NY Times math problem

Simple algebra is sufficient to determine that the agent's speed needs to be at least pi+1 times the rabbit's speed.

If the radius is r, the agent's speed is a and the rabbit's speed is b:

#1. The rabbit can remain diametrically opposed to the agent up to a distance of r*b/a from the centre, i.e. r-r*b/a = r*(1-b/a) from the shore.

#2. From there, the time it takes for the rabbit to reach the closest point on the shore is r*(1-b/a)/b = r*(1/b-1/a).

#3. The time it takes the agent to reach that point is pi*r/a.

#4. They will draw when the times in #2 and #3 are equal, i.e.

r*(1/b-1/a) = pi*r/a => 1/b-1/a = pi/a => a/b-1 = pi => a/b = pi+1

If the ratio is greater than pi+1, the rabbit can't win using this strategy. But proving that this strategy is optimal (i.e. it can't win using another strategy) may be rather more involved.

han

de.

I'd hate to tackle it analytically. But numerically, it seems perfectly tractable.

Assume that the rabbit isn't going for style points (swimming the backstroke in circles to taunt the agent) and is going for the maximum "lead" at the point where he exits the water.

Assume that the rabbit has reached his 1/4 r "jumping off" point. The rabbit will initially head straight for shore. But only long enough for the agent to pick a chase direction.

Once the agent has picked a direction, the rabbit chooses a new straight line course. This course can be chosen by differentiation -- how much additional distance must the agent run versus how much additional distance must the rabbit swim as a function of an incremental change in course angle.

If agent running distance is a and rabbit swimming distance is h and the rabbits selected swimming angle is theta then the hare is trying maximize

a/4 - h

so he needs to solve:

d(a)/4 - d(h) / d(theta) =3D 0

The rabbit may continue making this calculation as it approaches the shore. The result will remain unchanged as long as the agent does not turn around. The selected angle only depends on the angle made between the shoreline and the rabbit's course line at the point where they intersect. This means that the rabbit will swim a straight line course.

If one is trying to zero in on the optimal value for the agent's speed, one can can adjust the agent/rabbit speed ratio and until the critical value is found.

Some very basic calculus is helpful. A solution is given at

=20

Riddle:

There is a rabbit in the middle of a perfectly circular pond. An agent is trying to get the rabbit. The rabbit swims exactly away from the agent. After a few seconds, the agent's head explodes. Why?

Ya know, if the agent always seeks the closest path (with no underlying intelligence to escape the following scenario), the rabbit (if it were more intelligent) could follow a zig-zag path. As soon as it moves somewhat to the right, the agent sees this and moves in that direction. The rabbit, noticing the reduced distance, changes direction immediately. As it crosses the diameter the agent is standing on, the agent reverses direction. The opposite then happens, ad nauseum, until the rabbit reaches the shore safely.

Theorem 1: The rabbit can reach the shore regardless of the agent's relative speed. Theorem 2: Either the agent's head explodes, or the Church-Turing Theorem is false.

Theorem 2 follows from taking the limit as delta x approaches zero (that is, the width of the zig-zag). In the limit, the rabbit appears to proceed in a straight line, exactly opposite the agent (this also works if the rabbit simply moves in exactly this path, with no infinnitessimal shaking). The agent cannot decide which direction to go, because his distance-o-meter is saying both directions are equal. In terms of angle, sign(tangent(theta)) is undefined (where sign(x) is

Tim

In real life, the rabbit leaps from the pond and slaughters all concerned with his "big, nasty teeth," despite Tim the Enchanters' best efforts to warn them.

Tragic, really.

James Arthur

No. Got a link?

Thanks, Rich

Is the "agent" wading/swimming, or just walking around the perimeter of the pond? What's each of their respective speeds? How did the rabbit get to the middle of the pond in the first place? Was somebody using him for Muxkie bait? ;-)

Thanks, Rich

It doesn't have to. All it needs to do is float there until the agent dies of exhaustion from running around and around the pond forever. >:->

Hope This Helps! Rich

James Arthur wrote in news:odELl.2340$ snipped-for-privacy@nwrddc01.gnilink.net:

didn't Jimmy Carter get attacked by a killer rabbit in the water while he was canoeing?

_I_ would award your solution the prize, at any rate.

At least it forces us to make the agent's strategy explicit.

No, if the rabbit is far enough from the center of the circle it cannot swim fast enough to cross the diameter so the agent does not have to change direction. Indeed the agent's optimal strategy is to always seek to minimize the angular separation.

Nope. The agent knows that there are two optimal strategies. The agent knows that it is suboptimal not to choose between the two optimal strategies, but the method of choice is arbitrary. The agent will choose one of the optimal strategies (either at random or by some arbitrary rule, e.g. alphabetical order).

- William Hughes

han

de.

ed text -

I'm not convinced about the 'keep the agent on the opposite side' logic. Lets expand the problem to their being TWO agents, already on opposite sides. If the rabbit swims in any direction, off-center, then both agents will move together to be at the point the rabbit is apparently aiming for, then the rabbit moves slowly back towards center and the problem is reduced to the original one. So the rabbit can escape with TWO agents around the pool...or even an INFINITE number of them? (well, slightly less than infinite, in this case....)

--riverman

That incident was the back-story to the NYT math problem which is being discussed in this thread.

"Jimmy Carter's Killer Rabbit Puzzle"

When the rabbit moved back towards the centre of the pool, the agents would move back to positions oppoisite each other.

It appears that the rabbit cannot escape if there are two agents.

Sylvia.

No, that is not their best cooperative strategy even if it would be their best strategy individually. Though actually it is not even their best individual strategy, as the rabbit could then dictate which direction they run with arbitrarily little cost by "feinting" motion toward any arbitrary point on shore.

- Tim

They may not have time.

Assume that the two agents are unaware of each other's existence and choose the optimal strategy for a lone agent (always move so as to decrease the angular difference). The rabbit must very slightly modify his strategy, keep almost but not quite 180 degrees from the agents (otherwise they might split up). In this case the rabbit escapes.

However, assume that the two agents are aware of each other's existence but cannot communicate. They can stop the rabbit escaping. They divide the circle in two halves. Each agent adopts the strategy, move toward the point in my half that is closest to the rabbit.

- William Hughes

On May 5, 9:41=A0am, Tim Little wrote: =A0Though actually it is not even

The 'feinting' move would only work while the rabbit is extremely close to the center. The strategy for any agent is to determine the point that the rabbit is aiming for, and make haste toward that point. While the rabbit is very close to the center, 'feinting' merely moves it around a circle with a small radius, and changes the targetted landing zone immensely with each 'feint'. But its a moot point: the rabbit cannot make a break for the shore from the center...it needs to be somewhere along a circle of a larger radius, which is where feints become less productive.

The agent, meanwhile, would always race along the perimeter arc that is shortest from where he is to where the rabbit is headed (if the rabbitwere going directly along a radius). If the rabbit gains enough on the agent that the arclength 'behind' him is suddenly shorter than the one he is travelling along, he should turn around.

I think the rabbit could make use of this to follow a sinusoidal pattern to the shore, always jogging back across the 'opposite radius' to the position of the agent, causing the agent to reverse course.

In fact, if the rabbit were able to run in ANY circle until it were precisely opposite the agent, the agent would be faced with two equally desirous paths, and might even freeze in place instead. It would if it were a robot programmed to follow the shortest arc to the rabbit's landing zone...

It works everywhere, as you later post:

Yes, this is exactly why the "aim point" strategy fails. The rabbit can make the agent run back and forth like a cat chasing the dot from a laser pointer.

Though the reasoning is more general than that: at any time, the situation is completely determined by the distance between rabbit and shore, and the center angle between rabbit and agent. For a given angle, closer to shore is always better for the rabbit. For a given distance, a smaller angle is always better for the agent. If the agent ever voluntarily increases the angle, the rabbit's position is needlessly improved.

- Tim

It's possible they only expected the readership to be aware of 2-pi-r and to assume the rabbit will swim direct towards the shore away from the agent. In that case the general reader can say the rabbit cannot escape. Which was my first thought..

I don't like these spiraling paths and looked at the probability of escaping if the rabbit swam in a straight line, invisibly under water.

But I only made that odds of 50% of escaping, if the rabbit avoided the point on the shore +/- 50 degrees opposite the agent. Not so good. And that assumes the agent will start running around the pond.