Counting n-gons

[franktaw at netscape.net]

(First of all, you seem to have left off "non" a few times.  Plus, I 
think you mean to refer to "simple" and "nonsimple" polygons, rather 
than "convex" and "[non]convex".)

Allowing nonsimple polygons doesn't make any difference.  The sequence 
of side lengths for a nonsimple polygon are also the sequence of side 
lengths for a simple polygon.  So, unless you want to explicitly make a 
distinction between polygons based on their crossing properties, you 
get the same number of polygons.  (This might be an interesting 
question, but it does require some more clarification: do you allow 
crossings at vertices, or only in mid-edge?  In the former case, can 
two vertices coincide?  In the latter, can three or more edges cross at 
a single point?  More subtly, if we have, for example, a pentagon whose 
sides are different but sufficiently close to equal, we can create a 
figure with one pair of crossed sides, crossing any pair of 
non-adjacent sides.  Is this 5 different polygons, or only 1?)

Second,  note that the current question is the number of polygons with 
a given perimeter.  Such polygons might be similar to polygons with a 
smaller perimeter, but not to another with the same perimeter.  In any 
event, the reference to A005044 makes it clear: we are counting all 
such polygons, not just the irreducible ones.

Finally, I thought my earlier post was clear enough, but not everybody 
seems to have gotten the message.  You *can't* count polygons of more 
than 3 sides with integral sides up to congruence.  It isn't just a 
matter of flipping a pair of sides; as the rhombus example shows, the 
number of non-congruent polygons with 4 or more sides of specified 
(possible) lengths is infinite - of the order of the continuum, to be 
precise.  This is something every builder knows - triangles are rigid, 
larger polygons are not.

Franklin T. Adams-Watters


-----Original Message-----
From: jvospost3 at gmail.com

This problem does not come up for triangles, but for n-gons with n>3 
one can either restrict to convex n-gons, or allow convex n-gons, such 
as bowties and pentagrams.  In the former case, there is a triangle 
inequality applicable, which is not the case for convex n-gons.  I am 
not clear if this has been addressed.  Also, are not compounds 
degenerate cases of star-polygons? Also, are there not various ways to 
derive equivalence classes, considering the polygons "the same" under 
some set of rotations, reflections? Then, do we want to retrict to 
"primitive" n-gons, which are not similar to earlier ones in the list?

On 10/23/06, franktaw at netscape.net <franktaw at netscape.net > 
wrote:There's a problem with this formulation.  Triangles are 
determined up
to congruence by their side lenths, but polygons with more sides are
not.  (Consider rhombus vs. square.)

To make it work, you have to define it purely in terms of edge lengths:
sequences of n positive integers, totalling p, whose largest value is
less than the sum of the others (equivalently, less than p/2); up to
equivalence under rotation and reflection.

Franklin T. Adams-Watters


-----Original Message-----
From: davidwwilson at comcast.net

Let f(n,p) be the number of non-congruent integer-sidedn-gons with
perimeter p. Then f(3,p)= A005044.

Can we come up with a general formula/recurrence for f(n, p)?


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