RE: Comments on the new RDF Test Cases draft

> -----Original Message-----
> From: Dan Brickley [mailto:danbri@w3.org]
> Sent: Wednesday, May 08, 2002 2:16 AM
> To: Massimo Marchiori
> Cc: Www-Rdf-Comments@W3. Org
> Subject: Re: Comments on the new RDF Test Cases draft
>
>
> On Wed, 8 May 2002, Massimo Marchiori wrote:
>
> > [sent already, but it didn't seem it went thru... maybe just the thin
> > air of Hawaii... retrying now, sorry again for double postings]
> >
> > I just quickly read (yes, same flight... ;) the new RDF Test
> Cases as per
> > http://www.w3.org/TR/rdf-testcases/
> > In the main text (not time so far to read all the actual use cases... ;)
> > there is in Section 2:
> > <quote>
> > A parser is considered to pass the test if it produces a graph
> isomorphic
> > with the graph described by the N-triples output document.
> > </quote>
> >
> > This is wrong, according to the standard definition of graph isomorphism
> > (care when using words without accurate definitions...!).
>
> Could you give a citation for the 'standard definition', and outline how
> we deviate from that concept?
Citation: every book on graph theory, or just look on Google.
Definition: a graph isomorphism is an adjacency-preserving bijective map
between vertices
(bijection is clear, adjacency-preserving means that two vertices are
connected in one
of the graphs if and only if the corresponding, via the map, ones in the
other graph are
connected).

> > You'd define it using the RDF-MT semantical equivalence instead.
>
> Hmm, not so sure. RDF parsers aren't expected to exhibit knowledge of all
> the semantic equivalencies implied by RDF's MT.
Beep! (wrong button! ;)
Quiz: Why are you doing the MT...?
Answer to your specific question: correct, RDF parsers don't have to exhibit
knowledge of
the MT's semantic equivalence. But the above change just implies that a
parser,
to be considered RDF compliant, has to generate a graph that has the same
semantics as the one you provide in the test case. Rephrased: each test case
you put in the draft identifies one *equivalence class* of graphs, and a
parser is free to choose whatever representative of this equivalence class
it likes.
The moment you will *check* for compliance, then yes, you will have to
have knowledge of the MT semantics...

-M

Received on Thursday, 9 May 2002 02:50:16 UTC