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

From: Jan Hidders <jan.hidders_at_REMOVETHIS.pandora.be>
Date: Sat, 10 Apr 2004 08:30:29 GMT
Message-ID: <FgOdc.66173$337.4619398@phobos.telenet-ops.be>


Jonathan Leffler 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?

Oops. That's a mistake in the example relation. Sorry about that.

Received on Sat Apr 10 2004 - 03:30:29 CDT

Original text of this message

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