NY Times math problem
Mark-T
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?
--
Mark
Mensanator
Didn't see it.
>
> --
> Mark
Jake
I imagine this is what you're referring to:
http://tierneylab.blogs.nytimes.com/2009/04/13/jimmy-carters-killer-rabbit-puzzle/
Sue...
It conjures memories of rip tides, swimmers and lifeguards.
http://en.wikipedia.org/wiki/Fermat%27s_principle
Sue...
Sylvia Else
Yes.
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Since the agent can run four times as fast as the rabbit can swim, no
matter what the agent does, the rabbit can swim away from the centre,
and keep the agent on the far side as long as the rabbit is no more than
1/4 of the radius from the centre. So the rabbit can reach a point 3/4
of a radius from the edge while the agent is still on the opposite side.
From that point, if the rabbit swims directly to the edge, he has to
swim 3/4 of a radius. In that time, the agent can run at most 4 times as
far, or 3 * radius. But the distance to the rabbit's exit point is Pi *
radius, or a bit further, so the rabbit is free.
For the second part, instead of 4, write m, and let R be the radius.
The rabbit can reach a point R/m from the centre while keeping the agent
opposite. From that point, the rabbit needs to swim R * (m - 1) / m to
reach the edge. In the available time, the agent can run R * (m - 1). So
the break even point is where R * (m - 1) = R * Pi, or where m = Pi + 1.
Sylvia.
Tim Little
> I imagine this is what you're referring to:
>
> http://tierneylab.blogs.nytimes.com/2009/04/13/jimmy-carters-killer-rabbit-puzzle/
(To recap: 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. Can the
rabbit reach the shore without the agent getting to the same point
first? Extra credit: what is the lowest ratio of speeds for which the
agent can catch the rabbit?)
For the main problem, it's pretty easy without any calculus - just
simple geometry and arithmetic. Hint: how far from the center can the
rabbit keep the agent diametrically opposite?
The extra credit part does require more math background, but that's to
be expected.
- Tim
Tim Little
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> The rabbit can reach a point R/m from the centre while keeping the agent
> opposite. From that point, the rabbit needs to swim R * (m - 1) / m to
> reach the edge. In the available time, the agent can run R * (m - 1). So
> the break even point is where R * (m - 1) = R * Pi, or where m = Pi + 1.
No, the rabbit can do better than swimming in a straight line toward
the closest point on shore.
For example, suppose the shore were straight instead of circular, with
the agent starting distance pi from the closest point, while the
rabbit is at distance pi/m. If the rabbit swims in a straight line,
they will reach that point at exactly the same time.
If the rabbit swims at angle A, then the agent must run extra distance
(pi/m) tan A, while the rabbit must swim an extra (pi/m) (sec A - 1).
The rabbit is better off if tan A > m (sec A - 1), which has solutions
in A for any m. So for a straight shore, the rabbit is always better
off swimming at some nonzero angle.
The circular shore does change the distances, but the same pattern
holds: the rabbit is always better off swimming at an angle, even if a
small one.
- Tim
Dave
I solved this when it appeared in rec.puzzles in 1992 (using a duck
and a fox).
See http://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
M Rath
The rabbit is in the middle of a circular pond. The agent is on the edge of
the pond. The agent can run 4-times as fast as the rabbit can swim. Can the
rabbit get away ?
Okay the radius of the pond is 1 so the rabbit must swim 1 . The pond
circumference is Pi*radius*2 but the agent must run only half the
circumference for a distance of Pi . Then rabbit can swim at a rate of 1 so
the agent can run at a rate of 4.
Rate * Time = Distance
&
Time = Distance / Rate
So the rabbit can swim to the shore in
Time = 1 / 1 or 1 .
And the agent can run to the opposite shore in
Time = Pi / 4 or 0.785 .
And so I seem to have the wrong answer. But the second part of the problem
based on my answer would be
Pi / Rate = 1
Rate = 3.14
If the agent runs only 3.14 times as fast as the rabbit then they time...and
a tie goes to the base runner.
M Rath
>
> The rabbit is in the middle of a circular pond. The agent is on the edge
> of the pond. The agent can run 4-times as fast as the rabbit can swim. Can
> the rabbit get away ?
>
> Okay the radius of the pond is 1 so the rabbit must swim 1 . The pond
> circumference is Pi*radius*2 but the agent must run only half the
> circumference for a distance of Pi . Then rabbit can swim at a rate of 1
> so the agent can run at a rate of 4.
>
> Rate * Time = Distance
> &
> Time = Distance / Rate
>
> So the rabbit can swim to the shore in
>
> Time = 1 / 1 or 1 .
>
> And the agent can run to the opposite shore in
>
> Time = Pi / 4 or 0.785 .
>
> And so I seem to have the wrong answer. But the second part of the problem
> based on my answer would be
>
> Pi / Rate = 1
> Rate = 3.14
>
> If the agent runs only 3.14 times as fast as the rabbit then they
> time...and a tie goes to the base runner.
So the rabbit swims an arc. The agent is running a counterclockwise circular
curve while the rabbit is swimming a counterclockwise...outward spiral.
M Rath
See...in the extreme...at each instantaneous moment of the agents movement
the rabbit makes a new line from the agent to the rabbit to the shore.
Virgil
"M Rath" <mr...@notmail.com> wrote:
The rabbit initially moves to stay on the same diameter as the agent
while maximizing the distance between them.
If the agent acts optimally, the rabbit can get no more than 5/8 of that
diameter ( or 1 and 1/4 radii) away from the agent along that common
diameter.
From there the can rabbit race for the nearest point on shore.
The rabbit's distance from that point is (3/4)*r and the agent's is pi*r.
Since (3/4)*r is slightly less than 1/4 of pi*r, ...
Androcles
"M Rath" <mr...@notmail.com> wrote in message
news:uXaLl.36338$i9....@bignews8.bellsouth.net...
Dave
No. The rabbit wants to swim in a straight line. It is the shortest
distance from where he is to a point on the shore. If he swims in an
arc, it will take longer than on a line, meaning that the agent has
more time to reach him.
Dave
Virgil
<622b6f29-1f94-4a8f...@r34g2000vba.googlegroups.com>,
Dave <dave_an...@Juno.com> wrote:
By swimming so as to keep on the same diameter as the agent while moving
as fast as possible away from the agent, the rabbit can arrive at a
point that is only 3/4 of a radius from a point on the edge while the
agent is still half a circumference's running distance from that point.
Rath@notmail.com M Rath
Dave wrote:
No. The rabbit wants to swim in a straight line. It is the shortest
distance from where he is to a point on the shore. If he swims in an
arc, it will take longer than on a line, meaning that the agent has
more time to reach him.
M Rath wrote:
If the rabbit swims ONE straight line to the shore...then the agent is
waiting for him.
And the solution hereabouts says that the rabbit keeps turning away from the
agent until a point where he can break for shore. But how does the rabbit
know when he can break for shore ?
I'm saying that with each movement of the agent...the rabbit turns away from
the agent. So the rabbit swims an outward spiral that does prove that the
rabbit can get away. See...the rabbit stays in the water until the point
when he touches shore with the agent some distance away...
Tim Williams
> On May 2, 8:15 pm, Mark-T <MarkTanne...@gmail.com> wrote:
>
> > 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?
>
>
Seems clear to me. If the rabbit travels in a straight line,
regardless of which direction is taken, that line is always a radius.
Therefore, the point on the shore closest to the agent (without
reaching the agent) is just four radii of circumference (that is, 4
radians angle) from where he was standing.
However, the agent can reach the other side of the pond sooner than
3.14 radii, that is, the optimal point exactly opposite the agent. So
a straight-line solution is very easy: the ratio of speeds is pi, and
the rabbit cannot escape.
But that would be too easy, wouldn't it? If the rabbit is allowed a
curved path, then I would believe some of these complicated solutions
(a trancendental equation, eh?).
Tim
Sylvia Else
What in the original puzzle suggests that the rabbit is not allowd to
follow any arbitrary path?
Sylvia.
Nobody
>> So the rabbit swims an arc. The agent is running a counterclockwise circular
>> curve while the rabbit is swimming a counterclockwise...outward spiral.
>
> No. The rabbit wants to swim in a straight line. It is the shortest
> distance from where he is to a point on the shore. If he swims in an
> arc, it will take longer than on a line, meaning that the agent has
> more time to reach him.
If the rabbit swims in a straight line from the centre, the agent can beat
him to the shore. So he must get closer to the shore while remaining
diametrically opposite the agent.
So long as the rabbit is less than r/4 from the centre, he can maintain
diametric opposition. If the agent runs clockwise, the rabbit swims
anti-clockwise and vice-versa. The rabbit moves at 1/4 of the speed,
but is at less than 1/4 the radius, so his maximum angular speed is
higher than the agent's. He only needs to match the angular speed,
so whatever speed is left over can be used for outward motion.
Once he reaches r/4 from the center, the rabbit can maintain diametric
opposition, but not while moving outward. But at this point, he is only
3r/4 from the closest point on the shore, while the agent is pi*r from
that point. So he can head in a straight line to the shore and get there
before the agent does.
riverman
Only if the agent stands still until the last possible moment. I think
the agent will move continuously, and as soon as the rabbit starts
moving in any direction, the agent will move toward the point that is
the closest to the rabbit at that moment. Let's call the agent 'Xeno'.
Sylvia Else
No - until the rabbit is 1/4 of the radius away from the centre, he can
swim fast enough, in a suitable direction, to keep the agent exactly
opposite. So it doesn't matter if the agent moves.
Sylvia.
Nick
Which shows, I think (to return to the very original question about
whether this needs horrible maths to solve) that there is a simple proof
that pretty-well anyone can follow that shows the rabbit can escape in
the specific case (it goes to where it can just swim faster than the
agent, swims round in a circle until it gets to 180 degrees away from
the agent, and can then make it to the shore faster than the agent can
run round), but that the optimum strategy, and hence the answer to the
unasked question about which size of ponds or relative speeds allow the
rabbit to escape do require the horrid maths.
--
Online waterways route planner: http://canalplan.org.uk
development version: http://canalplan.eu
legg
At least it makes more 'sense' than the duck and the fox.
There's no reason for the duck to leave the pond.
The only problem I see with the elaborate solution is the assumption
that the obviously ill 'killer' rabbit will react in any way to the
presence of suits around the pond's periphery. After all, a rabbit
with any sense wouldn't be in the middle of a pond in the first place.
RL
Sylvia Else
It also seemed to me that issues relating to the agent moving are
totally hypothetical. Since the agent will know in advance that the
rabbit can escape, and will assume that the rabbit knows this too, the
agent won't even attempt to catch the rabbit, and will stay put.
Sylvia.
Richard Heathfield
<snip>
> It also seemed to me that issues relating to the agent moving are
> totally hypothetical. Since the agent will know in advance that
> the rabbit can escape, and will assume that the rabbit knows this
> too, the agent won't even attempt to catch the rabbit, and will
> stay put.
...and shoot the rabbit.
--
Richard Heathfield <http://www.cpax.org.uk>
Email: -http://www. +rjh@
Google users: <http://www.cpax.org.uk/prg/writings/googly.php>
"Usenet is a strange place" - dmr 29 July 1999
Nobody
>> No - until the rabbit is 1/4 of the radius away from the centre, he
>> can swim fast enough, in a suitable direction, to keep the agent
>> exactly opposite. So it doesn't matter if the agent moves.
>
> Which shows, I think (to return to the very original question about
> whether this needs horrible maths to solve) that there is a simple proof
> that pretty-well anyone can follow that shows the rabbit can escape in
> the specific case (it goes to where it can just swim faster than the
> agent, swims round in a circle until it gets to 180 degrees away from
> the agent, and can then make it to the shore faster than the agent can
> run round), but that the optimum strategy, and hence the answer to the
> unasked question about which size of ponds or relative speeds allow the
> rabbit to escape do require the horrid maths.
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.
jbrig...@gmail.com
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) = 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.
Anon
>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?
Some very basic calculus is helpful. A solution is given at
<http://www.mathrec.org/old/2003jul/solutions.html>
Tim Williams
> Some very basic calculus is helpful. A solution is given at
> <http://www.mathrec.org/old/2003jul/solutions.html>
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
+1 when x > 0, -1 when x < 0, and either 0 at x = 0, although sign(0)
may sometimes defined as +1). So now it's an undecidable problem, and
if the agent somehow succeeds, a lot of theorems (including those
about decidability) are wrong, or the agent's head simply
explodes. ;-)
Tim
James Arthur
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
Rich Grise
> DId anyone here see the problem presented in
> the Science section of NY Times last week?
No. Got a link?
Thanks,
Rich
Rich Grise
> On May 4, 9:56�am, Anon <a...@domain.invalid> wrote:
>> Some very basic calculus is helpful. �A solution is given at
>> <http://www.mathrec.org/old/2003jul/solutions.html>
>
> 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?
>
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
Rich the Cynic
>
>> 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?
>
> The rabbit is in the middle of a circular pond. The agent is on the edge of
> the pond. The agent can run 4-times as fast as the rabbit can swim. Can the
> rabbit get away ?
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
erg
Good question. Also, we may need to make some assumptions about the
strategy followed by the agent.
erg
_I_ would award your solution the prize, at any rate.
At least it forces us to make the agent's strategy explicit.
Virgilius
<1eb58bfd-eb18-4d34...@z19g2000vbz.googlegroups.com>,
erg <spamsp...@netzero.com> wrote:
An optimal agent strategy is:
The agent always stays on the edge of the circular pond.
While the rabbit is on the same radius (of the circular pond) as the
agent, the agent does not move.
While the rabbit is on the opposite radius (other half of the diameter)
from the agent, the agent moves at top speed in either direction along
the pond.
While the rabbit is not on the same diameter as the agent, the agent
moves at top speed towards the point on the edge closest to the rabbit.
An optimal rabbit strategy is:
Move to the center of the pond.
While staying on the same diameter as the agent move as far away from
the agent as possible. Even if the agent acts optimally, this will be no
more than 3/4 of a radius from the pond edge opposite the agent.
Then race for nearest point on the edge, which will, at that moment, be
opposite to the agent.
--
Virgil
William Hughes
>
> > Some very basic calculus is helpful. A solution is given at
> > <http://www.mathrec.org/old/2003jul/solutions.html>
>
> 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
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.
> 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.
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
Dave
> In article
> <1eb58bfd-eb18-4d34-a00f-04eec4122...@z19g2000vbz.googlegroups.com>,
This is the optimal strategy for the agent regardless of his speed,
but it is the optimal strategy for the rabbit only if the agent's
speed is less than (pi + 1) times the speed of the rabbit (assuming
appropriate adjustment of the preceding fractions). However, it is not
the rabbit's best strategy if the agent's speed, v, exceeds pi + 1. In
this case, once the rabbit aligns himself on the same diameter as the
agent 1/v of the radius from the center on the opposite radius, he
starts swimming toward the shore. Once the agent starts running, the
rabbit turns 90 degrees away from him and swims toward a new point on
the shore. (If the agent turns around, the rabbit turns to swim
outward on the radius until the agent reaches his diameter, and again
turns 90 degrees away from the agent. This is a losing strategy for
the agent, so he won't turn around.) This strategy allows the rabbit
to escape if the speed of the agent is less than
4.6033388487517003525565820291030165130674... times the speed of the
rabbit.
Dave
riverman
>
> - Show quoted 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
Sylvia Else
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.
Tim Little
> Seems clear to me. If the rabbit travels in a straight line,
> regardless of which direction is taken, that line is always a
> radius.
There are two regions of qualitatively different behaviour: close
enough to the center that the rabbit can always stay diametrically
opposite the agent, and outside that radius.
Within the center region, the rabbit's path depends upon what the
agent does - though the agent may as well just stand in one place
because the rabbit can counter any move made as it moves toward the
boundary.
Once the rabbit leaves that region, its optimal path is a straight
line that is *not* a radius. That is unless the agent changes
direction, which only helps the rabbit (and the new optimal path from
that point is again straight line that is not a radius).
- Tim
Tim Little
> 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
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
William Hughes
>
> >> - Show quoted 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....)
>
> When the rabbit moved back towards the centre of the pool, the agents
> would move back to positions oppoisite each other.
They may not have time.
>
> It appears that the rabbit cannot escape if there are two agents.
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
riverman
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
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...
Tim Little
> The 'feinting' move would only work while the rabbit is extremely
> close to the center.
It works everywhere, as you later post:
> 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.
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
BartC
"Mark-T" <MarkTa...@gmail.com> wrote in message
news:25a14f19-b336-44b6...@z16g2000prd.googlegroups.com...
> 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?
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.
--
bart
con...@email.rahul.net
There is an interesting book on this stuff:
"Chases and escapes: The Mathematics of Pursuit and Evasion," Paul
J. Nahin, Princeton University Press, 2007, ISBN-13:
978-0-691-12514-S, ISBN-10: 0-691-12514-7.
Pp. 78 references "Houghton's Problem: A circular Pursuit That Is
Solvable in Closed Form."
So, there, 8^).
John
--
John Conover, con...@email.rahul.net, http://www.johncon.com/
William Hughes
> On 2009-05-05, riverman <myronb...@yahoo.com> wrote:
>
> > The 'feinting' move would only work while the rabbit is extremely
> > close to the center.
>
> It works everywhere, as you later post:
>
> > 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.
>
> Yes, this is exactly why the "aim point" strategy fails.
It is not clear what the "aim point" is. If it is defined as the
intersection
of the tangent of the rabbit's path with the circle, then if the agent
always runs toward the aim point then the rabbit can escape
(even if the agent is very fast). But this
assumes a stupid agent (e.g. if the rabbit is close to shore it can
make the agent run away from it by swimming slowly toward the center).
If the agent chooses a simpler strategy, run in the direction that
decreases
the angular separation (if the angular separation is 0, do not move;
if the angular separation is 180 degrees, run clockwise)
then the rabbit cannot cause the agent to reverse
direction if the rabbit is more than 1/4 of the radius from the
center.
Indeed this second strategy is easily seen to be optimal for the
agent.
- William Hughes
legg
>
>"Mark-T" <MarkTa...@gmail.com> wrote in message
>news:25a14f19-b336-44b6...@z16g2000prd.googlegroups.com...
>> 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?
>
>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.
The cartoonist Sam Gross might suggest that the rabbit expire,
escaping to bunny heaven. That's what the befuddled scientist's
laboratory mice did, in one published work.
RL
Tim Little
> then if the agent always runs toward the aim point then the rabbit
> can escape (even if the agent is very fast). But this assumes a
> stupid agent
Yes it does, which is exactly why I was arguing that it was a poor
strategy for the agent to employ.
- Tim
CBFalconer
> "BartC" <ba...@freeuk.com> wrote:
>
... snip ...
>
>> 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.
>
> The cartoonist Sam Gross might suggest that the rabbit expire,
> escaping to bunny heaven. That's what the befuddled scientist's
> laboratory mice did, in one published work.
You have all missed the obvious. The rabbit teases the agent, who
spends his time running mightily to be on the same pond side as the
rabbit. This leads to mutual exhaustion and death by shriveling.
Now the only problem is "who shrivels first". Since the agent
travels 4 times as far, and does not have the benefit of cooling in
the pond, nor easily available drinking water, I contend the agent
shrivels first.
--
[mail]: Chuck F (cbfalconer at maineline dot net)
[page]: <http://cbfalconer.home.att.net>
Try the download section.
Bob Larter
Unless of course the agent is armed with a holy hand grenade. ;^)
--
W
. | ,. w , "Some people are alive only because
\|/ \|/ it is illegal to kill them." Perna condita delenda est
---^----^---------------------------------------------------------------
William Hughes
> James Arthur wrote:
> > legg wrote:
> >> On Sat, 2 May 2009 17:15:50 -0700 (PDT), Mark-T
> >> <MarkTanne...@gmail.com> wrote:
>
> >>> 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?
>
> >> At least it makes more 'sense' than the duck and the fox.
> >> There's no reason for the duck to leave the pond.
>
> >> The only problem I see with the elaborate solution is the assumption
> >> that the obviously ill 'killer' rabbit will react in any way to the
> >> presence of suits around the pond's periphery. After all, a rabbit
> >> with any sense wouldn't be in the middle of a pond in the first place.
>
> >> RL
>
> > 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.
>
> Unless of course the agent is armed with a holy hand grenade. ;^)
And can count to three.
- William Hughes
riverman
> >
> It is not clear what the "aim point" is.
Good point. I had an intuitive understanding for what I meant, but
upon closer introspection, I see that it is not clearly stated.
What about "the intersection of the radius containing the rabbit and
the edge of the pool, regardless of the direction of the rabbit's
motion"? Would aiming for that be synonymous to the strategy of the
agent moving to decrease the angular separation (between the rabbit
and the tangent line that defines the instantaneous direction of
motion of the agent?). What about just saying 'the agent moves to
minimize the distance between himself and the rabbit'?
--riverman
Richard Heathfield
<snip>
> What about just
> saying 'the agent moves to minimize the distance between himself
> and the rabbit'?
Splash!
--
Richard Heathfield <http://www.cpax.org.uk>
Email: -http://www. +rjh@
Google users: <http://www.cpax.org.uk/prg/writings/googly.php>
"Usenet is a strange place" - dmr 29 July 1999
Jim Thompson
>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've not done anything on paper, but isn't this simply a variation of
the pursuit curve problem... the rabbit swims along a vector defined
by the agent's position and the center-point of the pond?
...Jim Thompson
--
| James E.Thompson, P.E. | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona 85048 Skype: Contacts Only | |
| Voice:(480)460-2350 Fax: Available upon request | Brass Rat |
| E-mail Icon at http://www.analog-innovations.com | 1962 |
Stormy on the East Coast today... due to Bush's failed policies.
Michael A. Terrell
Jim Thompson wrote:
>
> On Sat, 2 May 2009 17:15:50 -0700 (PDT), Mark-T
> <MarkTa...@gmail.com> wrote:
>
> >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've not done anything on paper, but isn't this simply a variation of
> the pursuit curve problem... the rabbit swims along a vector defined
> by the agent's position and the center-point of the pond?
That must be a super rabbit, to keep swimming like that.
--
You can't have a sense of humor, if you have no sense!
Jim Thompson
We have a jackrabbit in the neighborhood. I see him every morning
running along the ridge as I'm pouring my coffee ;-)
William Hughes
> On May 5, 7:46 pm, William Hughes <wpihug...@hotmail.com> wrote:
>
>
>
> > It is not clear what the "aim point" is.
>
> Good point. I had an intuitive understanding for what I meant, but
> upon closer introspection, I see that it is not clearly stated.
>
> What about "the intersection of the radius containing the rabbit and
> the edge of the pool, regardless of the direction of the rabbit's
> motion"?
A useful point, but the name "aim point" seems odd
as the point has no dependence on the direction of the rabbit's
motion. (It is certainly not the meaning that Tim Little
ascribes to "aim point")
> Would aiming for that be synonymous to the strategy of the
> agent moving to decrease the angular separation (between the rabbit
> and the tangent line that defines the instantaneous direction of
> motion of the agent?).
Yes.
Note:
Angular separation: Take a polar coordinate system centered at the
circle.
The angular separation is the angular difference between the
rabbit's
angular coordinate and the agent's angular coordinate.
> What about just saying 'the agent moves to
> minimize the distance between himself and the rabbit'?
>
Equivalent to minimizing the angular separation
(noting that the agent's motion is constrained to be outside or on the
boundary of the disk of radius r, otherwise, as Richard Heathfield
points
out, splash).
- William Hughes
M Rath
> I've not done anything on paper, but isn't this simply a variation of
> the pursuit curve problem... the rabbit swims along a vector defined
> by the agent's position and the center-point of the pond?
>
M Rath wrote:
I writing a coordinate output computer program for an outward spiral...but
my compiler is acting up.
Delphi console mode is not recognizing "Readln" after a simple loop in the
main program...
M Rath
Somebody wrote this:
>> > 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.
>>
And then William Hughes wrote this:
>> Unless of course the agent is armed with a holy hand grenade. ;^)
> And can count to three.
>
But I don't know who wrote the following:
Well...you people don't understand Federal police work. Let the rabbit run.
Then coerce a witness into connecting the rabbit to the crime. Finally issue
an APB for the rabbit...and after that it's just a matter of time before
regional police are calling the Federal police with the capture.
(And a cable news columnist who is a former prosecutor sometimes points out
someone saying that they can "make the case" and that when knowing that the
case is not true. So the adversarial justice system sometimes gets confused
with the adversarial political system.)
Well...in the movie "Day of the Jackel"...they said "get a name...then we
have a passport...then we have a face...and after that it's just basic
police work..."
So they get you with your papers...and the politicians love systems that
make them look smart and capable.
So where did that rabbit get its shots ?
riverman
> riverman said:
>
> <snip>
>
> > What about just
> > saying 'the agent moves to minimize the distance between himself
> > and the rabbit'?
>
> Splash!
>
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
riverman
> If the agent chooses a simpler strategy, run in the direction that
> decreases
> the angular separation (if the angular separation is 0, do not move;
> if the angular separation is 180 degrees, run clockwise)
> then the rabbit cannot cause the agent to reverse
> direction if the rabbit is more than 1/4 of the radius from the
> center.
> Indeed this second strategy is easily seen to be optimal for the
> agent.
>
>
What do you mean by 'optimal for the agent', since it leads to the
rabbit escaping with the least travelled distance. I'd say its optimal
for the rabbit. Strange if it turned out to be optimal for both!
I can see that this strategy will work with any size pool, as the
distances and velocities are all relative. But what if the pool is
reduced to a POINT (as in a limit?) Who wins then?
--riverman
Bob Larter
It seems obvious to me that the rabbit will end up describing a spiral
path, as it will constantly be trying to stay 180 degrees away from the
agent.
The agent will move so as to to stay at the same angle as the rabbit,
& the rabbit will be trying to stay 180 degrees away from the agent. A
spiral is the obvious result, ending up with the rabbit hitting the
shore a whisker away from the agent. (Alternatively, the rabbit could
desribe a zig-zag path, but the end-result would be the same.)
Bob Larter
NOT FOUR!
Bob Larter
Well, that sure sucks!
William Hughes
> On May 5, 7:46 pm, William Hughes <wpihug...@hotmail.com> wrote:
>
> > If the agent chooses a simpler strategy, run in the direction that
> > decreases
> > the angular separation (if the angular separation is 0, do not move;
> > if the angular separation is 180 degrees, run clockwise)
> > then the rabbit cannot cause the agent to reverse
> > direction if the rabbit is more than 1/4 of the radius from the
> > center.
> > Indeed this second strategy is easily seen to be optimal for the
> > agent.
>
> >
>
> What do you mean by 'optimal for the agent',
Optimal in the sense that if there is any strategy that will
stop the rabbit from escaping this strategy will.
> since it leads to the
> rabbit escaping with the least travelled distance.
Not under any assumptions I can come up with
(If the rabbit uses a sure escape strategy, the
distance travelled will be identical as long as the
agent does not stand still).
What assumptions justify this claim?
What strategy would lead to the rabbit escaping with
a longer travelled distance?
> I'd say its optimal for the rabbit.
No, the optimal strategy for the rabbit is
for the agent not to move. (If the agent does
not move the rabbit swims
along a radius, if the agent moves
the rabbit does not swim along a radius).
- William Hughes
William Hughes
> It seems obvious to me that the rabbit will end up describing a spiral
> path, as it will constantly be trying to stay 180 degrees away from the
> agent.
Nope, when the rabbit is close enough to a point on shore
that he can reach it before the agent, the rabbit will
take a straight line path. It no longer matters if it
is 180 degrees from the agent.
- William Hughes
mich
> 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?
>
> Mark
Beacause the sped of agents is four times the rabbits speed, then the
rabbit get away.
The rabbit escapes if speed of agents is <= Pi+1.
Dave
Actually, the rabbit can escape if the agent's speed is less than
4.6033388487517003525565820291030165130674..., which exceeds pi + 1.
Dave
Dave
Please give the greatest agent speed for which the rabbit can escape
using this strategy.
Dave
Bob Larter
Point. But the path should be a spiral before that.
Michael Moroney
>On May 6, 2:35=A0pm, mich <micheletro...@graduate.org> wrote:
>> Beacause the sped of agents is four times the rabbits speed, then the
>> rabbit get away.
>>
>> The rabbit escapes if speed of agents is <= Pi+1.
>Actually, the rabbit can escape if the agent's speed is less than
>4.6033388487517003525565820291030165130674..., which exceeds pi + 1.
How is this figure derived?
mrda...@gmail.com
Missing a semicolon somewhere?
Michael
Nobody
Dave
wrote:
Assuming a pond's radius = 1 and the rabbit's speed = 1: If the
agent's speed is v, the rabbit can position himself 1/v from the
center on the opposite side of the pond from the agent. He then heads
radially toward the shore and watches to see which way the agent runs.
When the agent has committed himself to a direction, the rabbit turns
90 degrees in the opposite direction and heads in a straight line
toward the shore. If the agent reverses direction, the rabbit will
again head out on a radius until the agent, the rabbit, and the center
of the pond again are collinear or the agent reverses again. At this
point, the rabbit will be closer to the shore before and the agent
will find himself in a worse position; thus, once he commits to a
direction, the agent cannot afford to reverse it. The rabbit must swim
sqrt(1 - 1 / v^2) while the agent runs pi + arccos(1 / v). So solving
sqrt(1 - 1/v^2) = ( pi + arccos(1 / v) ) / v
for v gives the number in question.
Dave
Jim Thompson
That solution presumes the fox/agent doesn't move immediately.
Of course the problem is poorly stated, much like all requests for
circuit help here ;-)
...Jim Thompson
--
| James E.Thompson, P.E. | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona 85048 Skype: Contacts Only | |
| Voice:(480)460-2350 Fax: Available upon request | Brass Rat |
| E-mail Icon at http://www.analog-innovations.com | 1962 |
Buy a Fiarysler with your tax credit...
Ignore that grinding noise, that is normal :-)
Matthew Russotto
riverman <myro...@yahoo.com> wrote:
>
>I can see that this strategy will work with any size pool, as the
>distances and velocities are all relative. But what if the pool is
>reduced to a POINT (as in a limit?) Who wins then?
If the pool is a point, the agent wins, trivially. But in the limit as
the pool shrinks to radius zero, the rabbit wins.
(and this is why first-year calculus students get headaches, at least
until they learn to just push the symbols around and forget about it
:-) )
--
It's times like these which make me glad my bank is Dial-a-Mattress
CBFalconer
> Dave <dave_an...@Juno.com> writes:
>>
>>> Beacause the sped of agents is four times the rabbits speed,
>>> then the rabbit get away.
>>>
>>> The rabbit escapes if speed of agents is <= Pi+1.
>
>> Actually, the rabbit can escape if the agent's speed is less
>> than 4.6033388487517003525565820291030165130674..., which
>> exceeds pi + 1.
>
> How is this figure derived?
Laboriously.
--
[mail]: Chuck F (cbfalconer at maineline dot net)
[page]: <http://cbfalconer.home.att.net>
Try the download section.
Dave
Site.com> wrote:
> On Thu, 07 May 2009 00:12:56 +0100, Nobody <nob...@nowhere.com> wrote:
> >On Wed, 06 May 2009 22:32:23 +0000, Michael Moroney wrote:
>
>
> >>>On May 6, 2:35=A0pm, mich <micheletro...@graduate.org> wrote:
> >>>> Beacause the sped of agents is four times the rabbits speed, then the
> >>>> rabbit get away.
>
> >>>> The rabbit escapes if speed of agents is <= Pi+1.
>
> >>>Actually, the rabbit can escape if the agent's speed is less than
> >>>4.6033388487517003525565820291030165130674..., which exceeds pi + 1.
>
> >> How is this figure derived?
>
> >http://groups.google.com/group/rec.puzzles/msg/050273ed70e4a9f6
>
> That solution presumes the fox/agent doesn't move immediately.
No. It presumes that both the rabbit and the agent react instantly. So
the instant the rabbit gets more than 1/v from the center, the agent
chooses a direction and starts running. If he doesn't, the rabbit can
just swim straight to the shore.
Dave
M Rath
> M Rath wrote:
>
> I writing a coordinate output computer program for an outward spiral...but
> my compiler is acting up.
>
> Delphi console mode is not recognizing "Readln" after a simple loop in the
> main program...
mrdarrett wrote:
Missing a semicolon somewhere?
M Rath wrote:
No the compiler is not accepting "ads:= Pi / 2" as a test value to develop
formulas in a loop. Now the compiler might be saying "What's the point of
calculating the same result over and over in a loop ?" but it did not accept
"ads:= Pi/2" with the loop commented out. But in both cases it accepts the
numerical value...
James Arthur
Observation: I once chased a jackrabbit. I was on
a motorcycle, he was not. The rabbit got away.
Given his preternatural speed, the agent should fox
the rabbit into quitting the safety of the pond,
then give chase--he'll have the bunny in a few
strides.
Cheers,
James Arthur
KBH
10 , 10 and the radius is 10 units . The agent is at North and East
coordinates of 20 , 10 and moves clockwise in 1-degree steps. The rabbit at
the center of the pond moves away from the agent at each step on a line
between the agent and the rabbit. And the rabbit moves 1/4 the distance that
the agent moves.
The rabbit gets out at North and East coordinates of 1.3437 , 4.9694 when
the agent is at an azimuth of 232 degrees on the circle edge. But the rabbit
is out at an azimuth of 210 degrees or so. In other words the rabbit doubled
back while the agent kept going. Before the rabbit doubled back it looks
like a slow outward spiral but I will write a ScratchPlot file later.
Of course this is one systematic method with no answer for what the agent
does when the rabbit is within one unit of shore and near the agent.
Here is the KBH code:
Var
ads, nag, eag, rba, rds, sec, rbn, rbe, nd, ed, dgr, dst: double;
flg: integer;
Procedure D;
Begin
If (flg = 1) Then
Begin
{Procedure E}
rba:= rba + Pi;
Exit;
End;
rba:= rba + (Pi * 2);
{The structure of procedures D, E, & F are from the KBH Survey Progam for
the HP35S}
End;
Procedure F;
Begin
If (flg = 1) Then Exit;
rba:= rba + Pi;
End;
begin
{KBH Code}
flg:= 0;
sec:= 1 / 3600;
rbn:= 10;
rbe:= 10;
rds:= 0;
dgr:= 1;
While (dgr < 233) Do
Begin
ads:= (dgr * Pi) / 180;
nag:= 10 + Cos(ads);
eag:= 10 + Sin(ads);
nd:= rbn - nag;
If (nd = 0) Then nd:= 0.000000000001;
ed:= rbe - eag;
If (ed >= 0) Then flg:= 1;
rba:= ArcTan(ed / nd);
If (rba < 0.000000000001) Then D Else F;
rbn:= rbn + (Cos(rba) * 0.043633231);
rbe:= rbe + (Sin(rba) * 0.043633231);
dst:= Sqrt(Sqr(10 - rbn) + Sqr(10 - rbe));
WriteLn(rbn:15:4, rbe:15:4, dgr:15:4, dst:15:4);
dgr:= dgr + 1;
End;
ReadLn;
end.
KBH
Oh...that KBH Code will run in Delphi Console Mode.
KBH
The following is a file for KBH ScratchPlot. Copy to a text file and save
the file as "Name.plt" with the quotation marks and then the file will run
in ScratchPlot. And of course the file should contain text...not the html
being picked up here.
I didn't convert the entire coordinate file to a ScratchPlot file...I just
entered 15 points for the rabbit as lines. Then I laid out 15 points for the
agent from the radius point with the Fan and Azimuth features. Then I added
15 points to the rabbit line so that the points of the agent could be
matched up to the points of the rabbit. For instance both the agent and the
rabbit were at 220 degrees clockwise azimuth but at different times. I could
have added a full circle but that would make the file very large...
663&137@F+bkN0E0
41&759@S+bkN20E20
0
352&448@P+bkN10E10
353&447@P+rdN9.9564E9.9992
385&442@L+rdN8.9295E9.8114
416&429@L+rdN7.9231E9.3937
445&412@L+rdN6.9807E8.8454
473&392@L+rdN6.0868E8.2205
500&372@L+rdN5.2194E7.5591
527&351@L+rdN4.3545E6.8944
555&331@L+rdN3.4713E6.2543
583&312@L+rdN2.5578E5.6584
595&306@L+rdN2.1834E5.4343
600&302@L+rdN1.9944E5.3254
606&299@L+rdN1.8042E5.2184
612&296@L+rdN1.6130E5.1134
618&292@L+rdN1.4208E5.0102
621&291@L+rdN1.3437E4.9694
352&448@P+bkN10E10
352&448@P+bkN10E10
70&579@P+bkN19.0631E14.2262
352&448@+bkN10E10
41&453@P+bkN19.9985E10.1745
352&448@+bkN10E10
152&686@P+bkN16.4279E17.6604
352&448@+bkN10E10
271&748@P+bkN12.5882E19.6593
352&448@+bkN10E10
406&754@P+bkN8.2635E19.8481
352&448@+bkN10E10
530&702@P+bkN4.2642E18.1915
352&448@+bkN10E10
621&603@P+bkN1.3397E15.0000
352&448@+bkN10E10
661&475@P+bkN0.0381E10.8716
352&448@+bkN10E10
644&341@P+bkN0.6031E6.5798
352&448@+bkN10E10
621&292@P+bkN1.3397E5.0000
352&448@+bkN10E10
606&269@P+bkN1.8085E4.2642
352&448@+bkN10E10
590&248@P+bkN2.3396E3.5721
352&448@+bkN10E10
571&228@P+bkN2.9289E2.9289
352&448@+bkN10E10
551&209@P+bkN3.5721E2.3396
352&448@+bkN10E10
543&202@P+bkN3.8434E2.1199
352&448@+bkN10E10
352&448@P+bkN10E10
41&453@P+prN19.9985E10.1745
352&448@+prN10E10
41&453@P+bkN19.9985E10.1745
352&448@+bkN10E10
385&442@P+bkN8.9295E9.8114
416&429@P+bkN7.9231E9.3937
445&412@P+bkN6.9807E8.8454
473&392@P+bkN6.0868E8.2205
500&372@P+bkN5.2194E7.5591
527&351@P+bkN4.3545E6.8944
555&331@P+bkN3.4713E6.2543
583&312@P+bkN2.5578E5.6584
595&306@P+bkN2.1834E5.4343
600&302@P+bkN1.9944E5.3254
606&299@P+bkN1.8042E5.2184
612&296@P+bkN1.6130E5.1134
618&292@P+bkN1.4208E5.0102
621&291@P+bkN1.3437E4.9694
KBH
No...the plot didn't look right. And so in the previous code the coordinates
of the agent were based on a radius of 1 but should have been based on a
radius of 10.
And with this corrected code the Rabbit does not get out of the pond with
this systematic method ! But I'm sure the plot will make some nice
curves...which is what I was looking for.
Here is the corrected KBH code:
While (dgr < 360) Do
Begin
ads:= (dgr * Pi) / 180;
nag:= 10 + (Cos(ads) * 10);
eag:= 10 + (Sin(ads) * 10);
nd:= rbn - nag;
If (nd = 0) Then nd:= 0.000000000001;
ed:= rbe - eag;
If (ed >= 0) Then flg:= 1;
rba:= ArcTan(ed / nd);
If (rba < 0.000000000001) Then D Else F;
rbn:= rbn + (Cos(rba) * 0.043633231);
rbe:= rbe + (Sin(rba) * 0.043633231);
dst:= Sqrt(Sqr(10 - rbn) + Sqr(10 - rbe));
WriteLn(rbn:15:4, rbe:15:4, dgr:15:4, dst:15:4);
dgr:= dgr + 1;
End;
ReadLn;
end.
And that KBH Code will run in Delphi Console mode...
Then plot the coordinates in KBH Scratch Plot...
KBH
And the output of the KBH code is rabbit North coordinate, rabbit East
coordinate, clockwise azimuth of agent on circle, and rabbit distance from
circle radius of 10.
Baron
Just a guess, has Pi been defined before use ?
--
Best Regards:
Baron.
Dik T. Winter
...
> 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? ;-)
The rabbit had been caught by poachers in an helicopter, but he knew to
escape.
--
dik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
Dik T. Winter
> riverman wrote:
...
> > 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....)
>
> 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.
Of course the rabbit can escape if the agents use a stupid strategy.
Sylvia Else
> In article <0046a0fb$0$1968$c3e...@news.astraweb.com> Sylvia Else <syl...@not.at.this.address> writes:
> > riverman wrote:
> ...
> > > 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....)
> >
> > 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.
>
> Of course the rabbit can escape if the agents use a stupid strategy.
You mean like wandering off to get some donuts and coffee? I think it's
usual to assume in these puzzles that the participants will always use
the optimum strategy.
Sylvia.
Rob Johnson
Who says that the land speed of the rabbit is the same as its water
speed? My assumption is that once it makes it to shore, the rabbit
has no difficulty in escaping. However, the problem only requires
that the rabbit makes it to shore without being captured.
Rob Johnson <r...@trash.whim.org>
take out the trash before replying
to view any ASCII art, display article in a monospaced font
James Arthur
> In article <lLuMl.4916$b11....@nwrddc02.gnilink.net>,
> James Arthur <bogus...@verizon.net> wrote:
>> mich wrote:
>>> On 2 mayo, 20:15, Mark-T <MarkTanne...@gmail.com> wrote:
>>>> 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?
>>>>
>>>> --
>>>> Mark
>>> Beacause the sped of agents is four times the rabbits speed, then the
>>> rabbit get away.
>>>
>>> The rabbit escapes if speed of agents is <= Pi+1.
>>>
>> Observation: I once chased a jackrabbit. I was on
>> a motorcycle, he was not. The rabbit got away.
>>
>> Given his preternatural speed, the agent should fox
>> the rabbit into quitting the safety of the pond,
>> then give chase--he'll have the bunny in a few
>> strides.
>
> Who says that the land speed of the rabbit is the same as its water
> speed? My assumption is that once it makes it to shore, the rabbit
> has no difficulty in escaping.
I didn't assume that because the problem specifies each party's
speed. Otherwise I might've observed that people swim a lot
faster than rabbits, and the agent might have an easier time
with a running dive & a swim.
However, the problem only requires
> that the rabbit makes it to shore without being captured.
Yep. Two points.
Cheers,
James Arthur
ehsjr
C + x. Somewhere in the neighborhood of C the agent's perception
of the rabbit's position/speed/direction will be distorted.
Ed
Rob Johnson
The rabbit moves at a rate of 1, the agent at a rate of r. The edge
of the lake is the unit circle centered at (0,0).
As has been discussed, the rabbit can keep the agent on the opposite
side of the center of the lake until the rabbit gets to a distance
of 1/r from the center of the lake. So we assume that the rabbit
starts at (-1/r,0) and the agent starts at (1,0). Then the agent
moves counterclockwise and the rabbit head straight down until it
reaches the unit circle after time t (if it headed further left, it
would meet the agent sooner, if it headed further right, it would
decrease its distance to the center of the lake). The positions at
time t are
rabbit: ( -1/r , -t )
agent: ( cos(rt) , sin(rt) )
To solve for the extreme case, we need to solve the equations
cos(rt) + 1/r = 0
sin(rt) + t = 0
Eliminating the sin and cos using Pythagoras' Theorem, we get
rt = sqrt(r^2 - 1)
Thus, we need to solve
r cos(sqrt(r^2 - 1)) + 1 = 0 [1]
while making sure that
sin(sqrt(r^2 - 1)) < 0 [2]
The first two solutions of [1] are
2.261826334114651437542573924426
4.603338848751700352556582029103
For the first solution, inequality [2] fails to hold, so the
solution we are interested in is the second, for which [2] is
satisfied.
KBH
Okay...with the systematic method used here...the Rabbit does get out at
North and East coordinates of 10.5006 , 19.9976 where the radius point has
North and East coordinates of 10 , 10. And the rabbit gets out when the
agent is at a clockwise azimuth on the circle of 834 degrees !
Well...I have only plotted the first 360 degrees so far but here is the KBH
Scratch Plot file:
663&137@F+wtN0E0
41&759@S+wtN20E20
0
352&448@P+bkN10E10
353&447@L+rdN9.9564E9.9992
365&446@L+rdN9.5661E9.9588
378&443@L+rdN9.1452E9.8458
390&437@L+rdN8.7477E9.6669
402&430@L+rdN8.3826E9.4288
412&421@L+rdN8.0572E9.1387
421&410@L+rdN7.7774E8.8044
428&399@L+rdN7.5477E8.4338
433&386@L+rdN7.3716E8.0349
437&373@L+rdN7.2515E7.6158
439&360@L+rdN7.1886E7.1844
439&346@L+rdN7.1834E6.7484
437&333@L+rdN7.2357E6.3155
434&320@L+rdN7.3441E5.8932
429&307@L+rdN7.5067E5.4886
422&295@L+rdN7.7209E5.1089
414&285@L+rdN7.9834E4.7607
405&275@L+rdN8.2899E4.4506
394&267@L+rdN8.6357E4.1851
382&260@L+rdN9.0151E3.9703
369&255@L+rdN9.4215E3.8124
356&252@L+rdN9.8469E3.7173
343&251@L+rdN10.2819E3.6911
329&253@L+rdN10.7150E3.7398
316&257@L+rdN11.1311E3.8687
305&263@L+rdN11.5107E4.0819
295&273@L+rdN11.8285E4.3792
289&284@L+rdN12.00538E4.7515
285&297@L+rdN12.1525E5.1743
286&311@L+rdN12.1075E5.6063
292&323@L+rdN11.9231E5.9996
301&333@L+rdN11.6270E6.318
312&340@L+rdN11.2554E6.5445
325&344@L+rdN10.8406E6.6769
339&346@L+rdN10.4072E6.7212
325&347@L+rdN10.8414E6.7579
312&350@L+rdN11.2621E6.8715
352&448@P+bkN10E10
41&453@P+blN19.9985E10.1745
352&448@+blN10E10
45&502@P+blN19.8481E11.7365
352&448@+blN10E10
59&554@P+blN19.3969E13.4202
352&448@+blN10E10
82&603@P+blN18.6603E15.0000
352&448@+blN10E10
113&647@P+blN17.6604E16.4279
352&448@+blN10E10
152&686@P+blN16.4279E17.6604
352&448@+blN10E10
196&717@P+blN15.0000E18.6603
352&448@+blN10E10
245&740@P+blN13.4202E19.3969
352&448@+blN10E10
297&754@P+blN11.7365E19.8481
352&448@+blN10E10
352&759@P+blN10.0000E20.0000
352&448@+blN10E10
406&754@P+blN8.2635E19.8481
352&448@+blN10E10
458&740@P+blN6.5798E19.3969
352&448@+blN10E10
507&717@P+blN5.0000E18.6603
352&448@+blN10E10
551&686@P+blN3.5721E17.6604
352&448@+blN10E10
590&647@P+blN2.3396E16.4279
352&448@+blN10E10
621&603@P+blN1.3397E15.0000
352&448@+blN10E10
644&554@P+blN0.6031E13.4202
352&448@+blN10E10
658&502@P+blN0.1519E11.7365
352&448@+blN10E10
663&448@P+blN0.0000E10.0000
352&448@+blN10E10
658&393@P+blN0.1519E8.2635
352&448@+blN10E10
644&341@P+blN0.6031E6.5798
352&448@+blN10E10
621&292@P+blN1.3397E5.0000
352&448@+blN10E10
590&248@P+blN2.3396E3.5721
352&448@+blN10E10
551&209@P+blN3.5721E2.3396
352&448@+blN10E10
507&178@P+blN5.0000E1.3397
352&448@+blN10E10
458&155@P+blN6.5798E0.6031
352&448@+blN10E10
406&141@P+blN8.2635E0.1519
352&448@+blN10E10
352&137@P+blN10.0000E0.0000
352&448@+blN10E10
297&141@P+blN11.7365E0.1519
352&448@+blN10E10
245&155@P+blN13.4202E0.6031
352&448@+blN10E10
196&178@P+blN15.0000E1.3397
352&448@+blN10E10
152&209@P+blN16.4279E2.3396
352&448@+blN10E10
113&248@P+blN17.6604E3.5721
352&448@+blN10E10
82&292@P+blN18.6603E5.0000
352&448@+blN10E10
59&341@P+blN19.3969E6.5798
352&448@+blN10E10
45&393@P+blN19.8481E8.2635
352&448@+blN10E10
51&448@Agent->+blN19.65E10
352&463@Rabbit Begins Here+rdN10E10.5
Copy to a text file, save the text file as "Name.plt", and the file will run
in KBH Scratch Plot...
KBH
Copy to a text file, save the text file as "Name.plt" with quotation marks,
KBH
Looking at the plot...it seems that the rabbit path should continue to
spiral westward rather than turn back northward and eastward at the sharp
point.
But I don't yet see why the coordinate output of the KBH Code would be
correct at 340 degrees but incorrect at 350 degrees. Why I get this worked
out it will probably require a new subject posting...
KBH
All the above is voided...the agent coordinates should have been based on a
circle radius of 10 instead of a circle radius of 1 .
KBH
of the loop.
And according to this systematic method...the rabbit gets out at North and
East coordinates of 8.1332 , 0.1612 with the agent at a clockwise azimuth of
535 degrees on the circle edge.
The KBH Scratch Plot file below is correct until the sharp point...but is
edited below to stop before the sharp point.
>>>>> flg:= 0;
>>>>> End;
>>>>>
>>>>> ReadLn;
>>>>> end.
>>>>>
>>>>> And that KBH Code will run in Delphi Console mode...
>>>>>
>>>>
>>>> And the output of the KBH code is rabbit North coordinate, rabbit East
>>>> coordinate, clockwise azimuth of agent on circle, and rabbit distance
>>>> from circle radius of 10.
>>>>
>>>>> Then plot the coordinates in KBH Scratch Plot...
>>>>>
>>>
>>> Well...I have only plotted the first 340 degrees so far but here is the