From: Jeremy Carroll <jjc@hpl.hp.com>

Date: Fri, 13 Jun 2003 22:49:12 +0300

To: w3c-rdfcore-wg@w3.org

Message-Id: <200306132249.12048.jjc@hpl.hp.com>

Date: Fri, 13 Jun 2003 22:49:12 +0300

To: w3c-rdfcore-wg@w3.org

Message-Id: <200306132249.12048.jjc@hpl.hp.com>

Jeremy: >Let the nodes of R = V(G) U V(H) U { g, h, x } (all distinct blank nodes) > >Let the triples of R all have predicate rdf:value (which we will omit) > >Let the triples of R be E(G) U E(H) U { <g, g'> | g' in V(G) } > U { <h, h'> | h' in V(H) } U { <x, g>, <x, h> } Pat: > But not necessarily the reverse. (that is true ...) > For example let G be > <a b> <b c> <c a> > and H be > <d e> <e f> <d f> > then the instance a=d and b=e gives a redundancy (instance which is a > proper sub-RDFgraph) ???? R (on my construction is) x g x h g a g b g c h d h e h f a b b c c a d e e f d f R' (replacing a=d b=e) x g x h g a g b g c h a h b h f a b b c c a a b (duplicate) b f a f doesn't look anything like a subgraph. Even if you ignore all the triples I added it doesn't S a b b c c a d e e f d f S' a b b c c a b f a f S' contains a node (b) which has two outbound edges and one inbound edge, unlike S. irredunancy is not a local phenomenon - I need to work on my NP completeness proof but you've more work to do on your P proof! (Note to rest of group these pairs are in fact triples with a missing rdf:value in the middle; and all the nodes are blank - this might have something to do with RDF) JeremyReceived on Friday, 13 June 2003 16:49:17 EDT

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