[SPIN API] Constructor/constraint inheritance implementation with Jena rules?

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Martynas Jusevicius

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Sep 26, 2015, 10:51:21 AM9/26/15
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Hey Holger,

by inspecting SPINConstructors, I have noticed that constructors (and probably constraints as well?) are "inherited" by walking up the subclass chain and looking for them. Am I right?

I had a similar use case implemented as custom Jena rules. Properties can be inferred by constructing OntModelSpec with GenericRuleReasoner which parses the rules. They could look something like that:

  [constructors: (?class rdf:type rdfs:Class), (?class <http://spinrdf.org/spin#constructor> ?o), (?subClass rdfs:subClassOf ?class), noValue(?subClass <http://spinrdf.org/spin#constructor>) -> (?subClass <http://spinrdf.org/spin#constructor> ?o) ]
  [constraints: (?class rdf:type rdfs:Class), (?class <http://spinrdf.org/spin#constraint> ?o), (?subClass rdfs:subClassOf ?class), noValue(?subClass <http://spinrdf.org/spin#constraint>) -> (?subClass <http://spinrdf.org/spin#constraint> ?o) ]

I think that would be a cleaner approach and a better reuse of Jena's machinery. Would that work for SPIN API?

Martynas
graphityhq.com

Holger Knublauch

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Sep 27, 2015, 6:40:10 PM9/27/15
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Hi Martynas,

if you suggest to rely on Jena rules to walk the superclasses for SPIN then this is certainly an option but feels like an overkill to a relatively simple problem. It would require to activate inferencing, which has its own price tag and cannot be applied in all cases. Having said this, an alternative SPIN implementation could certainly use that machinery.

Holger
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