Re: A Normalization Question
From: Jan Hidders <jan.hidders_at_REMOVETHIS.pandora.be>
Date: Mon, 12 Jul 2004 21:17:44 GMT
Message-Id: <pan.2004.07.12.21.18.06.552156_at_REMOVETHIS.pandora.be>
>
> What I am saying is assuming tuple1 only and no FD A->B; that 1 isa
> person, 2 is a color and 3 isa street all named by the same string
> 'brown' thus
>
> R2(1, 'brown')
> R2(2, 'brown')
> R2(3, 'brown')
>
> which further normalizes to
>
> R2(1, 10)
> R2(2, 10)
> R2(3, 10)
>
> AND
>
> R3(10, 'b', 'r', 'o', 'w', 'n')
>
> and can be carried out one step further to each symbol.
Date: Mon, 12 Jul 2004 21:17:44 GMT
Message-Id: <pan.2004.07.12.21.18.06.552156_at_REMOVETHIS.pandora.be>
On Mon, 12 Jul 2004 13:28:42 -0700, Neo wrote:
>> Let's consider a small example. We take a relation R(A,B,C) with a >> functional dependency A->B. Now consider the following instance: >> >> A B C >> --------- >> 1 2 3 >> 1 2 4
>
> What I am saying is assuming tuple1 only and no FD A->B; that 1 isa
> person, 2 is a color and 3 isa street all named by the same string
> 'brown' thus
>
> R2(1, 'brown')
> R2(2, 'brown')
> R2(3, 'brown')
>
> which further normalizes to
>
> R2(1, 10)
> R2(2, 10)
> R2(3, 10)
>
> AND
>
> R3(10, 'b', 'r', 'o', 'w', 'n')
>
> and can be carried out one step further to each symbol.
Sure, you can represent it that way, even in the relational model, but you cannot call it normalization because normalization deals with removing logical redundancies and there were no logical redundancies in the original relation to begin with. I know you keep on denying this but that's what the definition of logical redundancy says.
- Jan Hidders