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.

  • 4
    $\begingroup$ 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 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 at 3:36

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.

  • 1
    $\begingroup$ 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 at 18:58

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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