Sparql - Traverse to top (broadest)

Question

I have an internal triples dataset.

I am trying to implement a typeahead feature for an application using the dataset.

I am trying to figure out how I can conditional traverse to the Top Concept

In this picture, if I searched for dreamworks, It would be 3 layers down (Business Organizations -> Private Company -> Dreamworks).

If I do something else, obviously it will be at different layers of the graph.

I am trying to get the value "Organization" or "Person" as the top level.

This very straightforward query works.

        SELECT DISTINCT  ?subjectPrefLabel ?a ?b ?o
    WHERE
        { 
            ?subject skosxl:prefLabel/skosxl:literalForm ?subjectPrefLabel .
    ?subject skos:broader/skos:broader/skos:broader/skosxl:prefLabel/skosxl:literalForm ?a


    FILTER regex(?subjectPrefLabel, "Dreamworks", 'i')
        }
    ORDER BY ?subjectPrefLabel

However, obviously this is unsustainable, since the developer would need to know how many levels down the hierarchy in order to issue the correct number of skos:broader.

Is there anyway I can conditionally discover the highest level of the hierarchy?

As this is an internal Ontology, I cannot share any data, so I know it will be hard to work with, any advice is appreciated.

Answer 1

Using the kleene star operator * incombination with a FILTER that there is no more broad concept should return the top concepts only:

SELECT DISTINCT  ?subjectPrefLabel ?a ?b ?o WHERE { 
    ?subject skosxl:prefLabel/skosxl:literalForm ?subjectPrefLabel .
    ?subject skos:broader* ?concept. 
    ?concept skosxl:prefLabel/skosxl:literalForm ?a
    FILTER NOT EXISTS { ?concept skos:broader ?supConcept } 
    FILTER regex(?subjectPrefLabel, "Dreamworks", 'i')
}
ORDER BY ?subjectPrefLabel