My number is bigger!

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Fejfo
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Re:

Postby Fejfo » Fri Mar 31, 2017 7:21 pm UTC

warriorness wrote:Define a function Q such that Q(x) is x→x→...x, where the number of arrows in that sequence is equal to x→x→...x, where there are x arrows in that second chain.

Now define W(x) = Q(Q(...Q(x)...)) so there are Q(x) number of "Q"s in there.

My number is:

W((g_64)!)

(edit: threw a factorial in there for good measure)


To make it easier to compare:

your Q function is
Q(x) = x→x→...x, where the number of arrows in that sequence is equal to x→x→...x, where there are x arrows in that second chain.

x→x→...x, where there are x arrows in that second chain so x→(2)x arrows in the first chain
so

Q(x) = x→(2) (x→(2)x) < x→(2) 3 →(2) 2

your W(x) function:
W(x) = Q^Q(x) (x)
~= x→(2) Q(x) →(2) 2
< x →(2) 3 →(2) 3

So I'm pretty sure your number is smaller than
3→(3)→4

< f_1(3) = 3→(4)→3 (the function defined in my last comment)

Daggoth
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Re: My number is bigger!

Postby Daggoth » Sat Apr 15, 2017 6:05 am UTC

Beaten on page 3, halfway through. Here

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Re: My number is bigger!

Postby Freddino18 » Mon Apr 17, 2017 8:23 pm UTC

Every time I try, I get beaten down, either by someone pointing out that it's too small, or by someone pointing out that it is incalculably large.

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Re: My number is bigger!

Postby Daggoth » Tue Apr 18, 2017 11:09 pm UTC

More like you haven't produced a number that wins while following the rules, it's understandable, the numbers on this thread are big

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Re: My number is bigger!

Postby Freddino18 » Wed Apr 19, 2017 7:55 pm UTC

The number of possible dick jokes.
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Re: My number is bigger!

Postby Daggoth » Sun Apr 23, 2017 8:42 pm UTC


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Re: My number is bigger!

Postby Freddino18 » Wed Apr 26, 2017 7:56 pm UTC

Cautiously pessimistic.
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Re: My number is bigger!

Postby Yet-One-More-Idiot » Sun Apr 30, 2017 1:31 am UTC

So, just wondering - what's the currently leading Biggest Number?

I just checked this thread and there's 40 pages of posts, I'm not archive-binging all of that at 3am on a Sunday. xD

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Re: My number is bigger!

Postby Thegeniusyoshi » Tue Jun 27, 2017 2:43 pm UTC

Imagine a function B, for which B(x) is 1 followed by x 0s.
Now imagine a function C for which C(x, y) is x factorialed y times. For example, C(10, 3) would ((10!)!)!.
My number is C(B(TREE(gA(g[sub]65,g65)[/sub]), B(TREE(gA(g[sub]65,g65)[/sub]))

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Re: My number is bigger!

Postby Yakk » Thu Jun 29, 2017 6:05 pm UTC

Thegeniusyoshi wrote:Imagine a function B, for which B(x) is 1 followed by x 0s.
Now imagine a function C for which C(x, y) is x factorialed y times. For example, C(10, 3) would ((10!)!)!.
My number is C(B(TREE(gA(g[sub]65,g65)[/sub]), B(TREE(gA(g[sub]65,g65)[/sub]))

Yakk wrote:I try to use the small/medium/large system to categorize numbers. Small is anything that doesn't use a particularly strong system. Medium uses something as powerful as the fast growing heirarchy and feeds it a reasonable cardinal. Large are some of the computability ones, or proof based ones, or large cardinal fast growing heirarchy ones.

B and C and A and g are "small" functions. They don't grow fast enough to really matter in this thread.

Tree appears to be a medium function.

So your value is "medium". And basically all of its size is because you used the word "TREE" and fed it a value bigger than 2.
One of the painful things about our time is that those who feel certainty are stupid, and those with any imagination and understanding are filled with doubt and indecision - BR

Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.

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Re: My number is bigger!

Postby Daggoth » Sat Jul 01, 2017 6:36 am UTC

I'll give it a tighter bound for gigs and shittles

// is for commentary
restrict x to positive integers

B(x) = 10^x

x! < x^x // for x=>2
since x! = x*y1*y2... where all yn are smaller than x and the amount of multiplicands is equal to x
and x^x = x*y1*y2... where all yn are not smaller than x and the amount of multiplicands is equal to x


x!! = (x!)! < (x^x)^(x^x) = x^(x*(x^x)) = x^((x^1)*(x^x)) = x^x^(x+1) which is roughly x^x^x or x^^3
x!!! < x^^4
C(x,y) < x^^y // for x,y=>2

C(x,x) > B(x) for x=>3

A(x,x) < A(x) = x↑xx (two argument ackermann vs ackermann number)
A(x) > C(x) for x=>3

g1 = 3^^^^3 = 3↑43
g2 = 3↑g13
gx > A(x) (no catch-up here)
gx+2 > A(A(x+1))

TREE(x) > gx for x=>3
TREE(x+1) > g...gx with x amount of g's (you are gonna have to trust me on this, TREE is ridiculously way faster than just this)

Now that we've placed the cutlery on the table, Lets put that whole salad on a plate and eat our way out from the inside , right to left
note, for those whom it might not be obvious, each subsequent number will be bigger and simpler
C(B(TREE(gA(g65,g65)), B(TREE(gA(g65,g65)))
A(g65,g65) < A(g65)

C(B(TREE(gA(g65)), B(TREE(gA(g65)))
g67 = 3↑g663 = 3↑3↑g6533 )
A(g65) = g65↑g65g65
g67 > A(g65)

C(B(TREE(g67), B(TREE(g67))
gx > A(x) > C(x,x) >B(x)

ggTREE(g67)
TREE(3)>ggg3>g67

ggTREE(TREE(3))
TREE(TREE(3)+1)>gg...TREE(TREE(3)) with TREE(TREE(3)) g's > ggTREE(TREE(3)+1)

Your number is less than TREE(TREE(3)+1), Beaten at page 12 by Deedlit's entry

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Re: My number is bigger!

Postby Rhombic » Sun Jul 02, 2017 10:48 am UTC

defining Primes(k) as the number of prime numbers < k;
Fn is the nth Fibonacci number

define Indigo(a, b, c, d) as the following

w = TREE(a)+c
for r in range(b):
w = (d^TREE((w^w)!)!
for t in range(d):
w = Primes(w!)^(w^Fw)
return n

My number is Indigo(2017, 2017^2017, 2017, 2017^F2017)

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Re: My number is bigger!

Postby Daggoth » Mon Jul 03, 2017 5:05 pm UTC

n has no value

unless you meant return w maybe?

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Re: My number is bigger!

Postby WarDaft » Fri Jul 28, 2017 1:58 pm UTC

The largest uncontested number is currently Deedlit's. There are two potentially larger numbers, but they are floating around at sizes where you can't hand-wave things that are being hand-waved, and I'll let someone with a sufficient math degree suss that out.

Instead, let's add more chaos. I'm going to propose a very large number, but I don't know how large. I solved a problem that was bugging me about an idea I had for a suggestion.

Worms come in orders now.
An order 0 worm is just a regular worm.
An order k worm is a list of order k-1 worms.
The local head of an order k worm is the highest indexed element of the order k worm, the global head is the local head chosen recursively until you have an integer.
The position of each integer in a worm can be specified as a 1-indexed vector based on the indicies from the groups it is a member of.
For example, in the worm 0101|1010|11 the bolded 0 has a position <2:2>
If the global head is 0, then to decrement an order k worm you:
1) Decrement the local head
- if you are in an order 0 worm, chop off the 0, skip the remaining steps, and return the new length l as the vector <l>
- if you are in an order k > 0 worm, decrement the local head to obtain an inspection vector v
2) Duplicate
- let l be the length of the current order k worm, form the inspection vector l:v, decrease l until you reach the first (IE highest) k+1:v such that k:v is an empty cell. Duplicate indicies k+1 through l n times. Return the vector k+1:v
If the global head is greater than 0, then to decrement an order k worm you:
1) Decrement the local head
- if you are in an order 0 worm, grow it like a normal worm, return the length d of the dead worm as the vector <d+1>, skip the remaining steps
- if you are in an order k worm, let l be the length of the current order k worm, decrement the local head to obtain inspection vector l:v.
2) Grow
- decrease l until you reach the first (IE highest) k+1:v such that element k:v is less than the current (IE post decrement) global head, or it's empty. Create l-k-1 copies of element l and append them to the current order k worm, which is now of length l2, let d be the first component of vector v, then, for elements k+1 through l-1, slice from element d through to the end of that order k-1 worm and append them pairwise to elements l through to l2-1, return the vector k+1:v

The goal was to embed a Buchholz hydra into a 2-d worm in a way that can be extended to higher dimensions.

An n-multiworm is an n-dimensional worm, where each dimension is of size n, and each cell contains n. I enter a (3-multiworm)-multiworm. :twisted: I have no idea what scale this is.
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Re: My number is bigger!

Postby phillip1882 » Wed Aug 09, 2017 1:14 am UTC

busybever(tree(5))
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Re: My number is bigger!

Postby Daggoth » Sat Aug 12, 2017 3:18 am UTC

I have no idea what scale this is.


any rough estimate / insight?

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Re: My number is bigger!

Postby Deedlit » Sat Aug 26, 2017 6:22 am UTC

Wardaft, a few questions:

1. Is the k in " decrease l until you reach the first (IE highest) k+1:v such that k:v is an empty cell" a different variable from the k in "order k worm"?

2. The "inspection vector" is just the vector for the current global head, right?

3. I don't understand the "Grow" procedure, in particular the "slice from element d through to the end of that order k-1 worm" part. Could you explain it more thoroughly, perhaps give a few examples showing how the procedure is implemented?

4. How does the Buchholz hydra embed into a 2D worm? (I assume you mean an order 2 worm here.) I wondered if the order 1 subworms of an order 2 worm were meant to be the branches of the Buchholz hydra tree for each leaf node, but the decrementing procedure you describe doesn't seem to match the Buchholz hydra procedure given this embedding.

5. Do you have any argument/intuition why this procedure for order k worms should terminate?

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Re: My number is bigger!

Postby fulvius » Thu Oct 12, 2017 5:55 pm UTC

Nice posts, deedlit et. al! I am out of my depth, but I'd like to comment and propose a number anyway. At first I thought I was going to be able to beat you by replacing the occurrences of 10 and 100 in the BEAF expression of the highest computable number at Bowers Infinity Scrapers, page, which is meameamealokkapoowa oompa. My thought was that instead of 10 and 100 we could use some other large number like SCG(13) or Loader's number in place of the 10s and 100 in the BEAF notation for oompa and get a bigger but still computable number... But then I noticed something at the end of the article at The Googology wiki for Loader's number. It said, "The final output of (Loader's number) is much larger than TREE(3), SCG(13), and likely anything that can be practically defined by BEAF. It is probably overpowered by finite promise games and greedy clique sequences." Oompa can be practically defined by BEAF so, assuming this is true, then oompa is not as good a starting point as Loader's number.

Next, I looked up Finite promise games and Greedy clique sequence in the Googology wiki and I couldn't understand those articles any better than Deedlit's posts on page 12... But I did notice a statement by Deedlit on page 12 where he says, "The limit of this notation, tsi_0(w_w), is an ordinal of some importance. It was shown to be the termination ordering of some decision problems. (I'm recalling this from one of Harvey Friedman's papers, I'm not sure which.) It's also the complexity of the Buchholz Hydra on finite labelled trees." And I also noticed this on the Bucholz hydra page: "The Buchholz hydra surpasses TREE(n) and SCG(n). It is likely weaker than loader.c as well as numbers from finite promise games."

So, unless I am misunderstanding, it sounds like Deedlit's numbers are on the order of numbers generated by the Buchholz Hydras, which are "likely??" smaller than the output of loader.c and finite promise games?

Here's my big number: Let's start wih loader.c and replace the "99" used as "n" to seed the D(n) call that generates Loader's number with meameamealokkapoowa oompa instead, just for good measure. More importantly, let's also embed that meamea seed in nested parentheses that are Loader's number + 2 levels deep, so we have D( . . . D(meameamealokkapoowa oompa) . . . ) where the elipsis represents a Loader's number of "D(" or close parentheses. Loader's number is D(D(D(D(D(99)))) where D is loader's "derive" function. That process of embedding Loader's derivation within a very deeply nested parenthetical statement generates a large number. Now let's step away from loader.c and instead use that very large number we just defined as the argument "a" in Friedman's (presumably Loader-beating) FLCI(a) described at http://googology.wikia.com/wiki/Finite_promise_game

I think this might beat other numbers in this thread, but I suspect deedlit would be a more qualified mathematician than I am to make that call... but if his iterations remain near the "level" of the Bucholz hydras, and if either loader.c or finite promise games are truly "stronger" than Bucholz hydras in the fast-growing heirarchy, then... my number is bigger!

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Re: My number is bigger!

Postby wabb2t » Tue Oct 17, 2017 7:26 pm UTC

fulvius wrote:Nice posts, deedlit et. al! I am out of my depth, but I'd like to comment and propose a number anyway. At first I thought I was going to be able to beat you by replacing the occurrences of 10 and 100 in the BEAF expression of the highest computable number at Bowers Infinity Scrapers, page, which is meameamealokkapoowa oompa. My thought was that instead of 10 and 100 we could use some other large number like SCG(13) or Loader's number in place of the 10s and 100 in the BEAF notation for oompa and get a bigger but still computable number... But then I noticed something at the end of the article at The Googology wiki for Loader's number. It said, "The final output of (Loader's number) is much larger than TREE(3), SCG(13), and likely anything that can be practically defined by BEAF. It is probably overpowered by finite promise games and greedy clique sequences." Oompa can be practically defined by BEAF so, assuming this is true, then oompa is not as good a starting point as Loader's number.

Next, I looked up Finite promise games and Greedy clique sequence in the Googology wiki and I couldn't understand those articles any better than Deedlit's posts on page 12... But I did notice a statement by Deedlit on page 12 where he says, "The limit of this notation, tsi_0(w_w), is an ordinal of some importance. It was shown to be the termination ordering of some decision problems. (I'm recalling this from one of Harvey Friedman's papers, I'm not sure which.) It's also the complexity of the Buchholz Hydra on finite labelled trees." And I also noticed this on the Bucholz hydra page: "The Buchholz hydra surpasses TREE(n) and SCG(n). It is likely weaker than loader.c as well as numbers from finite promise games."

So, unless I am misunderstanding, it sounds like Deedlit's numbers are on the order of numbers generated by the Buchholz Hydras, which are "likely??" smaller than the output of loader.c and finite promise games?

Here's my big number: Let's start wih loader.c and replace the "99" used as "n" to seed the D(n) call that generates Loader's number with meameamealokkapoowa oompa instead, just for good measure. More importantly, let's also embed that meamea seed in nested parentheses that are Loader's number + 2 levels deep, so we have D( . . . D(meameamealokkapoowa oompa) . . . ) where the elipsis represents a Loader's number of "D(" or close parentheses. Loader's number is D(D(D(D(D(99)))) where D is loader's "derive" function. That process of embedding Loader's derivation within a very deeply nested parenthetical statement generates a large number. Now let's step away from loader.c and instead use that very large number we just defined as the argument "a" in Friedman's (presumably Loader-beating) FLCI(a) described at http://googology.wikia.com/wiki/Finite_promise_game

I think this might beat other numbers in this thread, but I suspect deedlit would be a more qualified mathematician than I am to make that call... but if his iterations remain near the "level" of the Bucholz hydras, and if either loader.c or finite promise games are truly "stronger" than Bucholz hydras in the fast-growing heirarchy, then... my number is bigger!

Hello fulvius

By saying that Loader's D() is "likely" faster than Buchholz's BH(), it is meant that, assuming that the program does not contains any bug and indeed outputs what it is supposed to output, then it beats (i.e. eventually dominates) Buchholz's BH(). Similarly, Friedman's promise games are "likely" faster than D() means that IF these functions are well-defined as we believe and if Friedman's results about their speed are correct (if he didn't make any mistake) then FLCI()&co are faster than D().

tl;dr: Friedman's promise games beats Loader's D() which beats Buchholz's BH().
I think Deedlit's function on page 12 does reach BH() or near it, but I'm not sure. It most certainely does not go too far beyond that in any case.

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Re: My number is bigger!

Postby fulvius » Tue Oct 24, 2017 11:44 pm UTC

wabb2t wrote:Hello fulvius

By saying that Loader's D() is "likely" faster than Buchholz's BH(), it is meant that, assuming that the program does not contains any bug and indeed outputs what it is supposed to output, then it beats (i.e. eventually dominates) Buchholz's BH(). Similarly, Friedman's promise games are "likely" faster than D() means that IF these functions are well-defined as we believe and if Friedman's results about their speed are correct (if he didn't make any mistake) then FLCI()&co are faster than D().

tl;dr: Friedman's promise games beats Loader's D() which beats Buchholz's BH().
I think Deedlit's function on page 12 does reach BH() or near it, but I'm not sure. It most certainely does not go too far beyond that in any case.


Yep that sounds about right to me. I think there are two things to notice here: one is that it's more effective to start from a faster-growing function than it is to cleverly compound and manipulate a slower-growing one to make it seem like it's meaningfully larger. Two is that it's difficult to know for sure which of the known fast-growing functions actually grow the fastest. There is a formal proof, for example, that SCG(13) is larger than TREE(3), but when it comes to finite promise games, there doesn't seem to be as much certainty in the publications and wikis as to how those functions compare to Loader's, etc. It's very possible that these are still open problems, in which case it might not be feasible for us to know which number here is the largest without making all-new breakthroughs in mathematics.

The psychological "wow" factor from nesting and compounding functions seems less significant than the functions we choose to nest. By extension, it seems that putting the seed for D() within a Loader's-number-deep set of nested parentheses is probably a lot less meaningful than the fact that the result is subsequently used as the input for FLCI().

One thing I'd like to point out about loader.c is that it won't successfully output a number on any actual computer... the number grows too large for any available hardware. C is simply the mathematical syntax in which that function is expressed, and so it can be checked for self-consistency and accuracy just as any other problem, function or proof can be checked in any other notation. Still, without actually computing the number it brings us back to that uncertainty: which number is really bigger? We may never know.

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Re: My number is bigger!

Postby Daggoth » Fri Nov 03, 2017 1:07 am UTC

Still, without actually computing the number it brings us back to that uncertainty: which number is really bigger? We may never know.


But we may, analyzing the structures and methods upon which each number is built, we can draw a provable conclusion for either A or B being the biggest, even if neither can be actually computed in hardware. If we were to limit the topic to discuss only the (puny) numbers that could ever be computed, the thread would have ended since page 1 ( http://forums.xkcd.com/viewtopic.php?p=179672#p179672 ) . by a poster whose nickname is, perhaps ironically, Ended.
(10^^^10)!
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Re: My number is bigger!

Postby Yakk » Mon Nov 06, 2017 6:32 pm UTC

The psychological "wow" factor from nesting and compounding functions seems less significant than the functions we choose to nest.

In fact, the very act of compounding functions is wasteful to the reader.

Find the fastest growing function in the random compounding. Check that the other functions (A) feed it a value bigger than 2, and (B) don't map it back to 1. That is the only function that really matters in the noisy compounding mess.

If someone else has a "slightly" faster growing function, they'll beat it with a small input than yours does with a ridiculous input.

I tried to capture that in my 3 types of large number post above. We have small large numbers, medium ones and big ones.

Small ones are boring, like 10^^^^^^10. Medium ones are something reasonably strong, like the fast growing heirarchy powered by a reasonable ordinal. And then there are the Large ones.

Throwing around "we then pass it to" some other named algorithm is a waste of words. Almost all of the medium and large ones have ways to twerk their definition so they'll grow slightly faster that blow your "call A then call B" out of the water. And in turn, they blow each other out of the water, even when they are hard to work out which grows faster.
One of the painful things about our time is that those who feel certainty are stupid, and those with any imagination and understanding are filled with doubt and indecision - BR

Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.

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Re: My number is bigger!

Postby itaibn » Thu Nov 09, 2017 6:58 pm UTC

It's been a long time since I was active here, but I want to show you something I wrote. It's about a system for describing large numbers with precision, somewhat akin to scientific notation but extending to numbers in at the ordinal function regime. Here's a link:

https://itaibn.wordpress.com/2017/11/09/prefix-free-codes-and-ordinals/

The description is pretty bare at the moment, so I may want to write a clearer explanation in the future.
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