Timeline for Number of polytopes formed by connecting points on a hypercube
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Apr 24, 2020 at 19:54 | comment | added | skd | Joseph O'Rourke calculated that a(3,2) = 340. Searching the sequence 4, 56, 340 on the OEIS led to this: oeis.org/A255011. There doesn't seem to be a formula recorded there, though. | |
| Apr 24, 2020 at 7:12 | comment | added | ReverseFlowControl | Just realized, the question needs a little more precision, when you say partition into several distinct polytopes....are you counting polytopes that are solids, facets, ridges, planes, lines....is there any restriction to the kind of polytopes? I've been making assumptions. Also, for some reason the phrasing with lines instead of hyperplanes made a lot more sense. | |
| Apr 24, 2020 at 7:00 | answer | added | ReverseFlowControl | timeline score: 0 | |
| Apr 24, 2020 at 6:07 | comment | added | ReverseFlowControl | There is a combinatorial formula for this, it gets cumbersome to generalize for higher dimension. Its easy for a(2,2), it also validates 56 as the answer. Its a double sum, inner sum over outer lattice points - 2, and outer sum over increasingly fewer lattice points as the starting point. | |
| Apr 23, 2020 at 22:37 | answer | added | Gerhard Paseman | timeline score: 1 | |
| Apr 23, 2020 at 21:21 | answer | added | Joseph O'Rourke | timeline score: 0 | |
| Apr 23, 2020 at 20:39 | comment | added | skd | @JosephO'Rourke I just drew it and counted! I don't know how to code this up, but it's probably possible to do so and compute more values. It'd be way more efficient than drawing and counting... | |
| Apr 23, 2020 at 20:33 | comment | added | Joseph O'Rourke | May I ask: How did you calculate $a(2,2)=56$? | |
| Apr 23, 2020 at 16:35 | history | edited | skd | CC BY-SA 4.0 |
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| Apr 23, 2020 at 5:37 | history | edited | skd | CC BY-SA 4.0 |
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| Apr 23, 2020 at 4:08 | history | asked | skd | CC BY-SA 4.0 |