Posted 24 Nov 2013 · Report post A while back, we had fun rescuing a fair maiden in a boat on a circular lake from a dog.Here's another circular chase problem, but this time it's in a cage. A lion and its tamer are inside a circular cage of radius 1.They both can run effortlessly at a top speed of 1.If the lion gets hungry, does the tamer become his lunch?If so, what's the longest time the tamer has to live?Since this is a puzzle, they both can be represented as points. 0 Share this post Link to post Share on other sites

0 Posted 24 Nov 2013 · Report post The lion begins to pace; the tamer stands to his feet and looks nervously to his left. 0 Share this post Link to post Share on other sites

0 Posted 25 Nov 2013 · Report post Should we assume the lion will always run straight for the trainer, or can the lion do whatever it feels is an optimal strategy? (I get the feeling the answer might be "calculate it both ways.")And should we consider a starting position where the trainer is at the edge of the cage and the lion is in the center, or something else? (At least with this question you can't reasonably say "calculate it for all possible starting positions." ) 0 Share this post Link to post Share on other sites

0 Posted 25 Nov 2013 · Report post Lion is not constrained. He wants to eat. Tamer does not want to be eaten. Starting positions could be opposite ends of diameter. After one unit of time lion could be at the center; tamer maintains maximum separation by not moving. We can start the chase from there. 0 Share this post Link to post Share on other sites

0 Posted 25 Nov 2013 · Report post I be the Lion. After walking stealthily to the center of the cage, I draw an imaginary line from the center to the trainer. As the man runs sheepishly along some chord, the imaginary line rotates and I stick to it, while advancing towards my meal. There. Someone solve it for me. BTW, it was an Ogre last time, not a dog. 0 Share this post Link to post Share on other sites

0 Posted 25 Nov 2013 · Report post I be the Lion. After walking stealthily to the center of the cage, I draw an imaginary line from the center to the trainer. As the man runs sheepishly along some chord, the imaginary line rotates and I stick to it, while advancing towards my meal. There. Someone solve it for me. BTW, it was an Ogre last time, not a dog. Right. An ogre. It was a dog and rabbit. Does it have to be a chord? 0 Share this post Link to post Share on other sites

0 Posted 26 Nov 2013 (edited) · Report post I was thinking along the same lines as Prime as far as the lion's strategy. But based on this calculation, that might not be a viable approach.Let the cage be radius 1, and the lion be distance r from the center, and on a line from the center to the tamer who is at the edge of the cage.If the tamer moves a (very small) distance dtheta around the perimeter of the cage, the lion will have to move a distance r*dtheta to remain on a line connecting the center and the tamer, and could move a distance (1-r) dtheta closer to the tamer. So after that movement by the tamer of dtheta, the change in r will be (1-r) dtheta.dr = (1-r) dtheta1/(1-r) dr = dthetaIntegrate both sides, letting u = (1-r) and du = -drintegral -1/u du = integral dtheta-ln(abs(u)) = theta-ln(abs(1-r)) = thetaIf we have the lion start at the center, at theta=0 we have r=0, so no need to worry about adding constant terms to the integrals.We want to know how far the tamer can run (how large theta is) before r reaches 1. When r=1, that means theta = -ln(0), which is infinity.One possible flaw with my approach is that, in calculating how much closer the tamer could get to the lion, I calculated how far around a circle of radius r the lion would have to run to remain on a line connecting the center and tamer (r*dtheta) and then let it use the rest of the distance that it could travel over the time that the tamer moved dtheta to move outward along a cord ((1-r)dtheta). It might be more efficient if, instead of making the lion move in two orthogonal directions, you were to make it move along the hypotenuse of a right triangle. But it's been so long since I've done any tough calculus that I fear having to integrate trig functions. Edited 26 Nov 2013 by plasmid 0 Share this post Link to post Share on other sites

0 Posted 26 Nov 2013 · Report post Why should a hungry lion devise the best strategy of escape for a tasty tamer? Suppose, the man runs on some spiral curve approaching the middle of the cage. Then the lion has an inside track and must catch up with the tamer in the center of the cage at worst. And I smell, it would take a finite amount of time to reach the center. In a desperate attempt to escape retribution, the lion’s oppressor might try running along the perimeter of the cage. Then deviating from the imaginary radius could cause the target to change the direction and gain some distance. So cutting the prey off like you do with zebra may not work here. If you trace the lion’s path, it will look like a spiral with rings packed more and more densely as it approaches the perimeter. Until the lion comes within the pouncing distance... 0 Share this post Link to post Share on other sites

0 Posted 28 Nov 2013 · Report post I was thinking along the same lines as Prime as far as the lion's strategy. But based on this calculation, that might not be a viable approach. Let the cage be radius 1, and the lion be distance r from the center, and on a line from the center to the tamer who is at the edge of the cage. If the tamer moves a (very small) distance dtheta around the perimeter of the cage, the lion will have to move a distance r*dtheta to remain on a line connecting the center and the tamer, and could move a distance (1-r) dtheta closer to the tamer. So after that movement by the tamer of dtheta, the change in r will be (1-r) dtheta. dr = (1-r) dtheta 1/(1-r) dr = dtheta Integrate both sides, letting u = (1-r) and du = -dr integral -1/u du = integral dtheta -ln(abs(u)) = theta -ln(abs(1-r)) = theta If we have the lion start at the center, at theta=0 we have r=0, so no need to worry about adding constant terms to the integrals. We want to know how far the tamer can run (how large theta is) before r reaches 1. When r=1, that means theta = -ln(0), which is infinity. One possible flaw with my approach is that, in calculating how much closer the tamer could get to the lion, I calculated how far around a circle of radius r the lion would have to run to remain on a line connecting the center and tamer (r*dtheta) and then let it use the rest of the distance that it could travel over the time that the tamer moved dtheta to move outward along a cord ((1-r)dtheta). It might be more efficient if, instead of making the lion move in two orthogonal directions, you were to make it move along the hypotenuse of a right triangle. But it's been so long since I've done any tough calculus that I fear having to integrate trig functions. You are right about the lion's radial speed - it's faster than your approximation. Does the lion catch the tamer?. 0 Share this post Link to post Share on other sites

0 Posted 30 Nov 2013 · Report post If I remember well, if the tamer runs on the perimeter, the lion catches the tamer. If the lion follows the radius of the tamer, his trajectory is a 1/2 circle and not a spiral. He catches the tamer in a finite time. Then, he has an escape strategy that works for an infinite time. 0 Share this post Link to post Share on other sites

0 Posted 30 Nov 2013 · Report post The good news is that using equations for a more efficient approach strategy by the lion as the trainer runs around the perimeter doesn't require trig functions. The bad news is that I nonetheless end up with a differential equation that I can't see how to solve.If the tamer moves a distance dtheta along the perimeter, the "lateral" distance that the lion has to run to stay on a line connecting the center and the tamer is r*dtheta just like before. But this time, instead of having the lion run along an inner circle's path for r*dtheta and then outward for (1-r)dtheta, I'll ask how far outward the lion can go by running along a skewed arc of total length dtheta with a constantly increasing r as the lion goes from the start to the finish of this distance.Because r is changing as the lion is running we can't just use r*dtheta as the total radial distance that the lion moves to stay on a line connecting the center and tamer; we have to account for the fact that the circumference of the circle he's following grows larger as r increases. But since circumference is linearly proportional to radius, we can just use the average radius as the lion is moving outward: (r + (r+dr)) / 2 = r + dr/2. So the "lateral" distance the lion moves is dtheta*(r + dr/2). The "outward" distance the lion moves is dr. The "total" distance the lion moves is dtheta (the same total distance that the tamer moved). Then use the pythagorean theorem to find out how large dr can be in relation to dtheta for any given r.dtheta^{2} = [dtheta*(r + dr/2)]^{2} + dr^{2}Integrating that function with the initial conditions of r=0 and theta=0 should let you solve for how large theta is when r reaches 1, and thus how far the tamer runs before he gets eaten. I'm not sure if I would have been able to solve that differential equation back when I was taking calculus classes years ago, but I sure can't now.But I can show that the lion is able to catch the tamer, albeit with a much less optimal approach.With the first method I posted, the lion can get very close to the tamer but not quite catch him. Let the lion use the first strategy to reach a distance r from the center on a line toward the tamer. Then the lion's strategy will change, and the lion will now run perpendicular to the line connecting the center and the tamer. If the tamer runs along the perimeter for a very small distance dx, then the lion will run a distance r*dx perpendicular to the radius. After running for r*dx perpendicular to the radius, the lion's new distance from the center will be sqrt(r^{2} + (r*dx)^{2}) = sqrt(r^{2}*(1+dx^{2})) = r*sqrt(1+dx^{2}).Again, actually turning that into a differential equation and solving it is beyond me, but the key point to notice is that r is always being multiplied by sqrt(1+dx^{2}) and will be continuously increasing at a rate that does not approach zero, which is enough to show that the lion will catch the tamer. (Although it's funny that with this strategy the lion can't make it out of the center: the increase in r is r*sqrt(1+dx^{2}), so zero when r=0, hence why I made the lion use the first strategy to get out and then switch to this second strategy to reach the tamer.) 0 Share this post Link to post Share on other sites

0 Posted 2 Dec 2013 · Report post Then, he has an escape strategy that works for an infinite time. Hi Grimbal. Care to share the tamer's escape path that precludes being caught in finite time? 0 Share this post Link to post Share on other sites

0 Posted 2 Dec 2013 · Report post I have a Zeno-esque explanation why the tamer cannot be caught. The tamer runs full speed around the outside of the cage. To simplify, suppose the cage rotates about the origin clockwise at 1 radian/second. The tamer is on the x-axis where x=1 and y=0 and runs in place facing upward. The lion's task then is to navigate the x-axis from the origin to the tamer at x=1. Since the lion is never farther from the origin than the tamer is, he can stay on the x-axis. His initial direction is straight toward the tamer: his motion is pure radial. As soon as he leaves the origin his direction acquires a circumferential component: he faces slightly upward at an angle a. He must do this to stay on the x-axis because the cage is rotating clockwise. The farther from the center he goes, the larger that angle a that he faces becomes. In order to reach the tamer, the angle a must become 90^{o} But at 90^{o} his radial speed is zero. In order to retrace his steps and get back to the origin (the chase process must be reversible) the lion must have some radial component to his motion. Since there is no mechanism for the lion to acquire a radial component to his motion from the final state of capture, there is no way for him to have achieved the pure circumferential motion he needs to capture and eat the tamer. But I can prove mathematically that the tamer can be caught,Start the lion at the center of the cage and the tamer at the fence. along with a simple expression for the lion's path length (which is the tamer's (finite) expected survival time.) Which is correct? I'm interested to hear Grimbal's escape route. 0 Share this post Link to post Share on other sites

0 Posted 2 Dec 2013 (edited) · Report post I thought lions could pounce. But even if that is not allowed, the lion still must eat. Maybe, cutting the prey off, like you do it with zebra, will work for this hunt as well. After running directly towards the tamer for a bit (half a unit, or so,) the lion could just move directly to the cage wall. Whereupon, he will project the rendezvous point at the perimeter (assuming the trainer continues to walk along the wall) and sprint directly to that point in a straight line of a chord. E.g. if initially the central angle between the lion on the perimeter and the man on the perimeter is a, and the angle between lion’s starting position and projected meeting point is x, the lion could solve an equation to project the meeting place (all the while continuing to run): (x-a) = x*(2-2*cos x)^{1/2}. (Using the formula for the chord length and taking into account the unit radius.) If the quarry changes his path, the lion can recalculate meeting point for the new trajectory of the prey, head to it in a straight line and have his lunch even sooner. Edited 2 Dec 2013 by Prime 0 Share this post Link to post Share on other sites

0 Posted 2 Dec 2013 · Report post I thought lions could pounce. But even if that is not allowed, the lion still must eat.Maybe, cutting the prey off, like you do it with zebra, will work for this hunt as well. After running directly towards the tamer for a bit (half a unit, or so,) the lion could just move directly to the cage wall. Whereupon, he will project the rendezvous point at the perimeter (assuming the trainer continues to walk along the wall) and sprint directly to that point in a straight line of a chord. E.g. if initially the central angle between the lion on the perimeter and the man on the perimeter is a, and the angle between lion’s starting position and projected meeting point is x, the lion could solve an equation to project the meeting place (all the while continuing to run): (x-a) = x*(2-2*cos x)^{1/2}. (Using the formula for the chord length and taking into account the unit radius.) If the quarry changes his path, the lion can recalculate meeting point for the new trajectory of the prey, head to it in a straight line and have his lunch even sooner. Yes the lion can pounce, I.e. Feet can leave the ground. His top speed is still 1 radius/second. If the tamer's minutes, or seconds, are measured, how long does he have, with starting locations at center (lion) and perimeter (tamer)? 0 Share this post Link to post Share on other sites

0 Posted 3 Dec 2013 · Report post If I understand correctly, the Zeno argument says that (going in the reverse direction) when you are at the perimeter the change in the lion's distance from the center over time equals zero so there is no way to get out from the perimeter.Would this be the same as considering the function y = x^{2} and saying that dy/dx at x=y=0 is zero, so there is no way for the y value to increase and ever leave the x-axis? Since y is not changing and x is the square root of y, would that also imply that x cannot change and so the function can never even leave the origin at all? 0 Share this post Link to post Share on other sites

0 Posted 3 Dec 2013 · Report post If I understand correctly, the Zeno argument says that (going in the reverse direction) when you are at the perimeter the change in the lion's distance from the center over time equals zero so there is no way to get out from the perimeter. Would this be the same as considering the function y = x^{2} and saying that dy/dx at x=y=0 is zero, so there is no way for the y value to increase and ever leave the x-axis? Since y is not changing and x is the square root of y, would that also imply that x cannot change and so the function can never even leave the origin at all? Yes, sort of. I don't have the equations and I expected the lion would approach the edge of the cage asymptotically, never actually reaching it. But now I think that's wrong. It is more like a parabola, and the lion's path is simply tangent to the circular edge, touching it at a single, well-defined point. The path has finite length and is retraceable. 0 Share this post Link to post Share on other sites

0 Posted 3 Dec 2013 (edited) · Report post If I remember well, if the tamer runs on the perimeter, the lion catches the tamer. If the lion follows the radius of the tamer, his trajectory is a 1/2 circle and not a spiral. He catches the tamer in a finite time. Then, he has an escape strategy that works for an infinite time. [spoiler=Grimbal has the path shape] Starting the lion at (0,0) and the tamer at (1,0) at same speed. The paths coincide after the tamer has gone 1/4 circle, or pi/2 radii, which, at unit speed, takes pi/2 seconds. The lion's semicircular path has the same length as the quarter circle, having 1/2 the diameter of the latter circle. The interesting thing about this chase is that the lion runs directly toward the tamer at only one moment in time: at the very start. At the point of capture they are running in precisely the same direction, the tamer being at the right side of the lion. A ray from the origin intersects the two paths where they are occupied at the same point in time. Another way of saying that in order to stay on a common radius, the lion must face increasingly away from the tamer toward the eventual point of capture. If the lion runs straight toward the tamer, the tamer escapes. Or does he? Suppose they are both on the perimeter, separated by a very small distance. Does the lion catch up? Perhaps. By running straight toward the tamer, the lion decreases his radius slightly, allowing him to reduce the angular separation. Hmm. Subject for a different puzzle perhaps. Edit: By running straight toward the tamer, the lion traces a path different from the semicircle (a path whose radius is greater at each point in time than the corresponding point on the semicircle.) But, lacking a linear speed advantage, the lion does not want the tamer directly in front of him. Nevertheless, so long as the lion's radius < 1, he has an angular speed advantage, and eventually he captures the tamer. So this does define a new calculation: how much extra time does that chase strategy on the part of the lion give to the tamer? Summary: The tamer becomes the lion's lunch, and he has pi/2 seconds to live once the chase is on. I'm marking the puzzle solved but invite Grimbal to describe an off-perimeter escape strategy for the tamer. Edited 30 Jan 2014 by bonanova Describe another capture path and pose a new calculation 0 Share this post Link to post Share on other sites

0 Posted 3 Dec 2013 · Report post I like happy endings. And that uncomplicated trajectory is nice too. 0 Share this post Link to post Share on other sites

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A while back, we had fun rescuing a fair maiden in a boat on a circular lake from a dog.

Here's another circular chase problem, but this time it's in a cage.

A lion and its tamer are inside a circular cage of radius 1.

They both can run effortlessly at a top speed of 1.

- If the lion gets hungry, does the tamer become his lunch?
- If so, what's the longest time the tamer has to live?

Since this is a puzzle, they both can be represented as points.## Share this post

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