- From: Patrick Stickler <patrick.stickler@nokia.com>
- Date: Fri, 25 Jan 2002 12:16:02 +0200
- To: Pat Hayes <phayes@ai.uwf.edu>, Graham Klyne <Graham.Klyne@MIMEsweeper.com>
- CC: RDF Core <w3c-rdfcore-wg@w3.org>
On 2002-01-24 18:31, "ext Pat Hayes" <phayes@ai.uwf.edu> wrote:
>> At 10:19 AM 12/14/01 -0800, Sergey Melnik wrote:
>>>> Dan has raised an issue, rephrased by Pat as how many triples result from
>>>> merging:
>>>>
>>>> foo bar "10" .
>>>>
>>>> and
>>>>
>>>> foo bar "10" .
>>
>> Ha! I came up with exactly this issue in my review of the latest
>> model theory draft. Do we have a consensus yet?
>
> Well, let me suggest what the answer should be. This may seem odd, but....
>
> I think the answer depends on whether this is a single graph being
> 'merged' with itself, or whether it is two distinct, but identical
> (ie isomorphic) graphs. If the former, then the 'merge' is just the
> same graph, with 2 nodes and one arc: a single triple. If the latter,
> then it is a larger graph with three nodes (assuming foo is a uriref)
> and two arcs. This is because different occurrences of a 'bare'
> literal have to be treated as distinct nodes (in case one of them
> should get itself attached to a different datatyping scheme from the
> other...), unlike urirefs. Notice that if you did a similar exercise
> with two copies of an all-uriref triple:
>
> foo bar baz .
>
> then the merged graph would be the same in either case, ie it would
> simply be a single triple; since in this case, the merging would
> (re-)identify the two copies of each uriref.
>
> Pat
>
Thanks, Pat, for that excellent clarification.
This is what is presumed by the TDL proposal, that literal labeled
nodes do not participate in tidying, and that every occurrence of
a literal has a unique node in the graph.
Patrick
--
Patrick Stickler Phone: +358 50 483 9453
Senior Research Scientist Fax: +358 7180 35409
Nokia Research Center Email: patrick.stickler@nokia.com
Received on Friday, 25 January 2002 05:59:28 UTC