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Sir Col
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Re: Transfinite Subway  
« on: Oct 7th, 2003, 12:19pm »

::
If this is this is hard then I'm missing something?! (somehow I expect this to be quoted)  Wink
 
In transfinite arithmetic, aleph_0 * aleph_0 = aleph_0. So infinitely many embarkments of infinitely many passengers amounts to an infinity of passengers.
 
Actually, I don't know how to prove that aleph_02=aleph_0, which is a standard result in transfinite arithmetic. It is clear that aleph_0+aleph_0=aleph_0, by considering the sum of odds and evens. Therefore, any finite sum of aleph_0 will amount to aleph_0; that is, n *aleph_0 = aleph_0. How we extend this to squaring, I don't know?
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Re: Transfinite Subway  
« Reply #1 on: Oct 7th, 2003, 12:31pm »

Perhaps I'm missing something, but when I use the following definition:
 
aleph_0 = 1;
 
then the solution is easy...
 
To be serious now, aleph_0[sup2] = aleph_0 is easily shown from the countability of the rational numbers. We just number the pairs of natural numbers like this:
 
1 => 1,1
2 => 2,1
3 => 1,2
4 => 3,1
5 => 2,2
6 => 1,3
7 => 4,1
...
 
This lets us count two infinities with just one.
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Re: Transfinite Subway  
« Reply #2 on: Oct 7th, 2003, 4:24pm »

on Oct 7th, 2003, 12:19pm, Sir Col wrote:
::[hide]
If this is this is hard then I'm missing something?! (somehow I expect this to be quoted)  Wink

 
Happy to oblige!  Cool
 
So are you arguing that the train is empty at the Hilbert Hotel, or full, or somewhere in-between?
 
And by the way, you do too know the proof that [smiley=aleph.gif]02 = [smiley=aleph.gif]0. You've commented on it yourself in your attack on denumerability! It's the "snake back and forth over pairs" bit that is the first step to proving the countability of the rationals. (Knowing how you feel about mathematical development, I thought I should set the record straight that the denumerability of the Rationals follows from [smiley=aleph.gif]02 = [smiley=aleph.gif]0, not the other way around!)
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Re: Transfinite Subway  
« Reply #3 on: Oct 8th, 2003, 12:41am »

That's why I said I don't know how to prove it. I'm still unsure about using the assumption that the infinite grid is countable in the snaking fashion. Not that I doubt the process, and I accept its validity. However, I realise the we need to first assume that [smiley=aleph.gif]02 = [smiley=aleph.gif]0, and I don't know how to show this is true.
 
Hadn't I answered the question? [smiley=aleph.gif]0 passengers embark at each station, but we don't know the 'amount'. So infinitely many unknown, but infinite amounts, will leave the train with infinitely many ([smiley=aleph.gif]0) passengers on the train at Hilbert Hotel; that is, we do not enter a new degree of infinity (a higher aleph).
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Re: Transfinite Subway  
« Reply #4 on: Oct 8th, 2003, 10:08am »

I couldn't resist my own challenge, and I've just found a really cool way to prove that [smiley=aleph.gif]02 = [smiley=aleph.gif]0:
 
Suppose that n[in]{1,2,3,...} (set of natural numbers), and p[in]{2,3,5,,7,11,...} (set of primes).
 
Clearly pn is finite and each of these values created, by taking one element from n and one element from p, is unique. That is, 21, 22, 23, ..., 31, 32, 33, ... In other words, we have infinitely many lots of infinitely many values, hence we demonstrate that [smiley=aleph.gif]02 = [smiley=aleph.gif]0.
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Re: Transfinite Subway  
« Reply #5 on: Oct 8th, 2003, 8:30pm »

on Oct 8th, 2003, 12:41am, Sir Col wrote:
That's why I said I don't know how to prove it. I'm still unsure about using the assumption that the infinite grid is countable in the snaking fashion. Not that I doubt the process, and I accept its validity. However, I realise the we need to first assume that [smiley=aleph.gif]02 = [smiley=aleph.gif]0, and I don't know how to show this is true.

 
I'm not sure what you are saying here. It sounds like you believe that you need [smiley=aleph.gif]02 = [smiley=aleph.gif]0 in order for the "snaking" trick to work. If so, you've got it backwards. It is the "snaking" trick that shows that [smiley=aleph.gif]02 = [smiley=aleph.gif]0.
 
If you mean that [smiley=aleph.gif]02 = [smiley=aleph.gif]0 is needed to prove countability of the rationals, but you don't follow how the snaking trick shows [smiley=aleph.gif]02 = [smiley=aleph.gif]0, then:
 
A set is countable if it can be placed in 1-1 correspondence with the set of natural numbers. In the other thread, I gave you the formula for the nth pair of this correspondence for the variant of the snaking trick wherein instead of going back and forth, you always start at the bottom and go up the diagonals (this is easier to formulate). The formulas for this correspondance are:
 
(p, q) [mapsto] (p+q)(p+q+1)/2 + q
 
n(n+1)/2+k [mapsto] (n - k, k)   (n = 0, 1, 2, ...; k = 0, ..., n)
 
It is easy to verify that each of these maps is well-defined on all of [smiley=bbw.gif] or [smiley=bbw.gif] [times] [smiley=bbw.gif], and that they are inverses of each other. This proves the 1-1 correspondence.
 
Quote:
Hadn't I answered the question? [smiley=aleph.gif]0 passengers embark at each station, but we don't know the 'amount'.
Huh
 
This sentence makes as much sense as saying "2 passengers embark at each station, but we don't know the amount."
 
We do know the amount: [smiley=aleph.gif]0. This is a number, not a variable.
 
Quote:
So infinitely many unknown, but infinite amounts, will leave the train with infinitely many ([smiley=aleph.gif]0) passengers on the train at Hilbert Hotel; that is, we do not enter a new degree of infinity (a higher aleph).

 
But: One of the properties of cardinal multiplication is that if two cardinals are multiplied together and at least one of them is infinite, then the product is the larger of the two:
 
x * y = x    if x [ge] y and x [ge] [smiley=aleph.gif]0.

 In this case, we have [smiley=aleph.gif]0 passengers embarking at each of [smiley=aleph.gif]1 stations, for a total of [smiley=aleph.gif]0 * [smiley=aleph.gif]1 = [smiley=aleph.gif]1 passengers embarking. But we also have 1 * [smiley=aleph.gif]1 = [smiley=aleph.gif]1 passengers disembarking. So should we not have [smiley=aleph.gif]1 - [smiley=aleph.gif]1 = 0 passengers left on the train at the end?
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Re: Transfinite Subway  
« Reply #6 on: Oct 9th, 2003, 1:58am »

I should have expected this from T&B; he has a reputation for dropping a mathematical time bomb and running. Like a fool I walked over to it, and said, "Hmm... what's this nice present that T&B has left for us?"
 
I think I also warned myself before about playing with the big boys.  Cry
 
I forgot that [omega] was the first uncountable cardinal: in [smiley=aleph.gif]1.
 
Please excuse my utter naivity, but I'm going to continue my reputation as a fumbling enthusiast and give this a go...
 
I would answer your question in the affirmative, Icarus.  
 
The number of passengers embarking and disembarking, although different in a finite sense, are equivalent in a transfinite sense. As [omega] * 1 = [omega] * [smiley=aleph.gif]0, [omega]([smiley=aleph.gif]0–1)=0.
 
Actually this makes sense (even to me). On the grand scale of things, 1 and [smiley=aleph.gif]0 are dwarfed, in fact, they are nothing compared to [smiley=aleph.gif]1. It's a little like comparing a billion billion billion billion lots of a billion with one [smiley=aleph.gif]0; one remains finite, the other immediatley enters a new realm of size.
 
*prepares to be corrected, but looks forward to the lesson*
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Re: Transfinite Subway  
« Reply #7 on: Oct 9th, 2003, 10:58am »

on Oct 9th, 2003, 1:58am, Sir Col wrote:
Actually this makes sense (even to me).

Well you're the only one! Who is this idiot?  Roll Eyes
 
 
Actually, I've had a re-think...
 
We know that [smiley=aleph.gif]0–1=[smiley=aleph.gif]0, so it follows that [omega] * ([smiley=aleph.gif]0–1) = [omega] * [smiley=aleph.gif]0 = [smiley=aleph.gif]1.
 
But it is equally plausible that [omega] * ([smiley=aleph.gif]0–1) = [omega] * [smiley=aleph.gif]0 – [omega] * 1 = [smiley=aleph.gif]1 – [smiley=aleph.gif]1.
 
The question is, what is [smiley=aleph.gif]1 – [smiley=aleph.gif]1?
 
If we have set A, size [smiley=aleph.gif]1, and set B, size [smiley=aleph.gif]1, then A–B is null, iff A and B are identical; that is, each member of A maps in a one-to-one correspondence with B and vice versa.
 
As 1 and [smiley=aleph.gif]0 are not identical, their difference is not zero; cf. natural numbers minus the odd numbers = even numbers.
 
As I am losing track of what I think, I'm beginning to suspect that it is indeterminate. However, I'm swayed slightly to believe that it is [smiley=aleph.gif]1.
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Re: Transfinite Subway  
« Reply #8 on: Oct 9th, 2003, 11:09am »

Note that the impish pixie proof does not work here, because nowhere do we say that the people available are limited to the ones who get off the trains. In fact, I think you could show that there must exist people who get on the train and never get off. For instance, the people who board the train at the stop before the Hilbert Hotel. They don't all have a chance to get off. So the answer must be at least [smiley=aleph.gif]0.
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Re: Transfinite Subway  
« Reply #9 on: Oct 9th, 2003, 12:55pm »

Just when I thought I was getting somewhere...
 
Oh, I have an idea! All the people that get on have the names Abe XXX, Bert XXX, Cameron XXX, Diane XXX, ... , Albert Adam XXX, Art Bedelia XXX, ... and so on, where XXX is the number of the station where they got on. Abe is always the one that gets off. Therefore, there are [smiley=aleph.gif]1 people, with all sorts of names, but nobody named Abe.
 
Convincing, right?
 
Hmm ... but alternately, the people could get off in order, so Abe 0 gets off first, then Bert 0, etc., in which case there'd be nobody left.
 
So perhaps it's undecideable from the information given?
 
Whoa! TnB, what have you done to the question? There appears to be [smiley=aleph.gif]0aleph0 in step ii) of the question now.
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Re: Transfinite Subway  
« Reply #10 on: Oct 9th, 2003, 5:42pm »

on Oct 9th, 2003, 1:58am, Sir Col wrote:
I forgot that [omega] was the first uncountable cardinal: in [smiley=aleph.gif]1.

 
A small error here. [omega] represents the first infinite ordinal, not the first uncountable one. As such, [omega] is the ordinal representation of [smiley=aleph.gif]0, not [smiley=aleph.gif]1. I am not sure what convention holds for the first uncountable ordinal, but [omega]1 makes as much sense as anything.
 
By the way, T&B planted this trap, but I'm the one who set it. Wink
 
on Oct 9th, 2003, 12:55pm, James Fingas wrote:
So perhaps it's undecideable from the information given?

 
What was that last choice T&B offered?
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Re: Transfinite Subway  
« Reply #11 on: Oct 10th, 2003, 12:45am »

I can answer that question...
on Oct 9th, 2003, 10:58am, Sir Col wrote:
I'm beginning to suspect that it is indeterminate.

 
With regards the convention for use of omega, I was under the impression that [omega] was the original notation that Cantor devised for the first uncountable ordinal (position): 0,1,2,3,...,[omega],[omega]+1,..; but in terms of size (cardinality), [omega]=[omega]+1. Once he realised there were different degrees of infinity he switched to the use of subscript. I believe that convention now defines [omega]k to be the first uncountable ordinal with cardinality [smiley=aleph.gif]k.
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Re: Transfinite Subway  
« Reply #12 on: Oct 10th, 2003, 10:14am »

on Oct 9th, 2003, 12:55pm, James Fingas wrote:
So perhaps it's undecideable from the information given?

I agree with James. Also, I would like to quote the following post by Icarus from the "Impish Pixie" thread:
 
on Apr 13th, 2003, 7:28pm, Icarus wrote:
Change the imp's behavior to "The imp removes a ball completely at random", and you do not have enough information to determine the result completely. You can say that the probability of the basket being empty is 100% (with infinitely many cases, this is not the same as saying that the basket must be empty). - I read this result in another discussion of this puzzle, and have not checked myself if it is true, but it sounds right.

I remember that my friend and I once showed that it's true.  
 
Is it true also in the "uncountable" case? I.e., assuming that people get off randomly (do we need the Axiom of Choice here?), is it true that the probability of the train being empty is 1?
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Re: Transfinite Subway  
« Reply #13 on: Oct 10th, 2003, 8:38pm »

I don't know for certain. I never have bothered to try it. The probability has to be either 1 or 0. Of that much I am sure.
 
Concerning notation. I don't know what Cantor's original notation was, but I do know that nowhere in the literature have I ever seen [omega] used for anything other than the first infinite (but still countable) ordinal: [omega] = [smiley=aleph.gif]0.
 
And by the way, I should point out that when I wrote [smiley=aleph.gif]1 - [smiley=aleph.gif]1 = 0 earlier, I was pulling a fast one. Because, as I said in that same thread, when infinite cardinals are added together, the sum is always the larger of the two, subtraction cannot be defined for infinite cardinals. Like 0/0, [smiley=aleph.gif]1 - [smiley=aleph.gif]1 is a fundamentally undefinable quantity, because it has many answers that all work equally well.
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Re: Transfinite Subway  
« Reply #14 on: Oct 11th, 2003, 9:56pm »

Icarus, what is wrong with the following argument (which is not my own):
 
First, we prove that the subway must empty out at some positive countable ordinal. Suppose it doesn't. Then for each countable ordinal x > 0, someone gets out at stop x. He must have gotten on at some stop f(x) < x. f is a function from [omega]1 - {0} to [omega]1, and since only [smiley=aleph.gif]0 people get on at each stop, it is countable-to-1. Now for every countable ordinal x we have the sequence x > f(x) > f(f(x)) >... . This sequence cannot decrease forever, since ordinals are well-ordered. If it stops at fk(x) = 0, we can identify x by giving the value of k and saying which preimages of f we need to take to pass from fk(x) = 0 to f(k-1)(x), from f(k-1)(x) to f(k-2)(x), ..., and from f(x) to x. But this gives us a countable labelling of [omega]1, which can't exist. So the subway emptied out at some countable ordinal.
 
Now if the subway empties out at an ordinal y, exactly the same argument as above shows that it must empty out at y+z, for some countable ordinal z > 0.  So it empties out at some y' > y, and y' is countable if y is.
 
Finally, if the subway is empty at each of a set of stops S, it is empty at the supremum m of S; for, if there is someone on it, he got on at stop x < m; but there is some stop y in S with x < y, and he was not on the subway then.
 
It now follows by transfinite induction that the subway is empty when it reaches [omega]1.
 
« Last Edit: Oct 12th, 2003, 6:09pm by ThudnBlunder » IP Logged

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Re: Transfinite Subway  
« Reply #15 on: Oct 12th, 2003, 7:20am »

At each station people get on, after other people get off.. so at every station there should be at least the people that got on at the last station.
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Re: Transfinite Subway  
« Reply #16 on: Oct 12th, 2003, 9:24am »

on Oct 10th, 2003, 8:38pm, Icarus wrote:
Concerning notation. I don't know what Cantor's original notation was, but I do know that nowhere in the literature have I ever seen [omega] used for anything other than the first infinite (but still countable) ordinal: [omega] = [smiley=aleph.gif]0.

Icarus, you seem to be confusing finite ordinals with countable ordinals.
 
Ordinal refers to the numerical position: 1st, 2nd, 3rd, ... Whereas cardinality refers to the size of the set; cf. Peano's definition of the number system, giving us the cardinal numbers.
 
The cardinality of the set of natural numbers: 1,2,3,..., where each term is finite, is [smiley=aleph.gif]0, because each member can be put in a one-to-one correspondence with the finite ordinals.
 
Omega, [omega], is defined as the smallest infinite ordinal. Hence the order of numbers would be 1,2,3,...,[omega],[omega]+1,[omega]+2,...
 
The cardinality of the set of countable ordinals is [smiley=aleph.gif]1.
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Re: Transfinite Subway  
« Reply #17 on: Oct 12th, 2003, 3:05pm »

Sir Col, if you will read this post, I give an abbreviated account of the mathematical definitions of Cardinal and Ordinal numbers. While these objects get their names from the linguistic terminology you describe, their definition, particularly for ordinals, is considerably more rigorous. Both definitions are due to Cantor, I believe. Peano only provided a definition of Natural numbers (and zero), but no infinite numbers.
 
There is a natural identification between each cardinal number, and the smallest ordinal representing an ordering of that cardinality. Under this identification, the operations on ordinals restrict to the corresponding operations on cardinals. By it, we may take the cardinals to be a subclass of ordinals. It is by this identification that I said [omega] = [smiley=varaleph.gif]0.
 
By the definitions:
Every ordinal < [omega] = [smiley=varaleph.gif]0 is finite.
Every ordinal [ge] [omega] = [smiley=varaleph.gif]0  is infinite.
Every ordinal < [omega]1 = [smiley=varaleph.gif]1 is countable.
Every ordinal [ge] [omega]1 = [smiley=varaleph.gif]1 is uncountable.


I believe that the argument in T&B's last post is correct. Much to my chagrine, when I first saw this problem I made the same mistake (but with far less excuse) as James and Towr have. It was only upon reading that post that I went back over the argument and realized the error that should have been obvious to me.
 
James' "Abe gets off" and Towr's "the people who got on at the last stop will always be on" both run afoul of the same fact: not every ordinal number has an immediate predecessor. For first order limit ordinals, starting with [omega] itself, there is no Abe to get off, so one of the Berts must get the boot.
 
According to T&B's argument, it proves impossible to ever arrange leavings so that the train does not completely empty out infinitely many times before it gets to the Hotel.
 
I don't have time right now to give the argument the attention it deserves. I would very much like to recast it in a more mathematical form to make it clearer and better understand what it implies.
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Re: Transfinite Subway  
« Reply #18 on: Oct 12th, 2003, 4:08pm »

on Oct 12th, 2003, 9:24am, Sir Col wrote:

Icarus, you seem to be confusing finite ordinals with countable ordinals.

Sorry, Icarus, you're the last person who would need to have explained to them basic stuff like that; I was being very naughty making that post. I had a really tough week and I needed to cheer myself up. I had hilarious images of you reading it and exploding like something out of a cartoon: your face going red and ready to explode, accompanying klaxxons, screams of, "What!", the cat hurtling from the room to take cover, and steam venting from your ears. The beauty of the internet is that we have no idea what people look like so it is possible to employ our most wacky imaginations – the pictures I have of most people here are quite surreal.
 
If you were only slightly worried about my sanity before, you'd be really worried now!  Grin
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Re: Transfinite Subway  
« Reply #19 on: Oct 12th, 2003, 11:44pm »

on Oct 12th, 2003, 3:05pm, Icarus wrote:
Towr's "the people who got on at the last stop will always be on".
That's so not what I said.. I said they would still be on at the next station, nothing more.
Causality demands it.. (and the riddle states it)
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Re: Transfinite Subway  
« Reply #20 on: Oct 13th, 2003, 2:50am »

I also agree with THUDandBLUNDER's argument.
 
As this thread discussion surpasses the initial question, I would like to ask Icarus: do you agree with the following Sit Col's statement:
 
on Oct 12th, 2003, 9:24am, Sir Col wrote:
The cardinality of the set of countable ordinals is [smiley=aleph.gif]1.

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Re: Transfinite Subway  
« Reply #21 on: Oct 13th, 2003, 9:23pm »

on Oct 12th, 2003, 11:44pm, towr wrote:

That's so not what I said.. I said they would still be on at the next station, nothing more.
Causality demands it.. (and the riddle states it)

 
Pardon me for being somewhat inaccurate in referencing your argument - I did not mean to change it. However what I said still stands. Your argument was:
 
on Oct 12th, 2003, 7:20am, towr wrote:
At each station people get on, after other people get off.. so at every station there should be at least the people that got on at the last station.
(Emphasis is mine.)
 
It fails in that stations for limit ordinals do not have a "last station" before them.
 
on Oct 13th, 2003, 2:50am, Barukh wrote:
I would like to ask Icarus: do you agree with the following Sir Col's statement:
 
The cardinality of the set of countable ordinals is [smiley=aleph.gif]1.

 
Every ordinal represents an order-type. In particular, from the way the ordinals are defined, every ordinal [smiley=r.gif] represents the order-type of the set of all ordinals < [smiley=r.gif].
 
By definition, the least uncountable ordinal is the smallest ordinal representing an uncountable well-ordered set. This requires the collection of all ordinals less than it - i.e. all countable ordinals - to be uncountable. Since it is the first uncountable ordinal, it's corresponding cardinality can only be [smiley=varaleph.gif]1.
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Re: Transfinite Subway  
« Reply #22 on: Oct 14th, 2003, 12:28am »

on Oct 13th, 2003, 9:23pm, Icarus wrote:

It fails in that stations for limit ordinals do not have a "last station" before them.
True.. but then you can't get there going from one station to the next either..
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Re: Transfinite Subway  
« Reply #23 on: Oct 14th, 2003, 12:46am »

on Oct 13th, 2003, 9:23pm, Icarus wrote:
By definition, the least uncountable ordinal is the smallest ordinal representing an uncountable well-ordered set. This requires the collection of all ordinals less than it - i.e. all countable ordinals - to be uncountable. Since it is the first uncountable ordinal, it's corresponding cardinality can only be [smiley=varaleph.gif]1.

I wouldn't have put it quite so succinctly, and I trust both your judgement and the books from which I obtained the information originally. However, I do have a question which hasn't been answered. I accept that the cardinality is greater than [smiley=aleph.gif]0, but how can we be sure that it is not in a greater degree of infinity? If the ordinals are uncountable, why isn't it [smiley=aleph.gif]w?
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Re: Transfinite Subway  
« Reply #24 on: Oct 14th, 2003, 9:08pm »

Because by definition, [smiley=varaleph.gif]1 is the first cardinal after [smiley=varaleph.gif]0.
 
There is more to proving it rigorously, but it's late, and I am having trouble thinking about it, so I'll try to put it together tomorrow if noone beats me to it.
« Last Edit: Oct 14th, 2003, 9:09pm by Icarus » IP Logged

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