Re: How to normalize this?

From: Erwin <e.smout_at_myonline.be>
Date: Mon, 6 May 2013 12:27:19 -0700 (PDT)
Message-ID: <392e9eae-2482-4463-b5ac-35efe9131cc4_at_googlegroups.com>


Op maandag 6 mei 2013 17:19:36 UTC+2 schreef nvitac..._at_gmail.com het volgende:
> On Monday, May 6, 2013 12:52:44 AM UTC+2, Jan Hidders wrote:
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>
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> > >>>> 1. Apply the 3NF normalization procedure we discussed earlier
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> > >>>> 2. Include with each component only the FDs that generated that component
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> >
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> > >>> It's not clear to me what you are describing or how it relates to>> >
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> > >>> arguing against stopping at 3nf.
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> >
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> > >> I've described to you a normalization algorithm that seems to produce
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> > >> a>> schema in BCNF that is also lossless-join and dependency
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> > >> preserving,>> but also has no inter-component dependencies. That's not
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> > >> supposed to be>> possible according to the usual normalization
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> > >> explanation. So what went>> wrong here?
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> >
>
> > > That's simply not true is it ?
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> > > You are referring to the three-table decomposition, right ?
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> >
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> > No. I'm referring to the algorithm that is described by steps 1. and 2.
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> > above. It's an adapted variant of the usual algorithm that produced the
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> > three-table decomposition. That older algorithm does not always produce
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> > a schema in in BCNF. Mine seems to do so.
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>
>
> Hi, this is my first post in this group. I hope you won't mind if I dive into this very interesting discussion.
>
>
>
> One exercise that I usually give to my students seems to fit exactly the above. First, I ask them to prove that the schema R(A,B,C,D,E) with FDs F = {AB -> C, C -> D, D -> B, E ->D} is not in 3NF (which they usually do easily). Then I ask them to decompose it in 3NF and they typically come out with:
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>
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> R1(A,B,C), {AB -> C}
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> R2(C,D), {C -> D}
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> R3(D,B), {D -> B}
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> R4(D,E), {E -> D}
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> R5(A,E), {}
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>
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> Finally I ask them: does this decomposition preserve FDs? (“yes, by construction”). Is C -> B implied by F? (“yes, by transitivity”). Is C -> B preserved or not in the decomposition? (panic) :)
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>
>
> Nicola

If C->B applies, then how is R1 in 3NF ?

Looking at the relation schema for R1, sole candidate key is AB. The applicable FD's LHS is completely outside the candidate key. Received on Mon May 06 2013 - 21:27:19 CEST

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