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Re: First Impressions on Using Alphora's Dataphor

From: Laconic2 <laconic2_at_comcast.net>
Date: Fri, 27 Aug 2004 09:02:39 -0400
Message-ID: <Gcmdndgg16D0rbLcRVn-iA@comcast.com>

"mAsterdam" <mAsterdam_at_vrijdag.org> wrote in message news:412e5c87$0$65124$e4fe514c_at_news.xs4all.nl...

> It makes some intuitive sense to me. Would you care to elaborate?
> I would appreciate that, and I suspect more readers would.
>

Well, I'm not sure, but here goes.

It seems to me that, if you have a feature that has identity, but not order, then eliminating duplicates involves
comparisons on the order of (n*n/2). If you have a feature that has identity and order, then eliminating duplicates can be done with comparisons on the order of (n*log(n)). The latter is the order of magnitude of the comparisons needed to sort.

For large n, sorting is clearly preferable to comparing all pairs.

This is really just conjecture, but it sounds good to me. Received on Fri Aug 27 2004 - 08:02:39 CDT

Original text of this message

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