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Re: TEST: 5 of 7: was Re: Revisiting AllDisjoint

From: Ian Horrocks <horrocks@cs.man.ac.uk>
Date: Thu, 24 Jul 2003 10:31:01 +0100
Message-ID: <16159.42837.331750.773240@merlin.horrocks.net>
To: Jeremy Carroll <jjc@hpl.hp.com>
Cc: www-webont-wg@w3.org

On July 23, Jeremy Carroll writes:
> 
> Ian, Jim, Jonathan
> 
> I have include Ian's test from
> 
> http://lists.w3.org/Archives/Public/www-webont-wg/2003Jul/0157
> 
> as
> 
> http://www.w3.org/2002/03owlt/editors-draft/draft/proposedByIssue#I5.21-002

I wasn't able to run the test (I don't yet have the datatype reasoner
working), but it doesn't look like it would work, because
Amphisbaenians etc are not said to be reptiles and so would not
inherit the cardinality constraint on the reptile-name property.

Also, I'm not an expert on reptiles (not this kind of reptile anyway),
but I guess that "Sauria" is not a reptile, but a family or genus (or
some such?) of reptile. The proposed datatype idiom has to be used
carefully in such cases because if there is a hierarchy with
disjointness at each level, e.g., if there were sub-classes of
Serpentes such as "Viper" and "Cobra" (bad modelling, I know, but
let's ignore that for now), then to make these disjoint you must use a
different property than the one used to make Sauria and Serpentes
themselves disjoint (otherwise the ontology would be inconsistent).

I would suggest to either use reptile species in the example, or to
change the name of the property to something like "genus-name".

Ian


> 
> I could do with a few more reptile names. I currently have 6 (taken from Jim's 
> message)
> 
> The number of triples in Ian's idiom is 6 + 5N = 36
> 
> The number of triples in the O(N^2) version is N(N+1)/2 = 21
> 
> If we had N=11 (another five reptiles) we would have 61 verses 66 triples, 
> showing that Ian's idiom is better (with N>=11).
> 
> 
> Jeremy
> 
> 
> 
Received on Thursday, 24 July 2003 05:32:13 GMT

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