Timeline for Shortest path connecting two opposite points on a cube
Current License: CC BY-SA 3.0
18 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 17, 2017 at 11:17 | vote | accept | Arseniy Akopyan | ||
| Mar 17, 2017 at 11:03 | comment | added | Gro-Tsen | Maybe one way to explain this answer is to start by a simpler lower bound, $\pi/2$, which can be obtained by simply projecting centrally (=gnomonically) the unit cube to the (inscribed) sphere of radius $1/2$: this map is $1$-Lipschitz so we get a lower bound of $\pi/2$ on the cube from the sphere. The surprising fact that Anton shows is that by essentially patching together $2d$ azimuthal equidistant projections, one for each side of the cube, we can get a bound of $2$. | |
| Mar 17, 2017 at 10:59 | comment | added | Gro-Tsen | @მამუკაჯიბლაძე "Short" means $1$-Lipschitz (see other comments). The $2$ comes from the fact that for equator length $4$, the inverse exponential map (=azimuthal equidistant projection) of a spherical cube covering $1/(2d)$ of the $(d-1)$-sphere maps inside the Euclidean $(d-1)$-dimensional unit cube. (And because both have an inscribed sphere of radius $1/2$, this is optimal.) "Symmetric" means the maps for the faces can be glued together. I agree that, as written, the answer leaves a lot to be filled in, but it is impressive. | |
| Mar 17, 2017 at 7:46 | comment | added | Włodzimierz Holsztyński | Both the question and the answer were hard for me to digest. However, both are IMPRESSIVE. Excellent solution! | |
| Mar 17, 2017 at 6:07 | comment | added | მამუკა ჯიბლაძე | Sorry how do you conclude that the final length on the cube is bounded by $2$? Where will the number $2$ come from?? And what does "short" mean? Geodesic? And what is a symmetric map? | |
| Mar 16, 2017 at 23:08 | comment | added | Gro-Tsen | Ah, I see, you don't rescale the exponential map on each ray as I thought, you simply cut it off. Now I think I understand. | |
| Mar 16, 2017 at 22:47 | comment | added | Anton Petrunin | @Gro-Tsen Here is one way: note that inverse image $K'=\exp^{-1}K$ of the spherical cube $K$ is convex subset of the unit cube $Q$. The closest-point projection $Q\to K'$ is short and the exponent $K'\to K$ is short --- take the composition. | |
| Mar 16, 2017 at 22:40 | comment | added | Gro-Tsen | And your "modified" map from the (Euclidean) $(d-1)$-dimensional unit cube $C$ with center $o$ to the spherical cube $C'$ with center $o'$ is what, exactly? I would imagine something like: use a scaled exponential map on a half-line starting from $o$ in $C$ to the half-line with the same (i.e., tangent) direction from $o'$ in $C'$, scaled so that $o$ maps to $o'$ and the boundary of $C$ in that direction maps to the boundary of $C'$. But I don't find it obvious that this does cannot increase length. Or is it a different map? | |
| Mar 16, 2017 at 22:28 | comment | added | Anton Petrunin | @Gro-Tsen yes, all you say is correct. | |
| Mar 16, 2017 at 22:27 | comment | added | Anton Petrunin | @CarlSchildkraut "sphere with equator 4" means "sphere of radius $\tfrac2\pi$". | |
| Mar 16, 2017 at 22:22 | comment | added | Gro-Tsen | This is a bit hard to follow. Do I understand correctly that "sphere with equator $4$" means "$(d-1)$-dimensional sphere with equator of length $4$", that the "spherical cubes" are $(d-1)$-dimensional ones, corresponding to the facets of the cube in the original problem; and that a "short" map means one which never increases lengths? | |
| Mar 16, 2017 at 21:52 | comment | added | Carl Schildkraut | What does the "sphere with equator 4" mean? | |
| Mar 16, 2017 at 18:40 | vote | accept | Arseniy Akopyan | ||
| Mar 17, 2017 at 11:17 | |||||
| Mar 16, 2017 at 18:28 | vote | accept | Arseniy Akopyan | ||
| Mar 16, 2017 at 18:33 | |||||
| Mar 16, 2017 at 18:04 | comment | added | Arseniy Akopyan | It seams I've got it. Amazing proof. | |
| Mar 16, 2017 at 17:57 | comment | added | Anton Petrunin | @ArseniyAkopyan it is exponential map. | |
| Mar 16, 2017 at 17:50 | comment | added | Arseniy Akopyan | "Note also that if one maps a unit cube centered at the origin by the exponential map it will cover the spherical cube" — Sorry, I do not understand the description of the map. | |
| Mar 16, 2017 at 17:37 | history | answered | Anton Petrunin | CC BY-SA 3.0 |