flat assembler
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r22
(BB^G)(G)


09 Mar 2008, 17:42 

revolution
r22 wrote: (BB^G)(G) 

09 Mar 2008, 18:21 

victor
Even though I am the leader in the "Finites" section, I am here challenging the validity of the S function.
Refer to this. Quote: ...but there remain about 40 machines with nonregular behavior which are believed to never halt, but which have not yet been proven to run infinitely... Since S(n) is defined to be the largest number of steps taken by any nstate 2symbol Turing machine, S(n) may be infinity for some n! As infinity is not acceptable, it seems that we need to reconsider the validity of the S function and all BBrelated functions! 

10 Mar 2008, 03:33 

victor
Ok, I found this:
Quote: Lin, Shen and Radó, Tibor (1965), ''Computer Studies of Turing Machine Problems'', Journal of the Association for Computing Machinery, Vol. 12, No. 2 (April 1965), pp. 196212. Lin was a doctoral student under Radó. This paper appeared in part of Lin's thesis. Lin proves that Σ(3) = 6 and S(3) = 21 by proving that all 3state 2symbol Turing Machines which don't halt after 21 steps will never halt (Most are proven automatically by a computer program, however 40 are proven by human inspection). The definition seems to exclude all TMs that don't halt. Fine, I'm still the leader in the "Finites" section! 

10 Mar 2008, 04:35 

revolution
Yes indeed, creating a nonhalting machine is of course exceedingly easy and would be rather pointless trying to count the transitions.
No need to panic victor, I think your current position is quite safe. But I'm not guaranteeing it, 'cause ya just never know if/when someone will find a larger value that can be squeezed into 9 characters. 

10 Mar 2008, 06:57 

bitRAKE
Community Beaver, CB(n) is equal to the sum of all one bits generated by all halting nstate Turing machines. C(n) is the steps taken by all nstate Turing machines to reach CB(n).


10 Mar 2008, 15:54 

revolution
bitRAKE wrote: Community Beaver, CB(n) is equal to the sum of all one bits generated by all halting nstate Turing machines. C(n) is the steps taken by all nstate Turing machines to reach CB(n). Sounds interesting, what are the first three values? 

10 Mar 2008, 16:00 

r22
(CB^G)(G)
Will assume bitRAKE isn't making it up, but you know what they say about assuming 

10 Mar 2008, 17:25 

revolution
r22 wrote: (CB^G)(G) r22 wrote: Will assume bitRAKE isn't making it up, but you know what they say about assuming 

10 Mar 2008, 17:33 

bitRAKE
Guess it sounded better than I thought. Ya'd want to go right for (C^G$)(G) if it wasn't made up.


10 Mar 2008, 17:34 

r22
You can never be too sure with obscure math functions.
I don't think we spent enough time with Graham's number though Quote:
http://home.earthlink.net/~mrob/pub/math/largenum4.html Using gn() we can make a number much larger than Graham's number which is gn(64) So the largest number possible in 9 ASCII characters is... gn(gn(G)) 

10 Mar 2008, 20:09 

r22
BB(gn(G))
(gn^G)(G) S(gn(G$)) For completeness 

10 Mar 2008, 20:16 

bitRAKE
revolution wrote: Sounds interesting, what are the first three values? r22, unfortunately, all those values are smaller. The composite application of S() is just staggering. If a more compact notation can be found for composite then you'd have a winner  five characters are currently devoted to it: (^)() 

10 Mar 2008, 21:07 

r22
(gn^G)(G)
Is it greater than (S^G)(G$), I don't know. Is S(64) > G and as n > INF does this hold true? Quote: five characters are currently devoted to it: (^)() F^X_(Y) ^_() 4 characters? Using _ to negate the superscript to regular script. S^G$_(G$) gn^G_(G$) Last edited by r22 on 11 Mar 2008, 01:08; edited 1 time in total 

10 Mar 2008, 21:58 

bitRAKE
r22 wrote: Is S(64) > G and as n > INF does this hold true? 

11 Mar 2008, 00:23 

revolution
The BB function is that fastest growing function known to date. Even the graham's number function, while it looks like a candidate for huge numbers, will grow slower.
But, BB(X) > gn(X) where X = unknown_value. We don't know where the point is that BB will overtake gn. Maybe at X~10 or so will do it, but how to compute it? 

11 Mar 2008, 01:35 

r22
***EDIT*** will take the combined words of Rev & bitRake that BB() grows faster
re bitRAKE I have a hard time beleiving that S() > gn() S(1)=1; S(2)=6; S(3)=21; S(4)=107; S(5)=134,467 BB(1)=1; BB(2)=4; BB(3)=6; BB(4)=13 gn(1) >= 27; gn(2) >= 7,625,597,484,987; gn(3) >= 3^(7,625,597,484,987); hy(3, gn(n1)+2, 3) 

11 Mar 2008, 02:23 

r22
How about chained arrow notation
http://en.wikipedia.org/wiki/Conway_chained_arrow_notation G>G$$>G  G UpArrow^G$$ G 3→3→64→2 < Graham's number < 3→3→65→2 Not enough characters to get a 4 arrow chain in, I don't have a clue on how to check if 3 arrows using Graham's and superfactorials is enough. re: revolution Whats the ruling on my alternate composite function notation? (F^x)(y) == F^x_(y) == F^x (y) 5 chars 4 chars I think a subscript (_) or evena space ( ) is valid for negating the superscript (^) notation. 

11 Mar 2008, 02:53 

revolution
r22 wrote: How about chained arrow notation r22 wrote: Whats the ruling on my alternate composite function notation? Your point with the space (and even without the space) I have seen in things like tan²(x), and I guess it could be extended to tan^2(x). But rather than meaning tan(tan(x)) it means (tan(x))^2 But, just for interest sake, perhaps the construction S^n(x) can be accommodated to mean S(S(S...(x))). In which case a class of new leaders can be constructed. I don't mind keeping this open, so I have constructed a new top value, anyone care to make a new entry? 

11 Mar 2008, 03:38 

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