# Re: How is this collection called?

From: Mikito Harakiri <mikharakiri_at_iahu.com>
Date: Wed, 31 Mar 2004 11:05:25 -0800
Message-ID: <ANEac.32\$rc5.145_at_news.oracle.com>

"Paul" <pbrazier_at_cosmos-uk.co.uk> wrote in message news:51d64140.0403310835.cd5fa62_at_posting.google.com... "Paul" <pbrazier_at_cosmos-uk.co.uk> wrote in message news:51d64140.0403300727.2ef94b6_at_posting.google.com...
> Define a*b as "a U {b}" (where U is set union)
> Then a*(b*c) = aU{bU{c}} != (a*b)*c = aU{b}U{c}.
>
> But if you think in terms of physical boxes, imagine a small box
> inside a medium box, which in turn is inside a large box. For these
> things you could say the small box "is-in" the large one, even though
> it isn't directly inside.
>
> I think it's just the same (isomorphic to) as the tree example?

It is isomorphic indeed to a tree operation, but this is a different tree operation (and different tree structure?)

In case of binary trees with labeled ordered children the tree composition operation a*b is building a larger tree that has root connected to 2 branches a and b:

root
|
+--- a
|
+--- b

You, however, refer to a [nonbinary] tree with unordered children with composition being defined as attaching subtree b to the root of a (at level 1):

a
|
+----b
|
...<the rest of a>

In the first case the * operation could be defined commutative (unordered children) or not, but in your case

a*b != b*a

always. Received on Wed Mar 31 2004 - 21:05:25 CEST

Original text of this message