Timeline for Can the thief escape (from a smooth, simple closed curve)?
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Oct 13, 2021 at 19:35 | comment | added | Alessandro Della Corte | This is really nice! | |
| Oct 10, 2021 at 11:42 | vote | accept | Eric | ||
| Oct 10, 2021 at 10:51 | comment | added | Yaakov Baruch | @Eric Yes and no, respectively. It seems that a stretch of the curve with negative or $0$ curvature (plus finitely many singularities) can be guarded by a finite number of policemen, while any stretch with positive curvature (bounded below) cannot be guarded by even a dense, countable set. The concave square would be even easier to guard than the regular flat square. | |
| Oct 10, 2021 at 7:58 | comment | added | Eric | So it seems the key to the problem is positive curvature? If the thief is initially outside the circle, do you think he can get inside it? If the enclosure has negative curvature everywhere except at finite point, a concave square for example, does the thief get out? | |
| Oct 10, 2021 at 4:54 | comment | added | Yaakov Baruch | How different the situation if the curve isn't smooth! For a example a thief at the center of a square enclosure is always caught by 4 cops starting at the centers of the edges, but can escape 4 cops starting at the corners. | |
| Oct 9, 2021 at 22:56 | history | edited | Yaakov Baruch | CC BY-SA 4.0 |
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| Oct 9, 2021 at 19:52 | comment | added | Yaakov Baruch | @ChristianRemling Thank you for the thoughtful feedback. I'll edit the answer. Regarding internal segments, the problem is that their points with the property of the shadows moving faster, form subsegments that don't reach the circle. So one would need some messy pasting of infinitely many different segments and it may no longer be so easy to prove that they add up to an infinite path. The beauty of the 2 tangent circles is that because of the symmetry between left and right, the total length of the path is unaffected by the zigzagging. | |
| Oct 9, 2021 at 18:01 | comment | added | Christian Remling | By the way, I think you could also use straight line segments instead of the circular arcs, then you don't need the "tedious calculation" in item 3. | |
| Oct 9, 2021 at 18:00 | comment | added | Christian Remling | Thanks for clarifying, it's clear to me now. I probably didn't read it carefully enough. But maybe it would help to say explicitly that the structure of the argument is that you approach a point $p$ along a path that is divided into segments, and what we do on the $n$th segment will make sure that $d(P_n,P)\ge \epsilon_n$. | |
| Oct 9, 2021 at 17:04 | comment | added | Yaakov Baruch | @ChristianRemling. The definition of "inactive" uses a strict inequality (the word "more"), so for each policeman $P_i$ there a margin of $\epsilon_i$ that can never be closed no matter what $T$ does. Since $T$ (and $S$) clearly converge, each $P_i$ is at least $\epsilon_i$ away from their limit. Should I edit the answer to make that more clear? | |
| Oct 7, 2021 at 22:40 | history | edited | Yaakov Baruch | CC BY-SA 4.0 |
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| Oct 7, 2021 at 22:32 | history | answered | Yaakov Baruch | CC BY-SA 4.0 |