flat assembler
Message board for the users of flat assembler.

 Index > Heap > What is the best pie you can get with 9 digits? Goto page Previous  1, 2, 3 ... 13, 14, 15 ... 27, 28, 29  Next
Author
 Thread
bitRAKE

Joined: 21 Jul 2003
Posts: 2914
Location: [RSP+8*5]
bitRAKE
S(6) > 10^2876

gn() is just: x_0=3, x<=3^x

The three could be replaced by any number and S() would eventually be greater. Let us go one step further:

x_0=A, y<=x^x, x<=y^y

...for any constant A, S() will exceed the function value. Now construct any such function and S() will exceed it.

_________________
¯\(°_o)/¯ unlicense.org
11 Mar 2008, 04:00
victor

Joined: 31 Dec 2005
Posts: 126
Location: Utopia
victor
S^S(G)(G)
11 Mar 2008, 04:28
revolution
When all else fails, read the source

Joined: 24 Aug 2004
Posts: 17278
Location: In your JS exploiting you and your system
revolution
victor wrote:
S^S(G)(G)
Hehe, victor is very fast. Accept. But I think there is a larger value.
11 Mar 2008, 04:56
bitRAKE

Joined: 21 Jul 2003
Posts: 2914
Location: [RSP+8*5]
bitRAKE
S^^^^G(G)

_________________
¯\(°_o)/¯ unlicense.org
11 Mar 2008, 06:56
revolution
When all else fails, read the source

Joined: 24 Aug 2004
Posts: 17278
Location: In your JS exploiting you and your system
revolution
bitRAKE wrote:
S^^^^G(G)
How is this to be interpreted?
11 Mar 2008, 07:18
edfed

Joined: 20 Feb 2006
Posts: 4237
Location: 2018
edfed
ho, i feel so lonelly with this thread...

what means all these symbols? it is not numbers at all....
11 Mar 2008, 11:05
victor

Joined: 31 Dec 2005
Posts: 126
Location: Utopia
victor
Epsilon nought, E0, is a countable ordinal. So are Ew, EEw, and EEEw. Correct?

w_1, w with subscript 1, is the smallest uncountable ordinal. w_w is the limit of the w_n for natural numbers n.

So, under the "Transfinites" section, Beth w_w should be much, much bigger than Beth EEEw. Correct?
11 Mar 2008, 11:13
revolution
When all else fails, read the source

Joined: 24 Aug 2004
Posts: 17278
Location: In your JS exploiting you and your system
revolution
victor wrote:
Epsilon nought, E0, is a countable ordinal. So are Ew, EEw, and EEEw. Correct?

w_1, w with subscript 1, is the smallest uncountable ordinal. w_w is the limit of the w_n for natural numbers n.

So, under the "Transfinites" section, Beth w_w should be much, much bigger than Beth EEEw. Correct?
I'm not sure here, but I think that you are confusing countable/uncountable with smaller/larger.
11 Mar 2008, 11:31
revolution
When all else fails, read the source

Joined: 24 Aug 2004
Posts: 17278
Location: In your JS exploiting you and your system
revolution
edfed wrote:
it is not numbers at all....
Just for edfed, we will run a small competition with only numerals from '0' to '9' being allowed.

Here is my entry "999999999". Do I win?
11 Mar 2008, 11:33
edfed

Joined: 20 Feb 2006
Posts: 4237
Location: 2018
edfed
999*99999
i win.

sorry...
11 Mar 2008, 11:53
tom tobias

Joined: 09 Sep 2003
Posts: 1320
Location: usa
tom tobias
victor wrote:
w_1, w with subscript 1, is the smallest
where w here is really omega, right victor?
edfed wrote:
...what means all these symbols? it is not numbers at all....
no, you are correct dear friend, and now, at last, you understand how I feel, attempting to read your x86 asm code, filled up, as it is, with nonsense symbols: @@@@@@@
The symbols, like B, S, and omega and aleph, and so on, which revolution and MHajduk, and many others are so fond of, are inscrutable--> you are correct, in my opinion, edfed. Thank you for expressing your opinion.

I am still waiting for someone to explain, why this question requires NINE ascii characters????
Why not, for example THREE CHINESE HanZi??? Oh, you mean, some people don't know how to read HanZi? Yes, and some of us, don't know how to read "aleph", either.
Wouldn't it be interesting to know whether one could describe a number representing an even LARGER numeric quantity from using only three Chinese HanZi, rather than employing nine Roman letters? How about Three Chinese HanZi, versus Three Ascii characters?
I wonder how the Chinese define aleph???My guess, completely in ignorance, is that either (a) they don't bother, they simply employ a Chinese character representing approximately the same idea, else, (b) they use those HanZi which, in NORTHERN China would be pronounced (using PuTongHua) with a sound somewhat similar to the pronunciation of aleph, (c) They create a word (from two or three characters) corresonding exactly to the notion of aleph. This word would then be used EXCLUSIVELY for aleph, nothing else.
How are these questions "on topic"? Maybe they are not!!!! To me, they are relevant to the topic, only because I see the fundamental issue for this thread as very murky...i.e. why should the focus be on NINE ascii characters??? That is very arbitrary, in my opinion. As there was no justification, that I can detect, for Cantor's having introducing "aleph", a hebreic symbol, with no linguistic correlation whatsoever to Europe, where he lived, or European science, and mathematics, so too, there is little rationale for asking the largest value of 9 ascii characters (ASCII does not originate with Europe....) One of the arguments long championed by both Chinese, Japanese and Koreans, is that HanZi are more efficient than Roman letters, or any other phonetic alphabet, to describe MEANING. In other words, given the same group of folks, both educated from birth reading effortlessly both roman letters and HanZi, equally literate with both Chinese and some European language written with roman letters, and given the same passage to read, first in HanZi, then Roman letters for one half of the subjects, vice versa for the second group, one could measure the time needed to answer questions about the reading, and the HanZi written passage would in every case result in faster comprehension times, due to a more efficient representation of the meaning of the author. So, if this were accurate, then, it should be possible to represent the same large number in fewer HanZi compared to the quantity needed with Ascii characters. How many fewer symbols is my question. I am asking then, conversely to this thread, how many characters, whether Roman or BeiJing, are needed to describe a particular number. At least, if one demanded that the question be addressed using Chinese symbols, edfed and tom would be on equal footing with victor, revolution, and bitRAKE...i.e. same slippery slope of insufficent knowledge.
11 Mar 2008, 12:03
revolution
When all else fails, read the source

Joined: 24 Aug 2004
Posts: 17278
Location: In your JS exploiting you and your system
revolution
edfed wrote:
999*99999
i win.

sorry...
You didn't read the rules.
revolution wrote:
... only numerals from '0' to '9' being allowed
11 Mar 2008, 12:04
revolution
When all else fails, read the source

Joined: 24 Aug 2004
Posts: 17278
Location: In your JS exploiting you and your system
revolution
tom tobias wrote:
I am still waiting for someone to explain, why this question requires NINE ascii characters????
...
why should the focus be on NINE ascii characters??? That is very arbitrary, in my opinion.
It is arbitrary, I make no excuses, I chose nine because it seemed at the time to be the most interesting number of characters to use. I chose ASCII because everybody has it on their keyboard and can read it without needing to install some foreign language pack.

It has no real purpose or meaning except to provide a bit of an amusing distraction.

I like to think that some people have enjoyed partaking in this thread. And to me that is enough to make it worthwhile.
11 Mar 2008, 12:11
r22

Joined: 27 Dec 2004
Posts: 805
r22
S^G(S(G))

S^G\$(G\$)\$

hy(G,G,G)

hyG\$(G,G) == hy(G,G\$,G)

How does the hyper operator stack up ?

Quote:
ASCII notation a^^b Since the up-arrow is used identically to the caret (^), the tetration operator may be written as (^^).

http://en.wikipedia.org/wiki/Tetration
Maybe bitRAKE was trying for 4 UpArrows or some sort of super Tetration with ^^^^ ?

S^^G\$\$(G)

S(G)^^G\$\$

S^G^^G(G)
11 Mar 2008, 12:24
revolution
When all else fails, read the source

Joined: 24 Aug 2004
Posts: 17278
Location: In your JS exploiting you and your system
revolution
r22 wrote:
S^G(S(G))

S^G\$(G\$)\$
Alright, all seem to be in order. I have no idea how to compare them but I do think there is still one larger that has not yet been submitted.
r22 wrote:

hy(G,G,G)

hyG\$(G,G) == hy(G,G\$,G)

How does the hyper operator stack up ?
Not very well, smaller than BB()
r22 wrote:
Quote:
ASCII notation a^^b Since the up-arrow is used identically to the caret (^), the tetration operator may be written as (^^).

http://en.wikipedia.org/wiki/Tetration
Maybe bitRAKE was trying for 4 UpArrows or some sort of super Tetration with ^^^^ ?
Maybe. He didn't say what he intended.
r22 wrote:
S^^G\$\$(G)
The "S^" thing was defined as S(S(S(...))). How do you define "S^^"?
r22 wrote:
S(G)^^G\$\$
Okay, I'll accept as valid (assuming tetration), but not very big
r22 wrote:
S^G^^G(G)
What is the order of operations? Working left-to-right gives nothing sensible.
11 Mar 2008, 12:58
r22

Joined: 27 Dec 2004
Posts: 805
r22
Quote:
The "S^" thing was defined as S(S(S(...))). How do you define "S^^"?

The ~tetrated composite of a function
F^^x(y) == F^x...^x(F^x...^x(F^x...^x(...)))
where ... denotes x iterations
Sadly, I don't have a link to prove/verify this.

Quote:
What is the order of operations? Working left-to-right gives nothing sensible.

S^G^^G(G) , tetration not being an elementary function would have the higher order so it would be S^(G UpArrow G)(G)

*** Now for the real largest number ***
S(n,m) = max steps for an n states, m symbols for generalized busy beaver.
http://en.wikipedia.org/wiki/Busy_beaver#The_busy_beaver_function_.CE.A3.28n.29

S(x,y) < S(x,y+1)

S(x+1,x) < S(x,x+1)

SO (I believe these are the one's Revolution was alluding to when he stated there were larger)

S^G\$(G,G)

S(G\$,G\$)\$

S(S(G),G)

S(G,S(G))
11 Mar 2008, 14:24
revolution
When all else fails, read the source

Joined: 24 Aug 2004
Posts: 17278
Location: In your JS exploiting you and your system
revolution
r22 wrote:
Quote:
How do you define "S^^"?
The ~tetrated composite of a function
F^^x(y) == F^x...^x(F^x...^x(F^x...^x(...)))
where ... denotes x iterations
Sadly, I don't have a link to prove/verify this.
Hehe, your just making it all up. Good try, but I'll need that verification.
r22 wrote:
S^G^^G(G) , tetration not being an elementary function would have the higher order so it would be S^(G UpArrow G)(G)
Okay, sounds plausible, I'll allow. Things are getting larger.
r22 wrote:
*** Now for the real largest number ***
S(n,m) = max steps for an n states, m symbols for generalized busy beaver.
http://en.wikipedia.org/wiki/Busy_beaver#The_busy_beaver_function_.CE.A3.28n.29

S(x,y) < S(x,y+1)

S(x+1,x) < S(x,x+1)

SO (I believe these are the one's Revolution was alluding to when he stated there were larger)
Could be ... I'm not saying yet ...
r22 wrote:
S^G\$(G,G)
... it's bigger ...
r22 wrote:
S(G\$,G\$)\$
... it's small ...
r22 wrote:
S(S(G),G)
... it's small ...
r22 wrote:
S(G,S(G))
... it's small.

Current leader is now "S^G\$(G,G)"

But there be still larger available.
11 Mar 2008, 15:18
bitRAKE

Joined: 21 Jul 2003
Posts: 2914
Location: [RSP+8*5]
bitRAKE
If was thinking if: S^n() = S(S(S(...()...))), then S^^n() would be: s'(s'(s'(...()...))), where s' = S(S(S(...()...))).

Making S^^^^G(G) equal to: s''''(s''''(s''''(...(G)...))), where s''''= s'''(s'''(s'''(...()...))), s'''= s''(s''(s''(...()...))), s''= s'(s'(s'(...()...))), s'= S(S(S(...()...))). Think of a 4D hypercube of size G. Just an extension of what has already been agreed on?

_________________
¯\(°_o)/¯ unlicense.org
11 Mar 2008, 16:03
revolution
When all else fails, read the source

Joined: 24 Aug 2004
Posts: 17278
Location: In your JS exploiting you and your system
revolution
bitRAKE wrote:
If was thinking if: S^n() = S(S(S(...()...))), then S^^n() would be: s'(s'(s'(...()...))), where s' = S(S(S(...()...))).

Making S^^^^G(G) equal to: s''''(s''''(s''''(...(G)...))), where s''''= s'''(s'''(s'''(...()...))), s'''= s''(s''(s''(...()...))), s''= s'(s'(s'(...()...))), s'= S(S(S(...()...))). Think of a 4D hypercube of size G. Just an extension of what has already been agreed on?
Well that would certainly be a very big number. It would completely dwarf all other entries. What do others think, should it be allowed We would have to move away from the mathematical basis that has been followed (admittedly, weakly) up to now.

How about this: if you can find, in an external reference, somewhere that uses ^^^^ then I will consider it as more likely to be allowed.
11 Mar 2008, 16:16
bitRAKE

Joined: 21 Jul 2003
Posts: 2914
Location: [RSP+8*5]
bitRAKE
S^{G}G(G) - just seems wrong, lol. A G-cube of size G.

Only link I could find says ** and *** is used.
http://en.wikipedia.org/wiki/Knuth's_up-arrow_notation#Notation

S****G(G) would be a 3D-cube of size G, but this is exponentiation and not composition.

http://groups.google.com/group/sci.math/browse_thread/thread/dfc5506eff59adef/b0231dda6dd73968

x<y>z ...was trying to fanagle this notation in - have seen it several places - but only came up with smaller numbers.

_________________
¯\(°_o)/¯ unlicense.org
11 Mar 2008, 16:38
 Display posts from previous: All Posts1 Day7 Days2 Weeks1 Month3 Months6 Months1 Year Oldest FirstNewest First

 Jump to: Select a forum Official----------------AssemblyPeripheria General----------------MainDOSWindowsLinuxUnixMenuetOS Specific----------------MacroinstructionsCompiler InternalsIDE DevelopmentOS ConstructionNon-x86 architecturesHigh Level LanguagesProgramming Language DesignProjects and IdeasExamples and Tutorials Other----------------FeedbackHeapTest Area
Goto page Previous  1, 2, 3 ... 13, 14, 15 ... 27, 28, 29  Next

Forum Rules:
 You cannot post new topics in this forumYou cannot reply to topics in this forumYou cannot edit your posts in this forumYou cannot delete your posts in this forumYou cannot vote in polls in this forumYou can attach files in this forumYou can download files in this forum

Copyright © 1999-2020, Tomasz Grysztar.

Powered by rwasa.