James>http://faculty.uml.edu/jpropp/mathenchant/nonagon.pdf Nice. I think those are illegal in California. Use PossibleZeroQ. Tell us about any false positives. Alternatively, for n=9, e.g., In[478]:= I^Range[0, 4 - 4/9, 4/9] Out[478]= {1, (-1)^(2/9), (-1)^(4/9), (-1)^(2/3), (-1)^(8/9), -(-1)^(1/9), -(-1)^(1/3), -(-1)^(5/9), -(-1)^(7/9)} In[479]:= areaPolygon[%[[{1, 5, 3, 4, 8, 6, 7, 2, 9}]]] Out[479]= 1/2 (Cos[\[Pi]/18] + 1/2 Sqrt[3] Cos[\[Pi]/9] - 2 Cos[\[Pi]/18] Cos[\[Pi]/9] + 1/2 Sqrt[3] Cos[(2 \[Pi])/9] + 1/2 Sin[\[Pi]/9] - 2 Sin[\[Pi]/18] Sin[\[Pi]/9] + 1/2 Sin[(2 \[Pi])/9] - 2 Cos[(2 \[Pi])/9] Sin[(2 \[Pi])/9]) In[495]:= MinimalPolynomial@%479 // tim During evaluation of In[495]:= 0.241593,1 Out[495]= #1 & --rwg On 2016-07-30 04:34, James Propp wrote:

Tom,

I'm assuming you want the "area" as defined by the winding rule, which

...

means that some geometric areas may be counted multiple times. For instance, in a five-pointed star, the central pentagon, with winding number 2, counts double. Is this what you meant?

Yes, this is what I meant.

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Odd n=9 has the first solutions.

Odd n=15 has many; here are some:

0 1 2 3 4 6 5 11 7 14 13 10 12 9 8 0 1 2 3 4 6 5 12 8 7 14 11 13 10 9 0 1 2 3 4 6 5 12 11 8 10 7 14 13 9 0 1 2 3 4 6 5 12 14 11 7 13 10 9 8 0 1 2 3 4 6 5 12 14 11 10 7 13 9 8

Odd n=21 has many; here are some:

0 1 2 3 4 5 6 7 8 16 10 15 14 13 12 20 19 18 17 11 9 0 1 2 3 4 5 6 7 8 16 14 13 12 20 19 18 17 11 10 15 9 0 1 2 3 4 5 6 7 8 16 15 13 12 20 14 19 18 17 11 10 9 0 1 2 3 4 5 6 7 8 16 15 14 12 20 19 13 18 17 11 10 9 0 1 2 3 4 5 6 7 8 16 15 14 13 18 12 20 19 17 11 10 9

Thanks!

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So far I have not found any for any other odd values through n=31 (but I'll let them run overnight).

31 factorial is really large; it could be a very long night... :-)

He seems to be on n=33 !

One could mod out by a couple of dihedral group actions (one is geometric, based on the cyclic ordering of the n points, the other is combinatorial, based on the order in which the tour visits the points), but it's tricky because if I'm not mistaken they act "from opposite sides" (I never

learned

to like double cosets), and anyway, 31 factorial divided by 62 squared is still enormous...

Searching for symmetric solutions may well prove to

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find them more quickly than what I'm currently doing (just searching lexicographically).

Are any of your new solutions symmetrical?

I should mention that my program found some of the n=9 solutions but missed others because Mathematica wrongly computed their area to be nonzero (on the order of 10^-16); the problem disappeared when I stuck in a (somewhat time-consuming) FullSimplify. How do you try to avoid getting false positives and false negatives?

Jim