Re: MV Keys

"Jon Heggland" <heggland_at_idi.ntnu.no> wrote in message news:MPG.1e775ecac7857681989790_at_news.ntnu.no...
> In article <1141662881.772158.68960_at_j52g2000cwj.googlegroups.com>,
> jog_at_cs.nott.ac.uk says...
> > > > Because some appear to be missing this point, and hence confusing
> > > > actually what a compound type is. A comma seperated string, for
> > > > example, sitting in a relational database is hardly a compound type
as
> > > > far as the RM is concerned. It is merely a string.
> > >
> > > What difference does it make to the RM?
> >
> > I don't follow your question.

> Why should the RM be concerned whether something is a compound type or
> not?

Yesterday when I was going from downtown to home I noticed that the snowing became raining at some point.
I wondered if the raining replaced the snowing or not. Inquisitive minds like to know. Do you have an answer ?

> > > > > That doesn't make sense to me. Compound attributes are valid if
the
> > > > > system can decompose them, and if the system can't decompose them,
> > > > > they're not compound?
> > > >
> > > > Yes. What about that doesn't make sense to you?
> > >
> > > It is an overly complicated way of saying that compound attributes are
> > > by definition always "valid". What, then, is the point of making the
> > > distinction between simple and compound?
> >
> > It doesn't mean that at all.

> Perhaps that's not what you meant, but it is what you said. Below, don't
> you say that if there are no "decomposition operators", the type is not
> compound (and hence not a problem for the RM); and if there are such
> operators, it is also not a problem for the RM?

> > It is stating that one cannot talk about
> > items that we view in our heads as "compound" (sets say) being plugged
> > into a model without explaining how the type will be manipulated by the
> > algebra underpinning the system. Ignore this and as far as I can see
> > you no longer have a compound datatype at all.

> Well, yes. A type must have operators in order to be of much use---but
> the way I see it, the relational algebra (RA) doesn't care, as long as
> there exists an equality operator.

Equality is an operator or a relation ?
How do you compare two sets of relations ? Received on Tue Mar 07 2006 - 12:46:34 CET