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On Sat, 24 Jun 2006 13:02:42 +0200, mAsterdam wrote:
> Dmitry A. Kazakov wrote:
>>> Dmitry A. Kazakov wrote: >>>> mAsterdam wrote: >>> >>>>> These requirements establish K as a clean point type. >>>>> Aside: With the last condition deleted one can make a >>>>> circular K, having distance(k1, k2) <> distance(k2, k1). >>>> >>>> That won't be formally a distance, which is required to be symmetric. >>> >>> Why must it be symmetric? A to B and B to A may be different >>> distances if there are one-way routes involved.
That's a ring (modulo), which is not a metric space.
> d(Mon, Fri) = 4
> d(Fri, Mon) = 3
d(Fri, Mon) = |4 + 7n|, days, for any integer n. Which includes both cases: n=0 and n=-1.
One could consider classes of equivalence { k + 7n } = k* as "distances", which is again the same modulo. Unfortunately k" are incomparable, as it expected from distances.
If we wanted to make it a space, we should probably return back to the time space.
Fri - Mon (this week) = 4 days (a vector, with a magnitude of duration) Mon - Fri (adjacent week) = -3 days (another vector)
-- Regards, Dmitry A. Kazakov http://www.dmitry-kazakov.deReceived on Sat Jun 24 2006 - 14:04:42 CDT