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Topic: Re: Chew on This (Read 5640 times) |
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aero_guy
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Re: Chew on This
« on: Apr 2nd, 2003, 11:47am » |
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Argument 1 is correct, and the lowest numbered ball is numbered infinity, but half the infinity of the total number of balls. There are people here who know much more about the concept than I, but I am pretty sure that is the answer.
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wowbagger
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Re: Impish Pixie
« Reply #1 on: Apr 2nd, 2003, 12:05pm » |
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I agree that argument 1 is correct.
But let's hear what Icarus has to say. I'm sure he just has to comment on this one!
edited to make subject agree with new thread title
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| « Last Edit: Apr 3rd, 2003, 2:25am by wowbagger » |
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ThudnBlunder
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Impish Pixie
« Reply #2 on: Apr 2nd, 2003, 1:54pm » |
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on Apr 2nd, 2003, 11:47am, aero_guy wrote:| Argument 1 is correct, and the lowest numbered ball is numbered infinity... |
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And the second-lowest ball will be numbered infinity+1, I presume. 
At every stage, only balls with finite numbers on them are placed in the bucket.
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| « Last Edit: Apr 2nd, 2003, 9:02pm by ThudnBlunder » |
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LZJ
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Re: Chew on This
« Reply #3 on: Apr 2nd, 2003, 4:59pm » |
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The rate at which something reaches infinity counts too: for example, as x => infinity, e^x reaches infinity much faster than x does, and so does 2x.
(Sorry for stating the obvious)
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Icarus
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Re: Chew on This
« Reply #4 on: Apr 2nd, 2003, 7:14pm » |
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on Apr 2nd, 2003, 12:05pm, wowbagger wrote:| But let's hear what Icarus has to say. I'm sure he just has to comment on this one! |
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Of course! But I am going let you ponder it awhile longer before putting in my two cents. I will give the following hint:
What sets contain all numbers on balls in the final bucket? The intersection of all such sets is obviously the answer, since the answer is one of these sets.
As you might expect from two such rational sounding approaches giving different answers, the true answer reveals a nasty (ie anti-intuitive) fact about infinite processes.
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SWF
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Re: Chew on This
« Reply #5 on: Apr 2nd, 2003, 8:19pm » |
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on Apr 2nd, 2003, 12:05pm, wowbagger wrote:| But let's hear what Icarus has to say. I'm sure he just has to comment on this one! |
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I thought you meant Icarus would comment on the irrelevant subject for this thread. I'll take care of this one...
THUDandBLUNDER, please use a descriptive title when you start a thread. As a member, you can go back and edit the thread title, or maybe change the question so it fits the title. For example, make the balls, numbered gumballs being throw into a giant gumball machine. Posting of new puzzles is always welcome, but this one probably should have been put in the easy section.
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ThudnBlunder
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Impish Pixie
« Reply #6 on: Apr 2nd, 2003, 8:57pm » |
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> "THUDandBLUNDER, please use a descriptive title when you start a thread."
OK, I will do so in future.
> "As a member, you can go back and edit the thread title..."
Unfortunately, this does not rename the titles of the replies.
> "Posting of new puzzles is always welcome, but this one probably should have been put in the easy section."
If you think this puzzle belongs in the Easy section I would say that you are doubly mistaken. What is your Easy answer and the simple reasoning behind it? Now explain, so that an educated layperson can understand, what is wrong with the other answer(s). In my opinion, that's what makes this puzzle 'hard'. Getting the correct answer may not be hard, but being sure that it is right is surely not Easy, or even Relatively Medium. 
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| « Last Edit: Apr 3rd, 2003, 5:24am by ThudnBlunder » |
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LZJ
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Re: Impish Pixie
« Reply #7 on: Apr 3rd, 2003, 5:29am » |
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Just to end it off,
Increase = 2 * t
Decrease = 1 * t
Net change = (2 - 1) * t = t
Therefore, as t => infinity, net change => infinity.
Argument 1 is correct, but I think argument 2 is flawed (of course, I don't know much about infinities: perhaps Icarus would care to elucidate?)
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ThudnBlunder
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Re: Impish Pixie
« Reply #8 on: Apr 3rd, 2003, 5:55am » |
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> "perhaps Icarus would care to elucidate?"
Why not stick around for Daedulus while we are at it? After all, (unlike Icarus) we know for sure that Daedulus is on his way. 
As for Argument 1 being correct, I believe that SWF will either agree with you or be able to point out your mistake so that you can easily understand it.
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| « Last Edit: Apr 3rd, 2003, 9:57pm by ThudnBlunder » |
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Icarus
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Re: Impish Pixie
« Reply #9 on: Apr 3rd, 2003, 5:28pm » |
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on Apr 3rd, 2003, 5:55am, THUDandBLUNDER wrote:| Why not stick around for Daedulus while we are at it? After all, (unlike Icarus) we know for sure that Daedulus is on his way. |
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I'm afraid I don't understand this remark. How do we know "that Daedalus is on his way"?
They are expecting me to reply because when it comes to infinities, I have an infinite amount of bombast.
This is a good puzzle. It even fooled me at first glance, and I have enough experience with this that I should have known better. (I will say in my defense that it was only at first glance that I was fooled.) I am still holding off yet on giving my analysis.
And SWF is mistaken - this one definitely belongs in the hard section. Your only mistake was not giving it a good name from the start - a very common newbie blunder you have already corrected.
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| « Last Edit: Apr 3rd, 2003, 5:33pm by Icarus » |
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ThudnBlunder
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Re: Impish Pixie
« Reply #10 on: Apr 3rd, 2003, 9:56pm » |
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> '"I'm afraid I don't understand this remark. How do we know "that Daedalus is on his way"?'
I merely meant that your mythical namesake died when he flew too close to the sun - unlike his father, Daedalus .
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| « Last Edit: Apr 4th, 2003, 8:59am by ThudnBlunder » |
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wowbagger
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Re: Impish Pixie
« Reply #11 on: Apr 4th, 2003, 7:30am » |
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on Apr 3rd, 2003, 9:56pm, THUDandBLUNDER wrote:| I merely meant that your mythical namesake died when he flew too close to the sun - unlike his father, Daedulus . |
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I'm sure Icarus is aware of who his father is/was.
Not sure about Daedalus being on his way either, though.
Back to the riddle:
Based on Icarus's hint, I came up with the following:
Let Si be the set of the natural numbers without the first i numbers, so ("-" denotes set difference)
Si = N - {1,2,3,...,i}.
As all natural numbers will eventually be put in the bucket, and all will somewhen be taken out again, the set S of all finally remaining numbers has to be a subset of Si for all i.
On the other hand, it can't be a superset of S0 = N. This is why - as Icarus pointed out - the set S is the intersection of all Si:
S = interi=0oo Si
To me as a person without set-theoretical background, the set S seems to be empty.
So much for my initial answer!
To make sure I'm on the right track, consider the following alteration:
The impish pixie takes that ball back out which has the lowest odd number. So we add 1, 2, take out 1, add 3, 4, take out 3, add 5, 6, ...
This should lead to a set S consisting of all even natural numbers - and infinitely many balls in the bucket!
What do you all think of this?
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| « Last Edit: Apr 4th, 2003, 7:35am by wowbagger » |
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Ulkesh
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Re: Impish Pixie
« Reply #12 on: Apr 4th, 2003, 11:29am » |
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Didn't Cantor say that two sets are equal in magnitude if their elements can be put into one-to-one correspondence with each other?
wowbagger's alternative wording of the original problem is equivalent to it, I think. And, the set of integers can be put in one-to-one correspondence with the set of even integers. Doesn't this make the answer zero ?
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| « Last Edit: Apr 4th, 2003, 12:01pm by Ulkesh » |
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cho
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Re: Impish Pixie
« Reply #13 on: Apr 4th, 2003, 12:52pm » |
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Let's throw in one more variation. Throw in 2 balls and take the highest numbered ball out. 1,2 in; 2 out; 2,3 in, 3 out; 3,4 in, 4 out...
To summarize:
Infinity-infinity=infinity, according to argument one.
infinity-infinity=0, according to argument two.
infinity-infinity=1/2 infinity, leaving the even balls in.
infinity-infinity=infinity-1, if the balls that come out go back in.
And if you think an infinitely long discussion of infinity would be infinitely pointless and impractical, has anybody heard the news about a problem with quantum theories. If space and time are quantized, then space should be like a foam and photons passing through it will take different paths and slightly different speeds. That should make light traveling vast distances blurry, but Hubble photos of galaxies 5 billion light years away are razor sharp. So no quanta. If no quanta then time and space can be divided into infinitely small units. That should mean that the Big Bang began with an infinitely hot universe in an infinitely small space. Not possible.
So we'd better figure out this infinity quickly or the whole universe may be an impossibility and cease to exist.
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ThudnBlunder
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Re: Impish Pixie
« Reply #14 on: Apr 4th, 2003, 1:07pm » |
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Ulkesh, I believe Wowbagger's conclusion is correct.
As I see it, if after minute #N I take out ball #N there will be no balls left.
If instead I take out ball #2N (or, like Wowbagger, 2N-1) there will be an infinite number of balls left.
And for any natural number k, if I take out ball #N+k there will be k balls left. 
While it is true that the transfinite sets that you mentioned are the same size, the answer also depends very much on the ordering of the balls, strange though that may seem.
There are in fact three schools of thought: an infinite number of balls, no balls, or an indeterminate number of balls. (Balls, no balls, or bollocks, you might say.)
(Warning: discussions of this puzzle often generate more heat than light.)
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| « Last Edit: Apr 4th, 2003, 4:31pm by ThudnBlunder » |
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Icarus
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Re: Impish Pixie
« Reply #15 on: Apr 4th, 2003, 4:11pm » |
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Thanks, wowbagger! I was afraid nobody would get my hints that they really need to rethink the answer. My comments:
A technical way of approaching this problem is as a limit of sets:
Denote
[langle]n, k[rangle] = {m | m is an integer and n <= m <= k}.
[langle]n[rangle] = {m | m is an integer and n <= m}.
After the nth turn, the set of numbers corresponding to balls in the basket is [langle]n+1, 2n[rangle]. The problem essentially asks, what is Limn[to][subinfty] [langle]n+1, 2n[rangle] ?
How do you take the limit of a sequence of sets? The accepted method is this: Given a sequence {Sn} of sets, define
Limsup Sn = [bigcap]k=1[supinfty] [bigcup]n=k[supinfty] Sn
Liminf Sn = [bigcup]k=1[supinfty] [bigcap]n=k[supinfty] Sn
(The Unions in the Limsup definition contain every Sk for k >=n, so they "ought" to also contain any sort of limit of these sets. Since they all contain it, so does their intersection. So Limsup Sn is a "biggest possible" limit set. Similarly Liminf Sn is a "smallest possible" limit set.)
Define Lim Sn = A, if Limsup Sn = Liminf Sn = A.
Now, [bigcup]n=k[supinfty] [langle]n+1, 2n[rangle] = [langle]k+1[rangle], and [bigcap]n=k[supinfty] [langle]k+1[rangle] = [emptyset].
So, Limsup [langle]n+1,2n[rangle] = [emptyset]. The Liminf is always a subset of Limsup, so it must be that Liminf [langle]n+1,2n[rangle] = [emptyset] as well.
Hence the limit is empty. There are no balls in the basket when the job is finished.
What nasty thing about infinite processes does this teach? That cardinality is not well-behaved with respect to them. The number of balls in the basket is growing with each step, but when you move from a finite number of steps to an infinite number, instead of becoming infinite, the number of balls becomes zero! And as ThudandBlunder has indicated, you can vary the procedure so that the individual sets in the sequence stay the same size, but the limit set can have any number of balls from 0 to [smiley=varaleph.gif]0 (the countable cardinal infinity).
The fact is, cardinality is not continuous with respect to the limit expressed above.
This is the flaw with the first argument: It assumes that because the set sizes are growing, the Limit set must be infinite. Unfortunately this is not the case. But people are so wedded to the idea that even after ThudandBlunder pointed out the ridiculous conclusion that it leads to (what is the smallest ball still in the basket?), everyone was still sure it was correct, and hoped that infinite numbers would provide an answer. But the balls are labeled with Natural numbers, which are not infinite, so there is no ball labeled "infinity".
The second argument is a simple application of logic:
Claim: there are no balls in the basket at the end.
Proof: for every natural number k, ball k was removed from the basket at step k, so it cannot be in at the end. Since every ball has a natural number label, there can be no balls in the basket.
I have more to say on the subject, but I am out of time.
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| « Last Edit: Aug 18th, 2003, 5:31pm by Icarus » |
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towr
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Re: Impish Pixie
« Reply #16 on: Apr 6th, 2003, 7:10am » |
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One could also consider that the proces is simply impossible..
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ThudnBlunder
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Re: Impish Pixie
« Reply #17 on: Apr 6th, 2003, 10:13am » |
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on Apr 6th, 2003, 7:10am, towr wrote:| One could also consider that the proces is simply impossible.. |
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In the same way that 1/2 + 1/4 + 1/8 + 1/16 +......+ 1/2n +........ cannot possibly be equal to 1?
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| « Last Edit: Apr 6th, 2003, 10:28am by ThudnBlunder » |
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towr
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Re: Impish Pixie
« Reply #18 on: Apr 6th, 2003, 12:21pm » |
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yes :p
well there is a difference of course..
For every ball taken out, two are put in, so the bucket can't be empty. At the same time every ball with a label has been taken out, and since they're all labeled the bucket must be empty. That can't be true at the same time.
In short I do not think it makes sense for any real-world problem. So impish pixies that do this can't exist..
1/2 +1/4 + ... does make sense in the real world, but honoustly, you can only apply infinite processes by approximation in real life. Infinite processes would take infinite energy, if not infinite time.
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Chronos
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Re: Impish Pixie
« Reply #19 on: Apr 6th, 2003, 7:08pm » |
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I'm still a bit uneasy with this one. To explain why, consider a few simpler examples.
Case 1: I have the aforementioned balls with numbers on them. At each step, I take out the current ball, and place its successor ball in the bin. How many balls are left in the bin, after I'm done? By the same argument as in the original puzzle, there is no natural number left, so no balls.
Case 2: Instead of having an infinite number of balls at my disposal, I have just one. But the numbers are written in pencil. At each step, I take the ball out, erase the number on it, and write the successor number on it. This is equivalent to Case 1, is it not?
Case 3: Since I'm just erasing and re-writing the numbers, suppose that I just leave the ball in the basket, while I'm making the change. This is equivalent to Case 2, is it not?
Case 4: Suppose that I represent the numbers in such a way that I don't have to erase the old one in order to write the new one. For instance, I might represent 1 by a single dot, 2 by two dots, etc. In this case, the process of changing a number consists solely of adding a dot to the ball. So in each step, I add one dot to the ball. This is equivalent to Case 3, is it not? But on the other hand, if I have a ball in a basket and add a dot to it an infinite number of times, I do know what I have: A ball with a transfinite number (specifically, aleph-0) of dots. So now what I have is a ball in the basket, with markings on it. Those markings don't correspond to any natural number, but the ball is still in the basket. In what step did this become different from my Case 1?
A similar puzzle, by the way: Suppose I have a light bulb which is initially off. After half an hour, I flip it on. A quarter-hour after that, I flip it back off. An eighth of an hour after that, I flip it on again, and so on. Describe the state of the bulb after 1 hour. If you're going to dismiss this one as unrealistic, by the way, be prepared to explain why the balls-in-basket puzzle is not similarly unrealistic.
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Icarus
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Re: Impish Pixie
« Reply #20 on: Apr 6th, 2003, 8:03pm » |
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on Apr 6th, 2003, 7:08pm, Chronos wrote:I'm still a bit uneasy with this one. To explain why, consider a few simpler examples.
Case 1: I have the aforementioned balls with numbers on them. At each step, I take out the current ball, and place its successor ball in the bin. How many balls are left in the bin, after I'm done? By the same argument as in the original puzzle, there is no natural number left, so no balls. |
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Yes.
Quote:
Case 2: Instead of having an infinite number of balls at my disposal, I have just one. But the numbers are written in pencil. At each step, I take the ball out, erase the number on it, and write the successor number on it. This is equivalent to Case 1, is it not? |
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No. It LOOKS like Case 1 after any finite number of steps, but it is different. In Case 1, each ball makes two transitions: once into the basket, once out. Thus we can logically determine the location of all balls at the finish. In case 2, we have one ball going in and out infinitely many times. Case 2 is "[omega]-inconsistent". That is, there is no way to define it's final state.
Quote:
Case 3: Since I'm just erasing and re-writing the numbers, suppose that I just leave the ball in the basket, while I'm making the change. This is equivalent to Case 2, is it not? |
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No. Now the ball makes 1 transition only. Its final state is well-defined: in the basket. It is also not equivalent to Case 1, even though the contents of the basket look the same after any finite number of steps. There is something [omega]-inconsistent here, though. What label is on the ball at the end? This cannot be determined.
Quote:
Case 4: Suppose that I represent the numbers in such a way that I don't have to erase the old one in order to write the new one. For instance, I might represent 1 by a single dot, 2 by two dots, etc. In this case, the process of changing a number consists solely of adding a dot to the ball. So in each step, I add one dot to the ball. This is equivalent to Case 3, is it not? |
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For the ball, yes. For the label, no. The label on the final ball is now discernable, which of course is exactly why you introduced this case:
Quote:| But on the other hand, if I have a ball in a basket and add a dot to it an infinite number of times, I do know what I have: A ball with a transfinite number (specifically, aleph-0) of dots. So now what I have is a ball in the basket, with markings on it. Those markings don't correspond to any natural number, but the ball is still in the basket. In what step did this become different from my Case 1? |
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Case 2 and Case 3.
Quote:
A similar puzzle, by the way: Suppose I have a light bulb which is initially off. After half an hour, I flip it on. A quarter-hour after that, I flip it back off. An eighth of an hour after that, I flip it on again, and so on. Describe the state of the bulb after 1 hour. If you're going to dismiss this one as unrealistic, by the way, be prepared to explain why the balls-in-basket puzzle is not similarly unrealistic. |
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As in Case 2, the light changes states infinitely many times - so there is no way to decide its state at the finish.
Of course as towr points out, even the finite time does not make this problem realistic. To me though, the realism or unrealism of the puzzle is immaterial. The puzzle is not meant to be realistic. Neither is it realistic to believe 100 red-eyed monks will kill themselves 3+ months after some stranger mentions something they already knew, or that a bee/fly/bird would fly thousands of miles from one train to another AT CONSTANT SPEED, then turn around and head back WITHOUT ANY LOST TIME. Yet I have never heard anyone complain about the unreality of these puzzles.
The point of all such puzzles is: if such a thing were possible, could you arrive at a unique logically consistent answer? (Unique in the sense that to come up with a different answer requires assuming things not found in real life or suggested in some way by the puzzle.)
For this puzzle such an answer exists. Since each ball makes only two moves, its location at the finish is well-defined. And for all balls in the puzzle, that location is outside the basket, so the basket must be empty. The other analysis does not provide a second solution, since it depends on a false assumption: that a sequence of sets of increasing size must "converge" to an infinite set.
For Case 2 or the light, there is no answer. The position of the ball, and the state of the light, at the finish cannot be discerned from the given information.
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| « Last Edit: Aug 18th, 2003, 5:35pm by Icarus » |
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towr
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Re: Impish Pixie
« Reply #21 on: Apr 6th, 2003, 11:26pm » |
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I still don't buy it.
At every step k, k+1 is clearly in the bucket, so at every step there must be at least one ball in the bucket.
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Icarus
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Re: Impish Pixie
« Reply #22 on: Apr 7th, 2003, 5:06pm » |
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At every step after the first there are many balls in the bucket. And the number of balls is growing. If Bn is the contents of the bucket at step n, then
limn[to][subinfty] Card(Bn) = [infty] (actually, [smiley=varaleph.gif]0).
But this limit does not actually have anything to do with how many balls are in the bucket at the finish. Cardinality is not continuous here. It's similar to
limx[to]0 0x = 0
but 00 = 1.
It should be clear that every individual ball is not in the bucket at the finish. The ball is placed in the bucket at some point in the proceedings, and then removed at a later time. After that, it is never touched again. The only way for the bucket to still contain balls is if new ones magically appeared (one guy in a Usenet discussion of this puzzle actually proclaimed that this was a valid way of modelling the situation  ).
So we have the following situation:
Bn [to] B[subinfty]
Card(Bn) [to] [infty]
Card(B[subinfty]) = 0
Cardinality is not continuous.
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| « Last Edit: Aug 18th, 2003, 5:42pm by Icarus » |
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Re: Impish Pixie
« Reply #23 on: Apr 7th, 2003, 5:18pm » |
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Quote:
I'm not sure if that's a good example, as this is only true by definition.
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| « Last Edit: Apr 7th, 2003, 5:21pm by ThudnBlunder » |
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Re: Impish Pixie
« Reply #24 on: Apr 7th, 2003, 5:42pm » |
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Actually, it is an excellent example.
(Minor nitpick - of course it is only true by definition - everything that is true is only true by definition! What you meant was, it is only true by convention... but then, I suppose the same complaint can be raised with that!  )
We define 00 = 1 because it is a far more useful definition than the other candidate: 00 = 0. But even if the other definition were chosen, I would have given the example:
"limx[to]0 x0 = 1
but 00 = 0."
In any case, it is but one example of a discontinuous function. Cardinality is another.
One normally does not think of cardinality and continuity together. That is because most cardinals are discrete: they have no neighbors that approach them closer and closer... . But when you start tossing in infinite cardinals, this is no longer the case. Some cardinals are called "limit cardinals", because they have no immediate predecessors. Instead they are approached by an infinite number of cardinals below them. The first of these is [smiley=varaleph.gif]0 - the cardinality of the natural numbers.
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| « Last Edit: Aug 18th, 2003, 5:44pm by Icarus » |
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"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
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