An Effective Algorithm for the Membership Problem for Extended Regular Expressions
From FSL
This work has been published in a conference proceedings and in a technical report.
FOSSACS'07
- An Effective Algorithm for the Membership Problem for Extended Regular Expressions
- Grigore Rosu
- FOSSACS'07, LNCS 4423, pp 332-345, 2007
- Abstract. By adding the complement operator (
), extended regular expressions (ERE) can encode regular languages non-elementarily more succinctly than regular expressions. The ERE membership problem asks whether a word w of size n belongs to the language of an ERE R of size m. Unfortunately, the best known membership algorithms are either non-elementary in m or otherwise require space Ω(n2) and time Ω(n3); since in many practical applications n can be very large, these space and time requirements could be prohibitive. In this paper we present an ERE membership algorithm that runs in space 
and time 
. The presented algorithm outperforms the best known algorithms when n is exponentially larger than m.
- Abstract. By adding the complement operator (
- PDF, FOSSACS'07, BIB
Technical report
- An Effective Algorithm for the Membership Problem for Extended Regular Expressions
- Grigore Rosu
- Technical Report UIUCDCS-R-2005-2694, August 2005
- Abstract. By adding the complement operator ($\neg$), extended regular expressions ({\ERE}) can encode regular languages non-elementarily more succinctly than regular expressions. The {\ERE} membership problem asks whether a word $w$ of size $n$ belongs to the language of an {\ERE} $R$ of size $m$. Unfortunately, the best known membership algorithms are either non-elementary in $m$ or otherwise require space $\Omega(n^2)$ and time $\Omega(n^3)$; since in many practical applications $n$ can be very large (in the order of billions, e.g., in testing where $w$ represents the execution trace of some system), these space and time requirements could be prohibitive. In this paper we present a simple to implement {\ERE} membership algorithm that runs in space $O(n \cdot (m + \log n) \cdot 2^{m} \cdot k)$ and in time $O(n^2 \cdot (m + \log n)^2 \cdot 2^m \cdot k)$, where $k$ is the number of complement operators in $R$. The presented algorithm outperforms the best known algorithms when $n$ is large.
