SPARQL property paths based on a new property defined in a CONSTRUCT subquery

Question

Given the following schema, "driver-passenger" lineages can be easily seen:

tp:trip a owl:Class ;
    rdfs:label "trip"@en ;
    rdfs:comment "an 'asymmetric encounter' where someone is driving another person."@en .

tp:driver a owl:ObjectProperty ;
    rdfs:label "driver"@en ;
    rdfs:comment "has keys."@en ;
    rdfs:domain tp:trip ;
    rdfs:range tp:person .

tp:passenger a owl:ObjectProperty ;
    rdfs:label "passenger"@en ;
    rdfs:comment "has drinks."@en ;
    rdfs:domain tp:trip ;
    rdfs:range tp:person .

Consider the following data:

<alice> a tp:person .
<grace> a tp:person .
<tim> a tp:person .
<ruth> a tp:person .

<trip1> a tp:trip ;
    tp:participants   <alice> , <grace> ;
    tp:driver         <alice> ;
    tp:passenger         <grace> .

<trip2> a tp:trip ;
    tp:participants   <alice> , <tim> ;
    tp:driver         <alice> ;
    tp:passenger         <tim> .

<trip3> a tp:trip ;
    tp:participants   <tim> , <grace> ;
    tp:driver         <tim> ;
    tp:passenger         <grace> .

<trip4> a tp:trip ;
    tp:participants   <grace> , <ruth> ;
    tp:driver         <grace> ;
    tp:passenger         <ruth> .

<trip5> a tp:trip ;
    tp:participants   <grace> , <tim> ;
    tp:driver         <grace> ;
    tp:passenger         <tim> .

Now let a "driver-passenger descendent" be any tp:passenger at the end of a trip sequence where the tp:passenger of one trip is the tp:driver of the next trip

Ex. <ruth> is a descendent of <alice> according to the following sequence of trips:

<trip2> -> <trip3> -> <trip4>.

Question: How to get the (ancestor,descendent) pairs of all driver-passenger lineages?

Attempt 1: I initially tried the following CONSTRUCT subquery to define an object property: tp:drove, which can be easily used in a property path. However, this did not work on my actual data:

SELECT ?originalDriver ?passengerDescendent
WHERE {
    ?originalDriver tp:drove+ ?passengerDescendent .
    {
        CONSTRUCT { ?d tp:drove ?p . }
        WHERE { ?t a tp:trip .
                ?t tp:driver ?d . 
                ?t tp:passenger ?p .}        
    }
}

Attempt 2: I tried to create property path which expresses an ancestor as the driver of a passenger, but I don't think I've properly understood how this is supposed to work:

(tp:driver/^tp:passenger)+

Regarding MWE: Is there some kind of RDF sandbox that would allow me to create an MWE by defining a simple ontology like tp above, along with some sample data? The following "playgrounds" are available but none of them seem to support defining a toy ontology: SPARQL Playground, SPARQL Explorer.

Notes on related content:

This question is directly related to a previous question, but no longer requires saving the paths themselves, a feature not directly supported by SPARQL 1.1.

This answer by Joshua Taylor seems relevant, but doesn't address the identification of specific types of paths, such as the lineages defined above.

score 0 · Accepted Answer · answered Sep 24 '20 at 22:25

0

This one seems to do the trick:

select ?driver ?passenger where {
?driver (^tp:driver/tp:passenger)+ ?passenger .
filter( ?driver != ?passenger)
}

The filter condition can be removed if you want to also see relationships that lead back to the same person.

answered Sep 24 '20 at 22:25

karthik kotha

51
3

Thanks! However, on a query with actual data Virtuoso is complaining about lack of resources. I tried the following modifications (still got 42000): 1) Limit 10 results, 2) wrapping initial selection in a subquery, 3) specifying a single IRI for the ?driver – mkk Sep 24 '20 at 22:59
instead of a subquery, could you use VALUES? . Could you try using increasing the memory limit to say 4G ? You could also use a limit of 100 to begin with. – karthik kotha Sep 24 '20 at 23:02

SPARQL property paths based on a new property defined in a CONSTRUCT subquery

1 Answers1