On a tiny spherical planet there exist $N$ wolves and 1 hare.

The planet is so small any of these creatures can circle it in exactly 1 day. No creature needs to sleep or eat. The wolves communicate by howling. The hare doesn't understand. All creatures play perfectly.

As the hare has 4 very lucky feet, the original positions are the most favorable ones for the hare and it can always guess where the wolves are.

What is the minimum number $N$ that guarantees the hare is caught? How long does it take?

Consider the following cases:

  1. The wolves can always smell where the hare is.
  2. The wolves have stuffy noses and can only smell the hare if it is within 1 hour's run.
  • 1
    $\begingroup$ Are the wolves and hare just points? $\endgroup$
    – user88
    Apr 23, 2015 at 21:36
  • 8
    $\begingroup$ "Consider a spherical hare..." $\endgroup$ Apr 23, 2015 at 21:47
  • 10
    $\begingroup$ @IanMacDonald nature.ca/ukaliq/elem/pop/P0018-e.html $\endgroup$
    – kaine
    Apr 23, 2015 at 21:56
  • 6
    $\begingroup$ If no creatures need to sleep or eat, why are the wolves hunting the poor hare ;_; $\endgroup$
    – Lopsy
    Apr 23, 2015 at 22:20
  • 8
    $\begingroup$ Don't worry poor little @Lopsy. I'm sure they are just playing tag. That's what the hare keep telling herself too. $\endgroup$
    – kaine
    Apr 23, 2015 at 22:28

5 Answers 5


2 wolves

My thinking is thus: First Assume the wolves and hare are on the same line of longitude. The hare at the equator and equidistant from the wolves

However the hare moves the wolves can always stay at the same longitude as the hare by moving along their own line of latitude. Because the hare is between the wolves their latitude circles will be smaller than the hares. This gives them a free longitudinal component of motion they can use to close with the hare. Although the hare can move out of its equidistant point it will always be caught.. eventually

Now to consider out initial assumption. We can partially dispense with the equidistant requirement as the hare is able to move closer to one wolf or the other from its starting point without escaping or increasing its distance from either wolf.

We can also define our longitudinal axis as the plane of W-H-W. as long as the animals are on a plane they can be considered on a line of longitude

So, can the wolves move to position the hare between them and on a plane regardless of the hares movement? well if the wolves each move to a pole they form a plane with the rabbit wherever it is. The rabbit may not be in the same hemisphere, trapping one wolve behind the 'hump' of the planet. but this wolf is still able to maintain its longitudinal match with the hare and the nearer wolf can chase the hare towards the equator until they are all in the same hemisphere


NOW! max time to capture.


The best strategy from the hare is to move such that the wolves come closer at the slowest rate. This is achieved by having the maximum longitudinal speed.

Lets consider the case where the wolves are at the poles and the hare at the equator.

The hare runs along the equator and the wolves match its longitude but slowly approach. They cant run directly to the closest intersection point because the hare will reverse direction and escape. As they get closer and closer to the hare their courses become more and more parallel. But we are asked to consider them as points. Luckily they don't need to eat

1h smell radius. much harder. have to think

OK so, The wolves can detect the hare when they are 1h away from it. If they all happen to be on a plane as before they can catch it. this is the case where 2 wolves are 2hrs apart and the hare is directly between them.

So 12 wolves (24h/2) can start at one pole and, equally spaced, move to the other pole eventually trapping the hare

Say we have 1 less wolf, the ring of wolves almost reaches the equator and then MUST leave a gap. but can the rabbit squeeze though an infinitesimal gap? the wolves move at speed s same as the hare, the overlapped region of smell expands at sqrt(1-s^2)... no thats on a flat plane, for a sphere its.. arccos(cos(1)/cos(s)) ? which is always less than s. Ergo the hare can escape from between any two closing spherical caps

OK, Considering the strategy for 3 wolves.

Assuming the same best case (for the hare) starting point with wolves on poles and hare on equator but with one spare wolf at one pole. The 2 wolves move as before but the spare wolf moves to an arbitrary point on the equator to try and cut off the rabbit.

In this case the hare is still caught at infinity ( i think the same infinity, have to do the maths) as it can approach the third wolf but switch direction immediately before capture. As long as its speed remains constant the approach time of the original wolves is the same and the third wolf cannot catch it.

However what if the original two wolves vary their strategy, dropping back to prevent the reverse direction tactic? tbc..

  • $\begingroup$ This answer is very incomplete....did you accidentally submit it? $\endgroup$
    – kaine
    Apr 28, 2015 at 0:53
  • $\begingroup$ yes, although I was tempted to leave it at no parralell lines $\endgroup$
    – Ewan
    Apr 28, 2015 at 0:54
  • $\begingroup$ There is no reason in your answer to believe the hare must walk in a straight line or that paths that intersect must mean the hare and wolf must be there at the same time. $\endgroup$
    – kaine
    Apr 28, 2015 at 0:57
  • $\begingroup$ explaining makes the answer less poetic $\endgroup$
    – Ewan
    Apr 28, 2015 at 1:14
  • $\begingroup$ With the explanation, this answer is much better... i think there is a wolf strategy (i have one) for which the rabbit can be caught if finite time but otherwise I agree with everything. I really like the "same longitude" way of showing it. $\endgroup$
    – kaine
    Apr 28, 2015 at 16:03

I think case 1: 3 wolves (edited solution). Form an equilateral triangle around the hare. Let the hare run for an edge midpoint. Take the line between the hare's starting point and the edge midpoint, and double it. Let one wolf on that edge run to that point (the wolf and rabbit will get there at the same time). The other two wolves maintain a shrinking equilateral triangle with the first wolf. Continue to shrink until the hare is caught.

Regarding time: Let the wolves start at the north pole and the hare at the south. The wolves will take 24 hours to catch the hare if it just sits there, so it will take at least 24 hours with the hare playing optimally (since the starting position is optimized for the hare and the hare is capable of perfect play). I'm a little shaky on the timing of what happens in the shrinking capture plan (I think some calculus might be needed, and also I must play perfectly for the rabbit, which I have little confidence in my ability to do) but basically once the wolves cross the equator, the triangle is formed and the rabbit can run for the edge midpoint between two wolves. So we're at 12 hours now. As soon as the rabbit starts moving, the interceptor wolf starts running backwards for ((sqrt(3)/2)*8) - (((sqrt(3)/2)*8)/3) = 4.61880215352 hours (using equations off wikipedia to calculate height and height of center from edge midpoint when triangle side length = 8), which brings us to ~16.6 hours. However, the rabbit doesn't have to run willingly to meet the wolf, and could probably juke around for maybe 1-2 more hours. However, this calculation requires predicting perfect play by the rabbit, and I have no way of doing that. The hare is better off camping the south pole here, and maybe juking a little bit at the end to last a few more minutes.

Why 2 isn't enough in case 1: Let the 2 wolves come at the hare from different distances. The hare can position itself to be equidistant from either wolf by running towards the farther-away wolf. Now, let the wolves come at the hare from opposite directions, so that the three creatures form a line. The hare can draw a perpendicular line and follow it, and since the wolves will have to chase at an angle, they will fall behind. So let the wolves form an angle with the hare: the hare bisects the angle and follows this line, but in the direction away from the wolves. The wolves are even farther behind now! Since these are the only ways for the wolves to approach the hare, the hare always escapes.

Case 2: 8 wolves is enough for a naive comb: start them out at the north pole and send them southward, spread out so that by the time they get to the equator, there's 2 hours travel time between each wolf, then have them continue on to the south pole. Since the equator will be 24 hours travel time around, this guarantees finding the rabbit. Create a shrinking octagon around it. This should take, again, at least 24 hours, since the hare can camp at the south pole, and can take a tiny bit more time than that if the hare jukes a little, which again requires knowledge of perfect play to calculate, but since the wolves can reach the midpoint of the edge before the hare can, I think even with maximum juking the hare wouldn't make it more than 24 hours 15 minutes.

Best guess for the timing: I think the hare's best plan really might be to just sit on the opposite pole from the wolves' starting point and never move. This guarantees it 24 hours. Any fancy plays could cost it time because it's running to meet the wolves somewhat, so I'm guessing all methods with perfect play take exactly 24 hours.

I retract my hypothesized solution in 6; I found a flaw in it. 8 definitely works for case 2.

  • $\begingroup$ In an equilateral triangle, the center is closer to the edge midpoint than the vertices. So the hare could escape the trap. $\endgroup$
    – JS1
    Apr 24, 2015 at 3:25
  • 1
    $\begingroup$ It still might work, because the two closest wolves could intercept toward a point outside the triangle, while the farthest one follows directly. This results in the triangle shrinking on one side. I think you can prove that one wolf will always get closer no matter which direction the hare moves in. $\endgroup$
    – JS1
    Apr 24, 2015 at 3:40
  • $\begingroup$ The hare can reach the side-midpoint faster than the wolf can. I think you need to construct a hexagon to prove that your solution works. $\endgroup$
    – user88
    Apr 24, 2015 at 4:38
  • $\begingroup$ @CalebBernard Could you maybe provide a little more explanation? I am not convinced 2 can't catch the hare in part 1. I am not even convinced 6 can catch it in part 2. There is also no indication of how long it takes besides a guess. $\endgroup$
    – kaine
    Apr 24, 2015 at 13:07
  • 2
    $\begingroup$ Wolves are points. They don't take up the complete hour. It'd be more like 000|000|000|000|000|000|000|000| $\endgroup$
    – JonTheMon
    Apr 24, 2015 at 15:05

For scenario 1, I think it's possible for 2 wolves to get the rabbit. Take this case: 2 wolves on equator, opposite sides, rabbit at north pole. The rabbit moves perpendicular to the wolves towards the equator. In the simple case, if the wolves then move along the equator, they will hit the rabbit as it hits the equator. If the rabbit doesn't move, the wolves just go towards the pole until it does.

You could think of it like this, the wolves either move towards the rabbit, or parallel to the rabbit. But, parallel lines on a sphere always intersect. So, even when running parallel to the rabbit, they gain ground. More specifically, though, the wolves move so that the hare is always between them on a circumference. So in the first case, the wolves more towards the rabbit initially (so away from the equator) while still moving in the general equator direction.

If the rabbit is constantly moving, if the angle between them and the rabbit is B, they move $\frac{1}{12}\pi * (1-cos(B))$ towards the rabbit ©. They eventually need to make it $\frac{\pi}{2}$ radians towards the rabbit. A quick wolfram integration gives 5.06 radians, so a time of 19.33hours.

For scenario 2, I think the biggest issue is the wolves finding the rabbit. As such, it would take 12 wolves, since each the radius of the planet is 24hrs and each wolf can smell 2hrs. The wolves start 1 hour from a pole, then moves to the other pole (and at the equator are equally spaced). At some point, the rabbit will be found, and can then be caught. Max search time would be 10 hours.

© Imagine the sphere is divided up into 3 parallel planes: one through the poles (the rabbit's path) and two $\pi/3$ rads away (wolves' path). Since the wolves' path is 1/2 the length of the rabbits, they can use 1/2 their movement to keep it in line, and 1/2 to get closer.

  • $\begingroup$ This is guarenteed to work for either case and is a good upper bound but is it the minimum? $\endgroup$
    – kaine
    Apr 24, 2015 at 13:23
  • $\begingroup$ @kaine - I still think the second scenario is bound at 12 wolves, but the first one can go down to 2 $\endgroup$
    – JonTheMon
    Apr 24, 2015 at 15:20
  • $\begingroup$ I agree 2 is possible (though I would appreciate the expected amount of time and a more rigorous explanation). I think you can beat 12 wolves. $\endgroup$
    – kaine
    Apr 24, 2015 at 15:24
  • $\begingroup$ As an edge case for scenario 1, what is the hare keeps running back and forth perpendicular to the wolves and never stops or attempts to cross the border? The wolves would continue to run the equator and never approach the hare. I think you might need 3 wolves for your idea. :) $\endgroup$
    – Mark N
    Apr 24, 2015 at 16:23
  • $\begingroup$ @MarkN Staying on the equator is the simple example. Practically, they'd move in more to keep the hare between them. $\endgroup$
    – JonTheMon
    Apr 24, 2015 at 16:28

For my second approach on both the first/second case, I think the minimum is 3 wolves.

For my first approach on both the first/second case, I think the minimum is 6 wolves. Here's my approach:

Use X wolves to trap the hare, and 1 to chase it down.

2*pi*r*90/360 > 2*pirX/360. Or in other words, the hare's arc length must be larger than the arc length of the wolves, where X is the degrees between each wolf. This results in X < 90. Which means a minimum of 5 wolves (think 4 wolves will give 90 degrees each, +1 wolf will give < 90 each)

So now if the wolves equally spread out along the equator the hare will not be able to get past them to the other side. (As they can just rotate along their perimeter faster to block the hare)

[Imagine a circle with a dot in the middle trying to escape, except its a semi-sphere, This is the best possible distance for the hare]

Now we need the one wolf to chase the trapped hare! This wolf just needs to force the trapped hare into another wolf by chasing behind it for long enough. If the hare tries to run, he can only run closer and closer to another wolf, either a wolf on the boarder, or the one that is chasing it. Eventually, at least one wolf be able catch the hare.

Now you might be wondering why the hare would allow itself to be trapped as explained above, unfortunately it has no choice. Since the 5 wolves only have to place them self to cover their own arc length section with respect to the hare, they just have to run to where they should be. At first they will be trailing behind where they 'should' be in the setup to trap the hare, but once the hair reverses direction (or stops), the wolves will be able to catch up and match the location where they should be.

This same method will also work for the first part of the question with the same number of wolves! The only difference is that 1 wolf must start the chase first so that it can communicate the hares position to the other wolves so they can get into the proper positions. (at all times)

Second approach, you need 3 wolves. Influenced by the other answers, I think you could catch the hare with a minimum of 3 wolves (2 to trap, 1 to chase). The two wolves would have to be positioned on opposite sides of the sphere with the hare in the center (in the middle of a straight line drawn between the two wolves). Now these two wolves simply have to rotate around the axis to prevent the hare from escaping to the other half (essentially trapping it on one half). The third wolf chases the hare into the edge of the trap (and prevent it from out running/escaping the edge wolves). [This is also based on the same equation as above. The mistake I made earlier is that the wolves would only move it one direction along the sphere (covering 90 degrees), but if they moved both ways you could half the amount required (cover 90 degrees both ways or 180 degrees total)]

Setup: To set the positions of the wolves up, the first two need to initially get onto the opposites sides of the sphere. Now you can always draw a straight line between the wolves with the hare on it (think north pole and south pole with the hare in between). Now the issue is that the hare isn't exactly in the middle (or equator). This is where the third wolf comes into play, chasing the rabbit towards the 'equator'. Once the hare is there, the two edge wolves can chase around the sphere to follow the above tactic.

This might also be able to work for the second part given that one wolf initially finds the hare and communicates it position to help the other two wolves setup.


I don't want any credit, but Javier in a similar question gave the correct answer as 2 in $\infty$ time. The info can be found here. Basically 1 wolf mirrors the rabbits movement containing it to a hemisphere while the other has free movement, and can slowly work its way towards the rabbit.


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