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

"David Cressey" <cressey73_at_verizon.net> wrote in message news:e9v9j.3457$L91.2437_at_trndny05...
>
> "Hugo Kornelis" <hugo_at_perFact.REMOVETHIS.info.INVALID> wrote in message
> news:3rabm31r4vab1t5hn83qi9hslvvqk2kklr_at_4ax.com...
>> On Sun, 16 Dec 2007 02:57:36 GMT, Brian Selzer wrote:
>>
>> >
>> >"David Cressey" <cressey73_at_verizon.net> wrote in message
>> >news:ncR8j.287$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?
>
>

A relation R(a*,b*,c*) that has three candidate keys, a*, b* and c* is indeed in 5NF. Received on Mon Dec 17 2007 - 15:34:08 CET