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Re: reverse directions in property paths

From: Andy Seaborne <andy.seaborne@epimorphics.com>
Date: Fri, 07 Oct 2011 08:58:35 +0100
Message-ID: <4E8EB12B.9030405@epimorphics.com>
To: public-rdf-dawg-comments@w3.org

On 17/09/11 02:43, David Mizell wrote:
> Hi –
> I'm starting to look at property paths in SPARQL 1.1, and the spec
> doesn't make it clear how/whether reversal “distributes” over other
> operators. Suppose you have the triples
> y a o1
> o1 b x
> y b o2
> o2 a x
> in the database. Would { x ^(a/b) y } match with the first pair, or with
> the second pair of triples in the database?
> Thanks
> David Mizell
> Cray Inc., Seattle

David - section "9.1 Property Path Syntax" has list giving the 
precedence rules of the property path syntax forms. Groups using () has 
higher precedence than unary ^ which in turn is higher precedence than 
binary operator /

{ x ^(a/b) y } is the reverse path of a/b so "(reverse (seq :a :b))" so 
it matches the first case, reversing the whole of a/b. i.e. b then a

---- data
@prefix :        <http://www.example.org/> .

:y1  :a  :o1 .
:o1 :b  :x1 .

:y2  :b  :o2 .
:o2  :a  :x2 .

---- query
PREFIX :	<http://www.example.org/>

SELECT *
{  ?x   ^(:a/:b)   ?y  }

---- results
-------------
| x   | y   |
=============
| :x1 | :y1 |
-------------

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Andy (on behalf of the SPARQL WG)
Received on Friday, 7 October 2011 07:59:22 GMT

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