Re: Relational symmetric difference is well defined

On Jun 16, 7:16 am, "V.J. Kumar" <vjkm..._at_gmail.com> wrote:
> Vadim Tropashko <vadimtro_inva..._at_yahoo.com> wrote innews:1181951119.615693.101990_at_d30g2000prg.googlegroups.com:
>
>
>
> > On Jun 15, 4:21 pm, "V.J. Kumar" <vjkm..._at_gmail.com> wrote:
> >> Sure, second order lang. is more expressive, but its expressivity
> does
> >> not come for free !
>
> > Something is not right here. So, join, union, and difference can be
> > defined via first order formulas, while relational division don't?
> > Sure relational division is expressible via join, union, and
> > difference!
>
> Of course it is, e.g.(project def. project on some attributes A not in
> R2):
>
> R1 divide R2 = project(R1) - project(project(R1) times R2) - R1).

Given R(x,y) and Q(y,z), here is a RA expression for the set equality join R=\Q:

( project_x(R) /\ project_z(Q) ) \
project_xz( (project_x(R) /\ Q ) symm_diff (R /\ project_z(Q) ) )

It follows that perhaps the expression

(project_x(R) /\ Q ) symm_diff (R /\ project_z(Q) )

is more interesting variant of symmetric difference than symmetric difference of projections of both R and Q into y. Received on Mon Jun 18 2007 - 20:14:51 CEST