Dave,
"Normally you would have other rules which extract the
pieces of the functors that you want as normal triples (e.g. back into
an n-ary relationships using bNodes) for query or access them
programmatically.
“
I can’t find out how to implement a bNode. Can you give a short example,
please?
Is something like this:
(?x fam:executeP01 ?a) <-
(?x fam:p01 certainty(?y, ?c)),
(?a fam:object ?y),
(?a fam:confidence ?c),
makeTemp(?a).
Or this:
(?a fam:object ?y),
(?a fam:confidence ?c),
(?x fam:executeP01 ?a) <-
(?x fam:p01 certainty(?y, ?c)),
makeTemp(?a).
MBA
On 12/11/13 18:29, "Dave Reynolds" <[email protected]> wrote:
>On 12/11/13 15:13, Miguel Bento Alves wrote:
>> Dave,
>>
>> "Alternatively, Jena rules has the notion of a "functor" (bad name)
>>which
>> is essentially a structured literal. So you can represent your Y/C pairs
>> as certainty(Y,C) and write rules which generate and match triples whose
>> object is a certainty(?x,?y) pair"
>>
>>
>>
>> You are saying something like:
>> (?x exa:p1 certainty(?y, ?c)) <- (?x exa:p11 certainty(?y, ?c)).
>>
>> (?x exa:p1 certainty(?y, ?c)) <- (?x exa:p12 certainty(?y, ?c)).
>>
>> (?x exa:p11 certainty(exa:o1, 0.8)) <- (?x exa:qvalue ?v),
>>lessThan(?v,
>> 10).
>>
>> (?x exa:p12 certainty(exa:o2, 0.7)) <- (?x exa:qvalue ?v), ge(?v,
>>100).
>>
>>
>> ?
>>
>> Is this possible?
>
>Yes. For example the rules for the OWL rule reasoner use this for
>brevity and a bit of efficiency. They glue the OWL RDF triples that
>make up restrictions into a functor, then make deductions on the basis
>of those functors, the functors are an intermediate step only.
>
>> How can I invoke a "structured literal” in a Sparql
>> command?
>
>You can't (other than as a literal with an odd data type), they are a
>non-standard datatype, which is why by default they are hidden from
>external access. Normally you would have other rules which extract the
>pieces of the functors that you want as normal triples (e.g. back into
>an n-ary relationships using bNodes) for query or access them
>programmatically.
>
>[I'm not sure this whole mechanism was worth it in retrospect, and if
>designing a new rule engine I wouldn't include it but there are no plans
>to remove it.]
>
>Dave
>
>>
>> MBA
>>
>> On 12/11/13 14:31, "Dave Reynolds" <[email protected]> wrote:
>>
>>> On 12/11/13 11:54, Miguel Bento Alves wrote:
>>>> Dear Dave,
>>>>
>>>> Can I express myself in Prolog?
>>>>
>>>> I want something like:
>>>>
>>>> v(s1, 8).
>>>> v(s2, 120).
>>>>
>>>> p1(X, Y, C):-
>>>> p11(X, Y, C).
>>>>
>>>>
>>>> p1(X, Y, C):-
>>>> p12(X, Y, C).
>>>>
>>>>
>>>> p11(X, o1, 0.8):-
>>>> v(X, V),
>>>> V < 10.
>>>>
>>>>
>>>> p12(X, o1, 0.7):-
>>>> v(X, V),
>>>> V >= 100.
>>>>
>>>>
>>>> p1(s1, Y, C)?
>>>> Y = o1, C = 0.8
>>>>
>>>> p1(s2, Y, C)?
>>>> Y = o1, C = 0.7
>>>>
>>>>
>>>>
>>>> In best of my knowledge, is not possible what I want. But I¹m new in
>>>> Jena
>>>> and perhaps there are a solution that I can¹t figure out. I think that
>>>> is
>>>> not possible because I need a quad: Subject, Property, Object,
>>>> Confidence.
>>>> I tried with blank nodes but I didn¹t had success.
>>>
>>> If that's really what you want then that is more or less possible,
>>> however I think you'll find you need more.
>>>
>>> The issue is that in your example there is no overlap of rules. So p11
>>> and p12 never occur together, indeed for values of V between 10 and 99
>>> there's no value for p1. Normally once you have uncertainty measures
>>>you
>>> find cases where there are multiple routes to a conclusion, each of
>>> which contributes some confidence/certainty/belief/whatever. In which
>>> case you need to trace out all the possible routes and combine the
>>> confidence measures from each route, which in turn requires dependency
>>> assumptions. If that's where you are headed them I would start with a
>>> tool designed for that job, not start with a rule based system and hope
>>> to extend it.
>>>
>>> However, if there are *really* no overlaps, and only ever a single
>>> deduction path for a single conclusion, and that won't change, then you
>>> could do something in Jena.
>>>
>>> There are two options for handling n-ary relations in Jena rules.
>>>
>>> First you can represent them as groups of triples. To deduce an
>>>instance
>>> of an n-ary relationship from a rule then you need to create a bNode to
>>> represent the instance. For this you can use makeTemp (which a guard to
>>> prevent multiple instantiation) or makeSkolem. The latter is probably
>>> the better choice.
>>>
>>> Alternatively, Jena rules has the notion of a "functor" (bad name)
>>>which
>>> is essentially a structured literal. So you can represent your Y/C
>>>pairs
>>> as certainty(Y,C) and write rules which generate and match triples
>>>whose
>>> object is a certainty(?x,?y) pair. By default these literals are kept
>>> internal to the inference engine but there is an option to make them
>>> visible to external queries (PROPenableFunctorFiltering).
>>>
>>> Dave
>>>
>>>
>>>>
>>>> MBA
>>>>
>>>>
>>>> On 12/11/13 11:19, "Dave Reynolds" <[email protected]> wrote:
>>>>
>>>>> Depending on exactly what semantics you want for confidence
>>>>> coefficients
>>>>> then you probably want to look at a Bayesian network software.
>>>>>
>>>>> Dave
>>>>>
>>>>> On 12/11/13 09:04, Miguel Bento Alves wrote:
>>>>>>
>>>>>>
>>>>>> I want to introduce a confidence coefficient in my rules. For
>>>>>> instance,
>>>>>> in
>>>>>> the example below, let's consider that p11 has a confidence
>>>>>> coefficient
>>>>>> of
>>>>>> 0.8 while p12 has a confidence coefficient of 0.7. When ?x exa:p1 ?y
>>>>>> happens
>>>>>> I want to know the confidence of this conclusion. Any ideas? I have
>>>>>> been
>>>>>> study but I couldn't figure out a good solution.
>>>>>>
>>>>>> (?x exa:p1 ?y) <- (?x exa:p11 ?y).
>>>>>>
>>>>>> (?x exa:p1 ?y) <- (?x exa:p12 ?y).
>>>>>>
>>>>>> (?x exa:p11 exa:o1) <- (?x exa:qvalue ?v), lessThan(?v, 10).
>>>>>>
>>>>>> (?x exa:p12 exa:o2) <- (?x exa:qvalue ?v), ge(?v, 100).
>>>>>>
>>>>>> MBA
>>>>>>
>>>>>>
>>>>>>
>>>>>
>>>>
>>>>
>>>
>>
>>
>
