Thanks a lot!
I know that proving that one DFA is a full subset of another DFA is a
polinomial problem, therefore solveable.
Perl5 regular expressions: What you say is that Perl5 expressions are not
regular language?
Why?
The reason I directed this question here, is because I thought it would be a
nice feature in ORO.
And also, there might have been a provision for that in there.
But please, could you tell me whether you think Perl5 expressions are not
regular language?
Please and thanks,
Michael Borodiansky
----- Original Message -----
From: "Daniel F. Savarese" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Friday, March 30, 2001 12:45 AM
Subject: Re: Question:
>
> >Let R1 be the regular expression 1
> >Let R2 be the regular expression 2
> >
> >R1 < R2 iff all inputs that R1 can match R2 can, but not vice versa.
>
> Although an interesting question, it doesn't have much to do with the
> development of the jakarta-oro software and should probably be
> directed to a mailing list or newsgroup on lexical analysis. However,
> I will venture to say it is a solvable problem for DFA and even some
> NFA representations of regular expressions, but not for Perl5
> expressions. You have to be able to compute closures and show that
> the closure of R2 over all of its inputs is a superset of the closure
> of R1, which is straightforward for DFAs.
>
> daniel
>
>
