Friday Puzzler -- Fox and Duck

Re: Friday Puzzler -- Fox and Duck

bill earl

That would have been a good puzzle for yesterday.

Re: Friday Puzzler -- Fox and Duck

Alex Y

I thought it would have been a good one for today, but we were both wrong.

Re: Friday Puzzler -- Fox and Duck

Bill Earl

Hmmm. Haven't got a solid solution yet. But it looks like it might be a better question for next month.

Re: Friday Puzzler -- Fox and Duck

Larry Barrett

Assuming the duck is in the center of the pond, he heads to the shore 180* opposite to where the fox is. The duck must travel a distance or r, while the fox must travel r(pi), so if the fox travels at a rate of (pi) x the duck's rate the duck can not escape.

However, the duck can improve his odds by constantly changing his direction relative to the fox to maintain so that he is always heading 180* away. So the duck will swim in a spiral, slowly getting closer to the shore. The fox must run faster and faster since the distance the duck is from the shore is gradually approaching zero.

Seems to me that the duck will eventually escape. I am probably missing something.

Re: Friday Puzzler -- Fox and Duck

Jim Rutten

I will be happy to put the obvious answer out there because in my linear world the duck only swims in one direction (towards the shore opposite where the fox is) and the fox only runs around in half circles. In my (like I said very linear world) the fox would half to run over 3.1415926535897932384626433832795028841971693993751058209749445923078164062862089986280348253421170679821480865132823066470938446095505822317253594081284811174502841027019385211055596446229489549303819644288109756659334461284756482337867831652712019091456485669234603486104543266482133936072602491412737245870066063155881748815209209628292540917153643678925903600113305305488204665213841469519415116094330572703657595919530921861173819326117931051185480744623799627495673518857527248912279381830119491298336733624406566430860213949463952247371907021798609437027705392171762931767523846748184676694051320005681271452635608277857713427577896091736371787214684409012249534301465495853710507922796892589235420199561121290219608640344181598136297747713099605187072113499999983729780499510597317328160963185950244594553469083026425223082533446850352619311881710100031378387528865875332083814206171776691473035982534904287554687311595628638823537875937519577818577805321712268066130019278766111959092164201989 as fast as the duck swims to catch the duck.

Hint

Alex Y

The answer is somewhere between Jim's and Larry's.

Now wasn't that helpful?

Your ducks must be way smarter

Henry Higginbotham

. . . than the ones around here.

Re: Friday Puzzler -- Fox and Duck

Alex Y

However, the duck can improve his odds by constantly changing his direction relative to the fox to maintain so that he is always heading 180* away. So the duck will swim in a spiral, slowly getting closer to the shore.

That's a good strategy for a while. But if the fox can run around the pond in 5 minutes, where do you thin he will be when the duck is just 5 minutes paddling time from shore? Probably not on the opposite shore.

Re: Friday Puzzler -- Fox and Duck

Larry Barrett

Depends on how smart the fox is, but you do seem to have a point. More thought needed.

First two tries

Alex Y

The silence is deafening on this one, so let me provide a start.

Duck swims at velocity 1, pond radius 1, and fox runs at f.

The duck swims to the center, and notes where the fox is.

First attempt: Duck swims to the shore opposite the fox. However, since the fox's speed is greater than pi, the duck sees that the fox will get there before her. So back to the center to try again.

Second attempt: Duck swims toward the shore opposite the fox, but watches over her shoulder to see which way the fox goes. When the fox starts moving, she starts veering from her straight line, keeping the center of the pond directly between her and the fox. For a while, she can match the fox's angular velocity, with enough speed left over to progress toward the shore. But soon she reaches a distance 1/f from the center, and it takes all her swimming speed to stay 180 degrees away from the fox. At this point, she breaks for shore directly opposite the fox. As she is about to reach the shore at time 1-1/f after breaking for shore, she sees that the fox was running faster than pi+1, so he will have reached the opposite shore before her.

So, our exhausted little duck turns around and swims back to the center of the pond, awaiting your guidance on how she might escape.

I will tell you that I am looking for a strategy and numerical approximation, not an exact answer. If you get the right equation, I don't think you will be able to find the root (at least I couldn't) other than by numerical approximation methods. (back-solving in excel works well).

Re: First two tries

Larry Barrett

The first part of attempt 2 seems to be the best approach, up to the point where the duck is 1/f away from center and the fox is 180* opposite. The strategy beyond that is not obvious. As a guess, I think that the duck should watch which way the fox runs, then, instead of heading directly for the shore, swim toward the shore on an opposite heading. For example, suppose the fox starts at 360*, runs CCW, and is at 270* when the duck is at 1/f from the center. Then, if the fox continues CCW, the duck swims at 0* toward the shore. Shortly, the fox sees that a shorter path to the duck would be to reverse course; then duck also reverses course, thus swimming in a zig-zag path.

Even if this is generally correct, I don't see an equation in sight. Maybe another hint is in order.

I think you are there

Alex Y

except for the zig-zag.

We are in agreement on getting the duck to 1/f from center, 180* away from the fox. As you said, if the fox at 270* headed CCW, then the duck sitting at 9o* will head due north.

Now it may occur to our fox that it is better to run a little more than 1/4 way around the circle to the ~5* to 10* position that the duck is swimming toward rather than the long way. But here is where he has been outfoxed by our wily duck.

When the duck was at the 1/f position, she initially broke directly toward the east, and gave the fox a very small amount of time to commit to his direction before the duck made her 90-degree turn. Then, if the fox reverses direction, the duck immediately starts swimming toward the nearest shore until such time as the fox is 180* opposite her. At this point, they are in the same relative position as before, except that the duck is closer to shore. The fox figures this is not a pattern that will satisfy his appetite, so he realizes he needs to pick one direction and stick to it. Obviously, if we make the head start the duck gives the fox arbitrarily small (epsilon-delta, anyone?)we can just solve for an equation assuming the fox goes counterclockwise and the duck goes due north.

What remains, then, is to determine the f that gets the fox and duck to that point at the same time, and to determine WHY this is the duck's best strategy.

As a hint to the latter, once outside of the 1/f circle, whatever point they end up at is most quickly reached by the duck in a straight line, assuming that straight line doesn't put her back inside the 1/f circle.

Second hint: consider what their paths look like over the tiny portion of time at the end, when the fox can be considered to be approximately running along a tangent to the circle.

Re: I think you are there

Larry Barrett

Continuing with the same set-up (fox at 270*, duck at 1/f from center on a heading of 90*, then the duck makes a left turn to head north and the fox continues CCW), the first question is at what speed must the fox run to just meet the duck when the duck reaches the shore?

The fox must run almost 3/4 of the way around the pond. Almost, because the angle A from the pond center to where the duck meets the shore must be accounted for. This is determined as a function of f by (90* - inverse cosine of 1/f). The fox must therefore run a distance of (270 - A)/360 x 2 x pi around the pond, at a speed of f. The duck must swim from the 1/f point to the shore, a distance determined by sqrt(1 - (1/f)^2) at a speed of 1.

For example, if f = pi, then this angle is about 18*, the fox must run .7 of the circumference of the circle and it will take him 1.4 units of time. The duck swims a distance of .94 and it will take him .94 units of time. The duck easily escapes for this example.

Using a spreadsheet for various values of f, Time D for the Duck to get to shore, and Time F for the Fox to arrive at the same point:

f........D.........F

4.0.....968....1.11

4.2.....971....1.06

4.4.....974....1.02

4.6.....976......977

4.8.....978......938

So the best the duck can do to escape is to make the fox run at a speed of 4.6 times the duck's speed.

To address the second question, is this the best strategy for the duck?, we have already seen that if the duck heads directly for the shore, the fox can catch it by running at a speed of pi + 1 = 4.14, so by heading north the duck does much better. At the point where the fox has nearly reached the duck, the fox is moving nearly horizontally toward the duck; thus, without doing more calculations, it would seem that the duck can not improve on its strategy by aiming even more to the left, since the fox is moving faster in that direction.

Correct! *LINK*

Alex Y

Surprising, isn't it, that the fox will have to run a little over 4.6 times as fast to be able to catch the duck.

If you have a masochistic streak, or particularly like trig, here is a solution including why it is optimal:


http://www.mathrec.org/old/2003jul/solutions.html