[FT] FT word Distance exactly

• From: andrewc <andrew.cao@cisra.canon.com.au>
• Date: Thu, 31 Mar 2005 14:22:36 +1000
• To: public-qt-comments@w3.org
• Message-ID: <424B7B0C.1020604@cisra.canon.com.au>

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 Thursday, 31 March 2005 04:22:42 UTC