Re: [FT] FT word Distance exactly

Dear Andrew,

your observations are correct and exhibit indeed a systematic error in the 
distance semantics functions. The XQuery Full-Text task force is currently 
working on a proposal that fixes this.

Yours sincerely / Mit freundlichen Grüßen,
      Jochen Dörre
__________________________________________
IBM Germany Böblingen Laboratory
DB2 Information Management Software
Phone: +49-7031-16-2992,    Fax: -4891,   Email: doerre@de.ibm.com

> Dear editors,
>
> When I have a node: <Node>word1 word2 word3</Node>
>
> I apply the query[1]:
> /Node ftcontains ("word1" && "word2" && "word3") with distance exactly 0 

> words
> I will get the AllMatches[1] as:
> --- AllMatches
>       --- Match
>             --- StringInclude (pos = 1)
>             --- StringInclude (pos = 2)
>             --- StringInclude (pos = 3)
> The final result is True.
> 
> I apply the query[2]:
> /Node ftcontains ("word1" && "word2" && ! "word3") with distance exactly 

> 0 words
> I seem to get the AllMatches[2] as:
> --- AllMatches
>       --- Match
>             --- StringInclude (pos = 1)
>             --- StringInclude (pos = 2)
> The final result is also True.
> 
> The reason for AllMatches[2] is that the StringExclude (pos = 3) which 
> is generated by ! "word3" has been dropped, according to semantics of 
> ApplyFTWordDistanceExactly, because SE-3 does not have a word distance 0 

> with both SI-1 and SI-2.
> 
> Are my two results correct? If they are correct, would this be 
> inconsistent? Or what is the intuition when "word3" is a don't-care?
> Can I compare SE-3 to any one of SI-1 and SI-2, not to both of them?
> 
> Thanks,

Received on Tuesday, 12 April 2005 11:37:40 UTC