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Topic: Re: Chew on This (Read 5595 times) |
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Icarus
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Re: Impish Pixie
« Reply #50 on: May 18th, 2003, 1:20pm » |
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You don't need an uncountable number of balls to have balls with infinite labels come into play. All you need is to toss in one more ball after you have already taken infinitely many. Since the puzzle only specifies "after an infinite amount of time", this is actually allowable, since [infty] + 1 is still infinite.
To make this more understandable to those less aquainted with infinity, consider it according to the "increasing speed" variation: the first step takes 1/2 minute, the second step takes 1/4 minutes, the third 1/8 minute, etc. So all natural numbered balls have their turn within the first minute. Instead of stopping immediately after that minute, you toss in balls [omega] and [omega]+1 ([omega] is the symbol normally used to denote the least infinite ordinal), and the imp takes out [omega].
You have still executed an infinite number of steps. The balls are still all numbered starting with 1 and going up. It fits the wording of the puzzle (adjusted for the increasing speed variation), and leaves a single ball, labeled [omega]+1, in the basket.
Note that this does not violate the main point of this puzzle: after finishing with the integer balls, the basket is still empty. Rather, this approach makes use of a technical loophole to allow an extra step or steps which repopulate the basket. If you keep going, using ordinal-labeled balls, until you get every ball up to but not including 2[omega] (you worked twice as fast, so this took only an additional 1/2 minute), the basket is empty again. Keep going and the bucket repopulates, only to empty again at 3[omega]. Keep at it, and by the time you reach two minutes from the start you have arrived at (but not yet used) ball [omega]2. The basket is once again empty. Redouble the speed, and a minute later you've reached 2[omega]2. Redouble again... by 4 minutes you've reached [omega]3. ...
Only countable ordinals are reachable in a finite amount of time by this "increasing speed" method. If you accept that the wording of the puzzle implies a "well-orderedness" on the balls or process steps, then it is impossible to reach an uncountable number except after "an uncountably infinite time" by any model. And if you accept the "well-orderedness", you might as well use the ordinals to label the balls, as any other type of "well-ordered infinite numbers" are equivalent to the ordinals, as far as this puzzle goes. (This includes the use of cardinals.)
In this situation, AT NO TIME does the bucket ever contain an infinite number of balls, but the bucket could have any finite number of balls when you stop, depending on just when you choose to.
One can reasonably argue that using this scenario, or a similar one, violates the normal understanding people have of the word "infinity". But one can also argue that the normal understanding people have of the word "infinity" is not robust enough to make the puzzle sensible at all.
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| « Last Edit: Aug 18th, 2003, 6:15pm by Icarus » |
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Josh
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Re: Impish Pixies
« Reply #51 on: Jun 7th, 2003, 4:19pm » |
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The answer is that you will run out of balls to put into the basket.
You have an infinite number of them, and you add two at each step. An infinite number of steps requires twice as many balls.
Quote:| the balls are labeled with Natural numbers, which are not infinite, so there is no ball labeled "infinity". |
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Are you sure this is true?
If you can have an infinite number of numbered balls, and an infinite number of discrete steps, you can have an inifinite number on an infinite number of balls. I don't think Natural numbers are involved in this question at all.
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Josh
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Re: Impish Pixies
« Reply #52 on: Jun 7th, 2003, 4:21pm » |
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on Jun 7th, 2003, 4:19pm, Josh wrote:
Are you sure this is true?
If you can have an infinite number of numbered balls, and an infinite number of discrete steps, you can have an inifinite number on an infinite number of balls. I don't think Natural numbers are involved in this question at all. |
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Um, never mind this part... :-[
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Icarus
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Re: Impish Pixie
« Reply #53 on: Jun 8th, 2003, 8:21pm » |
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Josh: 2[infty] = [infty].
(More exactly, 2[smiley=varaleph.gif]0 = [smiley=varaleph.gif]0.)
An example: Compare the Natural numbers [bbn] with the even naturals [smiley=bbe.gif], and the odd naturals [smiley=bbo.gif].
Fact 1) There are no more or less natural numbers than there are Even natural numbers. This must be the case, since you can match up [smiley=bbe.gif] one-for-one with [bbn]:
1 [smiley=leftrightarrow.gif] 2
2 [smiley=leftrightarrow.gif] 4
3 [smiley=leftrightarrow.gif] 6
etc.
Hence the cardinality of [smiley=bbe.gif] is also [smiley=varaleph.gif]0.
Fact 2) There are no more or less natural numbers than there are Odd natural numbers. Same Argument:
1 [smiley=leftrightarrow.gif] 1
2 [smiley=leftrightarrow.gif] 3
3 [smiley=leftrightarrow.gif] 5
etc.
Hence the cardinality of [smiley=bbo.gif] is also [smiley=varaleph.gif]0.
Fact 3) The natural numbers are the disjoint combination of the Evens and the Odds: [bbn] = [smiley=bbe.gif] [smiley=uplus.gif] [smiley=bbo.gif]. So, [smiley=varaleph.gif]0 = [smiley=varaleph.gif]0 + [smiley=varaleph.gif]0 = 2[smiley=varaleph.gif]0.
Looked at another way, if you run out of balls, then you have placed an infinite number of balls in the basket. This cannot of happened after any finite number of steps, because on the Nth step you have only put in 2N balls, a finite number.
Putting it another way: when you do run out of balls, you are finished, and puzzle is to figure out the contents of the basket at this point.
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| « Last Edit: Aug 18th, 2003, 6:14pm by Icarus » |
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BNC
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Re: Impish Pixie
« Reply #54 on: Jun 9th, 2003, 2:05am » |
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I've been thinking about this problem for a long (though < oo) time now. I have read the thread several times very carefully. And I still have a problem with it. I will try to state my problem, and hopefully someone will be kind enough to response.
Disclaimer: I have never studied "discreet mathematics" or set theories, so my knowledge of infinities is limited at best (I did learn quite a lot of calculus at the time). The way I learned it, infinity is not a "number" but rather is a "concept" -- a definition of what is happening for very large or very low numbers (processes). Also, English is not my mother tongue so I may use the wrong terms; I hope that they would be understandable from the context.
Here goes:
As far as I remember, a number series an is defined to have a limit at infinity if: for every number M there exist a number N (N=N(M)), such that for every n>N, an > M.
Now, what is the number series of the number of balls in the basket?
a1 = 1
a2 = 2
.
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an = n
lim n->oo an = oo
So, number of balls is infinity.
Counter claims:
1)
The cardinality is not continuous. Thus, the limit has nothing to do with the actual value.
>> BUT: to prove the opposite claim of 0 balls, you still use a limit (a limit of sets, not numbers -- but still a limit).
2)
What is the lower numbered ball in the basket? It cannot be any finite number, as they are all out.
>> BUT: of course all finite number balls are out after a finite amount of time. But the same number s still in the basket. What is the number in the basket? Heck, I don't know. What is the number of the last ball that was removed before process end (thanks to rmsgrey)? And what is the "minute number" at which it happened? Similar questions may be asked regarding the alternative “shortening time slices” version.
To me, this reasoning could be used to achieve absurd results:
I flip a fair coin. Head: I keep quite. Tail: I start counting from 1 upward 1,2,3,… ad infinum, one number per minute, and thus am never quite. Question: What happened after an infinite time? Head: I'm quite. Tail: I must be quite too. Otherwise, what number am I saying? Every number was already said at a previous finite time. Thus, I'm quite. But Tail means I’m never quite. And therefore, the (fair) coin must have probability 1 for Head.
I will accept a reply of "that's the definition / convention for that case" (like in 00) -- but than I don't accept it as a valid riddle.
Thanks in advance for your time.
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rmsgrey
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Re: Impish Pixie
« Reply #55 on: Jun 9th, 2003, 8:50am » |
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Having consulted with some friends doing Maths doctorates at Cambridge University, I'm now of the opinion that, for infinite time (at least for small infinities), the value of the number of balls in the jar can be any natural number - basically, at time w (omega, the smallest infinity) the bucket is empty. At times after w the number of balls is the same as after the equivalent finite time.
The trouble with the arguments BNC highlighted (cardinality being discontinuous and not being clear on numbers on balls) is that they just show that the immediate attempts to answer are flawed, and don't show why the 'correct' method is.
At the moment, I'm accepting argument by authority, with a caveat that my friends are themselves not entirely comfortable playing with infinity in this way, so may have overlooked something subtle.
As to BNC's coin toss, there's a flaw in the argument - if you stop counting at time w on a tail, then it's not true that you're never quiet when the coin comes up tails. Since it was only specified that counting continues until infinity ("ad infinitum" in Latin) what happens after that is unspecifed, so you can't argue based on what you'll be doing then. In fact, the argument seems to show that you'll be quiet after infinite time whether you throw H or T
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Icarus
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Re: Impish Pixie
« Reply #56 on: Jun 9th, 2003, 7:11pm » |
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on Jun 9th, 2003, 2:05am, BNC wrote:| The way I learned it, infinity is not a "number" but rather is a "concept" -- a definition of what is happening for very large or very low numbers (processes). |
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Yes, infinity is a concept (actually a number of different concepts are all described by the same word "infinity" - that's why I've said elsewhere that there are several type of infinities), but "one" is also just a concept! All things mathematical are nothing more than concepts. To make these concepts useful we have to come up with consistent definitions. Then we can examine the properties of these things we have defined.
It is inaccurate to say "infinity is not a number". There is no fixed definition of the term "number" in mathematics. It is applied to many things you would not normally think of as numbers (matrices, for example ). There are several definitions of sets of "numbers" which include infinite numbers. What your instructors should of said (but didn't, because they didn't want to confuse their students further by explaining all this) is that infinity is not a REAL number (as in the set of real numbers [bbr]).
Quote:
lim n[to][subinfty] an = [subinfty]
So, number of balls is infinity.
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No - that only shows that the limit of the number of balls at finite times is infinity. To conclude that the number of balls in the basket at time [omega] (that is, when first an infinite number of steps is accomplished) is infinity, you must prove that this limit counts the balls in the basket AT [omega]. The whole basis of argument 1 is the hidden assumption that this is true. But in fact it is false.
Quote:The cardinality is not continuous. Thus, the limit has nothing to do with the actual value.
>> BUT: to prove the opposite claim of 0 balls, you still use a limit (a limit of sets, not numbers -- but still a limit).
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The problem with your limit approach is not the limit, but the hidden, never justified, assumption that the limit of the number of balls at finite times is the number at time [omega].
In the limit approach that I gave, there was no such assumption (other than a mistaken belief that the puzzle statement only allowed integer numbered balls - a rereading prompted by rmsgrey's last post on 5/16/03 revealed this to be false, which I admitted in my next post following).
But the problem does not need to be done by limits at all. With the following clarifications of the original problem, which I have also stated before, it can be deduced by simple set theory logic that the basket is empty:
(1) The only things available to be in the basket at any time are the balls described in the problem - one ball for each natural number.
(2) If a ball is never moved after time K - where K is finite, then its position at time [omega] is the same as at time K.
(I have said before, and still maintain, that these are reasonable assumptions about modeling the problem, and need not be specified in it, any more than one specifies any number of other assumptions. But there are apparently some who disagree).
From (1), the only possible contents are the integer balls, but by (2) all of them are outside the basket at time [omega]. Hence the basket is empty.
So what is the difference between your assumption that the limit represents the balls at time [omega], and my assumptions (1) and (2)?
My assumptions are about how to model the "physical" situation using mathematics. Modeling is by definition an assumption about how the world (real or imaginary) works. Your assumption is within the mathematics itself. And within the mathematics it can (and has) been shown false.
Quote:
2)
What is the lower numbered ball in the basket? It cannot be any finite number, as they are all out.
>> BUT: of course all finite number balls are out after a finite amount of time. But the same number s still in the basket. |
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 You seem to be saying here that the ball is in the basket and out of the basket at same time. Could you please clarify this?
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What is the number in the basket? Heck, I don't know. What is the number of the last ball that was removed before process end (thanks to rmsgrey)? And what is the "minute number" at which it happened? |
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There was no last ball removed. At time [omega], there terms "last" and "previous" are undefinable.
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Similar questions may be asked regarding the alternative “shortening time slices” version. |
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The "shortening time slices" statement is simply to allow people to get their heads around the concept of "time [omega]". It does not offer anything different from the regular version.
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To me, this reasoning could be used to achieve absurd results:
I flip a fair coin. Head: I keep quite. Tail: I start counting from 1 upward 1,2,3,… ad infinum, one number per minute, and thus am never quite. Question: What happened after an infinite time? Head: I'm quite. Tail: I must be quite too. Otherwise, what number am I saying? Every number was already said at a previous finite time. Thus, I'm quite. But Tail means I’m never quite. And therefore, the (fair) coin must have probability 1 for Head.
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"But Tail means I’m never quiet." - How so? You can easily prove this for finite times, but you are assuming it against all evidence after infinite times.
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I will accept a reply of "that's the definition / convention for that case" (like in 00) -- but than I don't accept it as a valid riddle.
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Like T&B, and I suppose most other people, you are not understanding my point with 00. I was NOT saying that the basket was empty because we define it to be. (Just to be clear, T&B's disagreement was entirely different.)
The basket must be empty, because (given the two modeling assumptions mentioned above) logic demands it. To say that it is not empty is to give rise to a contradiction.
Quote:
Thanks in advance for your time.
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Reasoned debate with someone who is intellectually honest is never a waste of my time.
on Jun 9th, 2003, 8:50am, rmsgrey wrote:| Having consulted with some friends doing Maths doctorates at Cambridge University, I'm now of the opinion that, for infinite time (at least for small infinities), the value of the number of balls in the jar can be any natural number - basically, at time [omega] (omega, the smallest infinity) the bucket is empty. At times after [omega] the number of balls is the same as after the equivalent finite time. |
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Yes, that was my conclusion too, after noting that the problem statement does not limit the balls to having finite labels. However, this is clearly an unintended oversight. As most people (not being familiar with the concept of infinite numbers) would understand the problem, only finite numbers are available. And with this added limitation, at time [omega] there are no more balls to put in, and as you have said, the bucket is empty.
Quote:
The trouble with the arguments BNC highlighted (cardinality being discontinuous and not being clear on numbers on balls) is that they just show that the immediate attempts to answer are flawed, and don't show why the 'correct' method is.
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What in the world are looking for in "showing why the 'correct' method is"?
In this thread, you have seen
Wowbagger's original solution
My limit of sets explanation and at the bottom of the same thread, a short logical proof.
This was followed by several posts raising supposed counter examples, all of which I answered, demonstrating why the counter-examples fail to apply.
I have also re-examined and made explicit the modeling assumptions made, and admitted the caveat that you were correct about the puzzle allowing transfinite numbers.
But with the two modeling assumptions as I have stated them in this post above (slightly changed from a previous post to limit to make explicit the limitation to finite numbers), I have demonstrated repeatedly that they logically lead to only one conclusion.
So what is it you require?
Quote:
At the moment, I'm accepting argument by authority, with a caveat that my friends are themselves not entirely comfortable playing with infinity in this way, so may have overlooked something subtle.
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The modeling side of this puzzle is (and will always be) a little shaky. This is because it does represent a fundamentally impossible scenario, so there is no way to demonstrate "what really happens". But the mathematical side is quite sound.
Before I say "yay or nay" concerning your refutation of BNC's argument, I would like to see his clarification of it, so that I can decide if this problem is in his argument or our understanding of his argument.
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| « Last Edit: Aug 18th, 2003, 6:26pm by Icarus » |
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BNC
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Re: Impish Pixie
« Reply #57 on: Jun 10th, 2003, 1:13am » |
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Thanks, guys!
I see now that my understanding of infinity (or infinities) is even more limited than I thought 
I didn't realize about "transinfinte" numbers, and the w thingi. I wouldn't amagine one may count "past infinity".
Oh, well. One more thing to add to my (too long) list of "things to read / learn / experiment at the sonnest available time".
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Icarus
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Re: Impish Pixie
« Reply #58 on: Jun 10th, 2003, 8:06pm » |
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*** Warning *** I'm about to go off on another of my rambles about infinite numbers, so all those given to drowsiness should stop reading now, less they enter a comatose state. 
One way that you can get a better idea about infinite or "transfinite numbers" is to look at models of them within the real numbers.
I gave a quick overview of the definition of the three main types of infinities in this post. It is the third type, ordinals, that we have been discussing here.
By definition, ordinals are "order types" - ways of "well-ordering" a set. A set is well-ordered if it has an ordering (<) for which every non-empty subset has a least element. It turns out that if you have any two well-ordered sets A and B, then either there is a K in B such that the set BK = {x | x is in B and x < K} can be matched up 1-1 with A by an order-preserving correspondance (if r,s in A and x,y in BK, with r ~ x and s ~ y, then r < s if and only if x < y), or there is a K in A for which the reverse is true.
One way of interpreting this result is that there is one "universal" well-ordering. Any particular well-ordering you come across is just a chunk off the front of the universal one. All the "chunks off the front" comprise the ordinal numbers.
If you have two disjoint well-ordered sets A and B, then you can define an ordering on A [cup] B as (x < y [bigleftrightarrow] [ (x,y [in] A or x,y [in] B) and x < y ] or (x is in A and y is in B) ). So every element of B is greater than every element of A. This is easily seen to be a well ordering, and addition of ordinals is defined by
ord(A) + ord(B) = ord(A [cup] B).
The cross product A [times] B also has a well-ordering (r, x) < (s, y) [bigleftrightarrow] (r < s or (r=s and x<y)). (This is sometimes called the "dictionary" ordering.) Ordinal multiplication is defined by
ord(A)*ord(B) = ord(A [times] B).
With these, we start building the ordinals. Because two ordered sets cannot have an order-preserving correspondence between them unless they have the same number of elements, ordered sets of different sizes (cardinality) necessarily have different ordinals. On the other hand, all well-orderings of a finite set are equivalent. So the first ordinals are just the Natural numbers [bbn] (for convenience, I am including 0 in the Natural numbers):
0 = ord( [emptyset])
1 = ord({0})
2 = ord({0,1})
etc.
The natural numbers are themselves well-ordered, so we define [omega] = ord([bbn]).
A little work shows that if R is an infinite subset of [bbn], then ord(R) = [omega] as well. Thus [omega] is the least infinite ordinal. We can build larger ordinals using the ordinal addition and multiplication:
[omega] + 1 = ord({0} [times] [bbn] [cup] {(1,0)}), where the order is defined by: [forall] n [in] [bbn], (0,n) < (1,0).
[omega] + 2 = ord({0} [times] [bbn] [cup] {(1,0), (1,1)})
etc.
After all [omega] + k, k [in] [bbn], we get
[omega] + [omega] = ord({0} [times] [bbn] [cup] {1} [times] [bbn]) = ord({0,1} [times] [bbn]) = 2[omega]
then 2[omega]+1, 2[omega]+2,...,3[omega],...4[omega],...[omega]*[omega] = [omega]2,...[omega]3... .
This allows the definition of P([omega]), for any polynomial P with integer coefficients. These comprise all the ordinals of countable sets. The least ordinal greater than all P([omega]) is the first uncountable ordinal.
So what was that about modeling them in the Real numbers [bbr]?
Consider the equivalence n ~ 1-1/(n+1) between [bbn] and N1 = {1-1/(n+1) | n [in] [bbn]}. This is an order preserving equivalence, so the latter set is well-ordered, and gives us an alternative way of viewing how the natural numbers are distributed. In particular, we can add another element to the "top" of N1 easily: N1* = N1 [cup] {1}. By extending the equivalence, we get [omega] ~ 1. We can continue extending: k*[omega] + n ~ k - 1/(n + 1); k,n [in] [bbn].
By this means we have a model for all ordinals less than [omega]2 within the real numbers.
This gives a better idea of the "topological" behavior of these ordinals (that is, their behavior with respect to order and limits). We can see the exact same behavior in a fairly simple set of real numbers. In particular, each ordinal exhibits one or the other of two "behaviors". Most are discrete - no other ordinals approach them beyond some fixed distance - these are the ordinals k*[omega] + n, where n > 0. The corresponding points on the real line are the non-integer elements in the set S={k-1/(n+1) | k,n [in] [bbn], k>0}. All non-integer elements of S are discrete - for each there is a distance d such that no other element is within d of them. However, the integer elements do not have this property. For any d, there are infinitely many points of S within d of each integer. These are called the "limit" or "cluster" points of S. The corresponding ordinals are the "limit ordinals" n*[omega], for natural n [ge] 1. Each is only approached on its lower side. On the upper side, there is a discrete distance (1/2 in S) between each limit ordinal and its immediate successor.
We can create correspondences between sets of Real numbers and larger collections of countable ordinals as well. The set of all ordinals < [omega]k+1 has the correspondence
ak[omega]k+ak-1[omega]k-1+ ... a0 ~ ak - 1/(ak-1+1) - 1/(ak-1+1)(ak-2+1) - ...
with a set Sk of Real numbers.
The limit points of Sk are exactly those for which a0 = 0. Points for which a1 = 0 as well are limit points of the limit points.
I do not believe it is possible to find a subset of the reals which corresponds to the set of all countable ordinals (I.e. no uncountable set of Real numbers is well-ordered.) But I don't recall for sure right now.
A countable ordinal P([omega]) = ak[omega]k+ak-1[omega]k-1+ ... a0 is of "limit order" n if al = 0 for all l < n, and an > 0. The degree of a countable ordinal is the degree of its polynomial.
To examine the arithmetic properties of ordinals requires a different tack, since under the correspondences above, addition and multiplication are poorly behaved. And the correspondence between the ordinal P([omega]) and the polynomial P is not much better. The problem is that ordinal addition and multiplication are not commutative. I.e., in general, x + y [ne] y + x, and xy [ne] yx.
To see this, look at [omega]+1. By definition this is the ordinal of the set [bbn] [cup] {a}, where a > n [forall] n [in] [bbn]. This ordering must be distinct from that of [bbn], since it has an element, a, with infinitely many things less than it, where as every element of [bbn] has only finitely many things less than it.
But 1+[omega] is the set {b} [cup] [bbn], where b < n [forall] n [in] [bbn]. The ordering on this set is equivalent to the ordering on [bbn]. In fact the correspondence
{b}[cup][bbn] [smiley=leftrightarrow.gif] [bbn]
b ~ 0
n ~ n+1 n [in] [bbn]
is order-preserving between the two sets. So 1+[omega] = [omega] < [omega]+1.
More generally, if x and y are two ordinals and deg(x) < deg(y), then x + y = y.
And if
x=ak[omega]k+ak-1[omega]k-1+ ... a0
y=bn[omega]n+bn-1[omega]n-1+ ... b0 (n [ge] k),
then y + x = bn[omega]n + bn-1[omega]n-1 + ... + bk+1[omega]k+1 + (bk + ak)[omega]k + ak-1[omega]k-1 + ... + a0.
Multiplication can be deduced from the rule if x is finite, then [omega]x = [omega], combined with associativity and distributivity, both of which hold for all ordinals.
Edit: corrected some typos. Thanks, Wowbagger, for pointing out some of the errors (yeah - only some, I found a couple more that you missed).
Edited again to make use of the new mathematical symbolry.
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| « Last Edit: Aug 18th, 2003, 7:01pm by Icarus » |
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rmsgrey
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Re: Impish Pixie
« Reply #59 on: Jun 13th, 2003, 7:11am » |
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I apologise for the unclear phrasing in my last post - I'm under time constraints in my internet access since I'm currently restricted to the local public library...
Anyway, I am happy with the arguments Icarus has presented for the correct answer under the assumption of termination at time w (as I have said before) but the point I was trying to address was that the specific arguments BNC listed were negative and non-constructive. The arguments which are constructive (and conclusive) weren't mentioned by BNC - possibly because he missed them or possibly because they are more subtle than those he did mention.
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xlh
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Re: Impish Pixie
« Reply #60 on: Jun 23rd, 2003, 8:29pm » |
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Sigh, lost my first post attempt for not filling out an email, I'll make this brief, especially since I'm disagreeing with everyone 
This problem looks to me like a tarted up version of the sum of an alternating series which doesn't converege absolutely.
Consider the model:
Define a the queue by the order with which the bucket will be emptied (so the next ball to leave the bucket is position 0, the second is position 1, etc). Now take the liminf of the set of filled positions in the queue and you get N. Thus the number of balls in the queue (and thus in bucket) goes to infinity not 0.
Why is this model any worse then the model which considers which balls end up in the bucket?
If the "proof" that the bucket winds up empty is valid, then what about this:
Take SUM( 2 -1 +2 -1 + 2 ...) and label the numbers with naturals with one label per unit and different labels for positive and negative.
{1,2} - {1} + {3,4} - {2} +{5,6} - {3} + ....
Thus SUM(2-1+2-1+2...) is apparently well defined and equal to 0.
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wowbagger
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Re: Impish Pixie
« Reply #61 on: Jun 24th, 2003, 4:24am » |
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on Jun 23rd, 2003, 8:29pm, xlh wrote:
This problem looks to me like a tarted up version of the sum of an alternating series which doesn't converege absolutely.
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xlh, a concept like absolute convergence doesn't make sense when considering sets, i think - but who knows what these mathematicians kill their time with?
Quote:Consider the model:
Define a the queue by the order with which the bucket will be emptied (so the next ball to leave the bucket is position 0, the second is position 1, etc). Now take the liminf of the set of filled positions in the queue and you get N. Thus the number of balls in the queue (and thus in bucket) goes to infinity not 0.
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I'm not sure I fully understand your model, but I think that's not even necessary. Did you read
Icarus's post? I don't want to stress the technical part, the most important bit is probably this:
on Apr 4th, 2003, 4:11pm, Icarus wrote:
The fact is, cardinality is not continuous with respect to the limit expressed above. |
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The limit of the number of balls after finitely many steps tends to infinity, yet after the process has ended, there are no balls left. This counter-intuitive result is what makes this riddle so intriguing.
Quote:If the "proof" that the bucket winds up empty is valid, then what about this:
Take SUM( 2 -1 +2 -1 + 2 ...) and label the numbers with naturals with one label per unit and different labels for positive and negative.
{1,2} - {1} + {3,4} - {2} +{5,6} - {3} + ....
Thus SUM(2-1+2-1+2...) is apparently well defined and equal to 0. |
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Please read Icarus's explanation. 
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| « Last Edit: Jun 24th, 2003, 4:27am by wowbagger » |
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xlh
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Re: Impish Pixie
« Reply #62 on: Jun 24th, 2003, 6:23am » |
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Quote:I'm not sure I fully understand your model, but I think that's not even necessary. Did you read
Icarus's post? I don't want to stress the technical part, the most important bit is probably this:
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You probably should attempt to understand my model before dismissing it, especially since my point is that you can get different answers with different equally valid models.
Quote:| The limit of the number of balls after finitely many steps tends to infinity, yet after the process has ended, there are no balls left. This counter-intuitive result is what makes this riddle so intriguing. |
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You can't point to a single ball thats left in the bucket, but yet you can't point to a space in the bucket that isn't filled by a ball. Why is one model right while the other is wrong?
Quote:| Please read Icarus's explanation. |
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Sorry to be so snarky but please read my explanation.
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wowbagger
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Re: Impish Pixie
« Reply #63 on: Jun 24th, 2003, 9:16am » |
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on Jun 24th, 2003, 6:23am, xlh wrote:
You probably should attempt to understand my model before dismissing it, especially since my point is that you can get different answers with different equally valid models.
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Well, actually I did attempt to understand it.
So what do you do in your model? You have a queue that is empty at the beginning, then we add numbers 1 and 2, then we remove number 1, then we add numbers 3 and 4, then we remove number 2, and so forth. Right?
In which way do you think this differs from what's happening in the bucket? Every number enters the queue exactly once, and every number leaves the queue exactly once. That's why the queue has to be empty after the whole process is done.
Quote:| You can't point to a single ball thats left in the bucket, but yet you can't point to a space in the bucket that isn't filled by a ball. |
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Yes, I can. Every point in the bucket that was occupied by ball number n is no longer occupied after removing that ball in step n. And after this removal, ball n is never touched again, it remains out of the bucket.
Quote:| Sorry to be so snarky but please read my explanation. |
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| « Last Edit: Jun 24th, 2003, 9:18am by wowbagger » |
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xlh
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Re: Impish Pixie
« Reply #64 on: Jun 24th, 2003, 4:21pm » |
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I guess I was too brief after my initial posting mishap. A queue works by adding at the end and taking from the front. A "position" in the queue corresponds to how far from the front of the queue an item is. Think of it like a soda rack: when you pull a bottle out all the others slide forward. This means the sets of positions in the queue are:
{} starts empty
{1,2} 2 bottles added
{1} imp takes one, other slides forward
{1,2,3} 2 bottles added
{1,2} imp takes one, others slide forward
etc.
The lim inf is Z+ so the queue never empties.
If it concerns you note that the we don't care about the "take from front add to back" aspect of the queue.
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Icarus
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Re: Impish Pixie
« Reply #65 on: Jun 24th, 2003, 5:27pm » |
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xlh - The problem with saying that the limit of the number of balls at finite times is infinite, so there must be an infinite number of balls in the basket, is that it's failing is not in the "model". Its failing is in the mathematics!
It assumes this "continuity of cardinality" - which can be shown (as I have above) to be false in a strict mathematical sense.
At step n, the contents of the bucket correspond to the set Sn = {n+1, n+2, ..., 2n}
lim card(Sn) = [infty]
lim Sn = [emptyset]
Just because the cardinality is growing doesn't mean you will have anything at the limit.
On the other hand, the only problem with the "what balls are in the bucket at the end" approach are strictly in the model - that is, the relationship between the "real" situation and the mathematics used to describe it. The logic and mathematics behind this approach are sound. The only question is whether there are "equally valid" models which give different answers.
Your "queue" model is a variation of the one I refered to in this post, wherein the placing of balls in the same position infinitely many times results in a "limit ball" in that location at step [omega].
This model may be considered to meet the criteria of the original puzzle, which is why I said there that the "canonical" solution is unique only with the following added assumptions (which I still say are reasonable to take for granted):
(1) That the only possible contents are the balls mentioned in the puzzle itself.
(1a) That the labeled balls are labeled only with natural numbers (thanks, rmsgrey, for pointing out that this is not actually specified in the puzzle statement).
(2) That the only way for the balls to change position is when you or the imp physically move them.
These assumptions disallow "limit ball" models such as yours, since there are no balls available to fill the limit positions.
Concerning your alternating sum "counterexample". I'm sorry, but the mathematics behind the "canonical solution" do not behave at all like you are claiming in it.
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| « Last Edit: Aug 18th, 2003, 7:05pm by Icarus » |
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xlh
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Re: Impish Pixie
« Reply #66 on: Jun 24th, 2003, 7:34pm » |
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Please explain why you think the queue model is prevented by your 3 assumptions. My explanation about soda bottles sliding around was simply to help describe the concept, there is obviously no reason we can't define our queue positions without physically moving the balls.
Repeating that cardinality is not necessarily continuous or arguing that your model leads to an empty bucket are not useful because I already think both of those things. My question remains "why is cardinality well defined here?"
I'm a little confused why you refer me to that link when its discussion ends with:
Quote:
Please note that is just
one way of extending our model to include super tasks. We have only
shown that for these experiments, in our model, we come to consistent
conclusions. It does not mean that there are no other models which lead
to different, but also, within that model, consistent solutions.
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and
Quote:I conclude by stating that the result of the super task depends on how
our standard models are enlarged to include the execution of
supertasks. We have given one extension which leads to consistent
results for the supertasks suggested by Ross. Other models may lead to
different, but also consistent, conclusions.
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This is exactly why I don't think the question has a unique answer.
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wowbagger
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Re: Impish Pixie
« Reply #67 on: Jun 25th, 2003, 4:03am » |
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xlh, I had no problem with your model because of it using a queue, I'm more or less familiar with that concept. My impression was just that your model doesn't add anything to simply counting the number of balls in the bucket. It seems to be an unnecessary comlication of your argument.
on Jun 24th, 2003, 7:34pm, xlh wrote:| Repeating that cardinality is not necessarily continuous or arguing that your model leads to an empty bucket are not useful because I already think both of those things. My question remains "why is cardinality well defined here?" |
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If you understand that cardinality is not continuous, why don't you accept the consequence? My (admittedly sometimes rather naive) grasp of mathematics tells me that the cardinality of the empty set is well-defined. Maybe the question you want to ask is why the limit of sets is well-defined.
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| « Last Edit: Jun 25th, 2003, 4:07am by wowbagger » |
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xlh
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Re: Impish Pixie
« Reply #68 on: Jun 25th, 2003, 6:13am » |
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Quote:| Maybe the question you want to ask is why the limit of sets is well-defined. |
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Sure. If you can tell me why the limit of the sequence of sets from the queue is really {} and not Z+ then I'll gladly concede. Alternatively, explain why the queue model is not legitimate for this problem.
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Icarus
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Re: Impish Pixie
« Reply #69 on: Jun 25th, 2003, 4:07pm » |
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Quote:| Please explain why you think the queue model is prevented by your 3 assumptions. |
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The 3 assumptions allow for nothing to ever be in the bucket except the integer-labeled balls. At the end time, after the task is completely finished, they are all demonstrably somewhere else. So the bucket must be empty.
Your queue model, like the one described in the link, works by setting up a situation with an infinite number of state changes (ie - the contents of each queue position). In order for it to be a complete model, you must define in some fashion the state at time [omega]. This is the extension to "supertasks" the link describes. There are several ways of doing this for the original problem, but they must be logically consistent with the rest of the puzzle. For instance, you cannot declare that position 1 contains ball #862 at time [omega], because ball #862 was removed from the bucket at step 862, and has never been replaced. And of course the same is true of any other integer ball.
What is not excluded by the original puzzle is the possibility of a new "limit ball" occuring in the spot, since it was always occupied. My additional 3 assumptions preclude the existance of any such ball, or anything else being in that position at time [omega]. Hence the only available state for position 1 at time [omega] is "empty".
Quote:| Alternatively, explain why the queue model is not legitimate for this problem. |
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It is only illegitimate for the problem given the 3 additional assumptions. If you are allowed to break either assumption 1 or 2, then it can work.
Quote:| My question remains "why is cardinality well defined here?" |
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No - YOUR question was "why ISN'T cardinality well-defined here"?
My argument was not based on cardinality. It is the other, false, argument that assumes that cardinality is not only well-defined here, but is "continuous". That is why Argument 1 is false, even for models such as yours that do end up with an infinite number of balls in the bucket. For those models it gets the right answer for the wrong reasons, just like "cancelling the 6s" in 64/16 to get 4/1 = 4.
Cardinality is well-defined for any set. The problem does not come in defining the cardinality, but rather the set corresponding to the contents of the bucket at the end time.
Quote:| If you can tell me why the limit of the sequence of sets from the queue is really {} and not Z+ |
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The sequence of sets is Sn = {n+1,n+2,...,2n}
The limit inferior of a sequence of sets is by definition the set {x | [exists] N [in] [bbn] such that [forall] n > N, x [in] Sn}
The limit superior of Sn is by definition the set {x | [forall] N [in] [bbn], [exists] n > N such that x [in] Sn}
The limit exists if and only if these two sets are the same. For Sn = {n+1,n+2,...,2n}, they are both empty.
Now the question of what model you use determines whether or not this particular definition of limit provides the balls found in the bucket at time [omega]. But my point in giving it was to show out why argument 1 is false reasoning. And that holds true no matter what model you use.
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| « Last Edit: Aug 18th, 2003, 7:11pm by Icarus » |
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