Re: foreign key constraint versus referential integrity constraint

From: Tegiri Nenashi <tegirinenashi_at_gmail.com>
Date: Wed, 28 Oct 2009 13:29:32 -0700 (PDT)
Message-ID: <c5b63b8e-045e-41d6-be5e-1ce753a15c0e_at_a37g2000prf.googlegroups.com>


On Oct 28, 10:41 am, paul c <toledobythe..._at_oohay.ac> wrote:
> Mr. Scott wrote:
> > "Marshall" <marshall.spi..._at_gmail.com> wrote in message
> >news:386975f9-472c-4184-8661-5c3d1e2f7621_at_r24g2000prf.googlegroups.com...
> >> On Oct 24, 10:53 am, Keith H Duggar <dug..._at_alum.mit.edu> wrote:
> >>> Anyhow, the question here is not one of our imagination but rather
> >>> simply this: if it makes sense for the RM to support constraints
> >>> on relational /values/ (taken on by variables) why does it not
> >>> make sense to support constraints on relational /expressions/?
> >>> That is a question of general principle not specific design.
> >> This question, it seems to me, is clear and to the point.
> >> And I would answer it by saying that we shouldn't really
> >> even make the distinction! (At least not formally.)
>
> > I think we should make the distinction, and formally.
>
> > (p /\ q) -> r   is not the same as   (p -> r) /\ (q -> r)
> > but  (p \/ q) -> r   is the same as   (p -> r) \/ (q -> r)
>
> > A view consisting of a natural join, for example, represents a set of
> > conjunctions.  Each row of the join represents a conjunction of
> > propositions, one for each operand.  A constraint defined on a join would be
> > of the form (p /\ q) -> r.  That is definitely not the same as constraints
> > defined on one or more tables, which would take the form (p \/ q) -> r.
> > ...
>
> Forgot to mention that I don't see that a "a constraint defined on a
> join" would necessarily be "of the form (p /\ q) -> r".  I had thought
> that many people think it could be any truth-valued expression such as
> "(p /\ q) = r".
>
> This leads me to think that most, if not all, view definitions can be
> interpreted as constraints.  It is interesting to me to then ask what
> makes a view different from a base.  Is it enough to say that a view
> always has one constraint (of possibly several) that is an equality and
> a view may be 'updated' without reference to the view?
>
> A more opaque way but perhaps less useful way of saying this is that a
> relation's definition in the first place amounts to nothing more than a
> constraint.

Is view definition a constraint? IMO it's purely terminological matter. Consider relations x and y defined by some algebraic identities. Is adding new view z (as a function of x and y) adding a constraint to the system?

Let's analyze a simpler example. Consider two real values constrained by the equality:

x + y = 5

Is introducing a new variable z, say

z = x - 2y

a new constraint imposed onto the system? Not really, because, variable z is redundant and can be eliminated, and it doesn't affect the formal property of the system of being under constrained. Received on Wed Oct 28 2009 - 21:29:32 CET

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