Re: a union is always a join!
- From: "Brian Selzer" <brian@xxxxxxxxxxxxxxxxxxx>
- Date: Tue, 6 Jan 2009 00:14:17 -0500
"paul c" <toledobythesea@xxxxxxxx> wrote in message
news:Pkv8l.42236$em5.22922@xxxxxxxxxxxxxxx
Criticisms, please (preferably ones based on some formal logic or other).
I think your reasoning is circular: A relation is a join because it can be
joined to a projection of itself over a subset of its own heading?
It is inescapable that every relation is a join (eg., Heath's theorem).
So every relvar points to a join. If we can't 'delete' through a join, we
can't delete from any relvar (my father, who thought a disk buffer was a
polishing material could have concluded this - also, I surmise, we can't
logically express some number of otherwise possible relations when we
have some purpose besides defining a 'view').
The reason for every relation being a join has nothing to do with the
expression that forms a join 'view'. I think it's a consequence of
Codd's algebra that every relation that has more than one subset of
attributes is a join of two or more relations. By Heath, there will
always be two heading subsets (at least one of them having the same
heading as the relation) that will determine the other possible pairs of
heading subsets in a join. Because the join on all of a relation's
attributes with any subset of them is algebraically equal to the relation,
it looks to me that the number of possible and irreducible join
expressions is at least equal to the number of subsets of attributes in a
relation heading. (This is implicit. Explicit constraints may have a
number of expressions that are fewer and still sufficient to express all
the possible expressions.)
For the same reason, every 'insert' is through a join, even if the
'insert' is also to a 'view' formed by union. One can equally say that
some relations involve union, but this doesn't change the fact that they
are also joins, in other words, not all joins are algebraically equal to
some union or other, ie., if a relation is formed by join, it is not
always also a union.
If so, it should be far more productive to design a language based on
axioms that are universal, the theorems that are always true, rather than
on statements that are true only sometimes.
.
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- From: paul c
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