n-gon sequence

[Brendan McKay]

If no two triangles can intersect in more than 6 points, then
three triangles can't have more than binomial(3,2)*6 = 18 points
of intersection altogether.  I can't understand your example at all.

Brendan.


* Jon Perry <perry at globalnet.co.uk> [021015 03:15]:
> With further research, I got 20 intersections from 3 triangles - I did say
> it was without proof, as is my 20 being maximal.
> 
> (Create a Star of David, and connect 2 opposite vertices up. The number of
> intersections here is 4).
> 
> I also got the sequence for maximal intersections of 2 n-gons wrong.
> 
> 3 and 4 are 6,8, but I got 18 and 20 intersection from two (non-self
> intersecting) 5- and 6- gons.
> 
> But again these are without proof, so the bound might rise again.
> 
> Jon Perry
> perry at globalnet.co.uk
> http://www.users.globalnet.co.uk/~perry/maths
> BrainBench MVP for HTML and JavaScript
> http://www.brainbench.com
> 
> 
> -----Original Message-----
> From: Brendan McKay [mailto:bdm at cs.anu.edu.au]
> Sent: 13 October 2002 14:53
> Cc: Seqfan at Ext.Jussieu.Fr
> Subject: Re: n-gon sequence
> 
> 
> * Jon Perry <perry at globalnet.co.uk> [021013 23:02]:
> > Is this sequence known? I've only worked out the first 2 terms of each.
> >
> > The maximum number of intersections of k n-gons.
> >
> > e.g. with triangles, the sequence opens 0,6 - my next term is 16, but this
> > is without proof.
> >
> > For the general n-gon, the sequence always opens 0,2n
> 
> Can you give a more precise definition please?  It is easy to
> draw 3 triangles so that each pair intersect in 6 points and
> all intersections are distinct, making 18 altogether.  So I
> guess you must mean something else.
> 
> Brendan.