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# Re: Canonical DB

From: Dmitry A. Kazakov <mailbox_at_dmitry-kazakov.de>
Date: Sat, 24 Jun 2006 21:04:42 +0200
Message-ID: <12lniumjtzvb9\$.1ceen1fyh0w2z.dlg@40tude.net>

On Sat, 24 Jun 2006 13:02:42 +0200, mAsterdam wrote:

```> Dmitry A. Kazakov wrote:
```

>> Gene Wirchenko 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.
```

>>
>> Rather when the space is anisotropic. If you just have many routes, you
>> still could get a metric distance by choosing the shortest path.
>>
>> ---------------
>> I found in Wikipedia that d(x,y) /= d(y,x) is called
>> quasimetric space (I never met such thing anywhere else).
>
> W = {Mon, Tue, Wed, Thu, Fri, Sat, Sun}

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.de
```
Received on Sat Jun 24 2006 - 14:04:42 CDT

Original text of this message

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