Re: thinking about UPDATE
Date: Wed, 28 Jul 2004 10:54:13 +0300
Message-ID: <41075b9e_at_post.usenet.com>
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> Of course, but I'm not starting with *any* relation. To show the type
> of completeness that we are dealing with here it is sufficient to show
> that for every set A of attributes that is not derived by the algorithm
> there is a relation that satisifies the initial set of candidate keys and
> in whose projection A is not a candidate key. Since the only restriction
> is that it satisfies the original set of candidate keys I'm free to assume
> that the only FDs that hold for it are those that are implied by these
> candidate keys.
Oh, that completeness.
It would have been easier if you stated that one need to know all the
constraints on the relation to determine the candidate keys (not a
superkey). Are you aware of any algorithm that solve this problem ? What is
the complexity of it ?
> You are right that I oversimplified the proof, but my goal was to
> get you thinking about it, and it seems to have worked. :-)
I'm glad.
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