# Re: Extending my question. Was: The relational model and relational algebra - why did SQL become the industry standard?

Date: 14 Feb 2003 12:25:32 -0800

Message-ID: <3c8cab4c.0302141225.50cf5f68_at_posting.google.com>

"Paul Vernon" <paul.vernon_at_ukk.ibmm.comm> wrote in message news:<b2gl27$nds$1_at_sp15at20.hursley.ibm.com>...

> "Jan Hidders" <jan.hidders_at_REMOVE.THIS.ua.ac.be> wrote in message

*> news:3e4bca2b.0_at_news.ruca.ua.ac.be...
**> >
**> > No. In fact, in theory, all optimizations that can be done in a set-based
**> > algebra can also be done in a bag-based algebra but not the other way
**> > around.
**>
**> Obviously, I guess. A bag algebra being a superset of a set alegbra.
*

Exactly. (Although the term "superset" is strictly speaking not correct because it is not necessarily so that the operators of the bag algebra are a superset of those of the set algebra.) My compliments for your insight. :-)

> However it does not follow that a dbms where users were exposed to a bag-based

*> algebra would be overall more efficient than one with users 'restricted' to a
**> set-based alegbra. Not by a long shot.
*

Yes, that by itself is not a sufficient argument. However, combined with the fact that you sometimes want internally a bag algebra anyway, and that if the user wants to do bag-like things these will be harder to recognize as such by the query optimizer and therefore also harder to match with the bag operations of its internal algebra, then this is suddenly not so clear anymore.

- Jan Hidders