Counting n-gons

David W. Wilson wilson at aprisma.com
Fri Nov 16 15:57:12 CET 2001

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Here is a question which would result in an interesting sequence.

Consider the space of possibly self-intersecting planar polygons in
which vertices are pairwise distinct, edge intersection points are
pairwise, distinct, and vertices do not lie interior to an edge.

Consider two such polygons equivalent if there is a continous
transformation from one to the other in this space.

Let f(n) be the number of nonequivalent n-gons according to this
definition (or the fixed/standard definition if it is faulty).

I am pretty sure that

f(3) = 1 (all triangles are equivalent)

f(4) = 2 (convex quadrilateral and bowtie)

I believe that f(5) = 6.  The six classes would be represented
by pentagons with vertex coordinates:

(0,0), (2,0), (2,1), (1,2), (0,1)
(0,0), (2,1), (1,2), (0,1), (2,0)
(0,0), (2,1), (0,1), (2,0), (1,2)
(0,0), (1,2), (2,0), (2,1), (0,1)
(0,0), (2,1), (1,1), (3,0), (1,3)
(0,0), (1,2), (2,0), (0,3), (2,3)

I wouldn't attempt the hexagons.

There are a couple of obvious generalizations.  In one, we make
mirror images equivalent.  Also, we could count "open n-gons"
(= n-edge polygonal arcs)

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