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The fugitive is at the origin. They move at a speed of $1$. There's a guard at $(i,j)$ for all $i,j\in \mathbb{Z}$ except the origin. A guard's speed is $\frac{1}{100}$. The fugitive and the guards move simultaneously and continuously. At any moment, all guards move towards the current position of the fugitive, i.e. a guard's trajectory is a pursuit curve. If they're within $\frac{1}{100}$ distance from a guard, the fugitive is caught. The game is played on $\mathbb{R}^2$.

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An animation with guards' speed $\frac{1}{4}$ looks something like this (source):

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Question: can the fugitive avoid capture forever?


What I know:

  1. The fugitive will be caught if they remain in a bounded area.

  2. The distance between two guards is strictly decreasing unless the fugitive and the guards remain collinear. But the further away the guards are, the slower that distance decreases.

  3. Even if there're only 2 guards, a straight-line dash into the gap between the pair will lead to capture, as long as they're sufficiently far away (see radiodrome).

  4. The fugitive can escape from arbitrarily large encirclement, provided the wall of guards is not too "thick" (4 or 5 layers are fine), such as this (3 layers):

enter image description here

The shape of the wall doesn't matter (doesn't have to be rectangular).


I asked the question sometime ago on math.stackexchange, where I received the cool animation above, but got no definite answer. I was inspired by a very similar problem here on MO, with additional complication of randomness.

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    $\begingroup$ Is there a specific reason that 1/100 is used here? Seems like making that a parameter might be more natural. Actually for that matter, you have it in two different places, since you use it as both the speed and the pursuit radius. These could be different numbers. $\endgroup$
    – JoshuaZ
    Jun 12, 2021 at 2:11
  • $\begingroup$ @JoshuaZ There is not. If guard's speed v is a significant fraction of the fugitive's, and the pursuit radius r is not too small, then the fugitive gets caught. Either the fugitive gets caught for all positive v and r, or they escape for small enough v and r. I don't know which is true. 1/100 is just there for ease of calculation. $\endgroup$
    – Eric
    Jun 12, 2021 at 3:36

3 Answers 3

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I expect the fugitive will be captured, but only after over $10^{200}$ (not sure about $10^{300}$) units of time.

As noted in my answer in the linked question, the fugitive will be captured for a random distribution of pursuers. However, we cannot just rely on the density of the pursuers: For some distributions of pursuers with at least one in each unit square, the fugitive can escape using a sinusoidal path.

What happens is that the distance between pursuers never increases. If the evader (i.e. fugitive) is far enough from a given pursuer $P$, the relative displacement of pursuers near $P$ is approximately linear, and the only way the evader can approach $P$ is to make the determinant (of the above linear transformation) very small, with a strong contraction in all but possibly one direction. Each time the evader gets twice as close to a faraway pursuer, the pursuer density increases in $1+Ω(v)$ times where $v$ is the pursuer speed ($Ω$ is per Big O Notation). Note that this permits $e^{Ω(v)}$ evasion time; for the above, $10^{200}<10^{300}<2500^{100}$, with $100=v^{-1}≈v^{-1}+1$ (this ignores the penalty for evasion maneuvers), and 2500 = 502 is the approximate pursuer density increase (depends on the pursuer configurations) to end the continuous gaps between pursuers.

However, large scale pursuer density increase does not suffice for capture if there are special local patterns of pursuers. Randomness rules out such patterns, while having pursuers at the integral coordinates allows some patterns, but apparently not enough to escape. For a lattice of pursuers, a faraway given area of pursuers will be impenetrable, expect possibly for one-dimensional movement (compared to fully impenetrable for random pursuer start). Now, for a straight line approach, the permitted movement direction is approximately perpendicular to the direction of approach, and so once you are confined long enough to one-dimensional movement, the pursuers should close in.

This should also apply to pursuer lattices in $k$ dimensions. After sufficient time, approaching an area confines fugitive movement to essentially a $k-1$ dimensional hyperplane (though one has to bound the nonlinearities). In turn, approaching a more local area after moving in the hyperplane should restrict movement to at most $k-2$ dimensions, and so on.

Moreover, if time $T$ is large enough, at some (large enough) scale $S$, the time-space movement up to $T$ will be close to linear (specifically, $E( |2p(t+S)-p(t)-p(t+2S)|) < εS$ where $E$ is expectation (i.e. average across time) and $p$ is position given time) since the speed is bounded and deviations from linearity cause inefficiency/slowdown that adds up across scales. And as noted above, with pursuers initially at the integral coordinates, linear movement will create an impenetrable wall requiring a turn (and thus large-scale nonlinearity), except possibly if we visited that area of pursuers before (which should not work for escape either).

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In the case that the pursuers have to actually catch the fugitive, this was answered in the article Escaping an infinitude of lions by Mikkel Abrahamsen, Jacob Holm, Eva Rotenberg, and Christian Wulff-Nilse.

They prove the following much more general theorem showing that the fugitive can always escape.

Theorem. For any $\epsilon >0$, a fugitive running at speed at most $1+\epsilon$, can always escape countably infinitely many pursuers running at speed at most $1$.

This holds for any starting position (provided no pursuer starts at the same point as the fugitive). The pursuers are even allowed to follow any pursuit strategy they like.

Note that in their version the fugitive and pursuers are allowed to slow down, but as pointed out by Alessandro Della Corte, this can be simulated by constant speed strategies.

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    $\begingroup$ In the paper that you mention the lions can be placed on any countable subset of the plane. This means that they can even be dense (!). Of course, in that case, in the OP’s version of the problem the man gets caught at once. $\endgroup$ Oct 2, 2021 at 18:58
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This is not an answer but an extended, elementary comment to argue in favor of a slight reformulation of the problem along the lines proposed in the comment by JoshuaZ.

Suppose that the fugitive escapes if the pursuers have speed $v$. Then, if the fugitive can slow down it would be immediate to see that he also escapes if the pursuers have any speed $v'$ less than $v$, because he can simply take speed $\frac{v'}{v}<1$. If the fugitive is obliged to proceed at speed 1, as you require, he can still emulate the process at speed $v$ making as many tiny oscillations (of amplitude less than $1/200$) as required along his way, and if you like he can do that in a $C^\infty$ way.

Since the fugitive is obviously caught if $v$ is large enough, then either there is a critical speed $v_c$ below which the fugitive always escapes and above which he is caught, or he never escapes (in which case you can set $v_c=0$). So a more natural version of the problem, for me, is to ask for the value of $v_c$. Of course that this also depends on the distance threshold of $d=1/100$ you fixed, so even more sensible would be to ask for $v_c$ as a function of $d$.

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