Re: Normalization by Composing, not just Decomposing
Date: Sat, 10 Apr 2004 12:13:55 +0200
> mAsterdam wrote:
>> Anyway, the irreducible normal form (one non-key attribute only) >> was recently mentioned again by Chris Date as the 6th normal form.
> Are you sure?
Your question got me to think about this again, and the only honest answer I can come up with is: no, I am not 100% sure. Either I read somewhere or at sometime I concluded that - for all practical purposes - 6NF is INF. Sorry for presenting this as a fact.
I could back out by rephrasing it so:
> Anyway, the irreducible normal form (one non-key attribute > only) was recently mentioned again by Chris Date in the > prelude to the 6th normal form.
... which would be correct, but would miss anything about 6NF and INF being the same or not. I think they are. Now I'm in an uneasy spot. I find that cannot reproduce how I got to that they are - provide a quote or proof.
I can just give you some of my thoughts about it. First: some nuance.
> Date's 6NF is a special normal form for when you have
> temporal data.
- Which is allways, *if* you look at it this way: "The database is not the database - the log is the database, and the database is just an optimized access path to the most recent version of the log." - B.-M Shueler, prominently quoted by Date, Darwen and Lorenzos in their recent book "Temporal Data and the relational model".
- Not just when you have temporal data. All data involving interval attributes in the key benefits from decomposing 5NF into 6NF. Temporal data, handled as proposed by Date, Darwen and Lorenzos is a special case of interval attributes in the key. (Temporal Data and the relational model, read e.g. pages [172:177]). So yes, you need 6NF when dealing with temporal data, but that is not it's only purpose.
> It's not uncontroversial, by the way.
> ... Actually finding out what
> the elementary facts are is essentially the same as normalizing to 5NF.
My take is that Date, Darwen and Lorenzos formulated 6NF the way they
did to make it fairly obvious that 6NF is more strict than PJNF (5NF)
(i.o.w. that every set of relations (relational variables) in 6NF is by
definition also in 5NF so 6NF is another step on the lossless
decomposition ladder). However, until I see an counterexample -
preferably pizza orders related - I'll look at 6NF as an alternative
predicate for the INF, the irreducable normal form (loose definition:
just one non-key attribute) (BTW great
acronym, don't you think? :-). Received on Sat Apr 10 2004 - 12:13:55 CEST