## Infinite Balls and Jugs [solution]

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dedalus
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### Re: Infinite Balls and Jugs [solution]

phlip wrote:
Octavian Starr wrote:i'm sorry for not being rigorous, but i simply pointed out the fact that just because any ball put in the jar will be removed eventually(which was the original argument) that does not mean that there are no balls in the jar in the end.

If there are any balls in the jar at the end, then there must be at least one example of a ball in the jug at the end. Name one.

The ball one greater then the last ball to be removed.
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### Re: Infinite Balls and Jugs [solution]

dedalus wrote:The ball one greater then the last ball to be removed.
That's clearly not well-defined, as there doesn't exist a "last ball to be removed" (the natural numbers do not contain a maximum).

Try again?

Code: Select all

enum ಠ_ಠ {°□°╰=1, °Д°╰, ಠ益ಠ╰};void ┻━┻︵​╰(ಠ_ಠ ⚠) {exit((int)⚠);}
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d0nk3y_k0n9
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### Re: Infinite Balls and Jugs [solution]

When did it become a requirement for something to exist that we be able to give it a specific name?

phlip
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### Re: Infinite Balls and Jugs [solution]

When we defined the puzzle so that we have a well-defined universe set, every element of which has a name?

Code: Select all

enum ಠ_ಠ {°□°╰=1, °Д°╰, ಠ益ಠ╰};void ┻━┻︵​╰(ಠ_ಠ ⚠) {exit((int)⚠);}
[he/him/his]

Goldstein
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### Re: Infinite Balls and Jugs [solution]

Moreover, if we assume "The ball one greater than the last ball to be removed" exists, you still have to show that it entered the jug before midnight if you're going to claim that it's in the jug at midnight. If you tell me when it entered, I'll tell you when it was removed. It's a contradiction.
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Malle
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### Re: Infinite Balls and Jugs [solution]

I have thought about this a bit, but before I go forth posting anything structured I want to ask some questions, to see if I can find any obvious flaw in my reasoning.

If it is justified to ask which ball was added but not later removed, why is it not justified to ask which ball was the one to be removed to leave the urn empty?

Also, consider the case where there is one and only one unlabeled ball in an urn at the start and at each time step we add, to the ball, a label that is numbered according to the time step, such that at time step n we add a label marked with the number n. What would be the end state if we a) remove any previous labels when we add the n'th label and b) if we simply cover the previous label with the n'th label such that the previous labels cannot be read? More specifically, Would the urn be there? If so, would the ball be in the urn? If so, would the ball have any labels on it? If so, how many labels would it have on it, how many could you read and what would those labels be marked.
Last edited by Malle on Tue Nov 17, 2009 8:30 pm UTC, edited 1 time in total.

jestingrabbit
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### Re: Infinite Balls and Jugs [solution]

Malle wrote:I have thought about this a bit, but before I go forth posting anything structured I want to ask some questions, to see if I can find any obvious flaw in my reasoning.

If it is justified to ask which ball was added but not later removed, why is it not justified to ask which ball was the one to be removed to leave the urn empty?

Its fair to ask, but it presupposes that there exists such a ball. The proposition that I and "the empty urn" proponents are advancing is that there is no such ball. Its exactly the same as asking what the largest whole number is. There is no last ball removed as there is no largest whole number.

Malle wrote:Also, consider the case where there is one and only one unlabeled ball in an urn at the start and at each time step we add a label that is numbered according to the time step, such that at time step n we add a label marked with the number n. What would be the end state if we a) remove any previous labels when we add the n'th label and b) if we simply cover the previous label with the n'th label such that the previous labels cannot be read? More specifically, Would the urn be there? If so, would the ball be in the urn? If so, would the ball have any labels on it? If so, how many labels would it have on it, how many could you read and what would those labels be marked.

To determine the disposition of this system I must ask whether the label tampering of the ball is done inside the urn or outside it. That is, does the ball move in and out of the urn during each step?
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Bubbles McCoy
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### Re: Infinite Balls and Jugs [solution]

Stricken.
Last edited by Bubbles McCoy on Tue Jan 25, 2011 8:57 am UTC, edited 1 time in total.

Malle
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### Re: Infinite Balls and Jugs [solution]

jestingrabbit wrote:
Malle wrote:Also, consider the case where there is one and only one unlabeled ball in an urn at the start and at each time step we add a label that is numbered according to the time step, such that at time step n we add a label marked with the number n. What would be the end state if we a) remove any previous labels when we add the n'th label and b) if we simply cover the previous label with the n'th label such that the previous labels cannot be read? More specifically, Would the urn be there? If so, would the ball be in the urn? If so, would the ball have any labels on it? If so, how many labels would it have on it, how many could you read and what would those labels be marked.

To determine the disposition of this system I must ask whether the label tampering of the ball is done inside the urn or outside it. That is, does the ball move in and out of the urn during each step?
The ball is never removed; it is only interacted with by adding the labels to it. The labels on the other hand are removed at each time step in scenario A while they are not removed in scenario B. I realize now that I never actually explicitly stated that the labels were added to the ball, so I'll go back and edit that.

Bubbles McCoy wrote:
jestingrabbit wrote:Its fair to ask, but it presupposes that there exists such a ball. The proposition that I and "the empty urn" proponents are advancing is that there is no such ball. Its exactly the same as asking what the largest whole number is. There is no last ball removed as there is no largest whole number.

I believe that's what Malle was trying to demonstrate, phlip was demanding the kind of evidence he couldn't give either.
I didn't so much ask it to prove anything, as to see the reasoning behind it.

jestingrabbit wrote:
Malle wrote:If it is justified to ask which ball was added but not later removed, why is it not justified to ask which ball was the one to be removed to leave the urn empty?

Its fair to ask, but it presupposes that there exists such a ball. The proposition that I and "the empty urn" proponents are advancing is that there is no such ball. Its exactly the same as asking what the largest whole number is. There is no last ball removed as there is no largest whole number.
But does that then not prove a contradiction in the statement that the urn is empty? If there is no ball that is not removed yet we never (in the time that we act) take a step that empties the urn, should the state of the urn not be undefined?

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### Re: Infinite Balls and Jugs [solution]

Malle wrote:But does that then not prove a contradiction in the statement that the urn is empty? If there is no ball that is not removed yet we never (in the time that we act) take a step that empties the urn, should the state of the urn not be undefined?

That's called Zeno's paradox. People have known how to resolve it since the time of Aristotle.
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Malle
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### Re: Infinite Balls and Jugs [solution]

Qaanol wrote:
Malle wrote:But does that then not prove a contradiction in the statement that the urn is empty? If there is no ball that is not removed yet we never (in the time that we act) take a step that empties the urn, should the state of the urn not be undefined?

That's called Zeno's paradox. People have known how to resolve it since the time of Aristotle.
Yes, that comparison seems valid. I can't believe I didn't see that.

Yakk
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### Re: Infinite Balls and Jugs [solution]

You have a supply of numbered balls.

At T_i, you place balls (10i+1, ..., 10i+9) into the jug. You then, in the jug, renumber the lowest numbered ball to 10i+10.

At midnight, there are no balls in the jug. No balls have been removed from the jug, and an infinite number of balls have been added to the jug.

This clearly, to me, indicates that whatever undefined axioms I used above to evaluate this "what happens at midnight" are bad axioms, as they don't do what we want them to do. Thus we should find better axioms for this kind of problem.

Axioms are tools to understand problems, not chains.

So the next step I'd take would be to formalize the problem (so all I'm doing to solve the problem is moving around symbols), see if there is anything crappy about the fromalization (ie, it should give me "undefined" in bad cases, instead of nonsense), and see if I can build an interesting formlization that does what I want.

As a side benefit, the axiom system that leads to possibly ridiculous results might be interesting to see if we can find an application for it.

Imagine if the balls are bills, balls removed from the jug are paid bills, midnight represents defaulting, and the exponentially decreasing interval between time periods represents a faster and faster (or larger and larger) business cycle. And then the problem becomes a description of how a company that is losing money might survive indefinitely (paying/dealing with the oldest bill), go under certainly (paying the most recent bill), or probably survive (paying a random bill). It is sketchy -- but under this model, you'll see that the (implicit) Axiom used by the empty-jar folk seem to make slightly more sense.

Hmm. I'm trying that inversion -- changing time-decrease to quantity-increase -- and am having problems.
One of the painful things about our time is that those who feel certainty are stupid, and those with any imagination and understanding are filled with doubt and indecision - BR

Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.

mike-l
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### Re: Infinite Balls and Jugs [solution]

Yakk wrote:You have a supply of numbered balls.

At T_i, you place balls (10i+1, ..., 10i+9) into the jug. You then, in the jug, renumber the lowest numbered ball to 10i+10.

At midnight, there are no balls in the jug. No balls have been removed from the jug, and an infinite number of balls have been added to the jug.

Why on earth would that be? At midnight in your situation, there ought to be infinitely many balls with undefined labels. If you change it slightly so that you start with ball 1, then at midnight there will be 1 ball for each natural number not terminating in 0, but each will bear that natural number followed by infinitely many 0s. (Or, be undefined, depending on how you are doing the labeling, but if you just append 0s as the modification, this is what you get). This solution fits perfectly with the theory that gives an empty jug in the original case - I look at each ball and sees what happens to it.
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Yakk
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### Re: Infinite Balls and Jugs [solution]

mike-l wrote:
Yakk wrote:You have a supply of numbered balls.

At T_i, you place balls (10i+1, ..., 10i+9) into the jug. You then, in the jug, renumber the lowest numbered ball to 10i+10.

At midnight, there are no balls in the jug. No balls have been removed from the jug, and an infinite number of balls have been added to the jug.

Why on earth would that be? At midnight in your situation, there ought to be infinitely many balls with undefined labels. If you change it slightly so that you start with ball 1, then at midnight there will be 1 ball for each natural number not terminating in 0, but each will bear that natural number followed by infinitely many 0s. (Or, be undefined, depending on how you are doing the labeling, but if you just append 0s as the modification, this is what you get). This solution fits perfectly with the theory that gives an empty jug in the original case - I look at each ball and sees what happens to it.

I was presuming reasonable (yet unspecified) axioms. To be explicit, if not formal:
Axiom of Inclusion:
A ball with label L is said to be in the jar if, for all n from N, for all a > n the ball is in the jar after T_a.

Axiom of Exclusion:
A ball with label L is said not to be in the jar if there exists an n from N such that for all a > n, no ball with the label L is in the jar after T_a.

Natural Axiom:
All balls in the jar have natural number axioms.

Axiom of Excluded Middle:
A ball is in the jar, or not in the jar, at each time, and at midnight.

Axiom of Count by Cardinality
The number of balls in the jar is the cardinality of the set of balls in the jar.

(And yes, more axioms are needed to finish the job).

These are all reasonable axioms -- and they lead directly to "yes, if you remove the lowest numbered ball while adding 10 higher numbered ones, you end up with zero balls in the jar at midnight". And similarly removing the highest numbered jar results in an infinite number of balls in the jar at midnight.

They are not, however, the only axioms you can choose for this problem. And your choice of axioms is both important, and not pre determined for this kind of problem. We pick reasonable axioms, and if there is a problem we examine why the problem is generated -- and see if we can find better axioms that are consistent, interesting, and don't generate the problem!
One of the painful things about our time is that those who feel certainty are stupid, and those with any imagination and understanding are filled with doubt and indecision - BR

Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.

mike-l
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### Re: Infinite Balls and Jugs [solution]

Yakk wrote:
Natural Axiom:
All balls in the jar have natural number axioms.

Axiom of Excluded Middle:
A ball is in the jar, or not in the jar, at each time, and at midnight.

Neither of these are reasonable. The second presupposes the limit exists, and the first has no reason to hold if you change the labels infinitely often (or rather, presupposes that such an occurrence has a limit existing and finite)
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### Re: Infinite Balls and Jugs [solution]

mike-l wrote:
Yakk wrote:
Natural Axiom:
All balls in the jar have natural number axioms.

Axiom of Excluded Middle:
A ball is in the jar, or not in the jar, at each time, and at midnight.

Neither of these are reasonable. The second presupposes the limit exists, and the first has no reason to hold if you change the labels infinitely often (or rather, presupposes that such an occurrence has a limit existing and finite)

Sorry, all balls in the jar have natural number labels (mistype).

And yes, I'm willing to accept that the answer to a given problem is "problem is inconsistent". If the axioms describing a given problem are inconsistent with my axioms, we just have an inconsistent system. That isn't a problem: it is information.

And sure, you can build a system of axioms in which labelled balls don't always have labels at midnight. I'm not saying my axiom system is the only reasonable one; I'm saying it is a reasonable one, and we can derive reasonable results from it. Throwing together an alternative one is perfectly ok as well!

The axiom of excluded middle would be stated formally in the form InJar( n, time ) | ~InJar( n, time ) for all n from N, with restrictions on what time is. You'll note that it makes no claim about you being able to determine if it is in the jar -- it just says that if you can prove that something is both in the jar and not, then you have shown the problem to be inconsistent. And, similarly, if you have proven that a ball is not in the jar, you have proven not (the ball is in the jar), and the converse. It is useful in a number of ways.
One of the painful things about our time is that those who feel certainty are stupid, and those with any imagination and understanding are filled with doubt and indecision - BR

Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.

joz
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### Re: Infinite Balls and Jugs [solution]

Sorry for bumping this, especially with my first post, but it seemed unresolved, and I believe I have resolved it.

Let us reform the problem mathematically, so we can do some analysis. People have done this with infinite sums, but others seem unsure about reordering, and the fact that the summation formulae don't distinguish the balls from each other. So I will present it as an infinite product, at each step we multiply by the next 10 primes in the list of all primes, and divide by the lowest prime left in the product, essentially the same as putting the next 10 numbered balls in a bucket and removing the lowest numbered ball. This is given by

$\prod_{n=1}^\infty \frac{p_{10n-9}\cdot p_{10n-8}\cdot\ldots\cdot p_{10n}}{p_n}$

where p1,p2,... is the ordered list of prime numbers.

There are 2 different intuitive arguments, which have been presented in the thread, translated into this new mathematical problem.
Argument 1: Eventually we multiply by every prime number, and then divide by it, so the final product is 1 (analogous to "all balls eventually removed" )
Argument 2: Each term in the sequence is greater than 2 and we multiply them together. How the hell can that give us an answer of 1! Obviously the answer is infinity. (analogous to "we keep on adding 9 balls each time, ffs, we arent taking anything away!! How is the answer 0!")

It turns out we can prove the second one using very basic analysis, thus disprove the first argument (which was kind of silly anyway). Notice if we try to determine the prime factorization of this "number" it tells us that there are no prime factors, yet every term in the sequence of partial products is a natural number. This still makes sense, since infinity doesnt need to have a prime factorisation. This is analogous to trying to determine which numbered ball remains.

The probability arguments were fundamentally flawed. It was another 0*infinity thing, e.g.
Fix m in N, and randomly, and independently choose n, in N. The probability m=n is zero, since there are an infinite number of natural numbers to choose from. => for all n in N, P(m=n)=0
=> p(m in N) = Πn=1 0 = 0
But we chose m in N, so p(m in N)=1

mike-l
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### Re: Infinite Balls and Jugs [solution]

joz wrote:Sorry for bumping this, especially with my first post, but it seemed unresolved, and I believe I have resolved it.

Let us reform the problem mathematically, so we can do some analysis. People have done this with infinite sums, but others seem unsure about reordering, and the fact that the summation formulae don't distinguish the balls from each other. So I will present it as an infinite product, at each step we multiply by the next 10 primes in the list of all primes, and divide by the lowest prime left in the product, essentially the same as putting the next 10 numbered balls in a bucket and removing the lowest numbered ball. This is given by

$\prod_{n=1}^\infty \frac{p_{10n-9}\cdot p_{10n-8}\cdot\ldots\cdot p_{10n}}{p_n}$

where p1,p2,... is the ordered list of prime numbers.

There are 2 different intuitive arguments, which have been presented in the thread, translated into this new mathematical problem.
Argument 1: Eventually we multiply by every prime number, and then divide by it, so the final product is 1 (analogous to "all balls eventually removed" )
Argument 2: Each term in the sequence is greater than 2 and we multiply them together. How the hell can that give us an answer of 1! Obviously the answer is infinity. (analogous to "we keep on adding 9 balls each time, ffs, we arent taking anything away!! How is the answer 0!")

It turns out we can prove the second one using very basic analysis, thus disprove the first argument (which was kind of silly anyway). Notice if we try to determine the prime factorization of this "number" it tells us that there are no prime factors, yet every term in the sequence of partial products is a natural number. This still makes sense, since infinity doesnt need to have a prime factorisation. This is analogous to trying to determine which numbered ball remains.

No, you've made the same mistake that numerous people in the thread have made. It's not true that the limit of your product is equal to the product of the limit. Which is exactly why this is an interesting problem. The intuition is 'I add balls every time, there must be some in there', but that is wrong. You're right that your product is infinity, but at that point it's not counting things, it's a limit of counting things, which is not counting the limit.

The probability arguments were fundamentally flawed. It was another 0*infinity thing, e.g.
Fix m in N, and randomly, and independently choose n, in N. The probability m=n is zero, since there are an infinite number of natural numbers to choose from. => for all n in N, P(m=n)=0
=> p(m in N) = Πn=1 0 = 0
But we chose m in N, so p(m in N)=1

All you've demonstrated here is that there isn't a uniform probability measure on the naturals. When you say 'pick a random number' you have to tell me 'how' to pick a random number. So when you say
The probability m=n is zero, since there are an infinite number of natural numbers to choose from
, you are suggesting that all of them should have equal probability, which you can't do, and you got a contradiction accordingly.
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joz
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### Re: Infinite Balls and Jugs [solution]

yes, i realise that this argument is completely wrong. The equivalent statement in terms of my product is "how many prime factors does infinity have", which you cant really answer. My probability lecturer also proved the lack of a uniform probability measure over the natural numbers this week *embarrassed*.

Anyway, I'd like to reform the problem into something simpler: at the first step, ball 1 is added, at the nth step, ball (n-1) is taken out, and ball n is added. Alternatively, ball (n-1) is relabelled as ball n.

I claim that, after midnight, both jars with contain a single ball labelled aleph-0

We can consider the label as tally marks. In the second case we have a single ball, and after every iteration, we add a tally mark to the ball, so after countable infinite iterations, the ball has countable infinite tally marks i.e. a countable infinite number, aleph 0.
In the second case, consider this: we prepare every ball that we are going to put in the jar, and line them up. In the first iteration, we put 1 mark on every ball, and put the first one in. In the nth iteration, we add a mark to each ball, and put the next one in. Since there is a countably infinite number of iterations, we eventually add a ball with a countably infinite number of marks on it, i.e. aleph 0.
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Qaanol
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### Re: Infinite Balls and Jugs [solution]

joz wrote:yes, i realise that this argument is completely wrong. The equivalent statement in terms of my product is "how many prime factors does infinity have", which you cant really answer. My probability lecturer also proved the lack of a uniform probability measure over the natural numbers this week *embarrassed*.

Anyway, I'd like to reform the problem into something simpler: at the first step, ball 1 is added, at the nth step, ball (n-1) is taken out, and ball n is added. Alternatively, ball (n-1) is relabelled as ball n.

I claim that, after midnight, both jars with contain a single ball labelled aleph-0

We can consider the label as tally marks. In the second case we have a single ball, and after every iteration, we add a tally mark to the ball, so after countable infinite iterations, the ball has countable infinite tally marks i.e. a countable infinite number, aleph 0.
In the second case, consider this: we prepare every ball that we are going to put in the jar, and line them up. In the first iteration, we put 1 mark on every ball, and put the first one in. In the nth iteration, we add a mark to each ball, and put the next one in. Since there is a countably infinite number of iterations, we eventually add a ball with a countably infinite number of marks on it, i.e. aleph 0.
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I claim you did not actually read this thread. Go back and read it from beginning to end. You should find several posts explicitly detailing the fallacies in your reasoning. Also several posts with intuitive explanations. I’m partial to this one.

If you look at your own post critically, you should be able to pick out exactly where your modifications to the problem statement change it enough to produce a different result.

Qaanol wrote:If you are the only person performing the steps (adding and removing balls from the jug) is your hand inside or outside the jug at midnight?

If infinitely many people are involved so person n only adds and removes balls at step n, where is each person's hand at midnight (inside or outside the jug)?

I’m still waiting for someone to respond to this.

I’ll even throw in a third scenario: you add and remove balls at the first time step. After removing the balls, you cease to exist, and a new person with identical particle structure—identical body, memories, etc.—comes into being. That person adds and removes the balls at the next time step, and the same thing happens. This continues until midnight. Is there a person at midnight?

Qaanol wrote:
skeptical scientist wrote:But they are not analogous, because in the lamp case, the switch itself is performing a supertask, and there is no way to deconstruct the setup into individual components which are not performing supertasks. In the balls+jug situation, however, while the entire system performs a supertask, the individual balls do not, and so a component-wise analysis is possible and completely explains the scenario. Apart from the superficial similarity that in both cases you have a lamp turning on and off an infinite number of times, the two setups are very different, because the mechanism controlling the lamp comes to a limit in the scenario with the balls, but not in the scenario with the switch.

What if we make them analogous?

We have a switch, and we toggle it at 11:50, 11:55, 11:57:30, etc.

The switch has two outputs. One goes to the room on the right and changes the state of Thompson's Lamp. The other goes to the room on the left and performs "Task a: place the next ten sequential balls into the urn, and remove the lowest-numbered ball from the urn."

These rooms are completely isolated. No information comes out of them: not from one room to the other, nor from either room to the switch room. Information only travels from the switch into the two rooms.

We are interested in the state of the switch at midnight if:

i) The machines in both rooms are functioning properly.
ii) Thompson's Lamp is disconnected from the switch, but the ball-and-urn device functions properly.
iii) The ball-and-urn device is disconnected, but Thompson's lamp works.
iv) Neither machine does anything.

And of course in each case, we at the switch do not and cannot know which (if either) machine or machines are actually connected to the switch.

And I’m still waiting for someone to respond to this.
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jestingrabbit
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### Re: Infinite Balls and Jugs [solution]

Qaanol wrote:
Qaanol wrote:If you are the only person performing the steps (adding and removing balls from the jug) is your hand inside or outside the jug at midnight?

If infinitely many people are involved so person n only adds and removes balls at step n, where is each person's hand at midnight (inside or outside the jug)?

I’m still waiting for someone to respond to this.

I’ll even throw in a third scenario: you add and remove balls at the first time step. After removing the balls, you cease to exist, and a new person with identical particle structure—identical body, memories, etc.—comes into being. That person adds and removes the balls at the next time step, and the same thing happens. This continues until midnight. Is there a person at midnight?

If you are the only person performing the steps (adding and removing balls from the jug) is your hand inside or outside the jug at midnight?
There is no sensible answer to this question imo.

If infinitely many people are involved so person n only adds and removes balls at step n, where is each person's hand at midnight (inside or outside the jug)?
They are all outside the jug.

Is there a person at midnight?
Well, you've added in the complexity of physics here, and I feel less confident in that realm, but would still say that there are no people at midnight.

The method that I use for answering all these questions is to simply ask "does it move in and out of a particular state an infinite number of times?" If so, I can't say which of those states it ends in, and consider that an answer cannot reasonably be arrived at. If it makes moves only a finite number of times, then we can determine where it is. This is the most sensible criteria that I can imagine, and it corresponds to the "limit of sets" idea that was brought up. To answer your final questions, I note that the switch has moved an infinite number of times regardless of what else is going on, so its position is indeterminate by any means that I can imagine.
ameretrifle wrote:Magic space feudalism is therefore a viable idea.

joz
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### Re: Infinite Balls and Jugs [solution]

I dont see which modifications make my first problem different. I add ball n, and take out ball (n-1). In the original, you add balls 10n +1 to 10n +10, or something, then take out ball n. Surely I can use the same reasoning as in my problem to state that it has remaining balls. There is a countably infinite number of balls taken out of the jar, everyone seems to agree on that, but we can use the same reasoning to show that there are balls that have a countably infinite "number" on them, and people seem to disagree.
Last edited by joz on Thu Feb 03, 2011 4:32 pm UTC, edited 1 time in total.

jestingrabbit
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### Re: Infinite Balls and Jugs [solution]

joz wrote:I dont see which modifications make my first problem different. I add ball n, and take out ball (n-1). In the original, you add balls 10n +1 to 10n +10, or something, then take out ball n. Surely I can use the same reasoning as in my problem to state that it has remaining balls.

The general consensus in the thread, which I thought was very nearly completely resolved more than a year ago, at least as far as I could be bothered with it, is that the standard jug has no balls remaining in the jug, and I would certainly contend that that is still the case in the "add n, remove (n-1)" version.

If there is only one ball, and you are putting tally marks on it, that is a very different question. The later examples put forward by ATCG are quite instructive regarding the subtleties.

joz wrote:There is a countably infinite number of balls taken out of the jar, everyone seems to agree on that, but we can use the same reasoning to show that there are balls that have a countably infinite "number" on them, and people seem to disagree.

I also disagree with you here. The only balls that are added to the jug have one and only one of the usual natural numbers written on them, using, I assume, the standard base 10 representation. How did this other ball come to be in the jug? At what instant was it added and not subsequently removed?

If you instead want to talk about balls that you put tally marks on, I suggest that there is indeed a ball with an infinity of tally marks on it, but that you cannot say whether it is inside or outside of the jug.

In your first post you claimed to have read the entire thread, and Qaanol expressed some doubt regarding this. I also find it hard to believe that you have read the thread. Its quite long, but there is discussion on the later pages that does not occur earlier, I assure you.
ameretrifle wrote:Magic space feudalism is therefore a viable idea.

joz
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### Re: Infinite Balls and Jugs [solution]

Ok I've finally understood it, there arent any balls (sorry for the huge bump, and thanks for not flaming me too hard). I was trying to show that there could be a non-natural number in the pot, but instead I showed that you only need natural number balls in the pot: If we have a countably infinite number of balls in the pot, (and we do since we add them in a countably infinite number of steps). Suppose you remove them all by going through ball 1, ball 2, to infinity, but always in the natural numbers. Then it is possible, as you are doing it, to "tally" each ball in the pot as you take one out. Since you take a countably infinite number of balls out, there are balls in the pot which have an infinite number of tallies on, which you supposedly take out No, there arent any balls left in the pot to tally. OK, too much time wasted. =/

WarDaft
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### Re: Infinite Balls and Jugs [solution]

I'm starting to think this question is simply not properly defined.

Suppose you have an ordinal click counter. It can display any computable ordinal in a straightforward and intuitive manner. The ordinal it displays at any given time is the number of times it has been clicked. You get a brand new one reading 0 right before you start your nightly jar assisted assault on the laws of reality. Every time you add a ball, you click the counter exactly once. After you have clicked it, it will read the same number as the highest numbered ball in the jar. There is no particular click after which the click counter reads ω, but after midnight has come and gone, it has been clicked ω times and must read ω or else it is faulty, if so we will demand our money back and repeat the experiment until we get a working one.

Now the next night, instead of writing on the balls, we have a nice box with infinitely many such (working) click counters to go with our box of infinitely many balls and the jar that can hold them all with room to spare. Now when we place a ball in the jar, we place a click counter beside it in a neat organized fashion, so that there will never be any question as to which ball and which counter go together, but they are not physically attached in any way.

Now, on each step, we remove the lowest counter and associated ball from the jar. We discard only the ball and click the counter up to the range of numbers we are adding in that round. The clicker is then replaced in the jar with the new ball. A counter is never, under any circumstances, placed in the jar without a ball. It cannot enter the jar without a ball accompanying it. No step must ever result in a counter being in the jar without it's ball. After any particular round of removing and adding is complete, every click counter that was in the jar before the round must still in the jar after the round.

After midnight... What on Earth is in the jar?
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jestingrabbit
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### Re: Infinite Balls and Jugs [solution]

The counters are put in the jar and then removed an infinite number of times, so its not well defined whether they are in the jar at the end of the night. The balls are not in the jar.

That said, these problems don't correspond to a physical reality, so its not really something that has an inarguable truth to it, but I do think the "follow individual objects to work out what happens" approach is internally consistent and the most sensible approach.
ameretrifle wrote:Magic space feudalism is therefore a viable idea.

WarDaft
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### Re: Infinite Balls and Jugs [solution]

But after each step, every counter that was in the jar is still in the jar. No step results in a permanent removal of a counter.

If Thompson's Lamp was turned on and then off again as a single explicit step, then infinitely many of those steps should result in it being off. This partitioning is important, because it prevents us from constructing any sequence of steps after which the lamp is on, which the base Thompson's Lamp problem does not. I see no way to explain how the lamp might get stuck on, but quite obvious ways for it to end up off. Without any arguments for why it might be on, and arguments for why it must be off, surely it must end up off must it not?
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Qaanol
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### Re: Infinite Balls and Jugs [solution]

WarDaft wrote:But after each step, every counter that was in the jar is still in the jar not empty. No step results in a permanent removal of a counter the jar being empty.

<snip>

I see no way to explain how the lamp might get stuck on jar might be empty, but quite obvious ways for it to end up off non-empty. Without any arguments for why it might be on empty, and arguments for why it must be off non-empty, surely it must end up off non-empty must it not?

wee free kings

jestingrabbit
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### Re: Infinite Balls and Jugs [solution]

WarDaft wrote:If Thompson's Lamp was turned on and then off again as a single explicit step, then infinitely many of those steps should result in it being off. This partitioning is important, because it prevents us from constructing any sequence of steps after which the lamp is on, which the base Thompson's Lamp problem does not.

I think that if the answer that you arrive at depends on whether you cut time up one way or another, then there is no reasonable way to assign an end state to what is going on.

One of the ideas that is fundamental to relativity is that no observer should be privileged over another. That is a good principle in general I think. If you need to specify one way of observing or describing events to get the answer you want, then your conclusions aren't all that sound imo.
ameretrifle wrote:Magic space feudalism is therefore a viable idea.

WarDaft
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### Re: Infinite Balls and Jugs [solution]

But by including both an on and off switching in the same step, we are describing a subtly different problem from where each step is only a single flip of the switch to it's opposite position. Breaking up these steps would be akin to interrupting someone mid flip in the original problem, thus leaving the state at midnight not only indeterminate between being on and off, but utterly indeterminate even in how close the switch is to being fully on or fully off.

If we presume as one of our axioms that no step is only partially completed, then surely the state of the lamp must be properly defined and in the off position.

Qaanol wrote:
WarDaft wrote:But after each step, every counter that was in the jar is still in the jar not empty. No step results in a permanent removal of a counter the jar being empty.

<snip>

I see no way to explain how the lamp might get stuck on jar might be empty, but quite obvious ways for it to end up off non-empty. Without any arguments for why it might be on empty, and arguments for why it must be off non-empty, surely it must end up off non-empty must it not?

That's a logical fallacy, they are not equivalent until proven so.

Given that we are assured the exit and reentry of the counters in the jar as a single discrete indivisible step, how can any count of wholly completed steps result in a counter that was once in the jar leaving the jar?
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cjdrum
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### Re: Infinite Balls and Jugs [solution]

Well, assuming that each action happens instantaneously:

If you keep halving the time until midnight, you will never reach 0. Therefore, there will be an infinite number of balls inside and outside of the jug at the precise moment of midnight. Is that right?

t1mm01994
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### Re: Infinite Balls and Jugs [solution]

cjdrum wrote:Well, assuming that each action happens instantaneously:

If you keep halving the time until midnight, you will never reach 0. Therefore, there will be an infinite number of balls inside and outside of the jug at the precise moment of midnight. Is that right?

If there are infinitely many balls inside the jug, you'll have no trouble at all naming one that's still in it, right?

Qaanol
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### Re: Infinite Balls and Jugs [solution]

WarDaft wrote:But by including both an on and off switching in the same step, we are describing a subtly different problem from scenario where each step gives a time strictly before midnight at which the state of things is undefined, as some objects are alleged to be both inside and outside the jug at the same time is only a single flip of the switch to it's opposite position. Breaking up these steps would be akin to interrupting someone mid flip in the original problem, thus leaving the state at midnight not only indeterminate on account of a supertask being performed, between being on and off, but utterly indeterminate even in how things were as time approached midnight, since things were not well defined beforehand close the switch is to being fully on or fully off.

If we presume as one of our axioms that no step is only partially completed, then surely the state of the lamp must be properly defined and in the off position. jar must be outside the scope of any classical description that ascribes a single position to an object at a single time.

WarDaft wrote:
Qaanol wrote:
WarDaft wrote:But after each step, every counter that was in the jar is still in the jar not empty. No step results in a permanent removal of a counter the jar being empty.

<snip>

I see no way to explain how the lamp might get stuck on jar might be empty, but quite obvious ways for it to end up off non-empty. Without any arguments for why it might be on empty, and arguments for why it must be off non-empty, surely it must end up off non-empty must it not?

That's a logical fallacy, they are not equivalent until proven so.

As I wrote here,

“No. Absolutely not. This is not how mathematics works. The answer exists independently of any of our abilities to comprehend or prove it. The answer does not change in time based on anything any of us says.”

WarDaft wrote:Given that we are assured the exit and reentry of the counters in the jar as a single discrete indivisible step, how can any count of wholly completed steps result in a counter that was once in the jar leaving makes any sense whatsoever, seeing as the counting itself was done at times where the counter was claimed to be both inside and outside the jar?

wee free kings

WarDaft
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### Re: Infinite Balls and Jugs [solution]

Qaanol wrote:As I wrote here,

“No. Absolutely not. This is not how mathematics works. The answer exists independently of any of our abilities to comprehend or prove it. The answer does not change in time based on anything any of us says.”

You applied your arguments against one problem to another. They have to be equivalent for that to not be a logical fallacy; unless they've already been proven equivalent somewhere in the thread and you can link me that post instead, my point stands.

You're still answering the question as if it's the series (1-1+1-1+1...), but the situation I presented is (1+(1-1)+(1-1)+(1-1)+...).

To put it another way, suppose we have a straight line. __________________

Then, we put a dent in it. ________/\__________

Then, we put another dent in it. ________/\____/\______
And another. ________/\____/\__/\____

I would keep going, but I can't make progressively narrower dents in ASCII. If all of our dents are made before midnight, how could the line look any different after midnight? Or to be more specific, given [imath]f(n,x) = 1-abs(x+\frac{1}{2^n})*2^{n+2}[/imath], and the sequence of functions [imath]g(0,x) = min(1,max(0,-x-1))[/imath] (putting the counter in the jug) [imath]g(n,x) = max(g(n-1,x),f(n,x))[/imath] (taking it out and putting it back in) are you really saying that [imath]\displaystyle\lim_{n→∞}g(n,1)[/imath] is undefined?
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cjdrum
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### Re: Infinite Balls and Jugs [solution]

t1mm01994 wrote:
cjdrum wrote:Well, assuming that each action happens instantaneously:

If you keep halving the time until midnight, you will never reach 0. Therefore, there will be an infinite number of balls inside and outside of the jug at the precise moment of midnight. Is that right?

If there are infinitely many balls inside the jug, you'll have no trouble at all naming one that's still in it, right?

I'm not entirely sure what you're asking... I'll give you all the possible answers.
"Sure, you just keep going up. Have more digits on the balls that need it."
"I read the OP as saying that all the balls are pre-numbered, so trying to name them doesn't need to happen."
"Um, all the balls whose number ends in 0 won't be in the jug, and all others will."

crayzeeepete
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### Re: Infinite Balls and Jugs [solution]

Surely the problem with the question is that they're imposing a finite time limit at the end of an infinite amount of time? The trick is that they've given 10 minutes, but have indicated that the addition/removal of the balls does not take an amount of time. As you iterate through the ball operations, you will never reach midnight.

The question isn't really "at midnight", the question is at t=[imath]\infty[/imath], which is an entirely different question.

Ralp
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### Re: Infinite Balls and Jugs [solution]

crayzeeepete wrote:Surely the problem with the question is that they're imposing a finite time limit at the end of an infinite amount of time? The trick is that they've given 10 minutes, but have indicated that the addition/removal of the balls does not take an amount of time. As you iterate through the ball operations, you will never reach midnight.

The question isn't really "at midnight", the question is at t=[imath]\infty[/imath], which is an entirely different question.
Nope. Though infinite, the sum of 5+2.5+1.25+... is a geometric series that converges to 10.

crayzeeepete
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### Re: Infinite Balls and Jugs [solution]

Ralp wrote:Nope. Though infinite, the sum of 5+2.5+1.25+... is a geometric series that converges to 10.

But, crucially, never equals 10. The moment you round to 10, you have a finite series, and thus it is possible to know how many balls are in the bag.

RonWessels
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### Re: Infinite Balls and Jugs [solution]

crayzeeepete wrote:
Ralp wrote:Nope. Though infinite, the sum of 5+2.5+1.25+... is a geometric series that converges to 10.

But, crucially, never equals 10. The moment you round to 10, you have a finite series, and thus it is possible to know how many balls are in the bag.

Oooh, I'm not sure I'd go that far. A limit is a well-understood concept, and the limit of an infinite series is a reasonable thing to talk about. However, you have touched on a key point, which is that the limit of a series need not obey the properties that all of the series members obey.

Ralp
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### Re: Infinite Balls and Jugs [solution]

crayzeeepete wrote:
Ralp wrote:Nope. Though infinite, the sum of 5+2.5+1.25+... is a geometric series that converges to 10.

But, crucially, never equals 10. The moment you round to 10, you have a finite series, and thus it is possible to know how many balls are in the bag.
I think you have it backwards; what you are saying sounds to me like Zeno's Paradox. But ok. The way the problem is stated, this doesn't matter: this series has nothing to do with how many balls are in any jug, it's simply the amount of time that has passed, and clearly it is possible for 10 minutes to elapse.

Rather, the problem describes the way to schedule the duration between one insertion/removal of balls, and the next insertion/removal of balls:
1. Start at 11:50pm
2. 5 minutes later, 11:55
3. 2.5 minutes later, 11:57:30
4. 1.25 minutes later, 11:58:45
and so on. So, you have three choices here:

• You believe that at midnight, every ball will have been inserted. This is the choice that everyone in this thread has been working with. (And I assure you, the only mathematically correct one, but ok ok I won't force you to choose it.)
• You believe that at some time before midnight, every ball will have been inserted. I suppose the problem doesn't define what to do from then until midnight if you finish inserting all infinity balls, but it's safe to assume we can just twiddle our thumbs and not move any more balls around. (It doesn't matter, since you definitely won't finish before midnight: name any time prior to midnight and I can tell you exactly how many steps you've taken and how many balls are in the jug, and both will be quite finite numbers.)
• You believe that by midnight, we won't be finished inserting balls -- maybe even that we will never be finished and that it would take an infinite amount of time, even though we keep speeding up by half the interval between steps. If this is the case, it should be easy for you to answer the question of how many balls are in the jug. Since at midnight we haven't yet inserted all of the balls (or we'd be done), there must be an infinite number of balls still to go*, and since we always insert them in order, we couldn't yet have already inserted an infinite number of balls*. So we must have inserted a finite number of them (and as the problem specifies, removed exactly one-tenth that number).
This choice can be shown to be false too though in a similar way to the previous one: Name any finite number of steps that you think we won't have time to take until after midnight, or the number on any ball which you think isn't scheduled to be inserted until after midnight, and I can tell you exactly what time that step will be taken or when that ball will be inserted: it will be a positive real duration before midnight.

* I assert both of these claims without reservation but I leave rigorous proof of them to those of you with better knowledge of formal properties of aleph null (or whatever!)