flat assembler
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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) 

08 Mar 2008, 21:09 

bitRAKE
Ackermann Numbers
A(S^Z)(Z) AAAAAAAAZ ? Edit: The Grahm numbers are much larger  didn't notice they were used. 

08 Mar 2008, 21:23 

tom tobias
revolution wrote: ...can I strongly suggest we try to stay a bit closer to the topic. 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 errorapparently 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!!!)
With regard to my question about the relationship between Goedel and Cantor: http://en.wikipedia.org/wiki/Controversy_over_Cantor's_theory wrote: 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. 

09 Mar 2008, 00:25 

revolution
bitRAKE wrote: (S^Z$)(Z) bitRAKE wrote: AAAAAAAAZ bitRAKE wrote: A(S^Z)(Z) 

09 Mar 2008, 01:31 

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 

09 Mar 2008, 04:17 

r22
(S^Z$)^Z$
S^Z$$$$$$ 

09 Mar 2008, 05:19 

r22
Biggest^Z = Its on the number line JUST after (S^Z$)(Z)
Max(*)^Z$ Max(I)^Z$ Max(R)^Z$ 

09 Mar 2008, 05:26 

revolution
r22 wrote: (S^Z$)^Z$ r22 wrote: S^Z$$$$$$ r22 wrote: Biggest^Z = Its on the number line JUST after (S^Z$)(Z) r22 wrote: Max(*)^Z$ r22 wrote: Max(I)^Z$ r22 wrote: Max(R)^Z$ 

09 Mar 2008, 05:50 

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, ... Goplat wrote: excluding those written by Goplat. 

09 Mar 2008, 05:55 

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. (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.)) 

09 Mar 2008, 07:00 

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. 

09 Mar 2008, 08:26 

victor


09 Mar 2008, 10:04 

revolution
victor wrote: How big is A(G$,G$)$ ? BTW: I thought the Ackermann function took only one parameter  A(x)? 

09 Mar 2008, 10:08 

victor
G = Graham's number
Obviously, G is much, much bigger than Z. Hence, (S^G$)(G) would be hard to beat! 

09 Mar 2008, 10:13 

revolution
victor wrote: (S^G$)(G) This was what I had in mind. Code: Finites: 5. BB(BB(9)) Tomasz Grysztar 4. (BB^9)(9) Tomasz Grysztar (belatedly accepted) 3. (BB^Z)(Z) victor 2. (S^Z$)(Z) bitRAKE 1. (S^G$)(G) victor Transfinites: 5. AlephZ$$ MHajduk 4. BethZ$$$ bitRAKE 3. beth_w^w Tomasz Grysztar 2. Beth_EEE0 MHajduk 1. Beth EEEw revolution (shamelessly ripped from MHajduk's Beth_EEE0) It is still an open problem, both sets might still be able to be improved. 

09 Mar 2008, 10:18 

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. 

09 Mar 2008, 10:30 

victor
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 ... 

09 Mar 2008, 10:49 

edfed
G^G^G^G^G


09 Mar 2008, 11:13 

revolution
victor wrote: Any formal proof that (S^G$)(G) is really bigger than [A(G$,G$)$]?


09 Mar 2008, 12:56 

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