NY Times math problem

[KBH]

If the rabbit also knows the direction to the nearest shore.
adopt the following strategy (which does not involve
calculation):


Step 1.

a: Keep "rotating with a radial to the agent"
b: if not using maximum speed for a:
turn toward the nearest shore until
using maximum speed.
c: continue until not approaching the nearest
shore

Step 2.

a: Swim directly toward the nearest shore
at maximum speed.

This will allow the rabbit to escape given any
agent strategy if the agent's top
speed is less than (pi+1) times the rabbit's
(a slightly more complex strategy for step 2, also
involving no calculation, can improve this a
little).

KBH wrote:

Okay the rabbit's name is Duck...

The speeds are fixed but since the agent can stop maybe I should allow Duck
to slow down. But when they sprint just say that both increase speed at the
same percentage so that it's not necessary to allow for increased speed of a
sprint...over their given rates.

But I'm not allowing Duck to know where 1/4 radius distance is as some point
in the water. I'm only allowing Duck to know whether or not he is rotating
with the agent...and that finds the break point.

 

[KBH]

KBH wrote:

Okay the rabbit's name is Duck...

The speeds are fixed but since the agent can stop maybe I should allow
Duck to slow down. But when they sprint just say that both increase speed
at the same percentage so that it's not necessary to allow for increased
speed of a sprint...over their given rates.

But I'm not allowing Duck to know where 1/4 radius distance is as some
point in the water. I'm only allowing Duck to know whether or not he is
rotating with the agent...and that finds the break point.

I think we should stay with the fixed speeds...

We just find them in motion. The agent can't stop but can reverse direction.
And don't worry about sprint speeds as both speeds could increase at the
same rate...

 

[KBH]

I think we should stay with the fixed speeds...

We just find them in motion. The agent can't stop but can reverse
direction. And don't worry about sprint speeds as both speeds could
increase at the same rate...

We just find them in motion...

And allowing the agent to reverse direction is just to eliminate the
strategy that I wrote the computer program for. Then Duck can find rotation
with the agent by spiraling outward but must also maneuver to be on the
point on the other side of the radius from the agent.

 

[Archimedes' Lever]

They may not have time.

They would eventually end up together as was originally claimed.

Just like a spinning cylinder in space will eventually develop a
wobble, and will eventually end up spinning end over end.

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

Not if the rabbit determines where their parting line is located.

 

[Archimedes' Lever]

Yes it does, which is exactly why I was arguing that it was a poor
strategy for the agent to employ.


- Tim

The rabbit will ALWAYS use his peripheral vision to determine the
agent's position, and will always swim away from that attempt at capture.

They have over 270 degrees of vision without even turning their head.

 

[Archimedes' Lever]

LOL. Within, of course, the constraints of the stated problem.
"...a rabbit in the middle of a pond, an agent who can run
around the shore four times as fast as the rabbit can swim."

Let's not let all this precision affect our accuracy, shall we?


--riverman

No. What he said is 100% true. Also, the rabbit is doing the exact
opposite. i.e Trying to maximize the distance between himself and the
agent.

The rabbit will win eventually, and he only needs to wait until the
agent is *near* the 180 degree point. He doesn't have to get him there
exactly.

 

[riverman]

We just find them in motion...

And allowing the agent to reverse direction is just to eliminate the
strategy that I wrote the computer program for. Then Duck can find rotation
with the agent by spiraling outward but must also maneuver to be on the
point on the other side of the radius from the agent.- Hide quoted text -

- Show quoted text -

Why are we constraining outselves to the rabbit moving in a spiral or
along the radius of a circle? Could there not be a solution where the
rabbit moves in an elliptical path, allowing the agent to gain and
lose ground relative to the rabbit until the rabbit finds itself
opposite the agent, but closer to the edge than a circular path would
bring it?

--riverman

 

[Jasen Betts]

The rabbit will ALWAYS use his peripheral vision to determine the
agent's position, and will always swim away from that attempt at capture.

They have over 270 degrees of vision without even turning their head.

the dumb bunny will never leave the pool.

 

[Dave]

The rabbit will win eventually, and he only needs to wait until the
agent is *near* the 180 degree point.  He doesn't have to get him there
exactly.

The rabbit does need to position himself exactly on the opposite side
of the diameter from the agent if he expects to escape when the agent
is running the fastest he can for the rabbit to escape, if both use
optimal strategies.

Assuming that the rabbit starts at the center and that the agent is
running at full speed, the rabbit can remain on the diameter through
the agent reach the furthest radius at which his angular rate equals
that of the agent while the agent travels 90 degrees around the pond.

Dave

 

[Dave]

Why are we constraining outselves to the rabbit moving in a spiral or
along the radius of a circle? Could there not be a solution where the
rabbit moves in an elliptical path, allowing the agent to gain and
lose ground relative to the rabbit until the rabbit finds itself
opposite the agent, but closer to the edge than a circular path would
bring it?

When the rabbit is further from the center than a circular path would
take it, its angular velocity is necessarily less than that of the
agent, so the agent is decreasing the angle, and therefore the rabbit
cannot catch up to the opposite side of the diameter through the
agent.

Dave

 

[Sanny]

DId anyone here see the problem presented in
the Science section of NY Times last week?
Quite startling, to see something so sophisticated
in a 'general readership' publication.

Is it solvable without a calculus of variations approach?

I solved it using brain No Calculus needed.

Bye
Sanny

Enjoy & Chat: http://www.GetClub.com

 

[Archimedes' Lever]

I solved it using brain No Calculus needed.

Bye
Sanny

Enjoy & Chat: http://www.GetClub.com

Stupid link for something I will NOT sign up for. Post your solution,
dingledorf. This IS the discussion group.

 

[Mark-T]

I solved it using brain No Calculus needed.

Whose brain did you use?

 

[[email protected]]

I solved this when it appeared in rec.puzzles in 1992 (using a duck
and a fox).

Seehttp://groups.google.com/group/rec.puzzles/msg/050273ed70e4a9f6

The bottom line is that there is a strategy by which the rabbit can
escape as long as the agent can run no more than v =
4.6033388487517003525565820291030165130674... times as fast as the
rabbit can swim. The number v is the reciprocal of the solution of the
transcendental equation sqrt(1-r*r) = r*(pi+arccos(r)), which can be
derived using trigonometry.

Dave

I computed the rabbit's fastest escape for various values of v (the
agent's maximal speed), assuming the rabbit's maximal speed is 1
and the pond's shore is the unit circle.

We may assume, without loss of generality, that the agent begins at
the point (1,0) running counterclockwise, and that, from the agent's
point of view, the rabbit never goes left of the center of the pond.
(If the rabbit did so, we could construct an equivalent solution in
which he did not.) We may also assume that the agent runs at full
speed the entire time (the solutions in which he slows down to stop
the rabbit from cutting behind him being of little interest).

The fastest escape for the rabbit involves a circle of diameter 1/v
whose top is at the pond's center. When the rabbit runs at full
speed counterclockwise on this circle, he keeps the pond's center
between him and the agent (until the agent gets to (-1,0)). For
any given spot on the pond's edge where the rabbit might try to
escape, his fastest route there (without going left of the pond's
center from the agent's point of view) is to run along this
circle until he can run straight to his exit point without
crossing through the circle. (The solution for the maximal value
v given by Dave is for the rabbit to run a semicircle, then to
continue straight (in the positive x direction).) Using paths of
this form for arbitrary v, one can solve numerically for the
first escape point that allows the rabbit to reach that point at
the same time as the agent (tie goes to the rabbit).

Here are some escape times t_e for various values of v:

v <= pi: t_e = 1
v = 3.5: t_e = 1.0011722480824568763140554691535330067350...
v = 4 : t_e = 1.0173266236654672461422542182088696225446...
v = 4.3: t_e = 1.0511345079733767600617899255156714598171...
v = 4.5: t_e = 1.1118849008932825665138398791684589180820...

For the critical value given by Dave,

v_c = 4.6033388487517003525565820291030165130673...
(last digit "3" not "4" because I'm truncating, not rounding),

the escape time is

t_e(v_c) = 1.3173494248307199060775592744716826605635...

Slightly smaller values of v yield significantly shorter escape
times:

t_e(v^c - 10^-2) = 1.2146176261268462636077456248661081871567...
t_e(v^c - 10^-3) = 1.2685499374063007502760823717840407058122...
t_e(v^c - 10^-4) = 1.2945402874735326613302923323908998597116...
t_e(v^c - 10^-5) = 1.3067373155331437760247440575734966228130...
t_e(v^c - 10^-6) = 1.3124192755216122410040697316416725441298...

Exploring this asymptotic behavior numerically yields

t_e(v^c - eps) = t_e(v_c) - k eps^(1/3) + O(eps^(2/3)), where

k = 0.4933500252621480311919413384450341753698...

-Jim Ferry
Metron, Inc.
f rr @m tsc .c m
e y e i o

 

[Rich Grise]

DId anyone here see the problem presented in
the Science section of NY Times last week?
Quite startling, to see something so sophisticated
in a 'general readership' publication.

Is it solvable without a calculus of variations approach?

I don't get the NY times here. Could you transcribe the problem
here? (I'm at s.e.design, but crossposting to all of the above.)

Thanks,
Rich

 

[Rich Grise]

SEE
http://tierneylab.blogs.nytimes.com/2009/04/13/jimmy-carters-killer-rabbit-puzzle/

Thanks!
Rich

 

[Spehro Pefhany]

Sounds like you need to switch to decaffeinated!

...a hookah-smoking caterpillar has given you the call

 

[Richard Cranium]

On Sat, 09 May 2009 12:43:03 -0700, Archimedes' Lever


Archie:

You can win and achieve the adulatrion you so desperately seek. You
simply take off at full speed running in a right handed helix until
your speed begins to unwind your DNA and you run through your asshole
and turn inside out. THEN, girls will talk to you ... maybe.

 

[Dave]

On Sat, 09 May 2009 12:43:03 -0700, Archimedes' Lever

Archie:

You can win and achieve the adulatrion you so desperately seek.  You
simply take off at full speed running in a right handed helix until
your speed begins to unwind your DNA and you run through your asshole
and turn inside out.  THEN, girls will talk to you ... maybe.

adulatrion? What's that?

 

[Richard Cranium]

adulatrion? What's that?

It's the confusion caused by the extra "r" from another post about
Archie's mental prowess where I spelled "hemmorhoid" (referencing a
potential cause for his frequent headaches) incorrectly. My sincere
apologies to the Mensa crew here. I trust no one has become so upset
that their solution to the world's problems is put on hold.