NY Times math problem

[KBH]

KBH Code:

{The rabbit at the center of a 10 radius circular pond and the chasing
agent
is at the edge of the pond.}

Var
ads, nag, eag, rba, 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;
rbn:= 10;
rbe:= 10;
dgr:= 1;

While (dgr < 361) Do {Change value for desired output range}
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));
rba:= (rba / Pi) * 180;
WriteLn(rbn:15:4, rbe:15:4, dgr:15:4, dst:15:4);
dgr:= dgr + 1;
flg:= 0;
End;

ReadLn;
end.

And the above KBH code runs in Delphi Console Mode...

Now the rabbit gets out at North and East coordinates of 8.1332 , 0.1612
while the agent is at an azimuth of 535 degrees on the edge of the
circular
pond and on his second trip around the pond. Of course this is a
systematic
method only...

And here is the KBH Scratch Plot file for the rabbit path:

663&137@F+wtN0E0
41&759@S+wtN20E20
0
352&448@P+bkN10E10
353&...@L+rdN9.9564E9.9992
365&...@L+rdN9.5661E9.9588
378&...@L+rdN9.1452E9.8458
390&...@L+rdN8.7477E9.6669
402&...@L+rdN8.3826E9.4288
412&...@L+rdN8.0572E9.1387
421&...@L+rdN7.7774E8.8044
428&...@L+rdN7.5477E8.4338
433&...@L+rdN7.3716E8.0349
437&...@L+rdN7.2515E7.6158
439&...@L+rdN7.1886E7.1844
439&...@L+rdN7.1834E6.7484
437&...@L+rdN7.2357E6.3155
434&...@L+rdN7.3441E5.8932
429&...@L+rdN7.5067E5.4886
422&...@L+rdN7.7209E5.1089
414&...@L+rdN7.9834E4.7607
405&...@L+rdN8.2899E4.4506
394&...@L+rdN8.6357E4.1851
382&...@L+rdN9.0151E3.9703
369&...@L+rdN9.4215E3.8124
356&...@L+rdN9.8469E3.7173
343&...@L+rdN10.2819E3.6911
329&...@L+rdN10.7150E3.7398
316&...@L+rdN11.1311E3.8687
305&...@L+rdN11.5107E4.0819
295&...@L+rdN11.8285E4.3792
289&...@L+rdN12.00538E4.7515
285&...@L+rdN12.1525E5.1743
286&...@L+rdN12.1075E5.6063
292&...@L+rdN11.9231E5.9996
301&...@L+rdN11.6270E6.318
312&...@L+rdN11.2554E6.5445
325&...@L+rdN10.8406E6.6769
339&...@L+rdN10.4072E6.7212
352&...@L+rdN9.9729E6.6856
365&...@L+rdN9.5502E6.5793
378&...@L+rdN9.1482E6.4106
390&...@L+rdN8.7736E6.1875
400&...@L+rdN8.4315E5.9171
410&...@L+rdN8.1258E5.6061
418&...@L+rdN7.8594E5.2609
425&...@L+rdN7.6345E4.8873
431&...@L+rdN7.4526E4.4909
435&...@L+rdN7.3148E4.0771
438&...@L+rdN7.2216E3.6511
439&...@L+rdN7.1730E3.2176
440&...@L+rdN7.1687E2.7815
438&...@L+rdN7.2079E2.3471
436&...@L+rdN7.2895E1.9186
432&...@L+rdN7.4120E1.5
427&...@L+rdN7.5736E1.0949
421&...@L+rdN7.7720E.7064
414&...@L+rdN8.0048E0.3376
410&...@L+rdN8.1332E0.1612
352&448@P+bkN10E10
41&...@P+blN19.9985E10.1745
352&448@+blN10E10
45&...@P+blN19.8481E11.7365
352&448@+blN10E10
59&...@P+blN19.3969E13.4202
352&448@+blN10E10
82&...@P+blN18.6603E15.0000
352&448@+blN10E10
113&...@P+blN17.6604E16.4279
352&448@+blN10E10
152&...@P+blN16.4279E17.6604
352&448@+blN10E10
196&...@P+blN15.0000E18.6603
352&448@+blN10E10
245&...@P+blN13.4202E19.3969
352&448@+blN10E10
297&...@P+blN11.7365E19.8481
352&448@+blN10E10
352&...@P+blN10.0000E20.0000
352&448@+blN10E10
406&...@P+blN8.2635E19.8481
352&448@+blN10E10
458&...@P+blN6.5798E19.3969
352&448@+blN10E10
507&...@P+blN5.0000E18.6603
352&448@+blN10E10
551&...@P+blN3.5721E17.6604
352&448@+blN10E10
590&...@P+blN2.3396E16.4279
352&448@+blN10E10
621&...@P+blN1.3397E15.0000
352&448@+blN10E10
644&...@P+blN0.6031E13.4202
352&448@+blN10E10
658&...@P+blN0.1519E11.7365
352&448@+blN10E10
663&...@P+blN0.0000E10.0000
352&448@+blN10E10
658&...@P+blN0.1519E8.2635
352&448@+blN10E10
644&...@P+blN0.6031E6.5798
352&448@+blN10E10
621&...@P+blN1.3397E5.0000
352&448@+blN10E10
590&...@P+blN2.3396E3.5721
352&448@+blN10E10
551&...@P+blN3.5721E2.3396
352&448@+blN10E10
507&...@P+blN5.0000E1.3397
352&448@+blN10E10
458&...@P+blN6.5798E0.6031
352&448@+blN10E10
406&...@P+blN8.2635E0.1519
352&448@+blN10E10
352&...@P+blN10.0000E0.0000
352&448@+blN10E10
297&...@P+blN11.7365E0.1519
352&448@+blN10E10
245&...@P+blN13.4202E0.6031
352&448@+blN10E10
196&...@P+blN15.0000E1.3397
352&448@+blN10E10
152&...@P+blN16.4279E2.3396
352&448@+blN10E10
113&...@P+blN17.6604E3.5721
352&448@+blN10E10
82&...@P+blN18.6603E5.0000
352&448@+blN10E10
59&...@P+blN19.3969E6.5798
352&448@+blN10E10
45&...@P+blN19.8481E8.2635
352&448@+blN10E10
51&448@Agent->+blN19.65E10
352&463@Rabbit Begins Here+rdN10E10.5
352&448@P+bkN10E10
661&...@P+blN0.0381E10.8716
599&...@L+blN2.0381E10.8716
597&475@Agent Stops Here on 2nd Loop+blN2.1E10.88

Copy to a text file, save the text file as "Name.plt" with quotation
marks,
and the file will run in KBH Scratch Plot...

And the rabbit path is a small-cap handwritten-style "e"...

Dave wrote:

I've totally lost track of what you are trying to do. Can you recap,
and possibly post the graph on the web and give us the link?

KBH wrote:

But you replied to the final version and thanks for that...

Now...with each step of the agent around the circle...the rabbit moves on a
line from the agent to the rabbit and away from the agent. So just say that
the rabbit turns away from each new position of the agent. And the rabbit
moves one-fourth the distance that the agent moves...each time that the
rabbit turns away.

And this is a systematic method. The agent keeps moving around the circle
and the rabbit keeps turning away...

The graphics that I have is a coordinate plot and not an animation movement.
But you can find Scratch Plot with google, download it in ten seconds,
install it, and make a file for it with the above data. And the rabbit path
is really nice looking...and I think it has a point to make.

Now with the rabbit path you can scale a tangent line off the path at any
point...knowing that the path runs away from the opposite side...and see
which agent point on the circle produced the rabbit path at that rabbit path
point.

 

[KBH]

Dave wrote:

I've totally lost track of what you are trying to do. Can you recap,
and possibly post the graph on the web and give us the link?

KBH wrote:

But you replied to the final version and thanks for that...

Now...with each step of the agent around the circle...the rabbit moves on
a line from the agent to the rabbit and away from the agent. So just say
that the rabbit turns away from each new position of the agent. And the
rabbit moves one-fourth the distance that the agent moves...each time that
the rabbit turns away.

And this is a systematic method. The agent keeps moving around the circle
and the rabbit keeps turning away...

The graphics that I have is a coordinate plot and not an animation
movement. But you can find Scratch Plot with google, download it in ten
seconds, install it, and make a file for it with the above data. And the
rabbit path is really nice looking...and I think it has a point to make.

Now with the rabbit path you can scale a tangent line off the path at any
point...knowing that the path runs away from the opposite side...and see
which agent point on the circle produced the rabbit path at that rabbit
path point.

Oh...to make a Scratch Plot file copy the graphics data from the 5/07/2009
6:48 post . Paste the graphics data to a text file, save the text file as
"Name.plt" with quotation marks, and Scratch Plot will run the file.

It's just that the replies to that 6:48 post are leaving out data...

 

[Jasen Betts]

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.

yes, this straight line may not be a radius, but it is unlikely to
differ greatly from a radius.

 

[Dave]

Dave wrote:

I've totally lost track of what you are trying to do. Can you recap,
and possibly post the graph on the web and give us the link?

KBH wrote:

But you replied to the final version and thanks for that...

Now...with each step of the agent around the circle...the rabbit moves ona
line from the agent to the rabbit and away from the agent. So just say that
the rabbit turns away from each new position of the agent. And the rabbit
moves one-fourth the distance that the agent moves...each time that the
rabbit turns away.

And this is a systematic method. The agent keeps moving around the circle
and the rabbit keeps turning away...

The graphics that I have is a coordinate plot and not an animation movement.
But you can find Scratch Plot with google, download it in ten seconds,
install it, and make a file for it with the above data. And the rabbit path
is really nice looking...and I think it has a point to make.

Now with the rabbit path you can scale a tangent line off the path at any
point...knowing that the path runs away from the opposite side...and see
which agent point on the circle produced the rabbit path at that rabbit path
point.

Using your strategy, what is the answer to part 2 of the puzzle?

Dave

 

[KBH]

Dave wrote:

Using your strategy, what is the answer to part 2 of the puzzle?

KBH wrote:

Part 2...what's the fastest speed the agent can run where the rabbit can
still escape ?

Oh...yeah thanks for that.

Well...I have the rabbit escape at an azimuth of 259.25649 when the agent is
at an azimuth of 535. And so the rabbit got out at a azimuth of 619.25649
relative to the agent .

So the agent got around 1.48611 loops but needed to get around 1.72016 loops
to catch the rabbit with this systematic method...

Okay...I have the radius at 10 so the arc distance around the circle is
62.832 . Divide by a agent rate of 4 and that's 15.708 time units for the
agent to go 360 degrees around the circle. So the agent went 535 degrees in
23.3438 time units. But the agent needs to go 619.25649 degrees (or 1.72016
loops) in 23.3438 time units. Okay...1.72016 loops * 62.832 circle arc
distance = a distance of 108.0811. Rate * Time = Distance so Rate = 108.0811
/ 23.3438 or a rate of 4.6300 . Tie goes to the runner...

But now does the rabbit make the same path with this systematic method ?

Since the agent is moving around the circle faster...the rabbit is making
directional changes at a quicker pace even as the rabbit speed stays the
same. Well the rabbit speed is slower relative to the agent at 21.5983%
verus the previous 25%...so I have to change the rabbit distance (per degree
of agent arc distance) in the KBH Code to 0.037696233 and run the code for
the answer.

And now the rabbit gets out at an agent azimuth of 786 degrees or 2.18333
loops...but the rabbit gets out at about the same spot as before.

So I'll just conclude that the rabbit always gets out no matter what the
agent speed ? Or try a slower agent speed ?

I change the rabbit distance (per degree of agent arc distance) to
0.049570229 and the rabbit gets out at an agent azimuth of 468 degrees. And
beginning here the exit coordinates are different than the previous two...

Okay change the rabbit distance (per degree of agent arc distance) to 0.06
and the rabbit gets out at an agent azimuth of 238 degrees.

Okay change the rabbit distance (per degree of agent arc distance) to 0.07
and the rabbit gets out at an agent azimuth of 171 degrees.

Change the rabbit distance (per degree of agent arc distance) to 0.10 and
the rabbit gets out at an agent azimuth of 108 degrees.

Change the rabbit distance (per degree of agent arc distance) to 0.20 and
the rabbit gets out at an agent azimuth of 51 degrees.

Change the rabbit distance (per degree of agent arc distance) to 0.30 and
the rabbit gets out at an agent azimuth of 34 degrees.

Change the rabbit distance (per degree of agent arc distance) to 0.40 and
the rabbit gets out at an agent azimuth of 26 degrees.

Change the rabbit distance (per degree of agent arc distance) to 0.50 and
the rabbit gets out at an agent azimuth of 21 degrees.

Change the rabbit distance (per degree of agent arc distance) to 0.60 and
the rabbit gets out at an agent azimuth of 17 degrees.

Change the rabbit distance (per degree of agent arc distance) to 0.70 and
the rabbit gets out at an agent azimuth of 15 degrees.

Change the rabbit distance (per degree of agent arc distance) to 1.00 and
the rabbit gets out at an agent azimuth of 11 degrees or less.

Change the rabbit distance (per degree of agent arc distance) to 10.00 and
the rabbit gets out at an agent azimuth of 1 degrees. So each time the agent
moves 0.174533333 the rabbit moves 10.00 ! So here the rabbit speed is
57.2956 times the agent speed instead of the given rabbit speed of 0.25
times the agent speed.

Part 2...what's the fastest speed the agent can run where the rabbit can
still escape ?

The fastest speed...not relative to the given speed...that the agent can
run...with the rabbit still escaping...is...the smallest possible speed
greater than zero ?

 

[Dave]

Dave wrote:

Using your strategy, what is the answer to part 2 of the puzzle?

KBH wrote:

Part 2...what's the fastest speed the agent can run where the rabbit can
still escape ?

Oh...yeah thanks for that.

Well...I have the rabbit escape at an azimuth of 259.25649 when the agentis
at an azimuth of 535. And so the rabbit got out at a azimuth of 619.25649
relative to the agent .

So the agent got around 1.48611 loops but needed to get around 1.72016 loops
to catch the rabbit with this systematic method...

Okay...I have the radius at 10 so the arc distance around the circle is
62.832 . Divide by a agent rate of 4 and that's 15.708 time units for the
agent to go 360 degrees around the circle. So the agent went 535 degrees in
23.3438 time units. But the agent needs to go 619.25649 degrees (or 1.72016
loops) in 23.3438 time units. Okay...1.72016 loops * 62.832 circle arc
distance = a distance of 108.0811. Rate * Time = Distance so Rate =108.0811
/ 23.3438 or a rate of 4.6300 . Tie goes to the runner...

But now does the rabbit make the same path with this systematic method ?

Since the agent is moving around the circle faster...the rabbit is making
directional changes at a quicker pace even as the rabbit speed stays the
same. Well the rabbit speed is slower relative to the agent at 21.5983%
verus the previous 25%...so I have to change the rabbit distance (per degree
of agent arc distance) in the KBH Code to 0.037696233 and run the code for
the answer.

And now the rabbit gets out at an agent azimuth of 786 degrees or 2.18333
loops...but the rabbit gets out at about the same spot as before.

So I'll just conclude that the rabbit always gets out no matter what the
agent speed ? Or try a slower agent speed ?

I change the rabbit distance (per degree of agent arc distance) to
0.049570229 and the rabbit gets out at an agent azimuth of 468 degrees. And
beginning here the exit coordinates are different than the previous two....

Okay change the rabbit distance (per degree of agent arc distance) to 0.06
and the rabbit gets out at an agent azimuth of 238 degrees.

Okay change the rabbit distance (per degree of agent arc distance) to 0.07
and the rabbit gets out at an agent azimuth of 171 degrees.

Change the rabbit distance (per degree of agent arc distance) to 0.10 and
the rabbit gets out at an agent azimuth of 108 degrees.

Change the rabbit distance (per degree of agent arc distance) to 0.20 and
the rabbit gets out at an agent azimuth of 51 degrees.

Change the rabbit distance (per degree of agent arc distance) to 0.30 and
the rabbit gets out at an agent azimuth of 34 degrees.

Change the rabbit distance (per degree of agent arc distance) to 0.40 and
the rabbit gets out at an agent azimuth of 26 degrees.

Change the rabbit distance (per degree of agent arc distance) to 0.50 and
the rabbit gets out at an agent azimuth of 21 degrees.

Change the rabbit distance (per degree of agent arc distance) to 0.60 and
the rabbit gets out at an agent azimuth of 17 degrees.

Change the rabbit distance (per degree of agent arc distance) to 0.70 and
the rabbit gets out at an agent azimuth of 15 degrees.

Change the rabbit distance (per degree of agent arc distance) to 1.00 and
the rabbit gets out at an agent azimuth of 11 degrees or less.

Change the rabbit distance (per degree of agent arc distance) to 10.00 and
the rabbit gets out at an agent azimuth of 1 degrees. So each time the agent
moves 0.174533333 the rabbit moves 10.00 ! So here the rabbit speed is
57.2956 times the agent speed instead of the given rabbit speed of 0.25
times the agent speed.

Part 2...what's the fastest speed the agent can run where the rabbit can
still escape ?

The fastest speed...not relative to the given speed...that the agent can
run...with the rabbit still escaping...is...the smallest possible speed
greater than zero ?

It looks like you have had the rabbit running faster and faster, but
you should be increasing the ratio of speed of the agent to the speed
of the rabbit, until the rabbit can't escape.

Now, if the agent can run 10 times as fast as the rabbit can swim, he
can position himself at the closest point on the shore to the rabbit.
The rabbit then will swim away from the shore. Thus the rabbit cannot
escape from the pond. We've already established that the rabbit can
escape when the agent runs 4 times as fast as the rabbit can swim.
What number between 4 and 10 is the fastest the agent can run and
still let the rabbit escape? We are assuming that the rabbit follows
your strategy and the agent follows his best strategy.

Dave

 

[KBH]

Using your strategy, what is the answer to part 2 of the puzzle?

KBH wrote:

Part 2...what's the fastest speed the agent can run where the rabbit can
still escape ?

Oh...yeah thanks for that.

Well...I have the rabbit escape at an azimuth of 259.25649 when the agent
is at an azimuth of 535. And so the rabbit got out at a azimuth of
619.25649 relative to the agent .

So the agent got around 1.48611 loops but needed to get around 1.72016
loops to catch the rabbit with this systematic method...

Okay...I have the radius at 10 so the arc distance around the circle is
62.832 . Divide by a agent rate of 4 and that's 15.708 time units for the
agent to go 360 degrees around the circle. So the agent went 535 degrees
in 23.3438 time units. But the agent needs to go 619.25649 degrees (or
1.72016 loops) in 23.3438 time units. Okay...1.72016 loops * 62.832 circle
arc distance = a distance of 108.0811. Rate * Time = Distance so Rate =
108.0811 / 23.3438 or a rate of 4.6300 . Tie goes to the runner...

But now does the rabbit make the same path with this systematic method ?

Since the agent is moving around the circle faster...the rabbit is making
directional changes at a quicker pace even as the rabbit speed stays the
same. Well the rabbit speed is slower relative to the agent at 21.5983%
verus the previous 25%...so I have to change the rabbit distance (per
degree of agent arc distance) in the KBH Code to 0.037696233 and run the
code for the answer.

And now the rabbit gets out at an agent azimuth of 786 degrees or 2.18333
loops...but the rabbit gets out at about the same spot as before.

Oh...at an agent rate of 4 the rabbit gets out at an azimuth of 619.25 with
the agent at an azimuth of 535 . But relate this as to one loop with a
rabbit azimuth of 259.25 and an agent azimuth of 175.

And at an agent rate of 4.63 the rabbit gets out at an azimuth (related as
to one loop) of 259.25 with the agent azimuth of 393 (related as going in to
the second loop).

So there might be some agent rate between 4 and 4.63 where the agent run and
the rabbit run catch...with this systematic method ?

So I'll just conclude that the rabbit always gets out no matter what the
agent speed ? Or try a slower agent speed ?

I change the rabbit distance (per degree of agent arc distance) to
0.049570229 and the rabbit gets out at an agent azimuth of 468 degrees.
And beginning here the exit coordinates are different than the previous
two...

Okay change the rabbit distance (per degree of agent arc distance) to 0.06
and the rabbit gets out at an agent azimuth of 238 degrees.

Okay change the rabbit distance (per degree of agent arc distance) to 0.07
and the rabbit gets out at an agent azimuth of 171 degrees.

Change the rabbit distance (per degree of agent arc distance) to 0.10 and
the rabbit gets out at an agent azimuth of 108 degrees.

Change the rabbit distance (per degree of agent arc distance) to 0.20 and
the rabbit gets out at an agent azimuth of 51 degrees.

Change the rabbit distance (per degree of agent arc distance) to 0.30 and
the rabbit gets out at an agent azimuth of 34 degrees.

Change the rabbit distance (per degree of agent arc distance) to 0.40 and
the rabbit gets out at an agent azimuth of 26 degrees.

Change the rabbit distance (per degree of agent arc distance) to 0.50 and
the rabbit gets out at an agent azimuth of 21 degrees.

Change the rabbit distance (per degree of agent arc distance) to 0.60 and
the rabbit gets out at an agent azimuth of 17 degrees.

Change the rabbit distance (per degree of agent arc distance) to 0.70 and
the rabbit gets out at an agent azimuth of 15 degrees.

Change the rabbit distance (per degree of agent arc distance) to 1.00 and
the rabbit gets out at an agent azimuth of 11 degrees or less.

Change the rabbit distance (per degree of agent arc distance) to 10.00 and
the rabbit gets out at an agent azimuth of 1 degrees. So each time the
agent moves 0.174533333 the rabbit moves 10.00 ! So here the rabbit speed
is 57.2956 times the agent speed instead of the given rabbit speed of 0.25
times the agent speed.

Part 2...what's the fastest speed the agent can run where the rabbit can
still escape ?

The fastest speed...not relative to the given speed...that the agent can
run...with the rabbit still escaping...is...the smallest possible speed
greater than zero ?

Okay...that's the slowest speed that the agent can run with a rabbit escape.
The fastest speed might be unlimited unless a note that I added above works
out...

 

[KBH]

Well I know the output of the systematic program...in the first case from
the plot...so I will go back to ultimate values as I expect them.

At an agent rate of 4 the rabbit gets out at an azimuth of 619.25 with the
agent at an azimuth of 535 ...

And at an agent rate of 4.63 the rabbit gets out at an azimuth of about
619.25 with the agent azimuth of 786...

So there might be some agent rate between 4 and 4.63 where the agent run and
the rabbit run catch...

 

[[email protected]]

yes, this straight line may not be a radius, but it is unlikely to
differ greatly from a radius.

If I were the rabbit, I'd sit in the center for a while to
consider the situation. Or maybe I'd run out near the edge,
taunt the Secret Service agent, then take a nap.

To discuss what the rabbit "will do" one needs to constrain
the problem further. I would add the following objective
function F (which the rabbit seeks to maximize and the agent
to minimize):

Let
a = 1 if the rabbit escapes, 0 otherwise
b = the time it takes the rabbit to reach the edge
c = the angle separating the rabbit from the agent when
the rabbit reaches the edge

Then let F = a x^2 - b x + c, where "x = infinity" (or
F = (a,-b,c) with dictionary ordering).

With these constraints, we can say, for example, that
when the agent's speed is less than pi times the rabbit's,
the rabbit will make a beeline for the point opposite the
agent's initial position.

When the agent/rabbit speed ratio is the critical value
given by Dave, I think that the family of optimal solutions
has the following form. Both agent and rabbit move at maximal
speed the entire time. The agent reverses direction arbitrarily
until the rabbit leaves the critical disc, and heads toward the
rabbit's exit point thereafter. The rabbit increases his
distance from the center while remaining opposite to the agent
until he leaves the critical disc, then travels along a tangent
to the disc thereafter.

 

[Dave]

KBH Code:

{The rabbit at the center of a 10 radius circular pond and the chasing agent
is at the edge of the pond.}

Var
ads, nag, eag, rba, 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;
rbn:= 10;
rbe:= 10;
dgr:= 1;

While (dgr < 361) Do  {Change value for desired output range}
 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));
 rba:= (rba / Pi) * 180;
 WriteLn(rbn:15:4, rbe:15:4, dgr:15:4, dst:15:4);
 dgr:= dgr + 1;
 flg:= 0;
 End;

ReadLn;
end.

And the above KBH code runs in Delphi Console Mode...

Now the rabbit gets out at North and East coordinates of 8.1332 , 0.1612
while the agent is at an azimuth of 535 degrees on the edge of the circular
pond and on his second trip around the pond. Of course this is a systematic
method only...

And here is the KBH Scratch Plot file for the rabbit path:

663&137@F+wtN0E0
41&759@S+wtN20E20
0
352&448@P+bkN10E10
353&...@L+rdN9.9564E9.9992
365&...@L+rdN9.5661E9.9588
378&...@L+rdN9.1452E9.8458
390&...@L+rdN8.7477E9.6669
402&...@L+rdN8.3826E9.4288
412&...@L+rdN8.0572E9.1387
421&...@L+rdN7.7774E8.8044
428&...@L+rdN7.5477E8.4338
433&...@L+rdN7.3716E8.0349
437&...@L+rdN7.2515E7.6158
439&...@L+rdN7.1886E7.1844
439&...@L+rdN7.1834E6.7484
437&...@L+rdN7.2357E6.3155
434&...@L+rdN7.3441E5.8932
429&...@L+rdN7.5067E5.4886
422&...@L+rdN7.7209E5.1089
414&...@L+rdN7.9834E4.7607
405&...@L+rdN8.2899E4.4506
394&...@L+rdN8.6357E4.1851
382&...@L+rdN9.0151E3.9703
369&...@L+rdN9.4215E3.8124
356&...@L+rdN9.8469E3.7173
343&...@L+rdN10.2819E3.6911
329&...@L+rdN10.7150E3.7398
316&...@L+rdN11.1311E3.8687
305&...@L+rdN11.5107E4.0819
295&...@L+rdN11.8285E4.3792
289&...@L+rdN12.00538E4.7515
285&...@L+rdN12.1525E5.1743
286&...@L+rdN12.1075E5.6063
292&...@L+rdN11.9231E5.9996
301&...@L+rdN11.6270E6.318
312&...@L+rdN11.2554E6.5445
325&...@L+rdN10.8406E6.6769
339&...@L+rdN10.4072E6.7212
352&...@L+rdN9.9729E6.6856
365&...@L+rdN9.5502E6.5793
378&...@L+rdN9.1482E6.4106
390&...@L+rdN8.7736E6.1875
400&...@L+rdN8.4315E5.9171
410&...@L+rdN8.1258E5.6061
418&...@L+rdN7.8594E5.2609
425&...@L+rdN7.6345E4.8873
431&...@L+rdN7.4526E4.4909
435&...@L+rdN7.3148E4.0771
438&...@L+rdN7.2216E3.6511
439&...@L+rdN7.1730E3.2176
440&...@L+rdN7.1687E2.7815
438&...@L+rdN7.2079E2.3471
436&...@L+rdN7.2895E1.9186
432&...@L+rdN7.4120E1.5
427&...@L+rdN7.5736E1.0949
421&...@L+rdN7.7720E.7064
414&...@L+rdN8.0048E0.3376
410&...@L+rdN8.1332E0.1612
352&448@P+bkN10E10
41&...@P+blN19.9985E10.1745
352&448@+blN10E10
45&...@P+blN19.8481E11.7365
352&448@+blN10E10
59&...@P+blN19.3969E13.4202
352&448@+blN10E10
82&...@P+blN18.6603E15.0000
352&448@+blN10E10
113&...@P+blN17.6604E16.4279
352&448@+blN10E10
152&...@P+blN16.4279E17.6604
352&448@+blN10E10
196&...@P+blN15.0000E18.6603
352&448@+blN10E10
245&...@P+blN13.4202E19.3969
352&448@+blN10E10
297&...@P+blN11.7365E19.8481
352&448@+blN10E10
352&...@P+blN10.0000E20.0000
352&448@+blN10E10
406&...@P+blN8.2635E19.8481
352&448@+blN10E10
458&...@P+blN6.5798E19.3969
352&448@+blN10E10
507&...@P+blN5.0000E18.6603
352&448@+blN10E10
551&...@P+blN3.5721E17.6604
352&448@+blN10E10
590&...@P+blN2.3396E16.4279
352&448@+blN10E10
621&...@P+blN1.3397E15.0000
352&448@+blN10E10
644&...@P+blN0.6031E13.4202
352&448@+blN10E10
658&...@P+blN0.1519E11.7365
352&448@+blN10E10
663&...@P+blN0.0000E10.0000
352&448@+blN10E10
658&...@P+blN0.1519E8.2635
352&448@+blN10E10
644&...@P+blN0.6031E6.5798
352&448@+blN10E10
621&...@P+blN1.3397E5.0000
352&448@+blN10E10
590&...@P+blN2.3396E3.5721
352&448@+blN10E10
551&...@P+blN3.5721E2.3396
352&448@+blN10E10
507&...@P+blN5.0000E1.3397
352&448@+blN10E10
458&...@P+blN6.5798E0.6031
352&448@+blN10E10
406&...@P+blN8.2635E0.1519
352&448@+blN10E10
352&...@P+blN10.0000E0.0000
352&448@+blN10E10
297&...@P+blN11.7365E0.1519
352&448@+blN10E10
245&...@P+blN13.4202E0.6031
352&448@+blN10E10
196&...@P+blN15.0000E1.3397
352&448@+blN10E10
152&...@P+blN16.4279E2.3396
352&448@+blN10E10
113&...@P+blN17.6604E3.5721
352&448@+blN10E10
82&...@P+blN18.6603E5.0000
352&448@+blN10E10
59&...@P+blN19.3969E6.5798
352&448@+blN10E10
45&...@P+blN19.8481E8.2635
352&448@+blN10E10
51&448@Agent->+blN19.65E10
352&463@Rabbit Begins Here+rdN10E10.5
352&448@P+bkN10E10
661&...@P+blN0.0381E10.8716
599&...@L+blN2.0381E10.8716
597&475@Agent Stops Here on 2nd Loop+blN2.1E10.88

Copy to a text file, save the text file as "Name.plt" with quotation marks,
and the file will run in KBH Scratch Plot...

And the rabbit path is a small-cap handwritten-style "e"...

I made a small spreadsheet to use your rabbit's strategy. Correct me
if I am wrong, but isn't it true that after the agent has gone about
295 degrees around the pond, he is approximately on the radius
containing the rabbit? If, at that point, he keeps running at full
speed, as in your program, he will be running away from the rabbit.
This is hardly an optimal strategy for the agent. In fact, once the
agent gets on the rabbit's radius, he should stop. The rabbit will
swim back through the center and the cycle will repeat. By your
strategy, the rabbit never will reach the shore even at a speed ratio
of only 4, assuming that the agent uses his best strategy.

Dave

 

[KBH]

Dave:

I made a small spreadsheet to use your rabbit's strategy. Correct me
if I am wrong, but isn't it true that after the agent has gone about
295 degrees around the pond, he is approximately on the radius
containing the rabbit? If, at that point, he keeps running at full
speed, as in your program, he will be running away from the rabbit.
This is hardly an optimal strategy for the agent. In fact, once the
agent gets on the rabbit's radius, he should stop. The rabbit will
swim back through the center and the cycle will repeat. By your
strategy, the rabbit never will reach the shore even at a speed ratio
of only 4, assuming that the agent uses his best strategy.


KBH wrote:

It's a systematic method only...

But we now know what happens if the rabbit takes one straight line to shore
and we know what happens if the agent only loops around the circle in one
direction...with the rabbit correcting direction with each movement of the
agent. In one case the rabbit has the wrong strategy and in the other case
the agent has the wrong strategy.

Really I was interested in the rabbit path plot with the agent looping
around the circle in one direction...and with the rabbit correcting
direction with each movement of the agent.

 

[KBH]

Dave wrote:

I made a small spreadsheet to use your rabbit's strategy. Correct me
if I am wrong, but isn't it true that after the agent has gone about
295 degrees around the pond, he is approximately on the radius
containing the rabbit? If, at that point, he keeps running at full
speed, as in your program, he will be running away from the rabbit.
This is hardly an optimal strategy for the agent. In fact, once the
agent gets on the rabbit's radius, he should stop. The rabbit will
swim back through the center and the cycle will repeat. By your
strategy, the rabbit never will reach the shore even at a speed ratio
of only 4, assuming that the agent uses his best strategy.


KBH wrote:

It's a systematic method only...

But we now know what happens if the rabbit takes one straight line to
shore and we know what happens if the agent only loops around the circle
in one direction...with the rabbit correcting direction with each movement
of the agent. In one case the rabbit has the wrong strategy and in the
other case the agent has the wrong strategy.

Really I was interested in the rabbit path plot with the agent looping
around the circle in one direction...and with the rabbit correcting
direction with each movement of the agent.

So the rabbit...and it should be a duck...is never caught if he keeps
turning away from the agent with each movement of the agent. Then assuming a
strategy of reversing direction by the agent...the rabbit can only get out
by picking some point where he breaks for shore. But the agent can't be sure
of where the rabbit curves are going to carry and the rabbit can't be sure
of his break point. It's merely possible for the rabbit to get out...

 

[Dave]

Dave:

I made a small spreadsheet to use your rabbit's strategy. Correct me
if I am wrong, but isn't it true that after the agent has gone about
295 degrees around the pond, he is approximately on the radius
containing the rabbit? If, at that point, he keeps running at full
speed, as in your program, he will be running away from the rabbit.
This is hardly an optimal strategy for the agent. In fact, once the
agent gets on the rabbit's radius, he should stop. The rabbit will
swim back through the center and the cycle will repeat. By your
strategy, the rabbit never will reach the shore even at a speed ratio
of only 4, assuming that the agent uses his best strategy.

KBH wrote:

It's a systematic method only...

But we now know what happens if the rabbit takes one straight line to shore
and we know what happens if the agent only loops around the circle in one
direction...with the rabbit correcting direction with each movement of the
agent. In one case the rabbit has the wrong strategy and in the other case
the agent has the wrong strategy.

In the second case, both the agent and the rabbit have the wrong
strategy.

Dave

 

[Dave]

The rabbit can always move to the center of the pond.

From the center, the rabbit can move away from the agent and stay on the
same diameter as the agent at least until the rabbit's angular velocity
about the pond center equals the agents angular velocity about the pond
center at full speed, i,e, to 1/4 of a radius from the center in the
direction opposite to the agent but on the same diameter.

The rabbit is now 3/4 of a radius from the nearest point on the
circumference and the agent must travel pi times that radius along the
edge of the pond to get to that same point.

This has been stated numerous times. Now, we're fighting down those
who have suboptimal strategies for either the rabbit or the agent,
especially when determining the answer to the second part of the
puzzle, the fastest speed of the agent that still allows the rabbit to
escape.

Dave

 

[KBH]

The rabbit can always move to the center of the pond.

From the center, the rabbit can move away from the agent and stay on the
same diameter as the agent at least until the rabbit's angular velocity
about the pond center equals the agents angular velocity about the pond
center at full speed, i,e, to 1/4 of a radius from the center in the
direction opposite to the agent but on the same diameter.

The rabbit is now 3/4 of a radius from the nearest point on the
circumference and the agent must travel pi times that radius along the
edge of the pond to get to that same point.

KBH wrote:

From the center of the pond the rabbit has to work away from the center
while the agent is moving. So I suppose the rabbit rotates with the moving
agent...from a postion on the far side of the radius point...while spiraling
outward. It's intricate...

Well...consider the graphics plot that I made. As the agent moves the rabbit
turns away from the agent and the rabbit carries the rabbit's distance of
movement. Now from any agent point on the circle a line can be scaled from
the agent point to a tangent point on the rabbit path and that by knowing
that the rabbit path direction...and the fact that the rabbit is moving away
from the agent. And from that strategies can be developed...

Then consider the computer code that I wrote for the same situation. The
code can be modified so that the agent systematically runs clockwise to a
point but then changes to counterclockwise movement. I don't have time to
work with it but that's interesting. For instance the agent can run
clockwise for 360 degrees and then run counterclockwise 90 degrees or so to
run the rabbit back toward the center.

 

[KBH]

Dave:

I made a small spreadsheet to use your rabbit's strategy. Correct me
if I am wrong, but isn't it true that after the agent has gone about
295 degrees around the pond, he is approximately on the radius
containing the rabbit? If, at that point, he keeps running at full
speed, as in your program, he will be running away from the rabbit.
This is hardly an optimal strategy for the agent. In fact, once the
agent gets on the rabbit's radius, he should stop. The rabbit will
swim back through the center and the cycle will repeat. By your
strategy, the rabbit never will reach the shore even at a speed ratio
of only 4, assuming that the agent uses his best strategy.

KBH wrote:

It's a systematic method only...

But we now know what happens if the rabbit takes one straight line to
shore
and we know what happens if the agent only loops around the circle in one
direction...with the rabbit correcting direction with each movement of the
agent. In one case the rabbit has the wrong strategy and in the other case
the agent has the wrong strategy.

Dave wrote:

In the second case, both the agent and the rabbit have the wrong
strategy.

KBH wrote:

No...in the second case if the agent is dumb enough to just systematically
run clockwise around the circle...then the rabbit's strategy in response is
good enough.

 

[KBH]

KBH wrote:

From the center of the pond the rabbit has to work away from the center
while the agent is moving. So I suppose the rabbit rotates with the moving
agent...from a postion on the far side of the radius point...while
spiraling outward. It's intricate...

Well...the rabbit must spiral out from the radius point as he works to get
on the same radial rotation with the agent but also position on the opposite
side of the radius point. If the rabbit is rotating too fast then he would
just keep working outward until the rotation is correct. But if the rotation
is correct but the positon is wrong then the rabbit would have to zig-zag to
work into position. When in correct position on the opposite side of the
radius point and on the same radial rotation with the agent...then break for
the opposite shore. But break for shore hoping that it will work because the
rabbit is not calculating...

 

[KBH]

Unless the agent is continually switching directions, the rabbit never
need zig nor zag until it is ready to run for the edge.

KBH wrote:

Well the speeds are fixed and the rabbit needs a particular postion on its
circle...as it hunts for that circle which rotates with the agent movement.

If the speeds are correct the rabbit has .75 radii to cover but the
agent has pi radii ( half a circumference) to cover, and the ratio of
those distances is less that the ratio of speeds, 1 to 4 ( or about
0.2387 ) so the rabbit has a bit of a cushion.

Relative distance ratio of .75/pi is roughly 5/21 to the relative speeds
5/20, so the rabbit has a bit of a cushion.

KBH wrote:

The rabbit...and it's really a duck...can't calculate anything. The rabbit
does not know how far to be on the other side of the radius. The rabbit only
knows if it is rotating with a radial to the agent and if it is on the other
side of the radius...but when so then break for it.

 

[KBH]

Unnecessarily complicated. All the rabbit has to do is stay on the
diameter thru the agent's position until it is 3/4 of a diameter from
the point opposite the agent, then run straight for that point.

KBH wrote:

The rabbit does not know anything except whether or not it is rotating with
a radial to the agent...and if it is on the other side of the radius.

 

[William Hughes]

KBH wrote:

The rabbit does not know anything except whether or not it is rotating with
a radial to the agent...and if it is on the other side of the radius.

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).


- William Hughes