# Re: boolean datatype ... wtf?

Date: Fri, 1 Oct 2010 16:16:57 -0700 (PDT)

Message-ID: <ced03efd-b4be-48cc-a6fc-abc9b3dfff73_at_j24g2000yqa.googlegroups.com>

On 2 okt, 00:39, Brian <br..._at_selzer-software.com> wrote:

> On Oct 1, 2:26 pm, Erwin <e.sm..._at_myonline.be> wrote:

*>
**>
**>
**>
**>
**> > On 1 okt, 19:54, Brian <br..._at_selzer-software.com> wrote:
**>
**> > > > Erwin wrote:
**>
**> > > > > Explain "cyclical referential constraint".
**>
**> > > > > And explain "additional attribute".
**>
**> > > The functional dependency AB -> CD is lost. If that dependency were
**> > > invalid in the 5NF schema, {A,B,C,D} KEY {A,B}, then it would not be
**> > > in 5NF because that dependency is implied by the key, so the correct
**> > > 5NF schema would be,
**>
**> > > {A,B,C} KEY {A,B}, {A,B,D} KEY {A,B},
**>
**> > > which just happens to also be in 6NF.
**>
**> > > In order for the functional dependency AB -> CD to be preserved, a
**> > > cyclical inclusion dependency
**>
**> > > {A,B,C}[A,B] = {A,B,D}[A,B]
**>
**> > If you care to explain, my question amounted to "why do you qualify
**> > this as 'cyclical' ?".
**>
**> > And you also spoke of "additional attribute". Please point out to me
**> > where this "additional attribute" is in your very own answer ...- Hide quoted text -
**>
**> > - Show quoted text -
**>
**> Suppose you have a 5NF schema R {A,B,C,D,E} KEY {A,B}
**>
**> The inclusion dependency equivalent 6NF schema,
**> Rc {A,B,C} KEY {A,B},
**> Rd {A,B,D} KEY {A,B},
**> Re {A,B,E} KEY {A,B},
**> Rc[AB] = Rd[AB] AND Rc[AB] = Re[AB]
**>
**> can also be written
**>
**> Rc[AB] IN Rd[AB] AND Rd[AB] IN Re[AB] AND Re[AB] IN Rc[AB]
**>
**> Notice the cyclical nature of the constraint:
*

So set equality is, by definition, "cyclic" ?

If A and B are sets, then A=B implies both A in B && B in A.

> When there are only two relation schemata, the cyclical nature of the

*> constraint doesn't stand out like it does when there are three or
**> more.
*

As I have shown above, it stands out EXACTLY AS OBVIOUS when there are less than three.

Oh yes, and of course A in A && A in A is also a "cyclic" construct ?

My God, man, shut the fuck up. Received on Sat Oct 02 2010 - 01:16:57 CEST