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

From: Brian Selzer <>
Date: Mon, 17 Dec 2007 09:43:35 GMT
Message-ID: <bxr9j.54127$>

"Hugo Kornelis" <> wrote in message
> On Sun, 16 Dec 2007 02:57:36 GMT, Brian Selzer wrote:
>>"David Cressey" <> wrote in message
>>> "Bob Badour" <> wrote in message
>>> news:4763dc87$0$5291$
>>> [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
>>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 the latter case, the decomposition is incorrect. R can't be
> decomposed.
> Best, Hugo
Received on Mon Dec 17 2007 - 10:43:35 CET

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