Re: Basic question?What 's the key if there 's no FD(Functional Dependencies)?

From: NENASHI, Tegiri <tnmail42_at_gmail.com>
Date: 9 Nov 2006 07:40:47 -0800
Message-ID: <1163086847.479747.63710_at_b28g2000cwb.googlegroups.com>


Cimode wrote:
> A second...
>
> Let R1 relation of RealPeople, R2 relation of Impostors, R3 relation of
> Impostors and RealPeople expressed as R3 = R1 UNION R2
> R1 and R2 are disjoint therefore:
> R1 INTERSECT R2 = Empty Set
> R1 MINUS R2 = R1
> R2 MINUS R1 = R2
>
> implies...
>
> R1 MINUS R2
> UNION
> R2 MINUS R1
> UNION
> R1 INTERSECT R2
>
> is the same as
>
> R1 UNION R2 therefore R3 (call it AllPeople). In english ? Can
> relation AllPeople defined as the union of Impostors AND RealPeople be
> updated? Answer YES. Is it updated as an Impostor NO Is it updated as
> RealPeople NO. It is simply updated as AllPeople. Period.
>
> Neither R1 nor R2 are updated in fact. A major advantage of RM is the
> ability to work in closure with sets (meaning independently from values
> included in the sets at the first place)
>
> For instance,
>
> R1:{A, B, C}
> R2:{D, E, F}
>
> R3 = R1 UNION R2 --> leads to R3 :{A, B, C, D, E, F}
> If inserting R4: {G}
> Therefore {A, B, C, D, E, F} UNION {G} you produce R5 ={A, B, C, D, E,
> F, G}
> Neither R1, R2, R4 are updated in the process just R3 and R5 are
> exrpressed
>
> A consequence of assimiliating relations to revalues sets instead of
> focusing on their transformational properties is a common error in RM
> theory. Operations between 2 relations necessarily produce a new
> relation. Even an update to such relation produces a new relation.

To produce a new relation it is a very good point if the computer is named "math". One knows that the "math" has infiite speed and infinite memory. Helas, with the real computer, one must think about efficiency so your idea will not work in the real world.

--
Tegi
Received on Thu Nov 09 2006 - 16:40:47 CET

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