Re: Newbie question about db normalization theory: redundant keys OK?
From: Bob Badour <bbadour_at_pei.sympatico.ca>
Date: Sat, 15 Dec 2007 09:54:11 -0400
Message-ID: <4763dc87$0$5291$9a566e8b_at_news.aliant.net>
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> As I already said, 6NF baffles me. Why is the relation above not in 6NF?
Date: Sat, 15 Dec 2007 09:54:11 -0400
Message-ID: <4763dc87$0$5291$9a566e8b_at_news.aliant.net>
David Cressey wrote:
> "David Portas" <dportas_at_gmail.com> wrote in message
> news:98a2073c-612a-4f98-bf79-a0d441b99313_at_d21g2000prf.googlegroups.com...
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>>On Dec 14, 7:49 pm, "David Cressey" <cresse..._at_verizon.net> wrote: >> >>> >>>Someone else questioned my categorization of DKNF as "final". What I
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>>>was this: there are no normal forms more restrictive than DKNF. This
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>>>been proven, although I have not seen the proof, and maybe wouldn't be >>>able to follow it if I did. The other participant pointed out that
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>>>table decomposition is possible. That's true, but that doesn't mean
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>>>the resulting schema is of a higher normal form. >>> >> >>Define "more restrictive". My interpretation is that you think a >>schema in DKNF must also be in 6NF, 5NF, ... etc. This is false >>though. Example: >> >>R{a*,b,c} >> >>where a* is the only candidate key. >> >>Given the dependencies {a}->{b} and {a}->{c} relation R IS in DKNF but >>NOT in 6NF. Does this mean 6NF is "more restrictive"? It is true to >>say that 6NF is not a proper subset of DKNF because there are some 6NF >>schemas that are not in DKNF.
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> As I already said, 6NF baffles me. Why is the relation above not in 6NF?
6NF:
6NF has at most one non-key attribute.
R1(a*,b)
R2(a*,c)