# Re: The "standard" way to get to 3NF

Date: Sat, 10 Apr 2004 15:10:12 +1000

Message-ID: <650f709mnj6r166utdr3i3l4dholqf2msb_at_4ax.com>

On Sat, 10 Apr 2004 04:31:06 GMT, Jonathan Leffler <jleffler_at_earthlink.net> wrote:

*>Jan Hidders wrote:
**>> Jan Hidders wrote:
*

>>> [...] The usual algorithm that gets you to 3NF in one step (the one

*>>> using the minimal cover) splits as little as possible. See for example
**>>> sheet 46 on:
**>>>
**>>> http://cs.ulb.ac.be/cours/info364/relnormnotes.pdf
**>>
**>> Did anyone notice that this algorithm is actually not correct? Take the
**>> following example of a relation R(A,B,C,D,E) with the set of FDs:
**>>
**>> { AB->C, AB->D, BC->D }
**>
**>You've lost E - was that a mistake in the FD's or in the example relation?
**>
**>> It is clear that the relation ABCD is not in 3NF. Since the set of FDs
**>> it is already a minimal cover the resulting decomposition is:
**>>
**>> { ABCD, BCD }
**>>
**>> But that gives us our old relation back (plus a projection) so this is
**>> definitely not in 3NF.
**>>
**>> The strange thing is that this algorithm appears as such in the Elmasri
**>> and Navathe and also in Date (but not Ullman). Surely these two major
**>> textbooks would not get the most fundamental algorithm in normalization
**>> theory wrong? Or would they? Reminds me a little of the
**>> misrepresentation of 5NF in many textbooks.
**>>
**>> -- Jan Hidders
*

See p321Table 10.1 in E&N 4th Edition & elswhere in the text.

It specifically mentions the need to preserve the Candidate Key (CK) as a separate relation in particular when the CK is not on the LHS of any FD in the minimal cover.

So you need an extra relation r3(A, B, E). Received on Sat Apr 10 2004 - 07:10:12 CEST