Re: constraints in algebra instead of calculus

"Jan Hidders" <hidders_at_gmail.com> wrote in message news:1182241089.155428.89490_at_q69g2000hsb.googlegroups.com...
> On 19 jun, 08:26, "David Cressey" <cresse..._at_verizon.net> wrote:
> > "paul c" <toledobythe..._at_oohay.ac> wrote in message
> >
> > news:vNFdi.37701$NV3.32514_at_pd7urf2no...
> >
> >
> >
> >
> > I can't see much use for grouping on all attributes. It seems to me
that
> > this has to be a null operation.
>
> That would actually make the math harder, which is often a bad sign.
> Right now the definition is quite simple:
>
> R GROUP A AS B = { t[H-A] + (B : { t'[A] | t' in R, t'[H-A]=t[H-A] })
> | t in R }
>
> where
> - H is the set of all attributes of R
> - t[X] is the projection of tuple t on the set of attributes X
> - + is tuple concatenation
> - (B : v) constructs a tuple with a single field B with value v
>
> If you let A be equal to H then
> - H-A is the empty set
> - t[H-A] is the empty tuple () for all t
> - t[A] = t for all t in R,
> so you get:
>
> { () + (B : { t' | t' in R, ()=() }) | t in R } =
> { (B : { t' | t' in R }) | t in R } =
> { (B : { t' | t' in R }) } =
> { (B : R) }
>
> Your suggestion would create an exception to that rule.

It wasn't my intent to make a suggestion. It was supposed to be an observation. An incorrect one, it would appear from your response.

It sounds like I'm mixed up on what GROUP does. In particular, it sounds like I was confusing

R GROUP A AS B with
R GROUP (H-A) AS B Where can I go for a definition of GROUP? Received on Tue Jun 19 2007 - 13:05:23 CEST