Re: Enforcing functional dependecy constraints

From: David Cressey <david.cressey_at_earthlink.net>
Date: Fri, 09 Dec 2005 12:40:31 GMT
Message-ID: <3Xemf.1296$nm.109@newsread2.news.atl.earthlink.net>

"paul c" <toledobythesea_at_oohay.ac> wrote in message news:eY6mf.71805$ki.39213_at_pd7tw2no...
> David Cressey wrote:
> > "paul c" <toledobythesea_at_oohay.ac> wrote in message
> > news:WwKlf.63515$ki.5786_at_pd7tw2no...
> >
> >>paul c wrote:
> >>
> >>>...
> >>>
> >>>I agree with David but also I guess that the above comes down to saying
> >>>that BCNF is possible either with S(A,C) and T(C,B) or with U(A,B) and
> >>>T(C,B), assuming that C is a key.
> >>>
> >>
> >>Oops, meant to say "assuming C is a key of T".
> >>
> >>
> >>p
> >
> >
> > I don't see how to reconstruct R (A,B,C) from U(A,B) and T(C,B)
> >
> >
> >
> >

>

> Thanks, David C. You're right and I'm wrong again. Just made another
> example for myself and using my earlier substitutes U (all-key) allows a
> teacher to use several books.

>
>

> But the FD AB->C says that a teacher can't use the same book for two
> different courses. Since U is many-to-one (ie. many C to one B), the
> join could result in rows where many C's show up for the same B and thus
> the same (A,B), allowing a teacher to use the same book for two
> different courses.

>
>

> Thinking about this, I wonder if it is because C, being the the
> determinant of the stricter FD, must be in both projections (whereas
> that's not so with U(A,B) and T(C,B).

>
>

> I guess I'll have to go study the theory again. It's not a waste of my
> time, but sorry for wasting everybody else's!

To tell the truth, when you get beyond 3NF, I'm dealing with theory rather than practice. And theory is not my strength.

The reason I waited for several days was to see what other people, maybe better trained than I, would do with the "elementary problem" that x posed.

All I was working on in this problem was "every determinant is a candidate key". That led me to T(C,B).
One I factored that out, All I was left with was S(A,C). As nearly as I can tell, the determinant of S is the entire relation. Received on Fri Dec 09 2005 - 06:40:31 CST