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Re: How is this collection called?

From: Michael Mendelsohn <keine.Werbung.1300_at_michael.mendelsohn.de>
Date: Wed, 31 Mar 2004 12:30:58 +0200
Message-ID: <406A9DE2.D88C80CC@michael.mendelsohn.de>


Paul schrieb:
> I think the problem is you've not defined exactly what you mean by
> "binary aggregation operator".

I thought that your "packing it in" approach was a bit unfair, counter to my (personal) intuition how aggregation should work.

I'd define aggregation to fulfil the axioms that

x is-in A => x is-in A*B
x is-in B => x is-in A*B

and that rules your approach out, because with

> Define a*b as "a U {b}" (where U is set union)

x is-in b => x is-in A*B breaks.

> There's no reason why a "union" of bags or lists has to be defined in
> the usual way, it's just an arbitrary choice that fits in with our
> vague conceptual ideas about what a union is.

With my aggregation constraint, I believe the collections defined by Mikito's set and bag axioms can only be set and bag, and that you cannot choose an aggregation operation that breaks these definitions.

I think that the list is not so useful because you can define aggregators to keep or break the third axiom; and you can (with ease) define the aggregator to always keep a*b=b*a if there's an ordering on the collections.
Now wait, I think I just contradicted half of what I wrote above! :-P

So could I have a collection and aggregation with

a*a!=a
a*b=b*a
a*(b*c)!=(a*b)*c

?

Michael

-- 
Feel the stare of my burning hamster and stop smoking!
Received on Wed Mar 31 2004 - 04:30:58 CST

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