Timeline for Walks that cannot hit the boundary
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 14, 2011 at 21:43 | vote | accept | Valerio Capraro | ||
| Oct 14, 2011 at 21:43 | comment | added | Valerio Capraro | Yes, on the squares everything get very intuitive. Actually my proof would work for more general subsets of $\mathbb Z^2$; intuitively those subsets without holes whose boundary has a hole. Well, maybe I will put it just as a remark. Many thanks for the clarifications. | |
| Oct 14, 2011 at 17:40 | comment | added | Ilya Bogdanov | Well, if your map is a bijection on the boundary, then some two points should come to opposite vertices of the square. Then they should be the vertices themselves, otherwise there would exist a path between them which is shorter than $4n$. Hence it is also a bijection on the vertices, and by the same reasons of the shotrest path the whole map is some symmetry of the square. | |
| Oct 14, 2011 at 17:32 | comment | added | Valerio Capraro | I was thinking that actually it would be enough for me that the points on the boundary are mapped bijectively in points on the boundary (veryfying the property in the definition). But probably also in this case the result remains trivial: it seems to me that in this case every square of the shape $[-k,k]^2$ is mapped in itself | |
| Oct 14, 2011 at 17:23 | comment | added | Valerio Capraro | Jesus! I'm doing a mess. It was enough to draw a stupid picture to realize it! | |
| Oct 14, 2011 at 17:06 | history | answered | Ilya Bogdanov | CC BY-SA 3.0 |