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

"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? Received on Mon Dec 17 2007 - 14:51:06 CET