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Generally, in pursuit-evasion games, there's one prey and one or many pursuers. I'd like to know how extending the food chain would change the dynamics of such games.

Specifically, let's consider a closed, circular shape arena in $\mathbb{R}^2$. Three wolves are uniformly distributed at the border. The sheep and his lion friend are at the center.

enter image description here

If $d(w(t),s(t))=0$, the wolf eats the sheep, if $d(l(t),w(t))=0$, the lion eats the wolf, where w(t), s(t) and l(t) are the trajectories of the animals, and $d$ measures euclidean distance. The pack of wolves work as a group, their goal being to eat the sheep. The goal of the lion-sheep team is to prevent the sheep from being eaten, indefinitely. All animals move continuously in time at equal speed and are intelligent.

Can the lion protect the sheep from the wolves? In general, how many lions are necessary to protect the sheep from a pack of $N$ wolves uniformly distributed at the border?

This is a puzzle I originally asked here on puzzling.stackexchange. I know already that

  • A single wolf doesn't catch the sheep.
  • Two wolves will catch the sheep.
  • $N-1$ lions are sufficient to fend off $N$ wolves.

See here for the proof of those claims. I'm interested in knowing whether $N-2$ or less lions can fend off $N$ wolves.

Eric
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    The extraneous problem with this nice question is that when the lions get hungry they may eat the sheep themselves. – Ethan Bolker Apr 06 '22 at 13:54
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    For lion = N-2, can they just go after the closest wolf first to reduce the problem to N-1? – Paichu Apr 06 '22 at 16:42
  • @Paichu I'm not sure. It takes at least 2 lions to work as a team to kill a wolf. Say N=4, then as the 2 lions are hunting down a wolf, the other 3 wolves are hunting down the sheep. Is there enough time for the lions to return and help the sheep? Seems unlikely to me. – Eric Apr 06 '22 at 17:20
  • @Eric If the arena is a bounded circle with wolf starting on the edge and lion in the center, you can prove that it takes at most $r/v$ for one lion to catch one wolf regardless of how the wolf chooses to run, where $r$ is the radius of the circle and $v$ is the speed (assuming they all have the same speed). – Paichu Apr 06 '22 at 18:34
  • @Paichu That can't be further from the truth. It's a well known result that one lion does not catch the wolf. There's a proof in the link in the question, too, if you wonder why. – Eric Apr 07 '22 at 00:18
  • @Eric If the circle is a bounded domain, it happens in $r/v$ time. – Paichu Apr 08 '22 at 04:20
  • @Paichu Prove it then. – Eric Apr 08 '22 at 04:29
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    On p.$135$ of his A Mathematician's Miscellany, J.E.Littlewood gives a plausible, but erroneous, argument purporting to show that a single lion can always catch a single wolf (which he calls a "man"). After noting that the argument is wrong, he gives A.S.Besicovitch's strategy for the wolf that enables it to avoid capture indefinitely – lonza leggiera Apr 08 '22 at 14:57
  • @Eric Think about it. The starting position between a lion at the center and a wolf at the circumference is the furthest. Regardless of how the wolf moves, the lion will move directly toward the wolf. The best scenario is for the wolf to move along the circumference, but even in that scenario, it will only take the lion $r/v$ time to get to the wolf. – Paichu Apr 10 '22 at 20:59
  • @Eric Actually, it's $\sqrt{2} r/v$. – Paichu Apr 10 '22 at 21:04
  • This doesn't make sense, why should the lion protect the wolf – Clemens Bartholdy Apr 12 '22 at 09:38
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    Four clarifications seem necessary to answer: (i) Is the "top speed" of each entity the same? (This may be a convention for pursuit problems, but it's advisable to be explicit.) (ii) Does each entity move differentiably, or just continuously? (iii) Can the sheep and lion be at the same point, as described verbally? (If so, my first guess would be that they should stay together, possibly modulo the fourth clarification.) (iv) What happens if the sheep, one or more wolves, and the lion meet at a point? (Hopefully not a universe-creating explosion.... :) – Andrew D. Hwang Apr 12 '22 at 13:45
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    @AndrewD.Hwang I've said in question that they move at the equal speed and continuously. As for your question (iii) and (iv), the sheep and lion can be at the same point, nothing happens. If all three at the same point then lion eats wolf and wolf eats sheep, wolves win. – Eric Apr 12 '22 at 14:38
  • My oops about (i) and (ii); I read right past that. <> It still looks to me that the answer depends on details of information-sharing: If the sheep and lion can stay at the same location, they can sit at the center of the circle waiting for a wolf to approach. Just before a wolf reaches them the sheep steps away in the opposite direction to the wolf's approach and the lion eats the wolf. (As necessary, the sheep/lion can move so as not to be equidistant from any two wolves, avoiding the need to handle two assailants at once.) – Andrew D. Hwang Apr 12 '22 at 14:49
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    @AndrewD.Hwang Reaction is instantaneous, and the situation is symmetric: there's no "just before" to which the sheep can react but the wolves can't. – Eric Apr 12 '22 at 15:46

1 Answers1

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I like your problem. Let me summarize.

Is it possible for a lion to guard a sheep on a 2D plane against three wolves working as a team? This is for a video game scenario where all five animals travel at the same speed continuously and can all share the same coordinate. The wolves begin equidistantly apart. If the lion, the sheep and at least one wolf share the same coordinate, then the wolves win. All the lions begin at some point equidistance from each other to begin (an equilateral triangle).

I think I might have a proof, but would like it checked to be sure.

Theorem. It is possible for the lion to defend the sheep in this scenario so long as the following conditions are met: 1) the lion and the sheep run in a rotating elliptical orbit, having width a and length b, such that 2) the rotation of the ellipse is in the opposite direction that the lion and sheep run around the ellipse, 3) the wolves begin at a distance from the sheep greater than 2 times the distance of that between the lion and sheep, and 4) given

S_half=2a ∫_0^(π/2) 1-e^2 sin^2⁡(θ) dθ,

where S_half is half the distance around the perimeter of the ellipse, θ is maintained to be equal or greater than 2pi/3 radians.

Proof of the Theorem. Let a wolf and sheep begin on both the perimeter and the semi-minor axis of an ellipse whose width is a and length b, but on opposite sides from each other. Let the wolves be on the points of an equilateral triangle to begin. Let the triangle encompass the area of the ellipse, but not inscribed. Instead, align the lion to be directly between two of the wolves (on the perimeter of the triangle) and the sheep directly center of the triangle. Let this center point be the center of the cartesian plane in order to work out the math.

To begin, the lion and sheep run in a counter-clockwise motion on the ellipse whose theta is 120 degrees or greater (2pi/3 radians) the wolves will run in a straight line toward where the sheep is headed. If they run in any curved motion, this will be farther a distance to travel and the result will be even more advantageous for the lion and sheep. Because the lion is running in a curved motion, he charges two lions at the same time, the first to his left and also the next to his left as his curved run continues. Since the wolves must move continuously, they will need to curve their trajectory away from the curved run of the lion, which pushes them in the opposite direction from where the sheep will soon follow on the opposite side of the ellipse.

The third lion is the only one that can run in a straight line, as the lion is running in the opposite direction from it. However, since the third wolf began more than 2 times the distance from where the sheep was from the lion, and the theta is 120 degrees or greater (the arc is slight), he will return to where the sheep was prior before the third wolf can get the sheep. Since he will be coming in at a certain trajectory, the wolf must curve away in the opposite direction.

The cycle repeats as the lion curves once more toward the first wolf, but this time at a 60 degree pivot clockwise (assuming theta equals 120 degrees), and then the second as before. Since they are always pushed a greater distance away from the sheep, their distance will converge around a boundary. The direction of the wolves attack doesn’t matter, as it is assumed that would at most only one wolf at a time will be able to maintain a straight line attack. ∎

Jeff
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