Re: Jan's well-defined view updates definition

"Jan Hidders" <jan.hidders_at_pandora.be> wrote in message news:d14ab.25480$rw3.1352807_at_phobos.telenet-ops.be...
> Mikito Harakiri wrote:
> >
> > We can't continue going round and round, so let's move on to the second
> > paragraph of your definition.
>
> Well, I would still be interested in your opinion on whether the
definition
> I gave makes sense or not, and what the intuition is behind the
definitions
> you gave. It's almost as if you are affraid of such discussions.
>
> > "Jan Hidders" <jan.hidders_at_pandora.be> wrote:
> >> Finally, we define a view V as *updatable* by a certain set of updates
U
> > if
> >> for all databases D that satisfy the schema all updates in U are
> >> well-defined. Additionally we say that V is *commutatively updatable*
by
> >> U if for all databases D that satisfy the schema it holds that if two
> >> series of updates from U have the same result when applied to Q(V) and
> >> perform only well-defined updates then they result in the same database
> >> when applied to D.
> >
> > Could you please provide a non-commutatively updatable example that we
can
> > discuss?
>
> Lets say I have base relations R(a,b) and S(b,c) with a foreign key R.b ->
> S.b and a view V that is defined by the natural join of R and S. The
> additions and deletions are both well-defined but if I add the tuple
> (a1,b1,c1) and then remove it then the end result is an additional tuple
> (b1,c1) in S, which is not the same as the end result of adding 0 tuples.
> So for the class of updates that consists of inserts and deletes it is not
> commutatively updatable.

Why would one have (b1,c1) in S instead of (a1,b1) in R or neither? Received on Thu Sep 18 2003 - 00:07:59 CEST