Re: Enforcing functional dependecy constraints

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

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!

Cheers,
p Received on Fri Dec 09 2005 - 04:35:38 CET