Timeline for Number of hypercube unfoldings
Current License: CC BY-SA 4.0
15 events
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| Apr 19, 2023 at 14:22 | comment | added | Dmitry Krachun | Note the following post which suggests that the ratio $N_d/F(d)$ tends to $c=e^{(e^{-2}+e^{-4})/2}\approx 1.07985$. It also contains the list of first 100 values. postchimpblog.wordpress.com/2022/05/29/… | |
| May 27, 2021 at 13:46 | history | edited | Ivan Aidun | CC BY-SA 4.0 |
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| May 21, 2021 at 20:40 | history | edited | Ivan Aidun | CC BY-SA 4.0 |
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| May 18, 2021 at 20:46 | comment | added | Ivan Aidun | I also agree that the real conjecture is probably something about compositions of reflections, since that's what my constructions look like so far. | |
| May 18, 2021 at 20:35 | comment | added | Ivan Aidun | @PeterTaylor I actually just made an edit saying that this conjecture is wrong. You can actually see it in $d=3$, as there are 4 nets fixed by a nontrivial rotation. | |
| May 18, 2021 at 20:34 | history | edited | Ivan Aidun | CC BY-SA 4.0 |
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| May 18, 2021 at 20:28 | comment | added | Peter Taylor | "I think (tho I haven't sat down to prove) that rotations will never fix a spanning tree." Counterexample when $d=5$: edge set $\{(-x,+z), (-y,+x), (-z,+y), (-v,-x), (-v,-y), (-v,-z), (-v,-w), (-w,+v), (+v,+w)\}$ is fixed by rotations which permute $(x, y, z)$. | |
| May 17, 2021 at 18:38 | comment | added | Peter Taylor | I brute forced all spanning trees up to $d=5$, grouped them by the automorphisms of the hyperoctahedral graph, and get (group size, frequency) data $[(24, 6), (48, 5)]$ ($d=3$); $[(48, 4), (96, 4), (192, 77), (384, 176)]$ ($d=4$); $[(80, 2), (240, 6), (320, 2), (480, 60), (640, 2), (960, 147), (1280, 2), (1920, 1971), (3840, 7502)]$ ($d=5$). That the higher frequencies occur for group sizes $2^{-j}$ times the size of the automorphism group suggests that a careful account of trees fixed by $j$ reflections might get a reasonably tight estimate. | |
| May 17, 2021 at 18:32 | comment | added | Moritz Firsching | That's cool! I was aware that the largest fixset of spanning trees has size $T_d(d-1)2$. If we look at the list (size of fixset, #elements in Aut with that size), we have for $d=3$ the list $[(16, 9), (384, 1)]$ and for $d=4$ the list $[(8, 24), (288, 12), (96, 12), (16, 12), (192, 12), (128, 6), (2304, 4), (82944, 1)]$. Those then give the exact number via Burnside after dividing by $48$ and $384$ resp. | |
| May 17, 2021 at 17:13 | history | edited | Ivan Aidun | CC BY-SA 4.0 |
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| May 17, 2021 at 17:00 | history | edited | Ivan Aidun | CC BY-SA 4.0 |
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| May 17, 2021 at 2:18 | comment | added | Brendan McKay | My money is still on a limit of 1 since the numbers are still small and it is highly unusual for this type of limit to be anything except 0 or 1. It would be nice to prove an exception, though. Try dividing the ratio by $1+1/n$, which doesn't change the limit but flattens the values a lot; I think you'll agree that 1 is looking good. | |
| May 16, 2021 at 20:01 | review | Late answers | |||
| May 16, 2021 at 21:13 | |||||
| May 16, 2021 at 19:47 | review | First posts | |||
| May 16, 2021 at 20:27 | |||||
| May 16, 2021 at 19:45 | history | answered | Ivan Aidun | CC BY-SA 4.0 |