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

From: Paul <paul_at_test.com>
Date: Sat, 19 Feb 2005 12:02:49 +0000
Message-ID: <42172aea$0$53482$ed2619ec@ptn-nntp-reader03.plus.net>

Mikito Harakiri wrote:
>>Surely the associativity of a domain operator is orthogonal to the
>>relation structure?

> 
> I disagree. Aggregate operator properties corelate with the relation
> agregation semantics.
>

>>All it means is that your aggregate operator needs
>>to have three parameters: (1) the column to aggregate, (2) an ordering
>>to apply to the relation for the non-commutativity, and (3) an ordering
>>to apply to the relation for the non-associativity. I think 2 & 3 would
>>be independent of each other?

> 
> In case 3 it's no longer ordering. Without associativity you have to use
> brackets. 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 brackets (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 as a binary tree though?

Paul. Received on Sat Feb 19 2005 - 06:02:49 CST