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revolution
When all else fails, read the source


Joined: 24 Aug 2004
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revolution
So, the current list in case some of you might have forgotten:
Code:
Finites:
5.  9($^9$$$)       Tomasz Grysztar
4.   G($^G$$$)       Tomasz Grysztar (belatedly accepted)
3.      BB(BB(9))       Tomasz Grysztar
2.   (BB^9)(9)       Tomasz Grysztar (belatedly accepted)
1.      (BB^Z)(Z)       victor

Transfinites:
5.   Aleph-Z$$       MHajduk
4.   Beth-Z$$$       bitRAKE
3.   beth_w^w        Tomasz Grysztar
2.   Beth_EEE0       MHajduk
1.   Beth EEEw       revolution (shamelessly ripped from MHajduk's Beth_EEE0)    


The "Finites" can still be improved. And maybe the Transfinites can also.

Let the entries roll on ...
Post 08 Mar 2008, 19:33
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bitRAKE



Joined: 21 Jul 2003
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bitRAKE
The function S() is the number of steps required for the BB() function.

(S^Z$)(Z)

(Sidebar: here is a sequence I contributed to: A036235)

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Post 08 Mar 2008, 21:09
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bitRAKE



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bitRAKE
Ackermann Numbers

A(S^Z)(Z)
AAAAAAAAZ ?

Edit: The Grahm numbers are much larger - didn't notice they were used.

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Post 08 Mar 2008, 21:23
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tom tobias



Joined: 09 Sep 2003
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tom tobias
revolution wrote:
...can I strongly suggest we try to stay a bit closer to the topic.
My question, perhaps too far off topic, was this:
Where does this terminology of aleph, beth, and gimel, originate? I did not imagine that it was Cantor himself who introduced these Hebreic alphabetical symbols. After some further reading, I recognize now, that I am in error--apparently Cantor himself DID specify the use of these letters from the Hebrew alphabet: (Maybe it is not so surprising to anyone but me, to learn that (true to his name!) Cantor, despite Danish nationality, birth in Russia, education in Germany, and ostensibly a practicing Lutheran, was of Jewish ancestry!!!)

Cantor first established cardinality as an instrument to compare finite sets; e.g. the sets {1,2,3} and {2,3,4} are not equal, but have the same cardinality, namely three. ... Naming this cardinal number, aleph-null, Cantor proved that any unbounded subset of N has the same cardinality as N, even if this might appear at first to run contrary to intuition.

With regard to my question about the relationship between Goedel and Cantor:
Since the original inception of Cantor's set theory, it has been rigorously formalised in a number of different ways, collectively referred to as Axiomatic set theory . The most common formulation of Cantor's theory is known as Zermelo–Fraenkel set theory (or "ZF"). The fact that this axiomatic theory cannot be proved consistent is today understood in terms of Gödel's incompleteness theorems and it is not generally regarded as reason to doubt the theory.

So, notwithstanding its dependence upon Goedel, ZF is not generally doubted....The more I read, the less I am convinced...
Revolution's goal is to display, using only 9 ascii characters, symbols representing the number with the largest value. I look forward to learning the answer.
Smile
Post 09 Mar 2008, 00:25
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revolution
When all else fails, read the source


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revolution
bitRAKE wrote:
(S^Z$)(Z)
Accept, a new number 1. We are still not there yet but getting very warm now.
bitRAKE wrote:
AAAAAAAAZ
I don't think the syntax is correct, but even if is was it would be smaller than BB() or S().
bitRAKE wrote:
A(S^Z)(Z)
You will need the A inside the first brackets to be consistent with the current entries.
Post 09 Mar 2008, 01:31
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Goplat



Joined: 15 Sep 2006
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Goplat
I'm seeing a lot of functions and operators here that aren't really standard mathematical notation. So I think I'll define one of my own: LNXC(x) is the Largest Number that can be written in X Characters, excluding those written by Goplat.

LNXC(9)+1
Post 09 Mar 2008, 04:17
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r22



Joined: 27 Dec 2004
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r22
(S^Z$)^Z$

S^Z$$$$$$
Post 09 Mar 2008, 05:19
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r22



Joined: 27 Dec 2004
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r22
Biggest^Z = Its on the number line JUST after (S^Z$)(Z)

Max(*)^Z$

Max(I)^Z$

Max(R)^Z$
Post 09 Mar 2008, 05:26
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revolution
When all else fails, read the source


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revolution
r22 wrote:
(S^Z$)^Z$
That makes no sense.
r22 wrote:
S^Z$$$$$$
That makes no sense, what is S here?
r22 wrote:
Biggest^Z = Its on the number line JUST after (S^Z$)(Z)
Biggest is of indeterminate value, the definition is also circular if we keep replacing "Biggest" with the new largest value we never end the sequence.
r22 wrote:
Max(*)^Z$
What is *?
r22 wrote:
Max(I)^Z$
What is I, the integer set? There is no maximum.
r22 wrote:
Max(R)^Z$
What is R, the real set? There is no maximum.
Post 09 Mar 2008, 05:50
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revolution
When all else fails, read the source


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revolution
Goplat wrote:
I'm seeing a lot of functions and operators here that aren't really standard mathematical notation. So I think I'll define one of my own: LNXC(x) is the Largest Number that can be written in X Characters, ...

LNXC(9)+1
But it has no defined value. And a little bit too non-standard to be acceptable.

Goplat wrote:
excluding those written by Goplat.
Haha, a very smart little additional condition.
Post 09 Mar 2008, 05:55
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bitRAKE



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bitRAKE
Goplat wrote:
I think I'll define one of my own: LNXC(x) is the Largest Number that can be written in X Characters, excluding those written by Goplat.
This puzzle has appeared in various forms before, and I was reading one account where the guy recommended publishing just such a formula in a peer reviewed journal - so he could cite it as an answer to this question, lol.

(Unfortunately, this is the way laws are absued. Some jerk wants to do something illegal, so he lobbies to create a law which makes it legal - at least for a little while. Or throw enough speculation in the area to take some very questionable action without consequences. (Oh man, I need to take a few deep breaths - felt myself getting carried away...sleep - almost forgot.))

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Post 09 Mar 2008, 07:00
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revolution
When all else fails, read the source


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revolution
Re: LNXC(x)

This can easily be made logically inconsistent. If Goplat defines LNXC(x) as "the Largest Number that can be written in X Characters, excluding those written by Goplat." AND the revolution defines CXNL(x) as "the Largest Number that can be written in X Characters, excluding those written by revolution", then we cannot resolve a value.

Say person ABC eventually comes up with ABCDEFGHI as the best possible answer, then Goplat invokes the LNXC clause and thus defines ABCDEFGHI+1, now revolution invokes the CXNL caluse and thus defines ABCDEFGHI+2, then again Goplat applies LNXC ... and so on.

So by me defining a similar clause then I can effectively cancel all future possible rewordings and variations. Because any such attempt will create an indeterminate value and thus make it invalid as an answer.

Therefore, not accept LNXC(x)+1. Because it is logically inconsistent with the similar CXNL(x)+1.
Post 09 Mar 2008, 08:26
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victor



Joined: 31 Dec 2005
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victor
Post 09 Mar 2008, 10:04
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revolution
When all else fails, read the source


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revolution
victor wrote:
How big is A(G$,G$)$ ?
It is tiny, the busy beaver function will completely dwarf it.

BTW: I thought the Ackermann function took only one parameter - A(x)?
Post 09 Mar 2008, 10:08
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victor



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victor
G = Graham's number
Obviously, G is much, much bigger than Z.

Hence, (S^G$)(G) would be hard to beat!

Razz
Post 09 Mar 2008, 10:13
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revolution
When all else fails, read the source


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revolution
victor wrote:
(S^G$)(G)
Bingo, I new leader.

This was what I had in mind.
Code:
Finites:
5. BB(BB(9))       Tomasz Grysztar
4.   (BB^9)(9)       Tomasz Grysztar (belatedly accepted)
3.      (BB^Z)(Z)       victor
2.    (S^Z$)(Z)       bitRAKE
1.   (S^G$)(G)       victor

Transfinites:
5.   Aleph-Z$$       MHajduk
4.   Beth-Z$$$       bitRAKE
3.   beth_w^w        Tomasz Grysztar
2.   Beth_EEE0       MHajduk
1.   Beth EEEw       revolution (shamelessly ripped from MHajduk's Beth_EEE0)    
But, can anybody do any better?

It is still an open problem, both sets might still be able to be improved.
Post 09 Mar 2008, 10:18
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revolution
When all else fails, read the source


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revolution
For my original inception of this little puzzle I had "Beth S(G)" as my original 'largest number'. Of course subsequent events of all you guys posting made me revisit the answer.

The first revisit came when I split the list into two parts. I then had to formulate a new finite result, which is the result that victor has just posted "S^G$)(G)".

The second blow came when Tomasz submitted "Beth_w^w", which of course completely killed my puny "Beth S(G)". It was only when MHajduk posted Beth_EEE0 that the transfinite leader was born.

But it sure has been interesting, I learned some new things. Smile
Post 09 Mar 2008, 10:30
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victor



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victor
Razz

I still want an estimate of the order of A(G$,G$)$ !

Any formal proof that (S^G$)(G) is really bigger than the above? Well, I know that the BB function is incomputable ...

Confused
Post 09 Mar 2008, 10:49
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edfed



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edfed
G^G^G^G^G
Post 09 Mar 2008, 11:13
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revolution
When all else fails, read the source


Joined: 24 Aug 2004
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Location: In your JS exploiting you and your system
revolution
victor wrote:
Any formal proof that (S^G$)(G) is really bigger than [A(G$,G$)$]?
Here is my attempt:

  1. change (S^G$)(G) into (S^G)(G$), I think we can agree this new form is smaller
  2. Take away the step of the S^G and put in one trailing '$' into this: S(G$)$, once again I think we can agree this is smaller.
  3. Now we can compare the two "A(G$,G$)$" and "S(G$)$". And we know that the BB function (and it's associated S() function) grow faster than the Ackermann function. Therefore we can say that A(G$,G$)$ < S(G$)$ < (S^G)(G$) < (S^G$)(G)
Post 09 Mar 2008, 12:56
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