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

Date: Sun, 16 Dec 2007 08:48:25 -0000

Message-ID: <78ydnSlTF7nFe_nanZ2dnUVZ8tChnZ2d_at_giganews.com>

"Hugo Kornelis" <hugo_at_perFact.REMOVETHIS.info.INVALID> wrote in message
news:bdl8m3tv1ucqf11fp4sak24chm226s7q0a_at_4ax.com...

> On Sat, 15 Dec 2007 09:54:11 -0400, Bob Badour wrote:

*>
**>>
**>>6NF:
**>>R1(a*,b)
**>>R2(a*,c)
**>>
**>>6NF has at most one non-key attribute.
**>
**> Hi Bob,
**>
**> Isn't the latter statement an oversimplification? Unless I am truly
**> misunderstanding 6NF, I'd say that the scema below is 6NF even though R1
**> has two non-key attributes:
**>
**> R1(a*,b,c)
**> R2(b*,c*,d)
**>
**> Best, Hugo
*

Your R1 is not in 6NF because it can be further decomposed as I already demonstrated in my example.

In R2 does "b*,c*" represent one candidate key or two? If two, then it can be further decomposed and it isn't in 6NF.

To repeat the definition I already quoted:

"Relvar R is in sixth normal form, 6NF, if and only if it can't be nonloss decomposed at all (other than into the identity projection of R). Observe, therefore, that 6NF is the ultimate normal form with respect to normalization as conventionally understood; in particular, every 6NF relvar is in 5NF."

-- David PortasReceived on Sun Dec 16 2007 - 09:48:25 CET