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Re: [PP] Duplicates with :p*

From: Lee Feigenbaum <lee@thefigtrees.net>
Date: Thu, 10 Feb 2011 11:33:57 -0500
Message-ID: <4D541375.1040500@thefigtrees.net>
To: Andy Seaborne <andy.seaborne@epimorphics.com>
CC: SPARQL Working Group <public-rdf-dawg@w3.org>
On 2/10/2011 11:08 AM, Andy Seaborne wrote:
> Summary:
>
> 1/ Proposal to have an explicit :p* operator
> instead of :p* = :p{0} UNION :p+
>
> 2/ path-* would return only one node for a possible path
> That does not path-* duplicate free if there are
> multiple ways to the same end.

Read as:

That does not make path-* duplicate free if there are multiple ways to 
the same endpoint.

(thanks to Andy on IRC for the clarification.)

Lee

> c.f. literals and aggregation example
> c.f. BGPs and projection.
>
> Details still to be worked on.
>
>
> Currently,
>
> :p* = :p{0} UNION :p+
>
> This is a source of duplicates because :p{0} may include a node and then
> :p+ will also include it if the path loops back to the start.
>
> Defining
> :p+ = :p/:p*
>
> would also lead to duplicates because of the hidden projection in the
> "/" so two operators are needed.
>
> Example 1:
>
> :a :p :b .
> :b :p :c .
> :c :p :a .
>
> which is:
>
> :c
> / \
> / \
> :a --> :b
>
> Paths:
> P1) :a :p* ?X
> P2) :a :p+ ?Y
>
> we want:
> P1) => ?X = :a :b :c
> P2) => ?Y = :a :b :c
>
> Example 2:
>
> :a :p :b .
> :b :p :c .
> :c :p :b .
>
> :c
> / \
> \ /
> :a --> :b
>
> we want
> P1) => ?X = :a :b :c
> P2) => ?Y = :b :c
>
>
> It should be possible to have the same algorithm, with different
> starting conditions, for :p+ and :p*.
>
> Andy
>
>
Received on Thursday, 10 February 2011 16:34:45 GMT

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