- 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