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

From: Bob Badour <bbadour_at_pei.sympatico.ca>
Date: Sat, 15 Dec 2007 09:54:11 -0400
Message-ID: <4763dc87$0$5291$9a566e8b@news.aliant.net>

David Cressey wrote:

> "David Portas" <dportas_at_gmail.com> wrote in message
> news: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.

>
> As I already said, 6NF baffles me. Why is the relation above not in 6NF?

6NF:
R1(a*,b)
R2(a*,c)

6NF has at most one non-key attribute. Received on Sat Dec 15 2007 - 07:54:11 CST