Re: Enforcing functional dependecy constraints

From: paul c <toledobythesea_at_oohay.ac>
Date: Fri, 09 Dec 2005 16:14:21 GMT
Message-ID: <x3imf.81318$Eq5.69103@pd7tw1no>

David Cressey wrote:

>>
>>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.
>

I believe it's Heath's Theorem that allows your original projection. According to Date it says "Let R{A,B,C} be a relvar, where A, B, and C are sets of attributes. If R satisfies the FD A->B, then R is equal to the jon of its projections on {A,B} and {A,C}." (In the OP's message, the FD was C->B). But Heath doesn't justify what I said.

cheers,
p Received on Fri Dec 09 2005 - 10:14:21 CST