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

From: Brian Selzer <brian_at_selzer-software.com>
Date: Sun, 16 Dec 2007 13:39:18 GMT
Message-ID: <aU99j.30510$lD6.18268@newssvr27.news.prodigy.net>

"David Portas" <REMOVE_BEFORE_REPLYING_dportas_at_acm.org> wrote in message news:roGdnS9k5tXOePnanZ2dnUVZ8h6dnZ2d_at_giganews.com...

> "Brian Selzer" <brian_at_selzer-software.com> wrote in message 
> news:Au09j.31688$JD.5283_at_newssvr21.news.prodigy.net...

>>
>> "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!
>>

>
> Wrong in what way? Your example doesn't contradict Bob's statement that: 
> "6NF has at most one non-key attribute".
>

The example shows a table that has at most one non-key attribute that is not in 6NF.

> -- 
> David Portas
>
> 

Received on Sun Dec 16 2007 - 07:39:18 CST