Re: How to normalize this?
Date: Mon, 6 May 2013 17:18:38 +0200
Message-ID: <5187c9ce$0$574$e4fe514c_at_dreader34.news.xs4all.nl>
On 2013-05-06 12:26:10 +0000, Erwin said:
> Op maandag 6 mei 2013 00:52:44 UTC+2 schreef Jan Hidders het volgende:
>> On 2013-05-05 19:32:35 +0000, Erwin said:
>>
>>> That one does have the inter-component rule that R1 JOIN R2 === R1
>>> JOIN>> > R3 (as I pointed out), no ?
>>
>> Only if you want to be information-equivalent but conventional>>
>> normalization does not require that. It only aims at obtaining a>>
>> lossless decomposition.
>
> Well there's the whole contradiction isn't it.
>
> How can deliberate admission of "information differences" coexist with
> "aims of being lossless" ?
>
> If "information differences" are deliberately allowed, this can mean
> either or both of two things : something may be added, something may be
> lost.
> I contend that the "lossless" in conventional normalization can be
> upheld only for such a perverse and narrow meaning of the term that the
> claim (of having "lossless" decompositions) becomes essentially
> meaningless/futile/irrelevant/...
That's an .. er .. interesting claim. :-)
> Lossless then seems to mean purely (exaggerating a little bit) that
> "all the original attributes are still in the design, somewhere,
> somehow". Relation schemas as mere sets of attributes, and
> normalization as manipulation/set-algebraic computation on sets of sets
> of attributes, and the only restriction being that the union of the
> final sets of attributes must be equal to the original set of
> attributes (/ union of sets of attributes).
Nope.That's a necessary condition, but certainly not a sufficient condition.
- Jan Hidders