# Re: blank nodes out the wazoo

From: Jeremy Carroll <jjc@hpl.hp.com>
Date: Fri, 13 Jun 2003 22:49:12 +0300

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)

Jeremy
```
Received on Friday, 13 June 2003 16:49:17 UTC

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