Re: I think that relational DBs are dead. See link to my article inside
Date: Fri, 21 Jul 2006 11:03:49 GMT
"David Portas" <REMOVE_BEFORE_REPLYING_dportas_at_acm.org> wrote in message
> Josip Almasi wrote:
> > >
> > > What multidimensional model? Kimball popularised some methodology and
> > > some jargon under the Dimensional banner. Some people find such
> > > terminology useful but it doesn't change the data model. It is still
> > > relational or SQL. Do you think relational is something other than
> > > multi-dimensional?
> > You really gave me some food for thought with this:)
> > Yes, you're right. Only difference is normalization.
> Not even that. The Kimball-styled "Dimensional Model" operates only at
> the logical level. Despite hype to the contrary it is orthogonal to
> normalization, not the antithesis of it.
Both can be useful, when used judiciously.
It's useful to distinguish between the dimensional model itself, and star schema design.
> Normalization is concerned
> only with base relations whereas Kimball's ideas can be implemented
> purely through views without disturbing NF at all.
Or materialized in derived tables, kept current by ETL processing.
> RK has done a lot of
> ill by not better explaining the nature of his ideas. His works contain
> plenty of confusion over logical versus physical.
> > > A recursive CTE. More generally speaking the query is just some
> > > restriction of a transitive closure (ie. it need not be defined
> > > recursively).
> > But needs to be iterated regardless, right?
> > If so, it's not relational.
> > Please explain; I just don't understand how it can be represented as
> The relational model does not exclude iteration. RM says nothing about
> how any particular operation is performed inside the DBMS. What matters
> is that the operation can be logically defined using only relational
> operators. The transitive closure of a graph is certainly a set. A
> subtree within that graph is just a subset of it. In the case of a
> hierarchy it is particularly easy to define a subset just by specifying
> the root of a subtree.
> [dropped comp.ai.neural-nets from my reply]
> David Portas
Received on Fri Jul 21 2006 - 13:03:49 CEST