Re: What is Aggregation? Re: grouping in tuple relational calculus

From: <>
Date: 19 Feb 2005 10:02:27 -0800
Message-ID: <>

Paul wrote:
> Mikito Harakiri wrote:
> > Brackets naturally give rise to binary trees.

That should read "Nested Brackets naturally give rise to binary trees."

> Are you sure that two separate orderings wouldn't suffice?
> If you think about how you would do an aggregate of a substraction
> operator for example. You would first write them all down in order so

> you get a-b and not b-a. Then you would pick an order to do the
> (actually only a partial order is required I think) so you get for
> example (a-b)-(c-d) and not a-(b-(c-d)). In the first case it doesn't

> matter if you do (a-b) or (c-d) first so that's why you don't need a
> full ordering. I suppose a partial ordering amounts to the same thing
> a binary tree though?

Hmm. Partial order is associated with a structure with transitive binary relation, which is not necessarily a tree. Received on Sat Feb 19 2005 - 19:02:27 CET

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