Re: What to call this operator?

From: Marshall Spight <marshall.spight_at_gmail.com>
Date: 2 Jul 2005 13:14:24 -0700
Message-ID: <1120335263.968045.238790@f14g2000cwb.googlegroups.com>

Jan Hidders wrote:
> Marshall Spight wrote:
>
> The rule is that if we take the natural join of R and S then we can
> derive a candidate key K for the result if K is a candidate key of both
> R and S. Is that what you wanted to hear?

That seems correct but incomplete. If R(a, b) and S(b, c) and b is non-empty, and (a) is a candidate key of R, then it seems that (R join S) will also have (a) as a candidate key. Wouldn't it? It will certainly be unique. Is saying that a set of columns in the result will be unique over the result relation the same as saying that it is a candidate key?

We also need some kind of fallback rule, such that if we cannot derive any keys for the relation, then the union of all columns is a key.

Marshall Received on Sat Jul 02 2005 - 15:14:24 CDT