7 Jan 2004 01:01:22 -0500

Related articles |
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What is the trick of languages being LL(k+1), but not LL(k)? oliver@zeigermann.de (Oliver Zeigermann) (2004-01-07) |

Re: What is the trick of languages being LL(k+1), but not LL(k)? cfc@shell01.TheWorld.com (Chris F Clark) (2004-01-09) |

Re: What is the trick of languages being LL(k+1), but not LL(k)? oliver@zeigermann.de (Oliver Zeigermann) (2004-01-12) |

Re: What is the trick of languages being LL(k+1), but not LL(k)? cfc@shell01.TheWorld.com (Chris F Clark) (2004-01-12) |

Re: What is the trick of languages being LL(k+1), but not LL(k)? oliver@zeigermann.de (Oliver Zeigermann) (2004-01-16) |

Re: What is the trick of languages being LL(k+1), but not LL(k)? cgweav@aol.com (2004-01-16) |

Re: What is the trick of languages being LL(k+1), but not LL(k)? cfc@world.std.com (Chris F Clark) (2004-01-17) |

[5 later articles] |

From: | Oliver Zeigermann <oliver@zeigermann.de> |

Newsgroups: | comp.compilers |

Date: | 7 Jan 2004 01:01:22 -0500 |

Organization: | T-Online |

Keywords: | LL(1), parse, question |

Posted-Date: | 07 Jan 2004 01:01:22 EST |

Experts!

Theory books tell me for every k there are languages that are LL(k+1),

but not LL(k). An example is the language described by this grammar

(upper case letters are terminals, lower case are non-terminals, A^k

means A repeated k times):

s : A s a | ;

a : A^k B s | C ;

I understand this grammar is LL(k+1), but what makes me sure there is no

way to rewrite it to be LL(k) or even less?

Cheers and thanks in advance,

Oliver

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