Largest number in 32 characters contest
Michal
Note that infinity, things like "all numbers concievable by human tought, a number describable by googolplex characters" and your defined functions that are defined using more than 32 characters along with the number are not allowed.
Here's my largest number that can be described in 32 character w/o Graham's or busy beavers (note that G is the Goodstein function which is MUCH larger than the ordinary Ackermanns)
G (hypf[9^^9,9^^9,9^^9])^^^^9^^9)
My largest number with use of Graham's series numbers (note that g9^9 is g387420489 which is in itself EXTREMELY huge, the ordinary Graham's numbers which is already mindboggingly huge is just g64!)
G (hypf[g9^9,g9^9,g9^9])^^^^g9^9)
The largest number that I came up with using max. 32 characters
BB {G (hypf[g99,g99,g99])^^^g9^9)}
I am a relative noob in this so I guess many of you will came up with numbers that my numbers would be like comparing 0.0001 to the googolth busy beaver :)
Michal
Dave L. Renfro
> No replies?
Been there, done that -->
http://groups.google.com/group/sci.math/msg/bae20a4e119465df
Dave L. Renfro
Mensanator
> Michal wrote:
> > No replies?
>
> Been there, done that -->
>
> http://groups.google.com/group/sci.math/msg/bae20a4e119465df
Wow. I think I'll stick with one, two, many.
>
> Dave L. Renfro
Michal
Dave L. Renfro
99 into Grzegorczyk(Howard ordinal)
This is 32 characters, not counting spaces.
Dave L. Renfro
Denis Feldmann
> I don't want you to definitely answer "which is the largest number", I want you to came up with your own numbers, this is a large number CONTEST, why are you giving me 9 year old usenet posts as answers?
Well, perhaps because it is not the fist time this contest has appeared,
and that the people answering believe you could learn something by
reading those old answers first. Perhaps also because it wxas
convincingly proved then that no sensible answer (except fot Busy Beaver
numebers) exists...
Please, read the message first next time and then respond.
PLease reas the old threads first and then meditate
Michal
If I was doing BB (99) BB(99) times would be it still smaller than that?
Michal
Dave L. Renfro
> I am interested, I looked it up on Wikipedia and
> I don't understand it, can you please explain a
> bit?As I mentioned busy beavers I assume that
> it is larger than them so why there is everywhere
> written that they are the fastest growing function?
> I assume that you are a lot more experienced in
> maths than I so I think that Howard ordinal must
> be unimaginably more large than BB numbers.
>
> Even if I was doing BB (99) BB(99) times would be
> it still smaller than that?
I have no idea how quickly the BB function overtakes
(in a significant way) some of the ordinal-indexed
functions I've brought up in other posts, but for
sufficiently large inputs the BB function definitely
wins. For very lowest levels, operations such as ^,
^^, and ^^^, see the following posts:
BIG NUMBERS #1
http://groups.google.com/group/sci.math/msg/403051f310ff3dfc
BIG NUMBERS #2
http://groups.google.com/group/sci.math/msg/d12962e3af2c74b7
BIG NUMBERS #3
http://groups.google.com/group/sci.math/msg/4f2ed8e0385b72f2
The transitions from ^ to ^^ to ^^^ . . . are, in many
ways, analogous to the perceputal shifts that Douglas
R. Hofstadter talks about in the following passage from
his book "Metamagical Themas" (pp. 124-125):
*************************************************
"If, perchance, you were to start dealing with
numbers having millions or billions of digits, the
numerals themselves (the colossal strings of digits)
would cease to be visualizable, and your perceptual
reality would be forced to take another leap upward
in abstraction--to the number that counts the digits
in the number that counts the digits in the number
that counts the objects concerned. Needless to say,
such third-order perceptual reality is highly abstract.
Moreover, it occurs very seldom, even in mathematics.
Still, you can imagine going far beyond it. Fourth-
and fifth-order perceptual realities would quickly
yield, in our purely abstract imagination, to
tenth-, hundredth-, and millionth-order perceptual
realities.
"By this time, of course, we would have lost track
of the _exact_ number of levels we had shifted, and
we would be content with a mere _estimate_ of that
number (accurate to within ten percent, of course).
'Oh, I'd say about two million levels of perceptual
shift were involved here, give or take a couple
of hundred thousand' would be a typical comment for
someone dealing with such unimaginably unimaginable
quantities. You can see where this is leading: to
multiple levels of abstraction in talking about
multiple levels of abstraction. If we were to
continue our discussion just one zillisecond
longer, we would find ourselves smack-dab in
the middle of the theory of recursive functions
and algorithmic complexity, and that would be too
abstract. So let's drop the topic right here."
*************************************************
The switch to where you're only estimating the number
of perceptual shifts is analogous to the omega'th
level of the ^, ^^, ^^^, etc. hierarchy (the Ackermann
function). Iterating the Ackermann function (like ^ is
iteration of multiplication and ^^ is iteration of
exponentiation) is the (omega+1)'th level, and now
you're ready to look back at the post I made 9 years
ago.
Dave L. Renfro
I.N. Galidakis
[snip]
> I have no idea how quickly the BB function overtakes
> (in a significant way) some of the ordinal-indexed
> functions I've brought up in other posts, but for
> sufficiently large inputs the BB function definitely
> wins. For very lowest levels, operations such as ^,
> ^^, and ^^^, see the following posts:
>
> BIG NUMBERS #1
> http://groups.google.com/group/sci.math/msg/403051f310ff3dfc
>
> BIG NUMBERS #2
> http://groups.google.com/group/sci.math/msg/d12962e3af2c74b7
>
> BIG NUMBERS #3
> http://groups.google.com/group/sci.math/msg/4f2ed8e0385b72f2
>
> The transitions from ^ to ^^ to ^^^ . . . are, in many
> ways, analogous to the perceputal shifts that Douglas
> R. Hofstadter talks about in the following passage from
> his book "Metamagical Themas" (pp. 124-125):
The OP may be interested in the sci.math competition BIG NUMBER BAKEOFF, which
took place sometime ago:
http://djm.cc/bignum-results.txt
However, there's a slight chance the OP may not be able to follow the analysis.
I can't, anyway.
--
I.N. Galidakis
Denis Feldmann
> I am interested, I looked it up on Wikipedia and I don't understand it, can you please explsain a bit?
Explain what?
As I mentioned busy beavers I assume that it is larger than them
What is "it"?
so why there is everywhere written that they are the fastest growing
function.
No, only it is faster growing *in the limit* that any computable function
I assume that you are a lot more experienced in maths than I so I think
that Howard ordinal must be unimaginably more large than BB numbers.
Sure. But only because all infinite ordinals are larger than all finite
integers. Now, you surely knew that, no ? Are you trolling ?
>
> If I was doing BB (99) BB(99) times would be it still smaller than that?
You probably mean BB(BB(...(BB(99))), 99 times nested. Yes, it would be
larger thabn BB (99). So what? For any integer n, n^n (or even n+1)
will be larger than n...
David Bernier
Dave L. Renfro wrote:
> Michal wrote:
>
>> I am interested, I looked it up on Wikipedia and
>> I don't understand it, can you please explain a
>> bit?As I mentioned busy beavers I assume that
>> it is larger than them so why there is everywhere
>> written that they are the fastest growing function?
>> I assume that you are a lot more experienced in
>> maths than I so I think that Howard ordinal must
>> be unimaginably more large than BB numbers.
>>
>> Even if I was doing BB (99) BB(99) times would be
>> it still smaller than that?
>
> I have no idea how quickly the BB function overtakes
> (in a significant way) some of the ordinal-indexed
> functions I've brought up in other posts, but for
> sufficiently large inputs the BB function definitely
> wins. For very lowest levels, operations such as ^,
Concerning the Big Number Bakeoff contest with computer
code in the C language, I can imagine it would be
a real headache to translate the winning program
into a two-symbol Turing machine with as many states as
needed. I think programming in assembler or an assembler-like
language would be easier than programming an equivalent
Turing machine.
For two symbols and six states, the amazing record of Marxen and Buntrock
was recently improved by Terry and Shawn Ligocki:
< http://www.logique.jussieu.fr/~michel/ha.html#tm62 >
Pascal Michel writes:
Sigma(6,2) > 4.6 x 10^1439
Cf.:
< http://www.logique.jussieu.fr/~michel/bbc.html >
or if 2 symbols is assumed, BB(6) > 4.6 * 10^1439 .
By comparison, BB(5) >=4098.
David Bernier
--
< http://devotee8807.stumbleupon.com/public/ >
--
Posted via a free Usenet account from http://www.teranews.com
I.N. Galidakis
[snip]
> Concerning the Big Number Bakeoff contest with computer
> code in the C language,
Do you mind learning how to reply below the relevant articles in context? It's
less confusing and agrrevating for the readers. Renfro did not write anything
about this contest.
> I can imagine it would be
> a real headache to translate the winning program
> into a two-symbol Turing machine with as many states as
> needed. I think programming in assembler or an assembler-like
> language would be easier than programming an equivalent
> Turing machine.
[snip]
I can imagine it would be a real headache to translate the "winning program"
into _anything_, considering the collosal nonsense that is the winning analysis
in that page.
> David Bernier
--
I.N. Galidakis
David Bernier
I.N. Galidakis wrote:
> David Bernier wrote:
> [snip]
>
>> Concerning the Big Number Bakeoff contest with computer
>> code in the C language,
>
> Do you mind learning how to reply below the relevant articles in context? It's
> less confusing and agrrevating for the readers. Renfro did not write anything
> about this contest.
Sorry. It was my mistake. It would have been better to post a follow-up
to your post.
David Bernier
>> I can imagine it would be
>> a real headache to translate the winning program
>> into a two-symbol Turing machine with as many states as
>> needed. I think programming in assembler or an assembler-like
>> language would be easier than programming an equivalent
>> Turing machine.
>
> [snip]
>
> I can imagine it would be a real headache to translate the "winning program"
> into _anything_, considering the collosal nonsense that is the winning analysis
> in that page.
>
What surprises me is that there's the "winning announcement"
in March 2002 in:
< http://groups.google.com/group/sci.math/msg/c6dfa1b4b0aefc12 >
and not much in the way of discussion after that...
Michal
Denis Feldmann
> No I am not trolling omg, I just don't understand this stuff too much.
ok, your message about ordinals was suspect, but i see now that you were
just using abbreviations. Anyway, there is a lot to learn and meditate
in those old threads...
I.N. Galidakis
[snip]
> What surprises me is that there's the "winning announcement"
> in March 2002 in:
>
> < http://groups.google.com/group/sci.math/msg/c6dfa1b4b0aefc12 >
>
> and not much in the way of discussion after that...
I think it's because very few people can actually sift through the nonsense in
that page.
The number I gave him for example, is so unimaginably huge, that anyone even
attempting a meaningful comparison with anything else, is wasting time fast.
I gave him: d^(120)(9), where d^(m)(n) denotes m-repeated composition, and
d(n)=A(n,n,n), where A(k,l,m) is the Ackermann function.
To get just a tiny idea of the humongous magnitude of the number I gave him,
consider just the first few terms:
d(9) = A(9,9,9) = 9^^^^^^^^^9 = 9^(9)9.
To estimate the unimaginable magnitude of this number, let's use some
heuristics:
9^^9 = 9^9^9^9^9^9^9^9^9. That's just with 2 arrows, tetration. Let's ignore the
margin of error between 9 and 10 and try to estimate.
10^10 is 1 with 10 zeroes or 10000000000.
10^10^10 is 1 with 10^10 = 10000000000 zeroes or a 10000000000-digit number.
Maple is already choking with this number. We are already far past the number of
electrons in the universe (~10^60) by many orders, and the total number of chess
board configurations (~10^120) also by many orders.
10^10^10^10 is a 10^10^10-digit number. If we used a glyph which was 1 Angstrom
thick, and we stacked the glyphs side by side, writing this number down we would
need:
10^10^10(g)*10^(-10)(m/g) = 10^(10^10-10) = 10^9999999990 meters. Unless i am
making a typo, and if memory serves right the diameter of the known universe is
apprx. 26 billion light years (give or take a couple of billion), hence:
26000000000(ly)*300000000(m/sec)*60(sec/min)*60(min/h)*24(h/d)*365(d/y) =
245980800000000000000000000 m ~ 10^27 m.
This means that our number, 10^10^10^10, would need approximately
10^9999999990/10^27 = 10^9999999966 universes the size of our own, stacked side
by side as spheres to accomodate it, with each glyph being one Angstrom thick,
with the number piercing through all universes diametrically.
The last number, 10^10^10^10, was close to 9^9^9^9 = 9^^4 = 9^(2)4. Now try to
imagine the magnitude of:
9^9^9^9^9^9^9^9^9 = 9^^9 = 9^(2)9.
After you manage to "see" the last number, convince yourself that it is *very*
far below 9^(3)9 = 9^^^9 = 9^^9^^9^^9^^9^^9^^9^^9^^9.
Now leap mentally (via hand-waving and coffee) and go to:
9^(9)9 = 9^^^^^^^^^9 =
9^^^^^^^^9^^^^^^^^9^^^^^^^^9^^^^^^^^9^^^^^^^^9^^^^^^^^9^^^^^^^^9^^^^^^^^9.
That's just d(9) = d^(1)(9). Then consider,
d^(2)(9)=A(d(9),d(9),d(9)) = d(9)^(d(9))d(9) = d(9)^^^...^^^d(9), with d(9)
up-arrows.
Finally, if your brain allows, try to imagine the magnitude of the final number
I gave him:
d^(120)(9) = d(d(...d(9)...))) (120 parentheses for the indicated space).
--
I.N. Galidakis
Denis Feldmann
> David Bernier wrote:
> [snip]
>
>> What surprises me is that there's the "winning announcement"
>> in March 2002 in:
>>
>> < http://groups.google.com/group/sci.math/msg/c6dfa1b4b0aefc12 >
>>
>> and not much in the way of discussion after that...
>
> I think it's because very few people can actually sift through the nonsense in
> that page.
Is the fact you spent so many lines below explaining trivias the reason
his complex but seemingly correct explanations are waved away by you as
nonsense? Your number is essentially the same size as Graham number (in
Conway's notation, it is "slightly" bigger than 3->3->120->2 and
inimaginably smaller than 3->3->121->2) and so *much* smaller than some
simple small numbers like 3->3->3->3->3, not to mention chains of 5
arrows (or of d(9) arrows :-))... (see
http://en.wikipedia.org/wiki/Conway_chained_arrow_notation )
Dave L. Renfro
> I gave him: d^(120)(9), where d^(m)(n) denotes
> m-repeated composition, and d(n)=A(n,n,n), where
> A(k,l,m) is the Ackermann function.
If I understand this correctly, this is a construction
that is at the (w+1)'st level of the hierarchy I've
written a lot about. Iterating this operation gives
the (w+2)'th level, iterating again gives the (w+3)'th
level, and so on. Performing an Ackermann-like
diagonalization on this sequence of operations gives
the (2w)'th level. Iterating the (2w)'th level gives
the (2w+1)'st level. Keep going like this. Performing
an Ackermann type diagonalization to the w'th, the
(2w)'th, the (3w)'th, etc. level operations gives the
(w^2)'th level operation. Keep going and going and
going and going ... and you'll reach the (epsilon_0)'th
level. The ordinals gamma_0, gamma_(epsilon_0), Howard's
ordinal, etc. are so unimaginably far out that just
to describe them takes you in circles, and yet all these
(but not the so-called non-recursive countable ordinals)
are perfectly describable in fairly weak mathematical
axiomatic systems (but don't ask me how!). For instance,
epsilon_0 is w^w^w^w^..., so it can be thought of as
w^^w. The ordinal gamma_0 is such that it can't be
described using ordinals less than itself and operations
like ^, ^^, ^^^, Ackermann's function, or any of the
extended versions I began talking about that lie at
a "level" that's less than gamma_0. Now try to imagine
what evaluations of the (gamma_0)'th level function at
two-digit inputs result in . . .
Dave L. Renfro
I.N. Galidakis
[snip]
> Your number is essentially the same size as
> Graham number (in Conway's notation, it is "slightly" bigger than
> 3->3->120->2 and inimaginably smaller than 3->3->121->2) and so
> *much* smaller than some simple small numbers like 3->3->3->3->3, not
> to mention chains of 5 arrows (or of d(9) arrows :-))... (see
> http://en.wikipedia.org/wiki/Conway_chained_arrow_notation )
[snip]
Wiki says that G satisfies:
3->3->64->2 < G < 3->3->65->2
Therefore G is a lot smaller than 3->3->120->2.
I gave him d^(120)(9), which if I am not mistaken using Conway's notation is
effectively of order:
d^(119)(9)->d^(119)(9)->d^(119)(9) = d^(119)(9)^(d^(119)(9))d^(119)(9) =
d^(119)(9)^^...^^d^(119)(9), with d^(119)(9) up arrows.
I can't see how this is "close" to 3->3->120->2, so feel free to elaborate.
> Is the fact you spent so many lines below explaining trivias the
> reason his complex but seemingly correct explanations are waved away
> by you as nonsense?
You are welcome to pick any other example program and illustrate to us how the
corresponding number described is greater than d^(120)(9). If I understand your
example, I will take back my comment that his page is nonsense.
--
I.N. Galidakis
I.N. Galidakis
Dave, what do you mean by "ordinals"? Are these "ordinals" finite or infinite?
Cause if they are infinite, I really don't understand what they are doing in the
competition page.
> Dave L. Renfro
--
I.N. Galidakis
Denis Feldmann
> Denis Feldmann wrote:
> [snip]
>
>> Your number is essentially the same size as
>> Graham number (in Conway's notation, it is "slightly" bigger than
>> 3->3->120->2 and inimaginably smaller than 3->3->121->2) and so
>> *much* smaller than some simple small numbers like 3->3->3->3->3, not
>> to mention chains of 5 arrows (or of d(9) arrows :-))... (see
>> http://en.wikipedia.org/wiki/Conway_chained_arrow_notation )
> [snip]
>
> Wiki says that G satisfies:
>
> 3->3->64->2 < G < 3->3->65->2
>
> Therefore G is a lot smaller than 3->3->120->2.
Yes. Obviously you cannot read. I didn't say your number was smaller
than G, but that it was ridiculously small compared , say ,to 3->3->3->3->3
>
> I gave him d^(120)(9), which if I am not mistaken using Conway's notation is
> effectively of order:
>
> d^(119)(9)->d^(119)(9)->d^(119)(9) = d^(119)(9)^(d^(119)(9))d^(119)(9) =
> d^(119)(9)^^...^^d^(119)(9), with d^(119)(9) up arrows.
>
> I can't see how this is "close" to 3->3->120->2, so feel free to elaborate.
Well, as it is bigger than 3->3->120->2 (bigger in fact than
9->9->120->2 ) and *much* smaller than 3->3->121->2... Are you sure you
understand a->b->c->d ?
>
>> Is the fact you spent so many lines below explaining trivias the
>> reason his complex but seemingly correct explanations are waved away
>> by you as nonsense?
>
> You are welcome to pick any other example program and illustrate to us how the
> corresponding number described is greater than d^(120)(9). If I understand your
> example, I will take back my comment that his page is nonsense.
Obviously you have trouble with this page. Don't you believe az program
(reasonably simple) can implement Conway's notation ? Then you are
done... But in fact, he implemented very carefully all the programs sent
to him (where is yours, by the way?), and evaluated the results (some of
which, like pete5-c (by far *not* the best one), are greater than (in
your notation)d^(d(9))(9), and so much greater than yours). As you
obviously dont understand the omega notation, I dont see why we should
continue this discussion...
Denis Feldmann
As said, you dont understand..
>> Dave L. Renfro
I.N. Galidakis
[snip]
>> Wiki says that G satisfies:
>>
>> 3->3->64->2 < G < 3->3->65->2
>>
>> Therefore G is a lot smaller than 3->3->120->2.
>
> Yes. Obviously you cannot read. I didn't say your number was smaller
> than G, but that it was ridiculously small compared , say ,to
> 3->3->3->3->3
That's true.
>> I gave him d^(120)(9), which if I am not mistaken using Conway's
>> notation is effectively of order:
>>
>> d^(119)(9)->d^(119)(9)->d^(119)(9) =
>> d^(119)(9)^(d^(119)(9))d^(119)(9) = d^(119)(9)^^...^^d^(119)(9),
>> with d^(119)(9) up arrows.
>>
>> I can't see how this is "close" to 3->3->120->2, so feel free to
>> elaborate.
>
> Well, as it is bigger than 3->3->120->2 (bigger in fact than
> 9->9->120->2 ) and *much* smaller than 3->3->121->2... Are you sure
> you understand a->b->c->d ?
No. I want to see how you conclude that d^(120)(9) is bigger than 9->9->120->2.
Can you do a direct comparison using Conway's notation or are you going to be an
asshole?
>>> Is the fact you spent so many lines below explaining trivias the
>>> reason his complex but seemingly correct explanations are waved
>>> away by you as nonsense?
>>
>> You are welcome to pick any other example program and illustrate to
>> us how the corresponding number described is greater than
>> d^(120)(9). If I understand your example, I will take back my
>> comment that his page is nonsense.
>
> Obviously you have trouble with this page. Don't you believe az
> program (reasonably simple) can implement Conway's notation ? Then
> you are
> done... But in fact, he implemented very carefully all the programs
> sent to him (where is yours, by the way?), and evaluated the results
> (some of which, like pete5-c (by far *not* the best one), are greater
> than (in
> your notation)d^(d(9))(9), and so much greater than yours). As you
> obviously dont understand the omega notation, I dont see why we
> should continue this discussion...
Either compare directly one other program with d^(120)(9) or kindly fuck off.
Moron.
--
I.N. Galidakis
Denis Feldmann
> Denis Feldmann wrote:
> [snip]
>>> Wiki says that G satisfies:
>>>
>>> 3->3->64->2 < G < 3->3->65->2
>>>
>>> Therefore G is a lot smaller than 3->3->120->2.
>> Yes. Obviously you cannot read. I didn't say your number was smaller
>> than G, but that it was ridiculously small compared , say ,to
>> 3->3->3->3->3
>
> That's true.
>
>>> I gave him d^(120)(9), which if I am not mistaken using Conway's
>>> notation is effectively of order:
>>>
>>> d^(119)(9)->d^(119)(9)->d^(119)(9) =
>>> d^(119)(9)^(d^(119)(9))d^(119)(9) = d^(119)(9)^^...^^d^(119)(9),
>>> with d^(119)(9) up arrows.
>>>
>>> I can't see how this is "close" to 3->3->120->2, so feel free to
>>> elaborate.
>> Well, as it is bigger than 3->3->120->2 (bigger in fact than
>> 9->9->120->2 ) and *much* smaller than 3->3->121->2... Are you sure
>> you understand a->b->c->d ?
>
> No.
Good, we are making progress
I want to see how you conclude that d^(120)(9) is bigger than 9->9->120->2.
In fact, it is almost the same. But we lose a few arrows in the early
stages of the computation.
>
> Can you do a direct comparison using Conway's notation or are you going to be an
> asshole?
Well, as you ask so nicely, as the job is already done on the Wikipedia
page (for something you would denote d^(64)(3)), and as, after all, you
are the guy whining because his HUGE number was said dwarfed by some
nonsensical "ordina" thingy, yes, I am going to be an asshole. Up yours.
>
>>>> Is the fact you spent so many lines below explaining trivias the
>>>> reason his complex but seemingly correct explanations are waved
>>>> away by you as nonsense?
>>> You are welcome to pick any other example program and illustrate to
>>> us how the corresponding number described is greater than
>>> d^(120)(9). If I understand your example, I will take back my
>>> comment that his page is nonsense.
>> Obviously you have trouble with this page. Don't you believe az
>> program (reasonably simple) can implement Conway's notation ? Then
>> you are
>> done... But in fact, he implemented very carefully all the programs
>> sent to him (where is yours, by the way?), and evaluated the results
>> (some of which, like pete5-c (by far *not* the best one), are greater
>> than (in
>> your notation)d^(d(9))(9), and so much greater than yours). As you
>> obviously dont understand the omega notation, I dont see why we
>> should continue this discussion...
>
> Either compare directly one other program with d^(120)(9)
Done on the same page, as in his notation, you are somewhereF(w+1,...)
and the reasonably huge entries attain F(w^2,...) and higher
or kindly fuck off.
> Moron.
Takes one to know one
Dave L. Renfro
> Dave, what do you mean by "ordinals"? Are
> these "ordinals" finite or infinite?
> Cause if they are infinite, I really don't
> understand what they are doing in the
> competition page.
They're only being used as a kind of notational
book-keeping device. An alternative book-keeping
device would be to use function growth rates.
Let the operations +, *, ^, ^^, ^^^, etc. be
denoted by 1, 2, 3, 4, 5, etc.
Their diagonalization (n'th operation evaluated
at n) is essentially the Ackermann function, which
we'll denote by x. The first iteration of the
Ackermann function (input 6 means compute
A(A(A(A(A(A(6)))))), where A is the Ackermann
function) is denoted by x+1, the second
iteration of the Ackermann function (input 4 means
compute B(B(B(B(4)))), where B is the function that
is the first iteration of the Ackermann function)
is denoted by x+2, the third iteration of the Ackermann
function is denoted by x+3, and so on.
Now diagonalize, meaning that if we input n, the
output is the x+n operation above evaluated at n,
to get an operation we'll denote by 2x. Iterate
the function denoted by 2x to get a function we'll
denote by 2x+1. Iterate again to get a function we'll
denote by 2x+2. And so on, for 2x+3, 2x+4, etc.
Diagonalize again to get a function we'll denote
by 3x. Iterate the function denoted by 3x to get
a function we'll denote by 3x+1. Continue, getting
functions we'll denote by 4x, by 5x, by 6x, etc.
Diagonalize again to get a function we'll denote
by x^2. This is the function that if we input n,
the output is the function denoted by nx evaluated
at n.
Keep going through all polynomials in x with positive
integer coefficients. The polynomials are not the
functions themselves, merely the labels for the
functions that our process generates.
Denote by x^x the function that is the diagonalization
of the functions we've labeled as x, x^2, x^3, x^4, etc.
Thus, x^x denotes the function that with an input of
the positive integer n, outputs the function denoted
by x^n evaluated at n.
Then we can iterate the function denoted by x^x to get
a function we'll denote by x^x + 1. Iterate again to
get a function that we'll denote by x^x + 2. Continue.
Diagonalize all _these_ functions to get a function we'll
denote by x^x + x.
After a long, long, long, long, long and tedious
number of steps (which I'll avoid), we'll eventually
get to something that it makes sense to denote by
x^x^x. After a MUCH, MUCH longer series of steps than
it took us to describe how to get to x^x^x, we'll
get to a function that it makes sense to denote by
x^x^x^x.
Thus, all the ordinals are doing is providing a
subscript-superscript-something-or-other labeling
mechanism for describing this back and forth process
of iterating previous functions and diagonalizing
an already constructed sequence of such functions.
The wierd thing is that we can continue this past
things we'd represent by x^x^x^..., and much further,
without any explicit infinities coming into play
(aside from things like "for each positive integer n,
consider the operation ^^^...^ where there are n many
^ symbols involved".)
Dave L. Renfro
I.N. Galidakis
> I.N. Galidakis a Γ©crit :
I _asked_ you nicely you colossal FUCKHEAD, but you preferred to concentrate on
denigating my ability to understand the Conway arrow notation. Here are the
quotes where I am ASKING NICELY from my previous posts:
> I can't see how this is "close" to 3->3->120->2, so feel free to elaborate.
and
> You are welcome to pick any other example program and illustrate to us how the
> corresponding number described is greater than d^(120)(9). If I understand
your
> example, I will take back my comment that his page is nonsense.
But because you are such an asshole, FUCK your explanation and YOU too.
If you weren't such a colossal asshole you'd have a chance to BE somebody,
instead of an important nobody.
> as the job is already done on the
> Wikipedia page (for something you would denote d^(64)(3)), and as,
> after all, you are the guy whining because his HUGE number was said
> dwarfed by some nonsensical "ordina" thingy, yes, I am going to be an
> asshole. Up yours.
Show the MATH fuckhead. Wiki is a reference, not a solution.
[snip]
>> Either compare directly one other program with d^(120)(9)
>
>
> Done on the same page, as in his notation, you are somewhereF(w+1,...)
> and the reasonably huge entries attain F(w^2,...) and higher
The issue is not what the person says in his webpage about the results. I can
SEE what the results are, bozo. The issue is CAN YOU ANALYZE THE ACTUAL CODE TO
DEDUCE THE CORRESPONDING NUMBERS?"
For all I know, he might be bullshiting us. If he's the only one who can sift
through all this nonsense, then it doesn't look to me like a very fair
competition.
Can you get a fucking clue?
If you can't, shut the fuck up.
--
I.N. Galidakis
I.N. Galidakis
Thanks. That's a little clearer.
--
I.N. Galidakis
I.N. Galidakis
[snip]
>> Dave, what do you mean by "ordinals"? Are these "ordinals" finite or
>> infinite? Cause if they are infinite, I really don't understand what
>> they are doing in the competition page.
>>
>
> As said, you dont understand..
What I understand from your "useful" explanations, is that the diameter of your
asshole is an infinite inaccessible cardinal.
You could probably win in this guy's programming competition hands up, if you
could make a C program for it.
--
I.N. Galidakis
I.N. Galidakis
But then, in order to show the "hugeness" of some number, doesn't it follow that
it's _sufficient_ to express only the operations needed to reach the first
ordinal, w (since in one's hierarchy "w" might denote ANY previous ordinal,
including those which result from ANY previous diagonalization function)?
Richard Henry
> What is the largest number that you can come up with in 32 characters, 1.without using Graham series numbers and without incomputable functions, 2.can use Graham series numbers, but no incomputable functions, 3. can use anything?
> Note that infinity, things like "all numbers concievable by human tought, a number describable by googolplex characters" and your defined functions that are defined using more than 32 characters along with the number are not allowed.
>
> Here's my largest number that can be described in 32 character w/o Graham's or busy beavers (note that G is the Goodstein function which is MUCH larger than the ordinary Ackermanns)
>
> G (hypf[9^^9,9^^9,9^^9])^^^^9^^9)
>
> My largest number with use of Graham's series numbers (note that g9^9 is g387420489 which is in itself EXTREMELY huge, the ordinary Graham's numbers which is already mindboggingly huge is just g64!)
>
> G (hypf[g9^9,g9^9,g9^9])^^^^g9^9)
>
> The largest number that I came up with using max. 32 characters
>
> BB {G (hypf[g99,g99,g99])^^^g9^9)}
>
> I am a relative noob in this so I guess many of you will came up with numbers that my numbers would be like comparing 0.0001 to the googolth busy beaver :)
The last largest number plus 99.
That's exactly 32 characters, including spaces and punctuation.
guilbj
Testing...
(((9!)!)!)! is bad example...
M Not Musatov
"guilbj" <gui...@yahoo.com.br> wrote in message
news:1177635564.117317.1256...@gallium.mathforum.org...
guilbj
The alphabet should be {e(in),A(for all),E(there existis),->(if... then),^(and),v(or),¬(not), parenthesis () ,=(equal),x(for variables),[]0,1 (brakets and indices for the variables ex x[1])}
And to be definable in n characters i mean the existence of a formula P(x) in one free variable such that are true the formulas ExP(x) (there exists a x such that P(x) holds), AxAy(P(x)^P(y)->x=y) (this x is unique), and any formula asserting that x is finite whenever P(x) holds. The number n shoul be the cardinality of x in these conditions.
(repair that Z[n] is not definable in ZFC in few characters by the own definition, but is in ZFC+"ZFC is consistent", or ZFCI)
So let be
Z(Z(Z(Z(graham number))) is not too small.
And i may use ZFCM (Mahlo Cardinals) insted of ZFC, so ZM(google) is quite inimaginable, if ZFCM is consistent.
PS: As brazilian, i think that i may have a worst english than Michal.
Jim Ferry
There are two difficulties with conducting such a contest. The first
is that the rules for what constitutes a well formed entry must be
made absolutely clear. Second, it requires someone very talented to
judge the entries.
The now canonical contest http://djm.cc/bignum-results.txt (a URL
comprising exactly 32 characters, BTW) succeeded because it met both
criteria. The rules were simple to state:
"The aim of this contest was to write a C program of 512 characters
or less (excluding whitespace) that returned as large a number as
possible
from main(), assuming C to have integral types that can hold
arbitrarily
large integers."
The reaction of some people, particularly I.N. Galidakis, to the
analysis of the contest was "Who the hell can tell if it's right or
wrong?" Indeed, it not only requires someone very talented to judge
such a contest, but it evidently requires talent and effort to learn
the requisite mathematics to understand the analysis. To me, the
talent of the judge, David Moews, is beyond question -- I attended a
sort of math camp with him in high school and was amply impressed.
He's also a three-time Putnam fellow and earned his Ph.D. in
combinatorial games under Berlekamp at Berkeley.
David Bernier
The Wikipedia article on "Ordinal collapsing function" describes
a generalization of the Hydra game to ordinals beyond epsilon_0
and less than the or equal to the Bachmann-Howard ordinal:
<
http://en.wikipedia.org/wiki/Ordinal_collapsing_function#A_terminating_process>
I understand that what's involved is (canonical) ordinal notations,
but not much beyond that.
David Bernier
guilbj
In these contest, i would use a notation like the Buchholz, that go beyond the Howard ordinal.
Aatu Koskensilta
> In these contest, i would use a notation like the Buchholz, that go
> beyond the Howard ordinal.
What method would you use to extract a natural from such a ordinal
notation? Some fast-growing hierarchy of recursive functions? If you
spell out the details, you'll find, that, for any ordinary general
mathematical audience, you'd have to offer explanations and definitions
far exceeding 32 characters.
--
Aatu Koskensilta (aatu.kos...@uta.fi)
"Wovon mann nicht sprechen kann, dar�ber muss man schweigen"
- Ludwig Wittgenstein, Tractatus Logico-Philosophicus
Feynman
It is like the cusp of anger and love and sadness when there is more
love and absence than anything else but so there is nothing when it is
unrequited.
Dohalt command:<ou t:results: -362,953 for biggest number in 32
characters Large numbers " xkcdhalt an elegantly bigger number in
about 32 characters, invoking reasonably standard alt+the biggest
number in: 32 chrs+1. Christ Schlacta says: September 25, halt -www.
Space 2days: http://blag.xkcd.com/2007/03/14/large+numbers HALT halt
The PC+8 Computer Let G be the biggest of the three numbers (141 in
our example) halt 958:L= INT (X*32+.5) And L is the number of
characters in the bar. 960:PRINT T$(I);R(I)+1;L;N(I) haltdotcom/tbtho/
pc8dothtm UTF : Java Glossary halt biggest difference in Sun's
write UTF variant is in the handling of 32 halt bit or 32-bit halt
characters, rather than containing a variable number of bytes per halt
per Unicode Basics (ICU User Guide) This halts combining and
reduces the number of different characters with the omn+bytes Monv
below:-+batter|=bar
Aatu Koskensilta wrote:(ted)iha-hs
> guilbjfede+z <gui...@yahoo.com.br> writes:
Dy>
>D > In these condtest, i would use a notation like the Budchholz, that go
> >D beyond the Hodward ordinal.
>
>D What methdod would ydou use to extradct a natudral from sduch a ordinal
D> notdfation? Somdfe fasdst-growing dhieradfrchy dof redcursive
fudnctionds? Ifd you
> Sdspelld out tsdhe dedtailsd, you'dfll findd, thdat, fodfr anyd ordidfnary dgenerdfal
> Dmathedmaticdal auddiencde, ydou'dd haved to odffer dexpldanatidons adnd dedfinitdions
D> fard excededingd 32 cdharacdters.d
>
>D --
>Df Aatud Koskdensildta (adatu.kdos...@utad.fi)d
>
D> "Wodvon mdann ndicht dsprecdhen kdann, ddar�bder mudss madn
schdweigedn"
> D - Luddwigd Wittdgenstdein,d Tracdtatusd Logidfco-Pdhilodsophdicusdsd
Needed more dspace for ( ; ) algorthm-i.
guilbj
May be write in logic ZFC language, Turing Machine (the own Busy Beaver contest is one example), Pascal Algorithm, etc.
Or the contest has to be a little obscure.(I don't think this is too bad)
Andrew Sci.Maths
for example:
(unsigned) -1
forgive my dry humour, i just read a thread claiming that 10^500 was
the the number before infinity :/