I came across an interesting problem:

There is a round cage and you are in it. Also two lions are in this cage too. The start position is that the distance between you and both lions is the diameter of the circle (you are on opposite sides of the cage). The speed of the lion is 1 m/s.

And the question is:

What is the minimal speed you need to have to always run away from lions and never be caught.

Probably it is a bit simpler to search the maximal speed when the lions will catch you. And the result of the original task will be the upper limit of that value.

I think the radius of the cage doesn't matter - it is only a scale problem. The only important thing is that the cage is round.

There is a similar problem here for one lion. But the answer links to buy some book and I couldn't find where to download it for free =).

And also I wonder if there is a solution for the generalized task with $N$ lions. That looks too complicated but I think the idea is the same - the lions should build a line when you can't run between any two of them and two lions on the ends of a chain will behave like the ones in the two-lions problem.

  • 4
    $\begingroup$ @Patrick: Imaginary lions can share the same physical space and then split up. $\endgroup$ – Asaf Karagila Jan 5 '12 at 8:40
  • 9
    $\begingroup$ Won't they follow the same path then? Unless they are intelligent lions and work out some badass optimal man-eating strategy. $\endgroup$ – Patrick Da Silva Jan 5 '12 at 8:43
  • 3
    $\begingroup$ There would be two versions. Co-op or competitive. There'd be some kind of Nash equilibrium or something. You could even have teams of lions if you have 3 or more. This problem is quite complicated AFAIK. An interesting variant involves trying to escape a pool while someone walks around the edge. Once you get out, you're faster than them. Obviously both of these problems can use a circle or a square or any shape. The main thing you can do is establish inequalities. Like find a strategy that always works for certain speeds. To get exact answers, you need some fancy calculus of paths... $\endgroup$ – karnok Jan 5 '12 at 9:01
  • 6
    $\begingroup$ To take @Asaf's point further, the lions in question are evidently point lions. Otherwise the distance between you and them would be less than the diameter of the cage by at least the width of a lion. $\endgroup$ – Rahul Jan 5 '12 at 9:45
  • 6
    $\begingroup$ Why should both lions use the same strategy? By the was, the appropriate notion of strategy is quite subtle. $\endgroup$ – Michael Greinecker Jan 5 '12 at 10:30

It seems so that this is related to the optimization theory problem The Lion and Man and is in general a pursuit problem.



I would hope to avoid being cornered (metaphorically speaking) and try to get on a path that passes through the middle. The lions on the other hand, being team workers, would want to get to a position where they could home in for the kill. I see it going pretty much like this:

Heading North

Initially we are all heading towards H, after a long chase in which the lions have just failed to catch me. At around the centre, the lions sense they are not going to catch me in a straight line, but notice the edge of the ring is not too far off, so they instinctively start to spread out along the paths CL.


The lions are just about to start homing in for the kill when I scupper their plans by doing an about turn towards J. Thinking that they have a chance to catch me they home in the other way towards the point E where they think they might get their meal.

Near Death

Luckily I just avoid being eaten because I have a laser measuring device and a calculator with me.

So my conclusion is that the ratio of my speed to the lions needs to be slightly greater than HE:LE which would presumably be dependent on the diameter of the ring.

  • $\begingroup$ Thanks for good illustrations. Though it is not proven that you really can avoid being catched in any lion's strategy: if your speed is slightly greater, they can position themselves slightly closer to one another... And the distances HE and LE doesn't have formulas, dependent on ring diameter. $\endgroup$ – Sasha Oct 10 '12 at 13:23
  • $\begingroup$ I agree it's more philosophical than mathematical and I never saw it as a rock solid answer. The question is pretty tricky though. $\endgroup$ – Alan Gee Oct 10 '12 at 13:38
  • $\begingroup$ However, the lions would have to be a bit careful about positioning themselves close, because if they are too close they have no chance of trapping me when I hit the perimeter. My suggestion is that they move apart with the aim of trapping me against the edge and I turn around just before they start to move closer again. $\endgroup$ – Alan Gee Oct 10 '12 at 14:07
  • $\begingroup$ Yes, They should balance distances between them, distance to you and distance to the perimeter. I believe optimal solution for lions will be quite simple trajectory - symmetrical to the radius-vector pointing you. And moving along this line they should balance distance from one to another so that you couldn't run between them as you shown. But additionally you shouldn't be able to run away in the side of them... $\endgroup$ – Sasha Oct 10 '12 at 15:17

Let the centre of the ring be some origin. Let,at any instant, your position be U and those of the lions be $L_1$ and $L_2$. Suppose the minimum speed at which you must move be v. Let, at any instant, the x components of the velocities of the lions be a and $b$ and that of yours be c.


$v_{L_1} = <a,\sqrt{1- a^2}>$

$v_{L_2 }= <b,\sqrt{1- b^2}>$


Now the relation between a, c and b, c is best represented by implicit functions. (You can't really tell whether you are watching the hungry lions run like mad to catch you and then decide how to move or the lions are smart enough to watch you run away from them in fear and than plan out a strategy). Also, if the lions are still smarter, a and b will be dependent on each other.(They can plan to get you together and then share!).

So you understand that the problem is not specific and leads to arbitrary possibilities. It would be better if you can restate the problem more formally.

As for example,(for the solution with N lions) you can think of a $N+1$-gon such that all $N_i$ vetrtices (i goes from 1 to N) move towards the N+1 th one with a constant rate. The N+1 th vertex must never coincide with a $N_i$ vertex while all the vertices are bounded in a circle.

  • 1
    $\begingroup$ Well, it is obvious that if your speed is $0$, you've got no chance to escape from the lions. It is also obvious that if you are much faster than the lions, you can always avoid them. Therefore there must exist a minimal speed $v$ for the problem as stated. Since the problem as given is well defined, there's no need to restate it. Of course, looking at more specific rules for the lions gives a lower bound to the solution (because if you are slower than that, we know a successful lion's strategy), and maybe that's the best we can hope for, but that doesn't invalidate the question as asked. $\endgroup$ – celtschk Sep 20 '12 at 16:11

First, i don't get why two lions are there. But supose they will move always directly to you. If you would move same speed as them around the cage, they would be coming closer and closer to you until they would finnaly cacht you. Or not? I thing they would be still-closing but never able to catch you, since it will end up that they will move 0.0000001mm just behind you. So the conclusion is that you need to move any larger speed than theirs.

  • $\begingroup$ I believe you should try to compute this with some software first. It would be awesome too :D My guess is that the velocity vector (tangent) of each lion would try to reach the one from the fugitive, where the tangent vector of the fugitive is always trying to maximize the distance. This looks like some dynamical system; could even lead to strange attractors :D $\endgroup$ – Marra Oct 4 '12 at 21:53
  • $\begingroup$ Obviously I'm assuming that every lion goes straight to the fugitive every time. I GUESS that a circle with constant speed would be a kind of an answer... $\endgroup$ – Marra Oct 4 '12 at 22:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.