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edfed



Joined: 20 Feb 2006
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edfed
the dot is for 1/10=0.1
Post 02 Jan 2010, 17:45
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Fanael



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Fanael
revolution wrote:
edfed wrote:
∞^∞^∞^∞.1
What does the dot mean?
Function composition. Razz
Post 02 Jan 2010, 17:50
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revolution
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revolution
edfed wrote:
the dot is for 1/10=0.1
Wouldn't a plus (+) be better than a decimal point? Wink
Post 02 Jan 2010, 18:49
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edfed



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edfed
jmp $

; assume a timer interrupt is set and generate a verybig number on a 1tera byte drive.
a 8terabits value. something like 2^3^2^40 bits value. it is little bit bigger than any existing number writen on thçs planet.
Post 03 Jan 2010, 00:54
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revolution
When all else fails, read the source


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revolution
edfed wrote:
jmp $

; assume a timer interrupt is set and generate a verybig number on a 1tera byte drive.
?
edfed wrote:

a 8terabits value. something like 2^3^2^40 bits value. it is little bit bigger than any existing number writen on thçs planet.
Googolplex?
Post 03 Jan 2010, 00:59
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edfed



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edfed
Quote:

?


of course, i assume jmp $ is less than 9 characters (5 chars, 6 if we count 0 ending, and 7 if we ad a tabulation at the start.
9 if we add 10,13 before 0 ending char., then, it is OK! Smile.
Post 03 Jan 2010, 02:02
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revolution
When all else fails, read the source


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revolution
edfed wrote:
of course, i assume jmp $ is less than 9 characters (5 chars, 6 if we count 0 ending, and 7 if we ad a tabulation at the start.
9 if we add 10,13 before 0 ending char., then, it is OK! Smile.
Laughing Well, yes, 'jmp $' certainly fits within 9 characters. But it ain't a number, or a representation of a number. It is assembly code dude.

And 256^40 ain't very big anyway.

And we have 2TB drives now. But sorry, 256^41 also ain't very big.

Wanna try again?
Post 03 Jan 2010, 02:30
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Borsuc



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Borsuc
revolution wrote:
Laughing Well, yes, 'jmp $' certainly fits within 9 characters. But it ain't a number, or a representation of a number. It is assembly code dude.
In that case all that abstract math is not numbers either, just representation of values. Wink

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Post 03 Jan 2010, 23:39
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bitRAKE



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bitRAKE
S(G,G)(G) or S(G,G)^G$, probably S^G$(G,G)

http://www.logique.jussieu.fr/~michel/tmi.html
Quote:
We denote by S(k,n) the time taken by the busy beaver with k states and n symbols which takes the most time to stop:
    S(k,n) = max {s(M) : M is a busy beaver with k states and n symbols}
...thank goodness for continued research establishing new notation.

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Post 30 Mar 2013, 05:18
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alessandro95



Joined: 24 Mar 2013
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alessandro95
http://en.wikipedia.org/wiki/Steinhaus%E2%80%93Moser_notation

we can get pretty big number with a single character, what about 9?

edit:
didn't notice the ASCII restriction, well, a megiston is the biggest named number (and not a googleplex, see link above) so

megiston!

should be an unthinkably big number
Post 30 Mar 2013, 05:46
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revolution
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revolution
megiston! does indeed sound large, but how to compare it to G or S(G)?
Post 30 Mar 2013, 17:32
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l_inc



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l_inc
revolution
Just look at the last paragraph of the linked article:
Quote:
Therefore Moser's number, although incomprehensibly large, is vanishingly small compared to Graham's number

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Post 30 Mar 2013, 21:08
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revolution
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revolution
l_inc wrote:
revolution
Just look at the last paragraph of the linked article:
Quote:
Therefore Moser's number, although incomprehensibly large, is vanishingly small compared to Graham's number
I saw that. But what does it say about megiston! (megiston factorial)?
Post 31 Mar 2013, 02:21
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alessandro95



Joined: 24 Mar 2013
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alessandro95
Uh, I didn't notice this line at the end, well, we can say S(G)! which is surely bigger than megiston! and solve the problem ^^
Post 31 Mar 2013, 02:26
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bitRAKE



Joined: 21 Jul 2003
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bitRAKE
The immenseness of S(k,n) is difficult to comprehend. It grows faster than any conceivable algorithm (excluding algorithms including S(k,n) or greater order S(), of course). The only question is where the cross-over point is for some of these larger numbers.

There might be a way to examine the equivalence by compressing the steps to reach these other numbers. Read something about a low-bound algorithm, but it's very far off even for small S().

For example, if we can say that G can be emulated by a 3,14 TM.

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Post 31 Mar 2013, 02:51
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uart777



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uart777
Code:
n*Infinite ; outcome will always be the same    
Post 31 Mar 2013, 03:30
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l_inc



Joined: 23 Oct 2009
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l_inc
revolution
According to the definition of megiston, it's obviously far less, than the Moser's number:
moser = M(2,1,mega) = M(2,2,mega-1) = M(M(2,1,mega-1),1,mega-1) >> M(10,1,5)

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Post 31 Mar 2013, 11:41
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revolution
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revolution
l_inc wrote:
revolution
According to the definition of megiston, it's obviously far less, than the Moser's number:
moser = M(2,1,mega) = M(2,2,mega-1) = M(M(2,1,mega-1),1,mega-1) >> M(10,1,5)
Okay. And megiston! (don't forget the factorial sign) is ... ?
Post 31 Mar 2013, 11:54
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l_inc



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l_inc
revolution
First of all, I don't think it makes any sense to apply a factorial in the light of presence of the superfactorial's definition.
Secondly, I don't think it makes any sense to apply anything to the megiston in the light of presence of the Graham's number expressible in a single character.
Thirdly, obviously megiston! is also far less, than the Moser's number, since factorial grows much slower than even M(n,1,3). Therefore:

M(2,1,mega-1) >> M(10,1,5) (analogously to my previous post)
consequently:
moser = M(M(2,1,mega-1),1,mega-1) >> M(M(10,1,5),1,3) >> M(10,1,5)!

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Post 31 Mar 2013, 12:31
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revolution
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revolution
l_inc wrote:
revolution
First of all, I don't think it makes any sense to apply a factorial in the light of presence of the superfactorial's definition.
I just wanted to give proper analysis to alessandro95'a suggestion.
l_inc wrote:
Secondly, I don't think it makes any sense to apply anything to the megiston in the light of presence of the Graham's number expressible in a single character.
Thirdly, obviously megiston! is also far less, than the Moser's number, since factorial grows much slower than even M(n,1,3). Therefore:

M(2,1,mega-1) >> M(10,1,5) (analogously to my previous post)
consequently:
moser = M(M(2,1,mega-1),1,mega-1) >> M(M(10,1,5),1,3) >> M(10,1,5)!
Thanks. Makes sense. So we can say that G is still the king as far as it applies to this topic.
Post 31 Mar 2013, 12:43
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