Re: Demo: Modelling Cost of Travel Paths Between Towns

From: Laconic2 <laconic2_at_comcast.net>
Date: Fri, 12 Nov 2004 10:03:46 -0500
Message-ID: <fKydnZV5S_xLTQncRVn-pw_at_comcast.com>


"Rok Debeljak" <rok.debeljak_at_milenij.si.remove> wrote in message news:Ru2ld.5529$F6.1282471_at_news.siol.net...
> > Here are the two times:
> >
> > 2004.11.11 24:00
> > 2005.11.12 00:00
> > ...
> > If a computation happened to result in 2004.11.11 24:00, I would
> want it to
> > convert it to 2005.11.12 00:00. And I would call this final step
>
> One year up or down does not really make any difference for a comparator
> that really understands what we are talking about ;-)
>
> Rokson

Ouch!!!! Of all the typos I've ever made, this is the worst. I don't know how I missed it. My proofreading skills are going down the drain!

Let me try the whole thing over again, corrected:

>According to ISO, we can assign the value 0000 or 2400 to midnight. Let's
>look at another time, say 2 PM, or 1400 . Do we assign two values to 1400?

 Believe it or not, this is a problem in "normalization". Not "data  normalization" as we ordinarily speak of it in database discussions, but  normalization nonetheless.

 Let's say that we are given two times, and asked whether they are equal or  not.

 Here are the two times:

 2004.11.11 24:00
2004.11.12 00:00

 The above form is for illustration purposes only.

 Now a naive comparator would answer the question "no". But a comparator  that really understands what we are talking about  might take a closer look, and come up with the answer: "yes". I like the  answer "yes", myself, better.

 In fact, I would like the "time type engine" whenever it's done some  calculation on time, (like adding an "interval" to a "base time") to  perform one final step, before delivering the result to the outside world.  If a computation happened to result in 2004.11.11 24:00, I would want it to
 convert it to 2004.11.12 00:00. And I would call this final step  "normalizing the result".

Once the results of a computation have been "normalized" , a naive comparator
 can test for equality by just comparing the representations. Right? Received on Fri Nov 12 2004 - 16:03:46 CET

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