Re: [math-fun] Reentrant polygons with area zero
Tom, can you please explain what the 2nd number on each line of your table means? And what it means when the leftmost number (which roots of unity you are using) is repeated on multiple lines? Dan -----Original Message-----
From: Tom Rokicki <rokicki@gmail.com> Sent: Jul 30, 2016 11:03 AM To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] Reentrant polygons with area zero
Here are the rotational symmetry searches I've done. The 'no's are for exhaustive complete searches. Searches that have not yet found solutions, but are still running, are not listed.
So far it looks like, for prime m, n has a rotationally symmetrical solution of order m if n is divisible by m^2.
n sym solution found? 9 3 yes 15 1 yes 15 3 no 15 5 no 21 1 yes 21 3 no 21 7 no 25 5 yes 27 3 yes 27 9 yes 33 11 no 35 5 no 35 7 no 39 13 no 45 9 no 45 15 yes 49 7 yes 51 17 no 57 19 no 63 21 yes 69 23 no 75 15 yes 75 25 no 81 27 yes
(Sorry for spamming the list; I'll stop now.)
-tom
On Sat, Jul 30, 2016 at 10:28 AM, Tom Rokicki <rokicki@gmail.com> wrote:
Oddly, there appear to be a lot of n=15 solutions, but none of them have rotational symmetry (like the n=9 case did).
-tom
On Sat, Jul 30, 2016 at 9:42 AM, Tom Rokicki <rokicki@gmail.com> wrote:
31 factorial is really large; it could be a very long night... :-)
My CPU runs at 2.4 GHz.
time-consuming) FullSimplify. How do you try to avoid getting false
positives and false negatives?
My philosophy is don't worry, be happy. Every additional vertex introduces an error of at most 2^-51 or something like that, so I check the final area against 0 with an epsilon of about 1e-8, and then print the computed final area with the solution. If it's on the order of 1e-15, I'm not certain it's zero, but close enough to report. I'll let Bill and his posse do the hard bits of proving the result. (Or I'll throw multiprecision arithmetic at it. Anyone want to write a Mathematica procedure that takes a permutation, applies it to the roots of unity and computes the area, using high-precision math?) If multiprecision math says the area is < 1e-10000, I'm inclined to view it as zero, at least provisionally.
-tom
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Sure; the first number is the number of points; the second is the rotational symmetry degree (i.e., m=3 means you can rotate the solution by 2*pi/3 and it will be unchanged). So 45 9 no 45 15 yes means there is a solution for n=45 that is symmetric with respect to rotation by 2*pi/15 but not one that is symmetric with respect to rotation by 2*pi/9. On Sat, Jul 30, 2016 at 12:03 PM, Dan Asimov <dasimov@earthlink.net> wrote:
Tom, can you please explain what the 2nd number on each line of your table means? And what it means when the leftmost number (which roots of unity you are using) is repeated on multiple lines?
Dan
-----Original Message-----
From: Tom Rokicki <rokicki@gmail.com> Sent: Jul 30, 2016 11:03 AM To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] Reentrant polygons with area zero
Here are the rotational symmetry searches I've done. The 'no's are for exhaustive complete searches. Searches that have not yet found solutions, but are still running, are not listed.
So far it looks like, for prime m, n has a rotationally symmetrical solution of order m if n is divisible by m^2.
n sym solution found? 9 3 yes 15 1 yes 15 3 no 15 5 no 21 1 yes 21 3 no 21 7 no 25 5 yes 27 3 yes 27 9 yes 33 11 no 35 5 no 35 7 no 39 13 no 45 9 no 45 15 yes 49 7 yes 51 17 no 57 19 no 63 21 yes 69 23 no 75 15 yes 75 25 no 81 27 yes
(Sorry for spamming the list; I'll stop now.)
-tom
On Sat, Jul 30, 2016 at 10:28 AM, Tom Rokicki <rokicki@gmail.com> wrote:
Oddly, there appear to be a lot of n=15 solutions, but none of them have rotational symmetry (like the n=9 case did).
-tom
On Sat, Jul 30, 2016 at 9:42 AM, Tom Rokicki <rokicki@gmail.com> wrote:
31 factorial is really large; it could be a very long night... :-)
My CPU runs at 2.4 GHz.
time-consuming) FullSimplify. How do you try to avoid getting false
positives and false negatives?
My philosophy is don't worry, be happy. Every additional vertex introduces an error of at most 2^-51 or something like that, so I check the final area against 0 with an epsilon of about 1e-8, and then print the computed final area with the solution. If it's on the order of 1e-15, I'm not certain it's zero, but close enough to report. I'll let Bill and his posse do the hard bits of proving the result. (Or I'll throw multiprecision arithmetic at it. Anyone want to write a Mathematica procedure that takes a permutation, applies it to the roots of unity and computes the area, using high-precision math?) If multiprecision math says the area is < 1e-10000, I'm inclined to view it as zero, at least provisionally.
-tom
-- -- http://cube20.org/ -- [ <http://golly.sf.net/>Golly link suppressed; ask me why] --
-- -- http://cube20.org/ -- [ <http://golly.sf.net/>Golly link suppressed; ask me why] --
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Tom, Can you give us more details on n=27 and n=81? Given the high amount of symmetry involved, my guess is that we'll be able to guess a pattern and show that for n=3^k there's a solution with 3^(k-1)-fold rotational symmetry, for all k>1. Jim On Sat, Jul 30, 2016 at 3:18 PM, Tom Rokicki <rokicki@gmail.com> wrote:
Sure; the first number is the number of points; the second is the rotational symmetry degree (i.e., m=3 means you can rotate the solution by 2*pi/3 and it will be unchanged). So
45 9 no 45 15 yes
means there is a solution for n=45 that is symmetric with respect to rotation by 2*pi/15 but not one that is symmetric with respect to rotation by 2*pi/9.
On Sat, Jul 30, 2016 at 12:03 PM, Dan Asimov <dasimov@earthlink.net> wrote:
Tom, can you please explain what the 2nd number on each line of your table means? And what it means when the leftmost number (which roots of unity you are using) is repeated on multiple lines?
Dan
-----Original Message-----
From: Tom Rokicki <rokicki@gmail.com> Sent: Jul 30, 2016 11:03 AM To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] Reentrant polygons with area zero
Here are the rotational symmetry searches I've done. The 'no's are for exhaustive complete searches. Searches that have not yet found solutions, but are still running, are not listed.
So far it looks like, for prime m, n has a rotationally symmetrical solution of order m if n is divisible by m^2.
n sym solution found? 9 3 yes 15 1 yes 15 3 no 15 5 no 21 1 yes 21 3 no 21 7 no 25 5 yes 27 3 yes 27 9 yes 33 11 no 35 5 no 35 7 no 39 13 no 45 9 no 45 15 yes 49 7 yes 51 17 no 57 19 no 63 21 yes 69 23 no 75 15 yes 75 25 no 81 27 yes
(Sorry for spamming the list; I'll stop now.)
-tom
On Sat, Jul 30, 2016 at 10:28 AM, Tom Rokicki <rokicki@gmail.com> wrote:
Oddly, there appear to be a lot of n=15 solutions, but none of them have rotational symmetry (like the n=9 case did).
-tom
On Sat, Jul 30, 2016 at 9:42 AM, Tom Rokicki <rokicki@gmail.com> wrote:
31 factorial is really large; it could be a very long night... :-)
My CPU runs at 2.4 GHz.
time-consuming) FullSimplify. How do you try to avoid getting false
positives and false negatives?
My philosophy is don't worry, be happy. Every additional vertex introduces an error of at most 2^-51 or something like that, so I check the final area against 0 with an epsilon of about 1e-8, and then print the computed final area with the solution. If it's on the order of 1e-15, I'm not certain it's zero, but close enough to report. I'll let Bill and his posse do the hard bits of proving the result. (Or I'll throw multiprecision arithmetic at it. Anyone want to write a Mathematica procedure that takes a permutation, applies it to the roots of unity and computes the area, using high-precision math?) If multiprecision math says the area is < 1e-10000, I'm inclined to view it as zero, at least provisionally.
-tom
-- -- http://cube20.org/ -- [ <http://golly.sf.net/>Golly link suppressed; ask me why] --
-- -- http://cube20.org/ -- [ <http://golly.sf.net/>Golly link suppressed; ask me why] --
-- -- http://cube20.org/ -- [ <http://golly.sf.net/>Golly link suppressed; ask me why] -- _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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participants (3)
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Dan Asimov -
James Propp -
Tom Rokicki