Re: SQL challenge

From: Jan Hidders <jan.hidders_at_REMOVETHIS.pandora.be>
Date: Thu, 15 Jul 2004 18:58:05 GMT
Message-Id: <pan.2004.07.15.18.58.36.38772_at_REMOVETHIS.pandora.be>


On Thu, 15 Jul 2004 11:45:51 -0700, Mikito Harakiri wrote:

> "Jan Hidders" <jan.hidders_at_REMOVETHIS.pandora.be> wrote in message
> news:pan.2004.07.15.17.52.05.184333_at_REMOVETHIS.pandora.be...

>> On Wed, 14 Jul 2004 18:35:49 -0700, Mikito Harakiri wrote:
>> > Consider tetrahedron defined by 4 vertices p1, p2, p3, p4. Then, for any
>> > point p it is easily decidable if p belongs to tetrahedron interior or
>> > not.
>> > Let's call this predicate as
>> >
>> > Inside(p, p1, p2, p3, p4)
>> >
>> > This predicate is expected to work in the "extreme" cases when p1=p2 (or
>> > even when p1=p2=p3). Finding closed form expression for Inside(p, p1,
>> > p2,
>> > p3, p4) is left as an exercise to the reader.
>>
>> Could you be a bit more precise here? What if p1 p2 p3 and p4 are on a
>> single line, what does it then exactly mean to be inside? I assume no node
>> would be inside then, right? Or is a node on the edges of the tetrahedron
>> also "inside". And the corner points, are these inside?
> 
> That is p belongs to convex hull of p1 p2 p3 and p4 but does not coincide
> with the vertices of that hull. This definition spans the cases where
> tetrahedron is collapsed into planar or linear figure. (I'm not totally
> happy being forced to exclude vertices from tetrahedron interior, though).

Ah, ok, then I agree with your solution. OK. So I hereby award you the unofficial Jim Gray SQL award of comp.databases.theory. :-)

  • Jan Hidders
Received on Thu Jul 15 2004 - 20:58:05 CEST

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