Re: Newbie question about db normalization theory: redundant keys OK?
Date: Mon, 17 Dec 2007 11:45:28 -0400
Message-ID: <47669999$0$5290$9a566e8b_at_news.aliant.net>
David Cressey wrote:
> "Hugo Kornelis" <hugo_at_perFact.REMOVETHIS.info.INVALID> wrote in message
> news:3rabm31r4vab1t5hn83qi9hslvvqk2kklr_at_4ax.com...
>
>>On Sun, 16 Dec 2007 02:57:36 GMT, Brian Selzer wrote:
>>
>>
>>>"David Cressey" <cressey73_at_verizon.net> wrote in message
>>>news:ncR8j.287$qv1.250_at_trndny01...
>>>
>>>>"Bob Badour" <bbadour_at_pei.sympatico.ca> wrote in message
>>>>news:4763dc87$0$5291$9a566e8b_at_news.aliant.net...
>>>>[snip]
>>>>
>>>>
>>>>>6NF:
>>>>>R1(a*,b)
>>>>>R2(a*,c)
>>>>>
>>>>>6NF has at most one non-key attribute.
>>>>
>>>>Thanks for the above definition. It's simple, and easily understood.
> > I >
>>>>actually "invented" this form on my own in my head, but didn't think
> > it >>>>
>>>>was
>>>>important enough to merit giving it a name.
>>>>
>>>
>>>It's also wrong. Consider,
>>>
>>>R(a*,b*,c*) where a*, b* and c* are each keys,
>>>
>>>R is not in 6NF because it can be decomposed into
>>>
>>>R1(a*,b*) and
>>>R2(a*,c*)
>>>even though R doesn't have any non-key attributes!
>>
>>Hi Brian,
>>
>>How should I interpret R(a*, b*, c*)? As a relation with three candidate
>>keys, each over one attribute? Or as a relation with a single candidate
>>key over three attributes?
>>
>>In the former case, R is indeed not in 6NF as I understand it. I guess
>>that Bob was thinking about the primary key only when writing down his
>>explanation for 6NF.
> > In this case, R is not in 5NF either, is it? While Bob didn't explicitly > say so isn't it presumed among us that the definition of 6NF includes the > restriction that it must also be in 5NF?
Bob just gave a simple and informal explanation for why one relation was not in 6NF. With the formal definition of 6NF, one can decompose R by arbitrarily choosing any 2 of the 3 combinations of 2 attributes from R. Received on Mon Dec 17 2007 - 16:45:28 CET