Timeline for Why are there 1024 Hamiltonian cycles on an icosahedron?
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
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| Mar 11, 2017 at 15:49 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
Fixed link.
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| Mar 11, 2017 at 14:54 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
Image links broken; now fixed.
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| Jun 18, 2011 at 15:02 | vote | accept | Joseph O'Rourke | ||
| Jan 16, 2011 at 14:03 | answer | added | Ed Wynn | timeline score: 17 | |
| Sep 5, 2010 at 20:50 | history | edited | Joseph O'Rourke | CC BY-SA 2.5 |
added 27 characters in body
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| Sep 5, 2010 at 20:43 | history | edited | Joseph O'Rourke | CC BY-SA 2.5 |
paths->cycles; added 4 characters in body; added 23 characters in body; added 7 characters in body
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| Sep 5, 2010 at 20:38 | history | edited | Joseph O'Rourke | CC BY-SA 2.5 |
directed vs. undirected; added 2 characters in body
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| Sep 5, 2010 at 17:10 | comment | added | Joseph O'Rourke | @Colin & Robin: Got it--cool! | |
| Sep 5, 2010 at 17:01 | comment | added | Colin Reid | To expand on Robin's comment, the group of rotational symmetries of the icosahedron has elements of order 3. If you consider the action of one such element on the set of Hamiltonian cycles, there must be a fixed point, otherwise the orbits would all have size 3 and so the whole set would have size a multiple of 3. | |
| Sep 5, 2010 at 14:26 | comment | added | BS. | The "translations" in terms of height of trees or paths in the square are not right. It is the distance between adjacent leaves which must be bounded (by 5), and also the distance to the root for the right- or left-most leaf. I don't understand the up/right path picture. | |
| Sep 5, 2010 at 14:17 | comment | added | Joseph O'Rourke | @Gil: Nice!! @Robin: Sorry to be slow---Could you expand on your logic a bit? @BS: That seems a promising approach! | |
| Sep 5, 2010 at 14:06 | comment | added | BS. | But there are also 12-2=10 triangles on each side of the cycle. This may incite to count pairs of triangulated 12-gons (without interior vertices) which "match up" to an icosahedron. Triangulated 12-gons are counted by the 10-th Catalan number, but there are far less triangulations with at most 4 triangles at each vertex, which is obviously necessary (and maybe sufficient) to have an icosahedral "complement". Translated in terms of binary trees, this amounts to a limitation to depth at most 4 trees, or up/right paths between the diagonal and some (3rd?) subdiagonal in a 10x10 square... | |
| Sep 5, 2010 at 12:53 | comment | added | Robin Chapman | As 2560 is not a multiple of 3, then there must be a Hamiltonian cycle with three-fold rotational symmetry. | |
| Sep 5, 2010 at 12:31 | comment | added | Gil Kalai | Joe, at least indeed 1024 x 30 / 12 = 2560. (Every edge belongs to 2/5 of all H. C. s) | |
| Sep 5, 2010 at 12:07 | history | asked | Joseph O'Rourke | CC BY-SA 2.5 |