Re: constraints in algebra instead of calculus

On May 26, 7:27 am, paul c <toledobythe..._at_oohay.ac> wrote:
> Vadim Tropashko wrote:
> > On May 20, 9:08 pm, paul c <toledobythe..._at_oohay.ac> wrote:
>
> >>Marshall wrote:
>
> >>>Okay, a while back we were talking about writing constraints
> >>>in a language with aspects of the relational calculus, specifically
> >>>the existential and universal quantifiers. The point was made
> >>>that that's unnecessary; the calculus is no more expressive
> >>>than the algebra.
>
> >>>So it ought to be possible to write any constraint from the
> >>>calculus in the algebra.
>
> >>>Well, I'm having a hard time figuring out how to do it. Can
> >>>anyone help?
>
> >>>How does one write a functional dependency in the algebra?
> >>>A foreign key?
>
> >>>Marshall
>
> >>I thought that one was easy, if FK is the set of 'foreign key'
> >>attributes and A is the 'referencing table' and using an op like D&D
> >><AND>, then it's something like A{FK} <AND> B{FK} = A{FK}. (I like this
> >>one because it's particularly easy to implement.)-
>
> > FK is easy indeed. Assuming that we rename attributes of A and B in
> > such a way that x becomes a common FK attribute, a FK constraint is
> > written in relational lattice as
>
> > A \/ X < B \/ X
>
> > where X is the empty relation with attribute x. Naturally, I'd suggest
> > too call the constraint between arbitrary 3 relations
>
> > A \/ C < B \/ C
>
> > as a generalized FK...
>
> Regarding the FD, or key for short, which hasn't been mentioned much so
> far, here's what I've seen so far, assuming one relvar A with
> non-overlapping attribute sets K and FK:
>
> 1) Use a COUNT function to test that the projection on K has the same
> cardinality as A. (I've seen this approach in a number of different
> places, so maybe it is most common.)
>
> 2) Take the PRODUCT of A and A with K and NK RENAME'd to RK and RNK and
> test that RNK = NK whenever K = RK.
>
> 3) Use TTM-style GROUP/UNGROUP to test that
> UNGROUP {NK}
> (GROUP ( (A GROUP {NK} as NK ))
> = A

Given a relation R(x,y,z,t), make the following 3 relations:

1. R'(x,y') = project R to (x,y) and rename y->y'
2. R"(x,y") = project R to (x,y) and rename y->y"
3. Id(y',y") -- identity relation y'=y"

The FD constraint is written as

Id(y',y") /\ R'(x,y') > R'(x,y') /\ R"(x,y")

Without clutter of attributes the inequality looks much cleaner:

Id /\ R' > R' /\ R"

Remarkably, it is shaped very similarly to FK constraint

A \/ X < B \/ X

that was mentioned earlier in the thread.

This is fine, but what are the implications? Can it be demonstrated that Armstrong axioms easily follow from this formulation? Received on Fri Jun 15 2007 - 18:23:33 CEST