Re: Newbie question about db normalization theory: redundant keys OK?

From: David Portas <dportas_at_gmail.com>
Date: Fri, 14 Dec 2007 12:25:25 -0800 (PST)
Message-ID: <98a2073c-612a-4f98-bf79-a0d441b99313_at_d21g2000prf.googlegroups.com>


On Dec 14, 7:49 pm, "David Cressey" <cresse..._at_verizon.net> wrote:
>
>

> Someone else questioned my categorization of DKNF as "final". What I meant
> was this: there are no normal forms more restrictive than DKNF. This has
> been proven, although I have not seen the proof, and maybe wouldn't be
> able to follow it if I did. The other participant pointed out that further
> table decomposition is possible. That's true, but that doesn't mean that
> the resulting schema is of a higher normal form.
>

Define "more restrictive". My interpretation is that you think a schema in DKNF must also be in 6NF, 5NF, ... etc. This is false though. Example:

R{a*,b,c}

where a* is the only candidate key.

Given the dependencies {a}->{b} and {a}->{c} relation R IS in DKNF but NOT in 6NF. Does this mean 6NF is "more restrictive"? It is true to say that 6NF is not a proper subset of DKNF because there are some 6NF schemas that are not in DKNF.

--
David Portas
Received on Fri Dec 14 2007 - 21:25:25 CET

Original text of this message