17 Jan 2003 20:03:38 -0500

Related articles |
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NFA to DFA question unmesh_joshi@hotmail.com (Unmesh joshi) (2003-01-12) |

Re: NFA to DFA question clint@0lsen.net (Clint Olsen) (2003-01-17) |

Re: NFA to DFA question thp@cs.ucr.edu (2003-01-17) |

Re: NFA to DFA question mchristoff@sympatico.ca (Michael N. Christoff) (2003-01-17) |

Re: NFA to DFA question lord@emf.emf.net (2003-01-21) |

Re: NFA to DFA question joachim_d@gmx.de (Joachim Durchholz) (2003-01-25) |

Re: NFA to DFA question rafe@cs.mu.oz.au (Ralph Becket) (2003-01-30) |

From: | thp@cs.ucr.edu |

Newsgroups: | comp.compilers |

Date: | 17 Jan 2003 20:03:38 -0500 |

Organization: | University of California, Riverside |

References: | 03-01-051 |

Keywords: | lex, DFA |

Posted-Date: | 17 Jan 2003 20:03:38 EST |

Unmesh joshi <unmesh_joshi@hotmail.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

string.

Tom Payne

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