# Re: An example LL(K) language that is not LL(K-1) ?

## Kaz Kylheku <kkylheku@gmail.com>

Wed, 3 Feb 2010 18:18:56 +0000 (UTC)

*From comp.compilers*

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**From: ** | Kaz Kylheku <kkylheku@gmail.com> |

**Newsgroups: ** | comp.compilers |

**Date: ** | Wed, 3 Feb 2010 18:18:56 +0000 (UTC) |

**Organization: ** | A noiseless patient Spider |

**References: ** | 10-02-009 10-02-015 |

**Keywords: ** | LL(1) |

**Posted-Date: ** | 05 Feb 2010 17:32:14 EST |

On 2010-02-02, klyjikoo <klyjikoo@gmail.com> wrote:

*>*

*>> I don't think your assumption that any LL(k) can be transformed into*

*>> an LL(k-1) is correct. The 'k' in LL(k) is assumed to be the supremum*

*>> of lookahead symbols that you need in order to parse your input. So,*

*>> suppose you have an LL(2) grammar, then you cannot convert it to an*

*>> LL(1) since the LL(1) equivalent won't have disjoint FIRST/FOLLOW sets!*

*>*

*>> I am not yet very experienced when it comes to compilers, so, if my*

*>> answer is wrong correct me please! :-)n*

*>*

*> Thanks to Hans, Consider this example :*

*>*

*> 1) Z := X*

*> 2) X := Y*

*> 3) X := bYa*

*> 4) Y := c*

*> 5) Y := ca*

This grammar generates only a finite set of strings. So it gives a

regular language, which can be described by the regular expression

(cca|bca|bcaa).

It would be astonishing if a regular language could not be described by

a LL(1) grammar.

*:)*

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