Obtaining the full property path from SPARQL query - sparql

Question

I have the following sample data:
#prefix hr: <http://ex.com/> .
#prefix rdf: <http://www.w3.org/1999/02/22-rdf-syntax-ns#> .
#prefix rdfs: <http://www.w3.org/2000/01/rdf-schema#> .
#prefix sch: <http://schema.org/> .
#prefix xml: <http://www.w3.org/XML/1998/namespace> .
#prefix xsd: <http://www.w3.org/2001/XMLSchema#> .
hr:AAAA a rdfs:Class .
hr:BBBB a rdfs:Class .
hr:ClassA a rdfs:Class ;
rdfs:subClassOf hr:AAAA ;
rdfs:subClassOf hr:BBBB .
hr:ClassB a rdfs:Class ;
rdfs:subClassOf hr:ClassA .
hr:ClassC a rdfs:Class ;
rdfs:subClassOf hr:ClassB .
hr:ClassD a rdfs:Class ;
rdfs:subClassOf hr:AAAA .
hr:ClassE a rdfs:Class ;
rdfs:subClassOf hr:ClassD .
What I would like to get as a series of strings are the full rdfs:subClassOf property paths leading from one node to another. In this case, the set of strings I would like to get back would look like or be similar to:
hr:ClassC -> hr:ClassB -> hr:ClassA -> hr:AAAA
hr:ClassC -> hr:ClassB -> hr:ClassA -> hr:BBBB
hr:ClassB -> hr:ClassA -> hr:AAAA
hr:ClassB -> hr:ClassA -> hr:BBBB
hr:ClassA -> hr:AAAA
hr:ClassA -> hr:BBBB
hr:ClassD -> hr:AAAA
hr:ClassE -> hr:ClassD -> hr:AAAA
Is this possible with SPARQL? If so, what would the query look like?
The following query seems like it is close:
SELECT ( GROUP_CONCAT( ?subclass; SEPARATOR = " -> " ) AS ?hierarchy )
WHERE {
?class rdfs:subClassOf+ ?subclass .
}
GROUP BY ?class
but the results are:
http://ex.com/AAAA
http://ex.com/AAAA -> http://ex.com/BBBB
http://ex.com/ClassA -> http://ex.com/AAAA -> http://ex.com/BBBB
http://ex.com/ClassB -> http://ex.com/ClassA -> http://ex.com/AAAA -> http://ex.com/BBBB
http://ex.com/ClassD -> http://ex.com/AAAA
I cannot explain where:
http://ex.com/AAAA -> http://ex.com/BBBB
is coming from or why hr:ClassE is missing from the results.

Answer 1

This is not possible with SPARQL.