Duck and Fox ... and other Great Puzzles

The Game Room
21

My argument in previous post was, I think, essentially correct. You procede North out of the “critical point”. Fox can’t trick you … because you’re ignoring which way he goes. This increases your necessary speed by f(epsilon) … but epsilon can be made arbitrarily small.

Although the essential argument remains, the details I posted were muddled.

:confused: :confused:

I should have spoken of a single “critical circle” with radius 1, with a move to 1+epsilon allowed when Fox’s speed is less than 4.60334.

I’m at an age where my math skills are declining. :frowning: This puzzle and it’s solution were clearer to me a few decades ago! :smiley: :smack:

22

The trick is that when the fox doubles back, the duck does not go straight back east; he goes a slight bit south, in such a way that the two end up on the opposite side of the center of the pond at a point slightly farther away than the critical circle, at which point we’re back at the beginning. (This axis will not be precisely north-south, so the duck’s optimal direction will be slightly north of west or slightly south of east). If the fox continues to double back, they will continue to end up on the opposite sides of the center-point, but always just a little bit further away from the center.

23

Ah, but is that actually possible? That’s the key question. And for this method to truly be simple, we must have a simple way for the duck to determine the proper direction to achieve that.

24

Oh, sure, if we’re looking at real-world constraints a fox running is going to be able to turn around more quickly than a duck swimming, but on the other hand what the hell kind of duck can’t take off from the water?

The straight-west-to-shore thing is a bit less arbitrary than it looks if you consider the path the duck & fox would take to get to the critical circle if neither of them doubled back - the fox constantly circling around to try to get on the same side of the pond as the duck, the duck spiraling out away from the center; at the point where the duck can no longer keep the center-point between itself & the fox, it just keeps going in the same direction it was going before.

25

I’m assuming that all direction changes are instantaneous. That’s not the question. I’m asking if the duck is fast enough to be able to get to a point opposite the fox.

26

Pretty much by definition. The two start out with the duck at the point south of center. The fox goes east; the duck goes west. Since the fox at this point is capable of covering a larger arc-segment than the duck in a given period of time, the line connecting the two drifts toward the east. When the fox reverses itself, even if the duck sat still, there still would be a moment where the line connecting the two would go through the center point of the pond.

27

There’s no doubt that the duck can move infinitesimally outside of the ‘safe’ circle[sup]*[/sup] and stay opposite the fox. The problem is once the fox has changed direction and is now on the same side as the duck’s heading. If the duck is outside of the safe circle, then it should be fairly obvious that the duck cannot now reach a point opposite the fox without heading toward the circle’s center.

That the duck may be able to defeat Buridan’s fox, or use psychological tricks to possibly escape a faster fox, only implies a plausible escape. I think fairly you need to show a guaranteed escape for the duck.

Consider a magical fox that can predict the future. With fox speed <= pi+1, then he can foresee the duck escaping, no matter what he does. Meeting that standard is not necessary, but it would be sufficient for an improved speed minimum.

(I should also point out that I mistakenly left in the first line of my ‘proof’ post, which is worded improperly. I don’t consider it a complete proof of the speed - it incompletely shows that the duck needs to cross a critical threshold to escape, which is a fairly weak condition. I’m open to possible alternative routes.)

[sup]*[/sup] I’m using ‘safe’ only to be consistent with my earlier use of ‘critical’. I’d consider the critical circle to be one of radius 1- ang[sub]f[/sub]/f, which is the one the duck can escape from if the fox’s position is ang[sub]f[/sub].

28

:cool: I’m glad Dopers find this good puzzle fun and challenging.

:mad: Of course I’m shocked my epsilon-based proof outline was ignored or disbelieved.

A few simple lemmas lead to the R=4.603 solution. (A further trick is needed to prove Duck can do no better.)

[del]a bit[/del] much less arbitrary…

29

All it appears to prove is that the duck can (marginally) move out of a circle of radius 1 (1/f using a unit circle).

The main problem I see is that at one point you say that epsilon gives the duck ‘margin for correction’. I’m not even sure what that means (considering that the duck has already moved the distance epsilon from the center), but regardless if you make epsilon arbitrarily small, that removes whatever correctional margin there was.

I think it seems possible that there is a duck strategy that may defeat whatever fox strategy there is, but I doubt that it describes a simple path. I can’t quite quantify it, but I don’t see why the fox is forced to commit to one direction over another.

I’d also like to see more details on the duck’s route. What exact direction does the duck travel when ‘switching’? Always tangent to the circle it is on?

30

In the following assume that Duck has already exited the “critical circle.”

Let t denote the angle (in degrees) formed by Duck-Center-Fox. Wlog, t <= 180.

If t = 180, Duck heads away from Fox, i.e. straight towards the shore.

If t < 180, Fox can do no better than to move so as to decrease t. He’ll never want to reverse course. Duck heads in a straight line for any point on the shore that ensures success. Below we demonstrate there will always be such a point.

The feint or zigzag situation arises only when t is within some epsilon of 180. Then Duck simply heads straight towards the shore, conceding a negligible (f(epsilon)) speed advantage.

In the critical position, t = 180, the Duck is R+1 distant from the Fox, and he does have a target point on the shore which gives him success.

It should be clear that if t = 180 and the Duck is more than R+1 distant from the Fox, he has successful target points. (For example, he can just proceed anti-parallel to the Fox’s initial chosen direction as before.)

Now, if t < 180 and Fox reverses direction (i.e. seeks a new t > 180), he will have to pass through t = 180. At that point, the Duck-Fox distance will exceed R+1 and Duck wins as we’ve seen.

The objection to all this is, presumably, that Fox can force some zigzagging in which the Duck’s progress will be some sum of infinitesimals which sum only to an infinitesimal. But Fox can force such zigzagging only by remaining near t = 180 and in that case Duck simply travels at maximum speed toward the shore!

31

I think I see how this works, but why is the ‘critical’ condition required? Shouldn’t the answer just be 3pi/2 (~4.7)?

As I see it, the duck heads directly away from the fox out of the center. Once the fox starts around the pond, the duck heads off on the tangent. The fox sees this, says, “I can’t reach him if I continue in this direction; I’ll have to turn around.” Fox turns around, and she sees that the duck has now switched directions. And thus turns around again, while the duck manages to extend its distance from the center, right?

32

In the correct solution, at the critical point when Fox decides to go West and Duck reacts by going East, Fox will not reverse course: he’d then be getting even farther (instantaneously) from the Duck. This holds due to a property of the tangent to the critical circle.

Or, another way of looking at it, doing what you describe Duck and Fox will do a sort of “zigzag” dance which will result in Duck reaching the critical circle opposite the Fox.

33

I did realize that the 3*pi/2 doesn’t work because the fox can continue toward the duck in the direction it’s going.

The way I’d put it:
In order for the duck to make progress out of the center, it must at least be moving tangent to the circle it’s currently on. The time for the duck to travel to the edge in this case is:

Td = sqrt(1-Dr^2)

where Dr is the duck’s current radius from the center (unit circle and speed=1 for the duck).

If the fox starts opposite the duck (distance pi around the circle), then to get to the duck “the long way around” it needs to move pi + ang[sub]d[/sub], where ang[sub]d[/sub] is the angle formed by the duck, circle center, and the duck’s target point. The triangle formed by these points has a hypotenuse of 1, and one leg = Dr.

The time for the fox to travel this far is

Tf = (pi + arccos(Dr/1))/f

where f is again the speed of the duck.

Setting these times equal gives a minimum fox speed:

fmin = (pi + arccos(Dr)/sqrt(1-Dr^2)

Plotting this on [0,1] gave me the most insight. It has a minimum point (call it fc) in this region. That’s the first location the duck could escape using this plan. This is because on (0,fc) the function is decreasing, meaning the fox does not need to change direction if it has a higher speed. Past this point, the function increases, which implies the duck actually no longer needs to target the tangent to its circle once it’s further out.

To ensure that the duck can be opposite the fox, we must find where 1/f intersects this function. If it’s > fc then the duck could escape when the fox is going faster. I suspect it’s not a mere coincidence that 1/f intersects exactly at that point (the expression for fmin looks suspiciously like the answer to a differential equation as well). The answer being 4.603, of course.

Now, this still doesn’t prove that duck can’t escape if the fox is going faster, as it’s only good for when we know the duck can reach a point opposite the fox.
One other note: There is no requirement that the Fox must move in any particular direction. Which means that it’s just as accurate to say the “correct solution” involves the duck moving off the safe circle in a straight line away from the fox, flipping infinitely fast back and forth (at least until it reaches 1-pi/f). It’s a little unsatisfying, to me.