Re: constraints in algebra instead of calculus

From: Jan Hidders <hidders_at_gmail.com>
Date: Tue, 19 Jun 2007 01:18:09 -0700
Message-ID: <1182241089.155428.89490@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.

• Jan Hidders

Received on Tue Jun 19 2007 - 03:18:09 CDT