Re: 4NF is Where It Is At! [WAS Re: 1:1 relationships]

In article <96hjgh$ilp$1_at_news.tue.nl>, Jan Hidders says...
>
>which is indeed equivalent with
>
> ~R = (~R[CT] JN R[X]) + (~R[CX] JN R[T])
>
>and also with
>
> R = R[CT] JN R[CX]
>
The manner in which you've written this prompts a natural question: could this equivalence be demonstrated by pure algebraic means? Or we have to resort to point-by-point consideration?

>> > ~R = (~R[CT] JN R[X]) + (~R[CX] JN R[T])
>>
>> I write mine rather as
>>
>> ~R = (~R[CT] CP X) + (~R[CX] CP T)
>>
>> without joins.
>
>De gustibus non disputandum est :-) In the most straightforward
>definition of the join without artificial limitations the JN is
>equivalent to CP if there are no common attributes, which is a
>precondition for the CP anyway.
>So it is more elegant to have only a JN
>and no CP in your algebra; instead of two operators with preconditions
>you get one operator with no preconditions.
>
By two operators do you mean CP and ~?

Indeed, from algebraic perspective JN looks simpler than CP. Not from geometric one, however.

In 3D modelling packages they often draw a nut (or rather simplistic nut without internal screw thread). A nut is an example of an object that is unambiguously defined by 2 projections. Any point inside nuts hole is enclosed within some 0-cylinder. On the other hand, if you consider a cup, glass or any container, then for the points inside container no such cylinder exists. Received on Fri Feb 16 2001 - 00:41:32 CET