What Tony did is very nice indeed. What Franklin wrote is certainly correct.<br>
<br>
The partitions can be the same, but the order of edges encountered in a circuit can differ.<br>
<br>
By bow-tie I mean, for instance, for a nonsimple quadrilateral of
perimeter 6, the sequence of coordinates (0,0), (0,1), (sqrt 3, 0),
(sqrt 3, 1), (0,0). The crossing point, not a vertex, is ((sqrt
3)/2, 1/2). Yes, that can be traversed in other orders, and the
quadrilateral can be deformed. Linkages were much studied until
perhaps the 2nd half of the 20th century in American education.<br>
<br>
As well as what I'd call the (1, 2, 1, 2) rectangle (deformable into a
parallelogram), there is a (1, 1, 2, 2) kite and a a (1, 1, 2, 2)
which are not similar to each other or to the rectangle. Both
kite and dart share a vertex at (0, 0) , a vertex at (2, 0), and a
vertex at (7/4, (sqrt 15)/4). The coordinates of the other two
vertices are elementary and left to the reader.<br>
<br>
Yes, for a general nonsimple n-gon, any of the n edges can be
crossed. The crossing point has integral coordinates precisely
when Pythagorus would want it. This relates these sequences to
those about Pythagorean triplets.<br>
<br>
There are other sequences yet to be mined here, under the various
assumptions that I made, including a count of the number of nonsimple
n-gons of permineter n, under appropriate equivalence classes.
Similarly, the maximum areas for each edge sequence. <br>
<br>
For perimeter 7, we have in the given example (1, 2, 2, 2) and (1, 1,
2, 3). But the trapezoid I'd denote by edges encountered in
circuit (1, 2, 1, 3) is not similar to the irregular convext
quadrilateral I'd call (1, 1, 2, 3). And perimeter 7 can have
nonvertex crossings near edges of length 1, 2, or 3 as desired.<br>
<br>
More work here for all of us to share.<br>
<br>
-- Jonathan (proud that his 18-year-old son in 5th year of university
just scored 99th percentile on the LSAT, and thus likely to attend one
of America's very best law schools, such as Harvard, Yale, Stanford,
Berkeley, NYU, ... even though that means not immediately going to grad
school to follow on his pending double B.S. in Math and Computer
Science)<br><br><div><span class="gmail_quote">On 10/24/06, <b class="gmail_sendername">T. D. Noe</b> <<a href="mailto:noe@sspectra.com">noe@sspectra.com</a>> wrote:</span><blockquote class="gmail_quote" style="border-left: 1px solid rgb(204, 204, 204); margin: 0pt 0pt 0pt 0.8ex; padding-left: 1ex;">
At 5:13 PM -0400 10/23/06, <a href="mailto:franktaw@netscape.net">franktaw@netscape.net</a> wrote:<br>>There's a problem with this formulation. Triangles are determined up<br>>to congruence by their side lenths, but polygons with more sides are
<br>>not. (Consider rhombus vs. square.)<br>><br>>To make it work, you have to define it purely in terms of edge lengths:<br>>sequences of n positive integers, totalling p, whose largest value is<br>>less than the sum of the others (equivalently, less than p/2); up to
<br>>equivalence under rotation and reflection.<br>><br>>Franklin T. Adams-Watters<br>><br>><br>>-----Original Message-----<br>>From: <a href="mailto:davidwwilson@comcast.net">davidwwilson@comcast.net
</a><br>><br>>Let f(n,p) be the number of non-congruent integer-sided n-gons with<br>>perimeter p. Then f(3,p) = A005044.<br>><br>>Can we come up with a general formula/recurrence for f(n, p)?<br><br><br>I just submitted A124278, triangle of the number of k-gons having perimeter
<br>n. Included is a generating function for each column of the triangle.<br>There is also a b-file for rows n=3..102 of the triangle. Here is the<br>internal format:<br><br>%S A124278 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 1, 1, 1, 3, 2, 2, 1, 1,
<br>3, 4, 4, 3, 2, 1, 1, 2, 5, 5, 4, 3, 2, 1, 1, 4, 7, 8, 6, 5, 3, 2, 1, 1, 3,<br>8, 9, 9, 6, 5, 3, 2, 1, 1, 5, 11, 14, 12, 10, 7, 5, 3, 2, 1, 1, 4, 12, 16,<br>16, 13, 10, 7, 5, 3, 2, 1, 1, 7, 16, 23, 22, 19, 14, 11, 7, 5, 3, 2, 1, 1,
<br>5, 18, 25, 28, 24, 20, 14, 11, 7, 5, 3, 2, 1, 1, 8, 23, 35, 37, 34, 27, 21,<br>15, 11, 7, 5, 3, 2, 1, 1<br>%N A124278 Triangle of the number of nondegenerate k-gons having perimeter<br>n, for k=3..n.<br>%C A124278 For a k-gon to be nondegenerate, the longest side must be
<br>shorter than the sum of the remaining sides. Column k=3 is A005044, column<br>k=4 is A062890, column k=5 is A069906, and column k=6 is A069907.<br>%H A124278 G.E. Andrews, P. Paule, and A. Riese, <a<br>href="
<a href="http://www.risc.uni-linz.ac.at/publications/download/risc_163/PAIX.pdf">http://www.risc.uni-linz.ac.at/publications/download/risc_163/PAIX.pdf</a>">MacMahon's<br>Partition Analysis IX: k-Gon Partitions, Bull. Austral. Math. Soc. 64
<br>(2001), 321-329.</a><br>%H A124278 T. D. Noe, <a<br>href="<a href="http://www.research.att.com/~njas/sequences/b124278.txt">http://www.research.att.com/~njas/sequences/b124278.txt</a>">Rows
<br>n=3..102 of triangle, flattened</a><br>%F A124278 G.f. for column k is x^k/(product_{i=1..k} 1-x^i) -<br>x^(2k-2)/(1-x)/(product_{i=1..k-1} 1-x^(2i))<br>%e A124278 For polygons having perimeter 7: 2 triangles, 2 quadrilaterals,
<br>2 pentagons, 1 hexagon, and 1 heptagon. The triangle begins<br>1<br>0 1<br>1 1 1<br>1 1 1 1<br>2 2 2 1 1<br>1 3 2 2 1 1<br>%t A124278 Needs["DiscreteMath`Combinatorica`"]; Table[p=Partitions[n];<br>Length[Select[p, Length[#]==k && #[[1]]<Total[Rest[#]]&]], {n,3,30},
<br>{k,3,n}]<br>%O A124278 3<br><br>Tony<br><br></blockquote></div><br>