W3C home > Mailing lists > Public > public-sparql-dev@w3.org > October to December 2014

Pattern involving BIND from 18.2.3 Examples of Mapped Graph Patterns

From: Rory Rother <rrother@algebraixdata.com>
Date: Thu, 2 Oct 2014 23:41:43 +0000
To: "public-sparql-dev@w3.org" <public-sparql-dev@w3.org>
Message-ID: <8EDCF559E852FB45A9C0F6991C67BAA81FABBA@EXCHANGE8.win.zerolag.com>
Hello,

I was working through some of the spec to get a better understanding of translating queries to SPARQL algebra and had a question about the eleventh example in section 18.2.3 (http://www.w3.org/TR/2013/REC-sparql11-query-20130321/#sparqlAlgebraExamples).

The example is

{ ?s :p ?v . {} BIND (2*?v AS ?v2) }

and the given translation to SPARQL algebra is

Join(
   BGP(?s :p ?v), ?v2, 2*?v) ,
   Extend({}, ?v2, 2*?v)
)

I am having a hard time seeing how to get that translation.  From my understanding, I am starting with a  GroupGraphPattern so I look to the relevant box in section 18.2.2.6.  I believe my GroupGraphPattern has three elements: the triple pattern, the empty GroupGraphPattern, and the BIND clause.  I set G equal to the Join identity.  Now I enter the For loop by first setting E equal to the triple pattern.  At the end of the first loop, I just have G=BGP(?s :p ?v).  At the start of the second loop, I set E equal to {}.  That satisfies the last If statement so I set A equal to Translate({}).  That should give me A equal to the Join identity.  Then G=Join( BGP(?s :p ?v, (the join identity) ).  So, at the end of the second loop, I just have G=BGP(?s :p ?v).  At the start of the third loop, I set E equal to the BIND clause.  That satisfies the third If statement, so I set G equal to Extend(BGP(?s :p ?v), ?v2, 2*?v).  That's the end of the For loop so I end up with

Extend( BGP(?s :p ?v), ?v2, 2*?v )

as the result.  I believe the above translation gives solution mappings different from those coming from the translation provided in the spec.  In fact, I believe the translation found in the spec would give solution mappings to the pattern

{
   {}
   BIND(2*?v AS ?v2)
   ?s :p ?v
}

Any insight you can provide is much appreciated.  If this is not the proper forum for such a question, then please accept my apologies (and maybe let me know where I should be asking :)).

Rory
Received on Friday, 3 October 2014 07:52:58 UTC

This archive was generated by hypermail 2.3.1 : Tuesday, 6 January 2015 20:15:52 UTC