|NFA to DFA question firstname.lastname@example.org (Unmesh joshi) (2003-01-12)|
|Re: NFA to DFA question email@example.com (Clint Olsen) (2003-01-17)|
|Re: NFA to DFA question firstname.lastname@example.org (2003-01-17)|
|Re: NFA to DFA question email@example.com (Michael N. Christoff) (2003-01-17)|
|Re: NFA to DFA question firstname.lastname@example.org (2003-01-21)|
|Re: NFA to DFA question email@example.com (Joachim Durchholz) (2003-01-25)|
|Re: NFA to DFA question firstname.lastname@example.org (Ralph Becket) (2003-01-30)|
|Date:||17 Jan 2003 20:03:38 -0500|
|Organization:||University of California, Riverside|
|Posted-Date:||17 Jan 2003 20:03:38 EST|
Unmesh joshi <email@example.com> 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.
It's much simpler than that. At every step of the deterministic
computation you are in the set of all states that you might be in
under the non-deterministic computation had you scanned the same input
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