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Syntax NP Complete - Hamiltonian Paths from Jeremy Carroll on 2003-04-02 (www-webont-wg@w3.org from April 2003)
Syntax NP Complete - Hamiltonian Paths
- From: Jeremy Carroll <jjc@hpl.hp.com>
- Date: Wed, 2 Apr 2003 15:44:21 +0300
- To: www-webont-wg@w3.org
- Message-Id: <200304021544.21956.jjc@hpl.hp.com>
On Friday while working on my formal objection I noted that there might be a relationship to the NP complete problem of Hamiltonian paths (i.e. finding a path through a graph that visits every node) and the EquivalentClasses mapping rule. Unfortunately, there is. I am sorry for not noticing earlier. It is not hard to fix, but I am not sure where I send my observations. Proof: Given an undirected., unlabelled graph G over the integers 1, ... n, and a tool that can tell if an RDF/XML document is in OWL DL or not, we can determine if there is a Hamiltonian path in the following fashion: 1: Is G connected, (O(n^3) Roy's algorithm, often misattributed to Warshall) If NO then there is no Hamiltonian Path. Otherwise: Construct RDF Graph _:n<i> owl:equivalentClass _:n<j> . For all i, j s.t. {i,j} is in G. _:n<i> rdf:type owl:Class . i = 1 .... n _:n<i> owl:unionOf rdf:nil . i = 1 ... n Then this RDF Graph is in OWL DL iff there is a Hamiltonian path in G. ========== Possible fixes include one of: 1: use B.1, B.2; 2: change mapping rule to EquivalentClasses( d1, .... dn ) T(di) owl:equivalentClass T(dj) . or T(dj) owl:equivalentClass T(di) . For all i,j s.t {i,j} in G, an arbitrary connected graph over {1, 2, ... n } 3: delete optional part of mapping rule for EquivalentClasses
Received on Wednesday, 2 April 2003 08:44:00 UTC