Infinite balls

[bonanova]

I read this Zeno-like paradox [some call them super tasks] some years ago. I was first convinced of answer [1] until someone argued convincingly for answer [2]. Then I changed the conditions innocuously [I think] to destroy answer [2]. Now I'm wondering if the paradox can even be stated, meaningfully. One can distinguish between super tasks like the man who moves half the remaining distance in half the remaining time, which is understandable and possible, and those like Thompson's Lamp that switches on and off in a convergent, geometrically decreasing time sequence. Is it on or off, finally? The latter question requires a nonexistent highest integer to provide an answer. But the following task doesn't seem to be of that type. I'll describe it, then ask:

1. Is answer [2] correct?
2. Does the question itself require an impossible assumption?

Bin A contains Aleph-null ping-pong balls painted with the counting numbers.
Bin B is empty
Bin C is empty.

After 1 tick, two balls, those numbered 1 and 2, are moved from Bin A to Bin B, then the lowest numbered ball in Bin B is moved to Bin C. Then,
After 1/2 tick, two balls, numbered 3 and 4 are moved from A to B, then the lowest numbered ball in B is moved to C. Then.
After 1/4 tick, 5 and 6 move from A to B and B's lowest ball goes to C. Then,
After 1/8 tick ... rinse and repeat.

After 2 ticks, describe the contents of Bin B. [If we object that ping-pong balls might at some point have to break the speed of light limit, the balls can be regarded as mass-less integers, and the bins can be thought of as sets.]

Answer [1] B contains an infinite number of balls. Proof: Each step adds two balls and subtracts one ball.
Answer [2] B is empty. Proof: Every ball enters and leaves the bin.

What convinced me that [2] must be right was the question: If B contains even one ball, what is its number? Any number you give has a well-defined tick value at which it transferred to Bin C. But let's make an innocuous change. Instead of moving the lowest numbered ball from B to C, move the highest numbered ball from B to C. Now I can say B is not empty. Proof: ball # 1 entered B and stayed there.

Can the answer depend on the shape of paint marks?
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[bonanova]

Not to answer my own puzzles, but it seems that all of the integers enter set B and "half of them" leave. Since the even integers have the same cardinality as the integers, it's somewhat undetermined whether you remove them all. If you make the 1-1 correspondence between the integers and the integers, [moving the lowest integer from B to C aleph-null times] then you've moved them all. If you make the 1-1 correspondence between the integers and the evens [moving the highest integer from B to C aleph-null times] then you've left all the odd ones behind in B.

OK, I'm cool with that, logically and mathematically. But I'm an engineer, and want the question to have an answer - I still want the difference in procedure to be physically innocuous. Do I have to give that up? Does physics have to stay finite? Are Banach and Tarski lurking in the room somewhere? Uhh nah, they messed around with aleph-one.
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[L'lanmal]

Alice the Friendly Chowder Server and Bob the Chowder Nazi both have difficulties dealing with the lunch crowd. Every minute, 2 people arrive, and each Alice and Bob can only serve one.

Alice tFCS solves this problem by having her customers form a line, first-come, first-served.

Bob tCN instead picks one of the two to serve, and refuses to serve the other.

Does Alice tFCS deny chowder to anyone who waits long enough in line? No. But each may have to wait a long time, and the later you come, the longer you will have to wait.
Does Bob tCN deny chowder to anyone who waits long enough in line? Yes. Some of them.

The difference here depends not on the people's clothing, skin color, or markings (unless Bob is a bigot as well as a, erm, "nazi"), merely on the algorithms they have chosen for chowder distribution.

I hope the mappings are clear. n minutes from the start of lunch maps to 2-1/(2^n) ticks. Bin A contains all people outside who will enter the store. Bin B contains people who entered the store, but have not yet received chowder. Bin C contains people who received chowder. Alice is the queueing (lowest number) method. Bob is the higher number method. Ball number is order in line for Alice. Ball number is the preference function for Bob (where he serves the higher, not it is arbitrary which is 1 and which is 2, etc.).

So I agree with [2].

Some of the intuitive confusion comes from the implication from ending at time "2" that the process is finite, and the length of the "line" converges to 0, or even converges at all. It does not. The "line" only gets longer as more people come, but each person who comes can still get served in finite time. In this case, time "2' is a label essentially meaning "eventually", and has little mechanical metric meaning. You would have great physical difficulty designing a machine that could move two balls (or even, say, electrons) in an arbitrarily short time (such as 1/(2^n) for arbitrarily large n).

[Antrax]

What L'l said I think. Assume to the contrary there's a ball left in B. But by its index you know exactly when it moved to C and it's in the interval [1, 2). So B has to be empty.
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[bgg1996]

The question is, indeed, impossible. The ticks would never reach two. Using the same logic as answer two, name the fraction that would eventually make it to exactly two. You can't, because there isn't one.


The argument against this is that the ticks could be a seperate occurence, and just a way of measuring the events that are taking place. The problem is, that means that the balls in the bins are moving at an infinite speed.


P.S. Wasn't there already a GL puzzle on this exact subject. Something about an imp shoving marbles under rugs.

[Antrax]

Urns of Infinity is similar: http://www.greylabyrinth.com/puzzle/puzzle091
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After years of disappointment with get rich quick schemes, I know I'm gonna get rich with this scheme. And quick!

[JohnP]

Answer [3] The question is nonsense.

"I protest against the use of infinite magnitude as something completed" - C. F. Gauss

[bonanova]

[L'lanmal]

[bgg1996]

The question is what balls would be in B when the machine is finished finished, right?
It would never finish, as it has an infinite amount of work to do.

The question is impossible, even when you replace the machine with integers, because it's the same problem, with the same answer. Since none of the variables that changed the eventual answer, then the answer is the same.

[Thok*]

[bgg1996]

There is no answer to the sequence described, because the sequence will never finish. The eventual outcome of the sequence will not be evident until it is complete. Since the sequence never runs its full course, an answer can never be found.

Consider the similiar sequence:
1-1+1-1+1-1+1-1+1-1+1-1+1-1+1-1+1-1+1-1.........
What is the final sum?

One answer is 0, [1-1]+[1-1]=0
Another answer is 1, [1-1]+[1-1]+1=1

Because the sequence continues indefinitely (never ends), there will never be a true answer.

[Thok*]

[bgg1996]

If the values did not fluctuate (0-0+0-0+...), then a logical assumption could be made about the eventual answer, yes.

The only way a logical assumption about an answer can be made with an infinite series, however, is if the change either never occurs (0-0+0-0+...), or if the change eventually becomes zero, or extremely close to zero (1+1/2+1/4+1/8+...). That is why we can assume that the eventual end of the ball movement is two.

This problem, however, is more like: a + b - a + c + d - b + e + f - c + g + h - d...
There is always a change in the value, so an answer cannot be deduced.

[Thok*]

[quote="bgg1996"This problem, however, is more like: a + b - a + c + d - b + e + f - c + g + h - d...
There is always a change in the value, so an answer cannot be deduced.[/quote]

Again, when you start talking about change in value, you are really talking about a metric structure and measuring distances between sets. And if you use the second metric structure I provided (because a priori, there isn't a default notion of distance between sets), you get a well defined notion of convergence of the sets.

The situation you would be talking about is more like "Take two sets A and B". At tick whatever, switch all the balls from A to B or from B back to A. And in that situation, there's no set that anything converges to.

The difference between the two situations can be made a bit more obvious if you assume that ball i requires work 1/2^i (the balls are getting progressively smaller or something as the numbers go up). Then in the original problem, only finite work is done and only a finite amount of work is done to any individual ball; the amount of work done to a given ball goes down as the number increases. In the switch back and forth problem, infinite work is being done to every single ball because every ball is moved infinitely many times.

[bonanova]

[bgg1996]

[Thok*]

bgg1996, go back to your favorite calculus book, and figure out why 1-1/2+1/3-1/4+1/5-... is ln(2) rather than not defined. Once you can do that, then you can see what's wrong with your argument.

[L'lanmal]