|NFA to DFA question email@example.com (Unmesh joshi) (2003-01-12)|
|Re: NFA to DFA question firstname.lastname@example.org (Clint Olsen) (2003-01-17)|
|Re: NFA to DFA question email@example.com (2003-01-17)|
|Re: NFA to DFA question firstname.lastname@example.org (Michael N. Christoff) (2003-01-17)|
|Re: NFA to DFA question email@example.com (2003-01-21)|
|Re: NFA to DFA question firstname.lastname@example.org (Joachim Durchholz) (2003-01-25)|
|Re: NFA to DFA question email@example.com (Ralph Becket) (2003-01-30)|
|From:||Clint Olsen <firstname.lastname@example.org>|
|Date:||17 Jan 2003 19:40:42 -0500|
|Posted-Date:||17 Jan 2003 19:40:42 EST|
Unmesh joshi wrote:
> I am reading the compilers book by Aho ullman, and I have one doubt about
> NFA to DFA conversion. "Every state of DFA corresponds to 'set of
> states' in NFA". Can anybody explain to me this? Does anybody has a
> source code sample for NFA-DFA? May be if I implement the DFA algorithm I
> will understand what that means.
Since an NFA can have multiple state transitions per character of input,
and a DFA cannot, it follows that a state in the DFA corresponds to a set
of NFA states (possibly multiple states). The best analogy is given at the
end of that paragraph you quoted where they discuss simultating an NFA and
DFA for the same input string.
The source code from 'flex' should give you an implementation of converting
an NFA to a DFA.
Return to the
Search the comp.compilers archives again.