ as a shorthand for $\lambda v.\mbox{\sf case}\ v\ \mbox{\sf of}\ [pat\Rightarrow c]$.

We give meaning to coercions in terms of a standard big-step operational semantics. Given a coercion (function) $f$ and some (parse tree) value $u$, we write $f\ u\Downarrow v$ to denote the evaluation of $f$ for input $u$ with resulting (parse tree) value $v$. We often write $f(u)$ as a shorthand for $v$. We say a coercion $f$ is bijective if there exists a coercion $g$ such that for every $u,v$ where $f\ u\Downarrow v$ we have that $g\ v\Downarrow u$. We refer to $g$ as the inverse of $f$.

We examine the three elementary solving steps, Arden, normalization and substitution. For each solving step we introduce an appropriate (bijective) coercion to carry out the transformation among parse trees.

Lemma 19** (Arden Coercion)**

Let ${\cal E}$ be a set of regular equations where $R\approx s\cdot R+\alpha\in{\cal E}$ such that $R\not\in\alpha$ and ${\cal E}\,\vdash\,\textsc{Fold}\ v:R$ for some parse tree $v$. Then, there exists a bijective coercion $f_{A}$ such that ${\cal E}\,\vdash\,f_{A}(v):s^{*}\cdot\alpha$ where $|v|=|f_{A}(v)|$.

Proof

By assumption ${\cal E}\,\vdash\,v:s\cdot R+\alpha$. The following function $f_{A}$ satisfies ${\cal E}\,\vdash\,f_{A}(v):s^{*}\cdot\alpha$ where $|v|=|f_{A}(v)|$. For convenience we use symbols $v$ and $u$ as pattern variables.

[TABLE]

Function $f_{A}$ is bijective. Here is the inverse function.

[TABLE]

∎

Lemma 20** (Normalization Coercion)**

Let $\gamma_{1},\gamma_{2}$ be two expressions such that $\gamma_{1}\simeq\gamma_{2}$ and ${\cal E}\,\vdash\,v:\gamma_{1}$ for some set ${\cal E}$ and parse tree $v$. Then, there exists a bijective coercion f such that ${\cal E}\,\vdash\,f(v):\gamma_{2}$ where $|v|=|f(v)|$.

Proof

For each of the equivalence proof rules, we introduce an appropriate (bijective) coercion. For rule (E1) we employ

[TABLE]

where the inverse function is as follows.

[TABLE]

Coercions for rules (E2-5) can be defined similarly. Rule (E7) corresponds to function composition and rule (E6) requires to navigate to the respective hole position. Details are omitted for brevity. ∎

We will write $\gamma_{1}\stackrel{{\scriptstyle f}}{{\simeq}}\gamma_{2}$ to denote the coercion $f$ to carry out the transformation of $\gamma_{1}$’s parse tree into $\gamma_{2}$’s parse tree.

What remains is to define coercions to carry out substitution where we replace subterms.

Definition 21** (Substitution Context)**

We define expressions with multiple holes to characterize substitution of a subterm by another.

[TABLE]

We refer to $\delta\langle\rangle$ as a substitution context.

We define a set of functions indexed by the shape of a substitution context. For $\delta\langle\rangle$ we transform $\alpha\langle R\rangle$’s parse tree into $\alpha\langle\alpha\rangle$’s parse tree assuming the equation $R\approx\alpha$.

[TABLE]

[TABLE]

Functions $f_{\delta\langle\rangle}$ navigate to the to-be-replaced subterm and drop the Fold constructor if necessary. There are inverse functions which we omit for brevity.

Lemma 22** (Substitution Coercion)**

Let ${\cal E}$ be a set of equations, $R\approx\alpha\in{\cal E}$ such that ${\cal E}\,\vdash\,v:\delta\langle R\rangle$ for some parse tree $v$ and substitution context $\delta\langle\rangle$. Then, we find that ${\cal E}\,\vdash\,f_{\delta\langle\rangle}(v):\delta\langle\alpha\rangle$ where $|v|=|f_{\delta\langle\rangle}(v)|$.

Proof

Follows by induction over the structure of $\delta\langle\rangle$. ∎

We integrate the elementary coercions into the solving process. For this purpose, we assume that regular equations and substitutions are annotated with parse trees. For example, we write $\{v_{1}:R_{1}\approx\alpha_{1},\dots,v_{n}:R_{n}\approx\alpha_{n}\}$ to denote a set of regular equations ${\cal E}$ where for each $i$ we have that ${\cal E}\,\vdash\,v_{i}:R_{i}$. Similarly, we write $\{v_{1}:R_{1}\mapsto\gamma_{1},\dots,v_{n}:R_{n}\mapsto\gamma_{n}\}$ for substitutions.

Definition 23** (Coercive Solver)**

[TABLE]

In the coercive Arden rule, we apply the Arden coercion introduced in Lemma 19. During substitution we uniformly normalize right-hand sides of equations and the codomains of substitutions. Side condition $R\not\in\delta^{\prime}\langle\alpha\rangle$ guarantees that all occurrences of $R$ are replaced. Parse trees are transformed by first applying the substitution coercion followed by the normalization coercion. Thus, we can transform parse trees of regular equations into parse trees of solutions.

Proposition 24** (Coercive Solving)**

Let ${\cal E}=\{v_{1}:R_{1}\approx\alpha_{1},\dots,v_{n}:R_{n}\approx\alpha_{n}\}$ be a parse tree annotated set of regular equations in normal form where ${\cal E}\,\vdash\,v_{i}:R_{i}$ for $i=1,\dots,n$. Then, $\langle\{\},{\cal E}\rangle\Rightarrow^{*}\langle\psi,\{\}\rangle$ for some substitution $\psi$ where $\psi=\{u_{1}:R_{1}\mapsto s_{1},\dots,u_{n}:R_{n}\mapsto s_{n}\}$ such that $\,\vdash\,u_{i}:s_{i}$ and $|u_{i}|=|v_{i}|$ for $i=1,\dots,n$.

Proof

Follows immediately from Lemmas 19, 20 and 22. ∎

Theorem 25** (Unambiguous Solutions)**

Let ${\cal E}$ be a set of non-overlapping equations where $\langle\{\},{\cal E}\rangle\Rightarrow^{*}\langle\psi,\{\}\rangle$ for some substitution $\psi$. Then, for each $R\in{\it dom}({\cal E})$ we find that $\psi(R)$ is unambiguous.

Proof

Follows from Propositions 17 and 24 and the fact that coercions are bijective. ∎

5 Brzozowski’s algebraic method

We revisit Brzozowski’s algebraic method [5] to transform an automaton into a regular expression. Based on our results we can show that resulting regular expressions are always unambiguous.

Definition 26** (Deterministic Finite Automata (DFA))**

A deterministic finite automaton (DFA) is a 5-tuple $M=(Q,\Sigma,\delta,q_{0},F)$ consisting of a a finite set $Q$ of states, a finite set $\Sigma$ of symbols, a transition function $\delta:Q\times\Sigma\rightarrow Q$, an initial state $q_{0}\in Q$, and a set $F$ of accepting states. We say $M$ accepts word $w=x_{1}\dots x_{n}$ if there exists a sequence of states $p_{1},\dots,p_{n+1}$ such that $p_{i+1}=\delta(p_{i},x_{n})$ for $i=1,\dots,n$, $p_{1}=q_{0}$ and $p_{n+1}\in F$.

Brzozowski turns a DFA into an equivalent set of (characteristic) regular equations.

Definition 27** (Characteristic Equations)**

Let $M=(Q,\Sigma,\delta,q_{0},F)$ be a DFA. We define ${\cal E}_{M}=\{R_{q}\approx\sum_{x\in\Sigma}x\cdot R_{\delta(q,x)}+f(q)\mid q\in Q\}$ where $f(q)=\varepsilon$ if $q\in F$. Otherwise, $f(q)=\phi$. We refer to ${\cal E}_{M}$ as the characteristic equations obtained from $M$.

He suggests solving these equations via Arden’s Lemma but the exact details (e.g. normalization) are not specified. Assuming we use the solving method specified in Definition 10 we can conclude the following. By construction, characteristic equations are non-overlapping. From Theorem 25 we can derive the following result.

Corollary 1

Solutions obtained from characteristic equations are unambiguous.

Instead of a DFA we can also turn a non-deterministic automaton (NFA) into an equivalent regular expression. Each $\varepsilon$ transitions is represented by the component $\varepsilon\cdot R$. For two non-deterministic transitions via symbol $x$ to follow states $R_{1}$ and $R_{2}$, we generate the component $x\cdot R_{1}+x\cdot R_{2}$. Resulting characteristic equations will be overlapping in general. Hence, we can no longer guarantee unambiguity.

6 Subtraction and Shuffle

We introduce direct methods to subtract and shuffle regular expressions. Instead of turning the regular expressions into a DFA and carrying out the operation at the level of DFAs, we generate an appropriate set of equations by employing Brzozowski derivatives. Solving the equations yields then the desired result. In essence, our method based on solutions resulting from derivative-based equations is isomorphic to building a derivative-based DFA from expressions, applying the product automaton construction among DFAs and then turn the resulting DFA into an expression via Brzozowski’s algebraic method.

For subtraction, equations generated are non-overlapping. Hence, resulting expressions are also unambiguous. First, we recall the essential of derivatives before discussing each operation including some optimizations.

6.1 Brzozowski’s Derivatives

The derivative of a regular expression $r$ with respect to some symbol $x$, written $d_{x}(r)$, is a regular expression for the left quotient of ${\cal L}(r)$ with respect to $x$. That is, ${\cal L}(d_{x}(r))=\{w\in\Sigma^{*}\mid x\cdot w\in{\cal L}(r)\}$. A derivative $d_{x}(r)$ can be computed by recursion over the structure of the regular expression $r$.

Definition 28** (Brzozowski Derivatives [5])**

[TABLE]

Example 4

The derivative of $(x+y)^{*}$ with respect to symbol $x$ is $(\varepsilon+\phi)\cdot(x+y)^{*}$. The calculation steps are as follows:

[TABLE]

Theorem 29** (Expansion [5])**

Every regular expression $r$ can be represented as the sum of its derivatives with respect to all symbols. If $\Sigma=\{x_{1},\dots,x_{n}\}$, then

[TABLE]

Definition 30** (Descendants and Similarity)**

A descendant of $r$ is either $r$ itself or the derivative of a descendant. We say $r$ and $s$ are similar, written $r\sim s$, if one can be transformed into the other by finitely many applications of the rewrite rules (Idempotency) $r+r\sim r$, (Commutativity) $r+s\sim s+r$, (Associativity) $r+(s+t)\sim(r+s)+t$, (Elim1) $\varepsilon\cdot r\sim r$, (Elim2) $\phi\cdot r\sim\phi$, (Elim3) $\phi+r\sim r$, and (Elim4) $r+\phi\sim r$.

Lemma 31

Similarity is an equivalence relation that respects regular expression equivalence: $r\sim s$ implies $r\equiv s$.

Theorem 32** (Finiteness [5])**

The elements of the set of descendants of a regular expression belong to finitely many similarity equivalence classes.

Similarity rules (Idempotency), (Commutativity), and (Associativity) suffice to achieve finiteness. Elimination rules are added to obtain a compact canonical representative for equivalence class of similar regular expressions. The canonical form is obtained by systematic application of the similarity rules in Definition 30. We enforce right-associativity of concatenated expressions, sort alternative expressions according to their size and their first symbol, and concatenations lexicographically, assuming an arbitrary total order on $\Sigma$. We further remove duplicates and apply elimination rules exhaustively (the details are standard [10]).

Definition 33** (Canonical Representatives)**

For a regular expression $r$, we write $\mathit{cnf}(r)$ to denote the canonical representative among all expressions similar to $r$. We write ${\cal D}(r)$ for the set of canonical representatives of the finitely many dissimilar descendants of $r$.

Example 5

We find that $\mathit{cnf}((\varepsilon+\phi)\cdot(x+y)^{*})=(x+y)^{*}$ where $x<y$.

6.2 Subtraction

Definition 34** (Equations for Subtraction)**

Let $r,s$ be two regular expressions. For each pair $(r^{\prime},s^{\prime})\in{\cal D}(r)\times{\cal D}(s)$ we introduce a variable $R_{r^{\prime},s^{\prime}}$. For each such $R_{r^{\prime},s^{\prime}}$ we define an equation of the following form. If ${\cal L}(r^{\prime})=\emptyset$, we set $R_{r^{\prime},s^{\prime}}\approx\phi$. Otherwise, $R_{r^{\prime},s^{\prime}}\approx\sum_{x\in\Sigma}x\cdot R_{\mathit{cnf}(d_{x}(r^{\prime})),\mathit{cnf}(d_{x}(s^{\prime}))}+t$ where $t=\varepsilon$ if $\varepsilon\in{\cal L}(r^{\prime}),\varepsilon\not\in{\cal L}(s^{\prime})$, otherwise $t=\phi$. All equations are collected in a set ${\cal S}_{r,s}$.

Let $\psi=\mbox{\it solve}({\cal S}_{r,s})$. Then, we define $r-s=\psi(R_{r,s})$.

As the set of canonical derivatives is finite, the set $\mbox{\it solve}({\cal S}_{r,s})$ is finite as well. Hence, a solution must exist. Hence, $r-s$ is well-defined.

Lemma 35

Let $r,s$ be two regular expressions. Then, we find that

[TABLE]

where $T=\{\varepsilon\}$ if $\varepsilon\in{\cal L}(r),\varepsilon\not\in{\cal L}(s)$, otherwise $T=\emptyset$.

Proof

By the Expansion Theorem 29 and Lemma 31, we find that $r\equiv\sum_{x\in\Sigma}x\cdot\mathit{cnf}(d_{x}(r))+t$ and $s\equiv\sum_{x\in\Sigma}x\cdot\mathit{cnf}(d_{x}(s))+t^{\prime}$ where $t=\varepsilon$ if $r$ is nullable. Otherwise, $t=\phi$. For $t^{\prime}$ we find $t^{\prime}=\varepsilon$ if $s$ is nullable. Otherwise, $t^{\prime}=\phi$.

By associativity, commutativity of $+$ and some standard algebraic laws

[TABLE]

the result follows immediately. ∎

Theorem 36** (Subtraction)**

Let $r,s$ be two regular expressions. Then, we find that $r-s$ is unambiguous and ${\cal L}(r-s)\equiv{\cal L}(r)-{\cal L}(s)$.

Proof

By construction, equations are non-overlapping. Unambiguity follows from Theorem 25.

We prove the equivalence claim via a coalgebraic proof method [17]. We show that the relation $\{({\cal L}(\psi(R_{r^{\prime},s^{\prime}})),{\cal L}(r^{\prime})-{\cal L}(s^{\prime}))\mid(r^{\prime},s^{\prime})\in{\cal D}(r)\times{\cal D}(s)\}$ is a bisimulation where $\psi=\mbox{\it solve}({\cal S}_{r,s})$. For that to hold two elements are in relation if either (1) they are both nullable, or (2) their derivatives, i.e. taking away the same leading literal, are again in relation.

Consider a pair $({\cal L}(\psi(R_{r^{\prime},s^{\prime}})),{\cal L}(r^{\prime})-{\cal L}(s^{\prime}))$. For ${\cal L}(r^{\prime})=\emptyset$ we have that $R_{r^{\prime},s^{\prime}}\approx\phi$. The conditions imposed on a bisimulation follow immediately.

Otherwise, $R_{r^{\prime},s^{\prime}}$ is defined by the equation

[TABLE]

where $t=\varepsilon$ if $\varepsilon\in{\cal L}(r^{\prime}),\varepsilon\not\in{\cal L}(s^{\prime})$, otherwise $t=\phi$. From Lemma 35 we can conclude that

[TABLE]

where $T=\{\varepsilon\}$ if $\varepsilon\in{\cal L}(r),\varepsilon\not\in{\cal L}(s)$, otherwise $T=\emptyset$. Immediately, we find that if one component of the pair is nullable, the other one must be nullable as well.

We build the derivative for each component w.r.t. some literal $x$. Given that $\psi$ is a solution and via (E1) and (E2) the resulting derivatives are equal to ${\cal L}(\psi(R_{\mathit{cnf}(d_{x}(r^{\prime})),\mathit{cnf}(d_{x}(s^{\prime}))}))$ and ${\cal L}(\mathit{cnf}(d_{x}(r)))-{\cal L}(\mathit{cnf}(d_{x}(s)))$. Hence, derivatives are again in relation. This concludes the proof. ∎

Example 6

We consider $r_{1}=(x+y)^{*}$ and $r_{2}=(x\cdot x)^{*}$. Let us consider first the canonical descendants of both expressions.

[TABLE]

The resulting equations are as follows.

[TABLE]

Solving of the above proceeds as follows. We first apply $R_{1,4}=r_{1}$.

[TABLE]

Next, we remove the equation for $R_{1,3}$ and apply some simplifications.

[TABLE]

Via Arden’s Lemma we find that $R_{1,2}=(x\cdot x)^{*}\cdot(x\cdot y\cdot r_{1}+x+y\cdot r_{1})$ and we are done.

6.3 Shuffle

Definition 37** (Shuffle)**

The shuffle operator $\|::\Sigma^{*}\times\Sigma^{*}\rightarrow\wp(\Sigma^{*})$ is defined inductively as follows:

[TABLE]

We lift shuffling to languages by

[TABLE]

For example, we find that $x\cdot y\|z=\{x\cdot y\cdot z,x\cdot z\cdot y,z\cdot x\cdot y\}$.

Definition 38** (Equations for Shuffling)**

Let $r,s$ be two regular expressions. For each pair $(r^{\prime},s^{\prime})\in{\cal D}(r)\times{\cal D}(s)$ we introduce a variable $R_{r^{\prime},s^{\prime}}$. For each such $R_{r^{\prime},s^{\prime}}$ we define an equation of the following form. If ${\cal L}(r^{\prime})=\emptyset$, we set $R_{r^{\prime},s^{\prime}}\approx\phi$. Otherwise, $R_{r^{\prime},s^{\prime}}\approx\sum_{x\in\Sigma}(x\cdot R_{\mathit{cnf}(d_{x}(r^{\prime})),s^{\prime}}+x\cdot R_{r^{\prime},\mathit{cnf}(d_{x}(s^{\prime}))})+t$ where $t=t_{1}+t_{2}$. Expression $t_{1}=s^{\prime}$ if $\varepsilon\in{\cal L}(r^{\prime})$, otherwise $t_{1}=\phi$. Expression $t_{2}=r^{\prime}$ if $\varepsilon\in{\cal L}(s^{\prime})$, otherwise $t_{2}=\phi$. All equations are collected in a set ${\cal H}_{r,s}$.

Let $\psi=\mbox{\it solve}({\cal H}_{r,s})$. Then, we define $r\|s=\psi(R_{r,s})$.

Lemma 39

Let $r,s$ be two regular expressions. Then, we find that

[TABLE]

where $T=T_{1}+T_{2}$. $T_{1}=s$ if $\varepsilon\in{\cal L}(r)$, otherwise $T_{1}=\phi$. $T_{2}=r$ if $\varepsilon\in{\cal L}(s)$, otherwise $T_{2}=\phi$.

Theorem 40** (Shuffling)**

Let $r,s$ be two regular expressions. Then, we find that ${\cal L}(r\|s)\equiv{\cal L}(r)\|{\cal L}(s)$.

7 Related Works and Conclusion

Our work gives a precise description of solving of regular equations including a computational interpretation by means of parse tree transformations. Thus, we can characterize conditions under which regular equations and resulting regular expressions are unambiguous.

Earlier work by Gruber and Holzer [11] gives a comprehensive overview on the conversion of finite automaton to regular expressions and vice versa. Like many other works [16, 5], the algorithmic details of solving regular equations based on Arden’s Lemma are not specified in detail.

Brzozowski’s and McCluskey’s [6] state elimination method appears to be the more popular and more widespread method. For example, consider work by Han [12] and in collaboration with Ahn [1], as well as work by Moreira, Nabais and Reis [15] that discuss state elimination heuristics to achieve short regular expressions.

Sakarovitch [18, 19] shows that the state elimination and solving via regular equation methods are isomorphic and produce effectively the same result. Hence, our (unambiguity) results are transferable to the state elimination setting. The other way around, state elimination heuristics are applicable as demonstrated by our implementation.

It is well understood how to build the subtraction and intersection among DFAs via the product automaton construction [13]. If we wish to apply these operations among regular expressions we need to convert expressions back and forth to DFAs. For example, we can convert a regular expression into a DFA using Brzozowski’s derivatives [5] and then use Brzozowski’s algebraic method to convert back the product automaton to a regular expression.

To build the shuffle among two regular expressions, the standard method is to (1) build the shuffle derivative-based DFA, (2) turn this DFA into some regular equations and then (3) solve these regular equations. Step (1) relies on the property that the canonical derivatives for shuffle expressions are finite.

In our own work [22], we establish finiteness for several variations of the shuffle operator. Caron, Champarnaud and Mignot [7] and Thiemann [23] establish finiteness of derivatives for an even larger class of regular expression operators.

We propose direct methods to build the intersection and the shuffle among two regular expressions. For each operation we generate an appropriate set of equations by employing Brzozowski derivatives. We only rely on finiteness of canonical derivatives for standard regular expressions. Solving of these equations then yields the desired expression. Correctness follows via some simple (co)algebraic reasoning and we can guarantee that resulting expressions are unambiguous.

Acknowledgments

We thank referees for CIAA’18, ICTAC’18 and ICTAC’19 for their helpful comments on previous versions of this paper.

Appendix 0.A Proofs

0.A.1 Proof of Proposition 17

Proof

For non-overlapping equations there can be at most one $v$ such that ${\cal E}\,\vdash\,v:R$. Suppose ${\cal E}\,\vdash\,v:R$ where $R\approx x_{1}\cdot R_{1}+\dots+x_{n}\cdot R_{n}+t$. Recall that $+$ is right-associative. From ${\cal E}\,\vdash\,v:R$ we conclude that ${\cal E}\,\vdash\,v^{\prime}:x_{1}\cdot R_{1}+\dots+x_{n}\cdot R_{n}+t$ where $v=\textsc{Fold}\ v^{\prime}$ for some $v^{\prime}$. For $|v^{\prime}|=\varepsilon$ (empty word), $t$ must be nullable. Based on the choice for $t$, we must have that $t=\varepsilon$. Hence, the choice for $v^{\prime}$ is fixed. Consider $|v^{\prime}|=x\cdot w$ for some literal $x$ and word $w$. Again the choice for $v^{\prime}$ is fixed because due to non-overlapping there is at most one $i$ such that $x=x_{i}$. ∎

0.A.2 Proof of Lemma 39

Proof

We employ the following algebraic laws.

[TABLE]

∎

0.A.3 Proof of Theorem 40

Proof

Similar to the proof of Theorem 36.

Unambiguity may no longer hold because equations to compute $r\|s$ are overlapping. ∎

Appendix 0.B Parsing with Regular Equations

We build a parser following the scheme of a derivative-style regular expression parser. Soundness results reported in [20] carry over to the extended setting. That is, if a parse exists the parser will succeed. It is possible to compute all parse trees following the scheme outlined in [21]. For brevity, we omit the details.

Definition 41** (Regular Expressions and Equations Derivatives)**

We assume a fixed set ${\cal E}$ of equations.

[TABLE]

All standard results, e.g. expansion, carry over to the extended setting.

Definition 42** (Empty Parse Trees)**

*We assume a fixed set ${\cal E}$ of equations. The (partial) function ${\mathit{m}kEmpty}_{\cdot}$ computes an empty parse tree for regular expressions and equations.

$\mathit{{}{\mathit{m}kEmpty}_{\epsilon}\mskip 4.0mu plus 1.0mu=\mskip 4.0mu plus 1.0mu\textsc{Eps}}$

$\mathit{{}{\mathit{m}kEmpty}_{r^{*}}\mskip 4.0mu plus 1.0mu=\mskip 4.0mu plus 1.0mu[]}$

$\mathit{{}{\mathit{m}kEmpty}_{\gamma_{1}\cdot\gamma_{2}}\mskip 4.0mu plus 1.0mu=\mskip 4.0mu plus 1.0mu\textsc{Seq}\ ({\mathit{m}kEmpty}_{\gamma_{1}})\ ({\mathit{m}kEmpty}_{\gamma_{2}})}$

$\mathit{{}{\mathit{m}kEmpty}_{\alpha_{1}+\alpha_{2}}\mskip 4.0mu plus 1.0mu=\mskip 4.0mu plus 1.0mu\mskip 4.0mu plus 1.0mu\mbox{\sf if}\ \varepsilon\mskip 4.0mu plus 1.0mu\in\mskip 4.0mu plus 1.0mu{\cal L}(\alpha_{1})\mskip 4.0mu plus 1.0mu\ \mbox{\sf then}\mskip 4.0mu plus 1.0mu\textsc{Inl}\ ({\mathit{m}kEmpty}_{\alpha_{1}})\mskip 4.0mu plus 1.0mu\mbox{\sf else}\ \textsc{Inr}\ ({\mathit{m}kEmpty}_{\alpha_{2}})}$

$\mathit{{}{\mathit{m}kEmpty}_{R}\mskip 4.0mu plus 1.0mu=\mskip 4.0mu plus 1.0mu{\mathit{m}kEmpty}_{\alpha}\mskip 4.0mu plus 1.0mu\ \ \ \mbox{\sf where}\ R\mskip 4.0mu plus 1.0mu\approx\mskip 4.0mu plus 1.0mu\alpha\mskip 4.0mu plus 1.0mu\in\mskip 4.0mu plus 1.0mu{\cal E}}$ *

Definition 43** (Injection)**

[TABLE]

Definition 44** (Parsing)**