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Re: property paths three more test cases pp13, pp14, pp15

From: Steve Harris <steve.harris@garlik.com>
Date: Tue, 21 Dec 2010 12:51:55 +0000
Cc: Axel Polleres <axel.polleres@deri.org>, SPARQL Working Group <public-rdf-dawg@w3.org>
Message-Id: <7A1BDF70-EA09-4292-8EEB-BA7C5E2EFB5B@garlik.com>
To: Andy Seaborne <andy.seaborne@epimorphics.com>
On 2010-12-21, at 12:39, Andy Seaborne wrote:

> 
>>> 
>>> One example is:
>>> 
>>>   ?x rdf:type/rdfs:subClassOf* ?T
>> 
>> I see, otherwise you'd have to write ?x (rdf:type|rdf:type/rdfs:subClassOf*) ?T, which looks a bit tautological.
> 
> Yes.
> 
>> On the other hand, trying to find just the transitively closed subclass relationships with:
>> 
>> ?C rdfs:subClassOf* ?SC .
>> 
>> you will get many non-sensical results. I think you can write
>> 
>> [] rdfs:subClassOf ?SC .
>> ?C rdfs:subClassOf [] .
>> ?C rdfs:subClassOf* ?SC .
>> 
>> if you want just the transitively closed classes/superclasses, but that also looks odd.
> 
> Isn't transitive closure "+"
> 
> ?C rdfs:subClassOf+ ?SC .

If you have <C> rdfs:subClassOf <D> . then <C> is a subClass of <C>.

I don't know a way to write that that's clearer than what I have above.

>> I'm also concerned that the relatively innocuous-looking expression like { ?s rdfs:subClassOf* ?o } will have a higher cardinality than { ?s ?p ?o }, but I guess users will learn not to do that after the first few hundred times they try it :)
> 
> ?s rdfs:subClassOf* ?o  will have ?s = ?o, not the cross product of the subjects and objects.
> 
> { ?s ?p ?o } could have more cardinality.
> 
> :x :p1 :y .
> :x :p2 :y .
> :x :p3 :y .
> :x :p4 :y .
> 
> { ?s rdfs:subClassOf* ?o } => cardinality 2
> 
> ?s=:x, ?o=:x
> ?s=:y, ?o=:y
> 
> and
> 
> {?s ?p ?o } is 4 rows.
> 
> It can be odd that "?x :p* ?o" does not depend on :p but it's similar to string regex:
> 
> "xyz" is matched by "a*"

To be strict, only the substring "" is matched by it.

$ echo "xyz" > foo.txt
$ egrep -o 'a*' foo.txt
$

- Steve

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Received on Tuesday, 21 December 2010 12:52:29 GMT

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