Counting Self-intersecting n-gons
David Wilson
davidwwilson at comcast.net
Fri Apr 22 17:06:59 CEST 2005
Don't know enough about homeomorphisms to tell you.
The basic idea is that if two "good" polygons are equivalent, I can deform one
into the other continuously, with all the intermediates being good polygons as
well. Because we stay within the space of good polygons, the transformation
will not allow, say, a vertex to pass through an edge or three edges to
intersect at a point during the transformation. Is this a homeomorphism?
About the graph observation, I don't think the polygon equivalence is the same
as the graph equivalance. For example, consider the following two pentagons:
X: (0,0) (4,0) (2,3) (3,2) (0,4)
Y: (0,0) (4,0) (1,2) (2,1) (0,4)
I believe that X and Y are distinct polygons according to my definition, but
their graphs are isomorphic.
The six pentagons I found were the two above plus
(0,0) (1,0) (2,1) (1,2) (0,1) (simple pentagon)
(0,0) (2,1) (0,1) (1,0) (1,2) (pentagram)
(1,0) (1,1) (0,1) (3,0) (0,3)
(2,0) (2,2) (0,2) (0,3) (3,0)
----- Original Message -----
From: Sterten at aol.com
To: ham ; davidwwilson at comcast.net
Sent: Friday, April 22, 2005 9:15 AM
Subject: Re: Counting Self-intersecting n-gons
>define two polygons in P as equivalent if there is a continuous
>tranformation between them in P.
homeomorphism ?
>Assuming this definition means anything,
you mean: "means something" ?
>I am pretty sure that
>equivalent polygons will dissect the plane into the same number
>of regions and corresponding regions will have the same number
>of edges.
a polygon X in P makes a planar graph G(X) by adding all intersection-points
as new vertices. X-vertices have degree 2 in G(X), but there can also be
(maximum (n-2)*(n-3)/2, I think) other vertices of degree 4.
When polygons X and Y are equivalent, then G(X) and G(Y) are isomorphic ?
When G(X) and G(Y) are isomorphic, then polygons X and Y are equivalent ?
>The question is, according to this definition, how many distinct
>self-intersecting n-gons are there?
>
>Clearly, a(3) = 1, since all triangles are equivalent.
>
>I believe a(4) = 2, all self-intersecting quadrilaterals begin equivalent
>to either a square or a bowtie.
>
>I have found 6 inequivalent pentagons, so a(5) >= 6.
I only found 5, they have 0,1,2,3,5 crossings.
-Guenter.
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