Re: Oids

From: Mikito Harakiri <mikharakiri_nospaum_at_yahoo.com>
Date: 18 Aug 2005 09:32:31 -0700
Message-ID: <1124382751.351453.167880@z14g2000cwz.googlegroups.com>

Marshall Spight wrote:
> Mikito Harakiri wrote:
> > Marshall Spight wrote:
> >
> > > > Now, it seems that you hit the nail, and the rowid maps Relational
> > > > Lattice into a boolean algebra with just 2 elements 00 and 01. This is
> > > > dual to header lattice homomorhism of Relational Lattice into a boolean
> > > > algebra of sets. The symmetry is sligtly broken, but who would insist
> > > > that rows and columns are completely symmetrical?
> > >
> > > If we define 10 existentially, does that fix the symmetry? Instead
> > > of a single boolean algebra, we have one-per-type-of-A.
> >
> > I'm not able to follow this idea.
>
> For every A, we define an algebra specific to that A, for
> which the value of 10 is zero rows, columns-same-as-A.
> So it's finite and easily constructable.

Let me draw a template which you perhaps can follow to express your idea a little bit more formally. Also, since the algebra is finite, you can even illustrate your idea on example!

<template>

Take several relations, say A,B, and C and calculate more elements by applying joins and unions. The resulting lattice is finite and has the greatest element with header

header(A) join header(B) join header(C)

and 0 rows. This lattice also has header elements

header(A)
header(A) join header(B)
header(A) union header(B)

(header(A) join header(B)) union header(C) ...

It is easy to see that the algebra of headers 1. Is distributive
2. Has more than 2 elements

<end of template> Received on Thu Aug 18 2005 - 11:32:31 CDT