Re: 4NF is Where It Is At! [WAS Re: 1:1 relationships]

Barry wrote:
>
> I find myself forced to come full-circle. I originally offered my
> vaguely-recalled expression of 4NF as:
>
> Given key k, non-key attributes of a and b, and f(k)->a, then f(k)->b for
> {k,a,b} to be in 4NF.
>
> In other words, if g(k)->b (i.e., the relationship of attribute b to the
> key is different to that of a), then {k,a,b} must be decomposed to {k,a}
> and {k,b} to satisfy 4NF.

No, no, no! I am very sorry but this is *really* not correct. You seem to be talking aboout functional dependencies here and these are already dealt with by BCNF and lower. Moreover, it is also not correct that if you have two functional dependencies which are semantically different then you should split the table. And that is simply not the case; no normal form (not even 5NF, the strongest one, says that).

But I am also not really clear about what you are saying here exactly. Let me ask you a question to verify something. If I have a table R(name, address, phone-number) with key 'name', would you then split it because you have the different dependencies address(name) -> address and phone(name) -> pone-number?

> But I do remember struggling with the ... dare I say it? ... Date
> formulations of Normalisation, and then bumping into statements of
> each more in the terms I cite above, and finding it much easier to
> understand.

Well, to be completely honest, I have been teaching normalization for a couple of years from his book and I have also been struggling with his definitions, but for exactly the opposite reason. :-) Date often tries to take intuitive shortcuts where the real formal definitions are actually more clear (provided you willing and able to understand them). Also he tends to simplify things (assume only one key for instance) and then later gets into trouble when he has to explain higher normal forms.

Kind regards,

-- 
  Jan Hidders