Sum values in a recursion way ( n3 / n3logic )

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#1

Hi there,

I'm a newbie learning n3logic. For some interest, I'd like to sum a set of values, which are listed like this:

(:a :b :c) a rdf:List.
:a :hasValue 1.0.
:b :hasValue 2.0.
:c :hasValue 3.0.
:sum :hasValue 0.

I tried to sum them up in a recursive way:

# Recursion
{?L a rdf:List;
     rdf:first ?Z;
     rdf:rest ?R. 
   ?Z  :hasValue ?V.
   :sum :hasValue ?X.
   (?X ?V) math:sum ?Y. 
    }
=> {?Z a :selected.
    ?R a rdf:List.    
   :sum :hasValue ?Y.
}.

Recursion (final step)


{?L a rdf:List;

rdf:first ?Z;

rdf:rest rdf:nil.

?Z  :hasValue ?V.

:sum :hasValue ?X.

(?X ?V) math:sum ?Y.

}

=> {?Z a :final;

a :selected.

:sum :hasFinalValue ?Y.

}.

But it fails into a dead loop. I guess the problem occurs when updating :sum's value, but cannot find a way to make it work. Is there a way to compute this sum in n3, or is it just not suitable for n3 to do such thing? (Here is the full code: .n3 file)

Sorry if the problem is inappropriate here

#2

If you have a SPARQL 1.1 processor, you can do the following: 

SELECT sum(?elem)
WHERE 
{  ?list rdf:rest*/rdf:first ?elem
}
...where ?list is bound to the list node.

I think there have been warnings here and elsewhere not to get too tied up in N3 and/or n3logic. It's not a standard and its features aren't faring all that well in the RDF 2 working group. Standards compliance-speaking, it's best to stick with Turtle/SPARQL, at least in the short term.

The problem is quite appropriate here.

#3

math:sum will compute the sum of all elements, you don't need to do something else then: 

@prefix math: http://www.w3.org/2000/10/swap/math# .

@forAll :x .
{
(1.0 2.0 3.0 4.0) math:sum :x .
} => {
  [ :sum :x ] .
} .
this except if you don't need to compute partial sums or want to experiment new ways, :)

edit: my apologise, i didn't notice that you have statements as elements of the list. You need first to extract the values into a separate list.