Re: boolean datatype ... wtf?

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