Timeline for Can the thief escape (from a smooth, simple closed curve)?
Current License: CC BY-SA 4.0
21 events
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| Oct 7, 2021 at 17:26 | comment | added | Yaakov Baruch | @AlessandroDellaCorte In plan 2, yes it's irrelevant what strategies the cops use: one by one each is forever left out of reach by the thief (provided the correction the thief makes to his landing plan at each step is vanishingly smaller than the previous one, so that no already lost cop can get back within reach again). I think this may work and a simple, finite starting path that (with the corrections) does the job for a circle $C$ of radius $1$ is one tangent to $C$ on the inside, of radius $>1/2$. Plan 1 is vague: I just feel the speed $>1$ makes the thief potentially unpredictable. | |
| Oct 7, 2021 at 14:52 | comment | added | Eric | @PietroMajer Can you describe what such a strategy of the cops might look like, say for cops on a circle for example? | |
| Oct 7, 2021 at 7:22 | comment | added | Alessandro Della Corte | @YaakovBaruch Please can you specify whether your strategy 2 falls within my proposed setting of the problem? As for 1, I'm a bit lost, as I don't really understand what you mean by "random point which has probability 1 of being cop-free". Even if the thief moves randomly (whatever it means), the cops are not, since they will know exactly where the thief is, so how this exactly applies? | |
| Oct 6, 2021 at 17:09 | comment | added | Yaakov Baruch | @PietroMajer I'm thinking along 2 lines, both in favor of the thief: 1) the thief could make continuous, rapidly decreasing random adjustments to his landing target to make his landing an unpredictable random point which has probability 1 of being cop-free; 2) as the thief nears $C$ and $P$ always moves at $>1$ speed, more cops fall out of the reachability radius. For the first cop (according to a predetermined numbering) that is still within that radius, the thief can exclude him out by a combination of cutting his own distance from $C$ and outrunning the cop, and so for every cop. | |
| Oct 6, 2021 at 15:39 | comment | added | Pietro Majer | @YaakovBaruch I think there is a strategy for the cops, such that, even if the thief is able to avoid any cop and leave him behind, this forces him to delay the arrival time, so that if he wants to avoid them all , he will never escape (in which case he looses according to rule 2). He will then rotate forever, and some cops will move towards his shadow, as soon as he passes, for a while, in case he tries to land. | |
| Oct 6, 2021 at 13:07 | comment | added | Alessandro Della Corte | Ok, I see that there are more subtleties than I thought. But my feeling is that quantifiers should be better specified in the problem statement. One possibility is: everyone has perfect and instantaneous information about the others. The officers place themselves after knowing the starting point of the thief and then they must follow a countable list of strategies decided in advance and such that their motion will only depend on current position and velocity of the thief; on the other hand, the thief knows in advance all the countable strategies that officers adopt. Is this the setting? | |
| Oct 6, 2021 at 12:53 | comment | added | Yaakov Baruch | @AlessandroDellaCorte If the thief approaches the boundary (a circle) along a spiral which has radius of curvature always less than that of the circle and approaches it very gently, then his shadow on the circle moves faster than 1. Perhaps he can use that (and modulating the landing point) to outpace all the countably many policemen that approach him (from either direction). How gently needs the spiral to approach? I think one can solve a differential equation to find the exact formula, but I haven't worked it out. | |
| Oct 6, 2021 at 12:38 | comment | added | Pietro Majer | @AlessandroDellaCorte The problem is that the projection P has a velocity slightly larger than 1, so the thief can either avoid or pass each single cop before touching the curve, and in principle touch the curve in a point not occupied by a cop. | |
| Oct 6, 2021 at 12:32 | comment | added | Alessandro Della Corte | @Yaakov Baruch: I don't really get this. Can you elaborate a bit further? | |
| Oct 6, 2021 at 12:31 | comment | added | Alessandro Della Corte | @Pietro Majer: +1 for the one on the two policemen's theorem :) | |
| Oct 6, 2021 at 10:27 | comment | added | Eric | That would be truly amazing if true. As @PietroMajer commented, whether the officers are coded or not can make a big difference. I think if it turns out that the thief has a strategy to escape, the plan must be something like coding the officers 1,2,3,... and getting rid of the $n$th officer at time $t_n$, traveling along shortening arcs or chords that reach the boundary in finite time $T=t_{\infty}$. | |
| Oct 6, 2021 at 8:13 | comment | added | Yaakov Baruch | @PietroMajer The thief doesn't even need to dribble. Assume $C$ is a circle and the thief starts from the center and approaches the boundary along a fixed spiral path which crosses the circle tangentially. He can probably modulate his speed to cross the boundary without getting caught. And perhaps he can even do that in a $C^1$ or $C^{\infty}$ way. | |
| Oct 6, 2021 at 8:09 | comment | added | Pietro Majer | Maybe one can modify your strategy this way. Two cops start moving as soon the thief is has distance $r$ to the curve $C$, say $C$ is parametrised in arc length, and they are in positions $A<P<B$ with $B-A \approx r$. Then they take care not to be passed by $P$, so that $A<P<B$ holds for all times (this means running after P when he moves away, and moving away when he gets closer). Even if $P$ can move faster, I think $A$ and $B$ can make their distance vanish in finite time. (So at the end the principle should be, of course, the 2 policemen’s theorem) | |
| Oct 6, 2021 at 7:58 | history | edited | Yaakov Baruch | CC BY-SA 4.0 |
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| Oct 6, 2021 at 7:52 | comment | added | Yaakov Baruch | The speed of the shadow $P$ should be bounded by $r/(r-d_P)$, so as the thief gets closer to $C$ its ability outrun the cops decreases, but I don't see if there is a way to use that in favor of the cops. @Eric neatly nails this point with his comment. | |
| Oct 6, 2021 at 7:51 | comment | added | Eric | Suppose the police bureau can freely place countably infinite officers on the x-axis, but each officer's speed is at most $1-\epsilon$. Can the thief cross over the line from one side to the other? I think he can, but I can't prove. | |
| Oct 6, 2021 at 7:42 | comment | added | Yaakov Baruch | @PietroMajer You make a very good point. I don't have an answer to that. | |
| Oct 6, 2021 at 7:37 | history | edited | Yaakov Baruch | CC BY-SA 4.0 |
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| Oct 6, 2021 at 7:34 | comment | added | Pietro Majer | But does this guarantee that one cop catches the thief? The shadow on $P$ of the thief on $C$ has possibly velocity $>1$, while each cop running after it has speed $\le1$. Can’t the thief move so smartly that he reaches $C$ at time $T$, but and at time $T-1/n$ The point $P$ has dribbled or escaped eventually Cop_n ? | |
| Oct 6, 2021 at 7:14 | history | edited | Yaakov Baruch | CC BY-SA 4.0 |
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| Oct 6, 2021 at 7:02 | history | answered | Yaakov Baruch | CC BY-SA 4.0 |