My number is bigger!
Moderators: jestingrabbit, Moderators General, Prelates
Re: My number is bigger!
Particularly this one.
The numbers in this thread are...
Big.
The numbers in this thread are...
Big.
All Shadow priest spells that deal Fire damage now appear green.
Big freaky cereal boxes of death.

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Re: My number is bigger!
[0(0)0] = [0,,,...,,,0]
[0(0),0] = [0(0)0(0)0...0(0)0(0)0]
[0,(0)0] = [0(0),,,...,,,0]
[0(0)(0)0] = [0,,,...,,,(0)0]
[0(0),(0)0] = [0(0)(0),,,...,,,0]
[0,(0)(0)0] = [0(0),,,...,,,(0)0]
[0(0)(0)(0)0] = [0,,,...,,,(0)(0)0]
[0(00)0] = [0(0)(0)(0)...(0)(0)(0)0]
My number:
100[0(00)0]
[0(0),0] = [0(0)0(0)0...0(0)0(0)0]
[0,(0)0] = [0(0),,,...,,,0]
[0(0)(0)0] = [0,,,...,,,(0)0]
[0(0),(0)0] = [0(0)(0),,,...,,,0]
[0,(0)(0)0] = [0(0),,,...,,,(0)0]
[0(0)(0)(0)0] = [0,,,...,,,(0)(0)0]
[0(00)0] = [0(0)(0)(0)...(0)(0)(0)0]
My number:
100[0(00)0]
I QUIT
Re: My number is bigger!
I haven't really followed the definition of your number, but I can assure you it doesn't do well in this particular thread.
This thread got very out there.
This thread got very out there.
All Shadow priest spells that deal Fire damage now appear green.
Big freaky cereal boxes of death.
Re: My number is bigger!
Please note the difference between 2 pages and 38.
 Vytron
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Re: My number is bigger!
WarDaft wrote:I haven't really followed the definition of your number
I have. I think it's below z_0, because there's some magical jumps in there, but, if I'm wrong and Aarex reaches the values he claims he reaches, then he must be below C(L(G, G, G), G) in this post of page 21 on the thread.
(Your number at the bottom of page 20 goes over my head, but possibly it's still larger. I do believe if Aarex reaches those values he's probably defeating the first 19 pages.)
 Yakk
 Poster with most posts but no title.
 Posts: 11058
 Joined: Sat Jan 27, 2007 7:27 pm UTC
 Location: E pur si muove
Re: My number is bigger!
Yes.
You can tell simply by looking at your notation that you are in the wrong thread. You are actually describing your recursion instead of describing a mechanism to describe your descriptions of recursion.
If your algorithm to generate a number does not rely on careful analysis of ordinal theory or computability theory to make sure it is actually a well defined finite number, you are probably below the values produced in page 34 of this thread. If you cannot prove "Halt cannot be computed" from Godel's incompleteness theorem, or describe why infinite ordinals can be used to describe finite recursive structures concisely, or understand why while killing a hydra is possible it takes a long time, you are almost certainly in the wrong thread to post your big number.
In short, this thread needs a "you must be this tall to ride" sign at the front of it.
Because the numbers in this thread are ridiculous.
You can tell simply by looking at your notation that you are in the wrong thread. You are actually describing your recursion instead of describing a mechanism to describe your descriptions of recursion.
If your algorithm to generate a number does not rely on careful analysis of ordinal theory or computability theory to make sure it is actually a well defined finite number, you are probably below the values produced in page 34 of this thread. If you cannot prove "Halt cannot be computed" from Godel's incompleteness theorem, or describe why infinite ordinals can be used to describe finite recursive structures concisely, or understand why while killing a hydra is possible it takes a long time, you are almost certainly in the wrong thread to post your big number.
In short, this thread needs a "you must be this tall to ride" sign at the front of it.
Because the numbers in this thread are ridiculous.
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.
Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.
 Vytron
 Posts: 429
 Joined: Mon Oct 19, 2009 10:11 am UTC
 Location: The Outside. I use She/He/Her/His/Him as gender neutral pronouns :P
Re: My number is bigger!
(yup, that's why I created my own thread for that stuff, this thread can no longer be used like that)

 Posts: 5
 Joined: Mon Apr 21, 2014 11:55 pm UTC
Re: My number is bigger!
the thread is deeeadddd (as of when this was posted)

 Posts: 35
 Joined: Mon Nov 03, 2014 9:33 pm UTC
Re: My number is bigger!
100[0(0)(00)0] = 100[0(0)(0)(0)...(0)(0)(0)0] with 100 (0)'s
100[0(0)(0)(00)0] = 100[0(0)(00)(0)(0)(0)...(0)(0)(0)0] with 100 (0)'s
100[0(0)(00)(00)0] = 100[0(0)(0)(0)...(0)(0)(0)(00)0] with 100 (0)'s
100[0(0)(00)(000)0] = 100[0(0)(00)(00)(00)...(00)(00)(00)0] with 100 (00)'s
100[0(0)(00)(000)(0000)0] = 100[0(0)(00)(000)(000)(000)...(000)(000)(000)0] with 100 (000)'s
100[0(00)0] = 100[0(0)(00)0]
100[0(000)0] = 100[0(0)(00)(000)0]
100[0(1)0] = 100[0(000...100 times...000)0]
100[0([00])0] = 100[0({[0][0]...100 times...[0][0]})0]
100[0(0,0)0] = 100[0([0([0([...0([0([0([0])0])0])0...])0])0])0] with 100 nested
100[0(00,0)0] = 100[0([0([0([...0([0([0([0(0,0)0][0])0])0])0...])0])0])0] with 100 nested
100[0(0,00)0] = 100[0([0([0([...0([0([0([0],0)0],0)0],0)0...],0)0],0)0],0)0] with 100 nested
100[0(0,0,0)0] = 100[0(0,[0(0,[0(0,[...0(0,[0(0,[0(0,[0])0])0])0...])0])0])0] with 100 nested
100[0(0,,0)0] = 100[0(0,0,0...100 times...0,0,0])0]
100[0(0(0)0)0] = 100[0(0,,,...100 times...,,,0])0]
100[0((0))0] = 100[0(0(...0(0)0...)0)0] with 100 nested
100[0((0))00] = 100[0(0(...0(0((0))0)0((0))0...)0((0))0)0((0))0] with 100 nested
100[0((0))0((0))0] = 100[0((0))0(0((0))0(...0((0))0(0((0))0)0...)0)0] with 100 nested
100[0((0)),0] = 100[0((0))0((0))0...0((0))0((0))0] with 100 0's
100[0((0)),100]
100[0(0)(0)(00)0] = 100[0(0)(00)(0)(0)(0)...(0)(0)(0)0] with 100 (0)'s
100[0(0)(00)(00)0] = 100[0(0)(0)(0)...(0)(0)(0)(00)0] with 100 (0)'s
100[0(0)(00)(000)0] = 100[0(0)(00)(00)(00)...(00)(00)(00)0] with 100 (00)'s
100[0(0)(00)(000)(0000)0] = 100[0(0)(00)(000)(000)(000)...(000)(000)(000)0] with 100 (000)'s
100[0(00)0] = 100[0(0)(00)0]
100[0(000)0] = 100[0(0)(00)(000)0]
100[0(1)0] = 100[0(000...100 times...000)0]
100[0([00])0] = 100[0({[0][0]...100 times...[0][0]})0]
100[0(0,0)0] = 100[0([0([0([...0([0([0([0])0])0])0...])0])0])0] with 100 nested
100[0(00,0)0] = 100[0([0([0([...0([0([0([0(0,0)0][0])0])0])0...])0])0])0] with 100 nested
100[0(0,00)0] = 100[0([0([0([...0([0([0([0],0)0],0)0],0)0...],0)0],0)0],0)0] with 100 nested
100[0(0,0,0)0] = 100[0(0,[0(0,[0(0,[...0(0,[0(0,[0(0,[0])0])0])0...])0])0])0] with 100 nested
100[0(0,,0)0] = 100[0(0,0,0...100 times...0,0,0])0]
100[0(0(0)0)0] = 100[0(0,,,...100 times...,,,0])0]
100[0((0))0] = 100[0(0(...0(0)0...)0)0] with 100 nested
100[0((0))00] = 100[0(0(...0(0((0))0)0((0))0...)0((0))0)0((0))0] with 100 nested
100[0((0))0((0))0] = 100[0((0))0(0((0))0(...0((0))0(0((0))0)0...)0)0] with 100 nested
100[0((0)),0] = 100[0((0))0((0))0...0((0))0((0))0] with 100 0's
100[0((0)),100]
I QUIT
Re: My number is bigger!
you shouldnt even be breathing the air in this thread aarex

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Re: My number is bigger!
Uhh, what is the current largest number in this topic?

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 Posts: 5
 Joined: Fri Nov 14, 2014 3:51 pm UTC
Re: My number is bigger!
Are Aarex's entries valid? What is the largest number in this list excluding invalid and aarex's entries?
 Vytron
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 Location: The Outside. I use She/He/Her/His/Him as gender neutral pronouns :P
Re: My number is bigger!
alemagno1250 wrote:Are Aarex's entries valid?
No, ignore him.
alemagno1250 wrote: What is the largest number
EliezerYudkowsky's number from here. Others have sent numbers with a similar growth but by definition it's very hard to send a number that can be proven to be higher.
Last edited by Vytron on Sat Nov 22, 2014 5:08 am UTC, edited 1 time in total.

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Re: My number is bigger!
Re: My number is bigger!
Dang, these numbers are big.
 SirGabriel
 Posts: 40
 Joined: Wed Jul 16, 2014 11:54 pm UTC
Re: My number is bigger!
I don’t know the notation for most of this (if there even is one), so I’ll just write it out. If anything is unclear, please let me know. Also, I can’t prove it’s greater than the previous winner, but I would be very impressed with that winner if someone can prove it’s not.
A is the sequence of all welldefined, finite numbers greater than 2 posted in this thread so far (whether or not they were the largest at the time of posting), excluding the previous winner, each rounded to the nearest integer. m is the number of numbers in A. The elements in A are arranged by size; A_{1} is the smallest number in A and A_{m} is the largest.
f(x) converts the input number x into a string equal to x written out in base 2. Example: f(3) = “11”
g(x) converts the input numeric string x into a number by interpreting it as a number written out in base one googolplex. Example: g(“11”) = 10^(10^100) + 1
h(x) = g(f(x)) Example: h(3) = 10^(10^100) + 1
j(x) = h(h(h(h(h(h(h(h(h(h(x))))))))))
k(x) is the function defined in the Matlab code below (where j(x) is the function defined above)
function y = k(x)
y = x
for a=1:(x^x)
y = j(y)
end
Example : k(3) = j(j(j(j(j(j(j(j(j(j(j(j(j(j(j(j(j(j(j(j(j(j(j(j(j(j(j(3))))))))))))))))))))))))))) = a really big number
l(x) = k(k(k(k(k(k(k(k(k(k(x))))))))))
p(X) applies l(x) to each element in sequence X (So if C = p(B), then C_{1} = l(B_{1}), C_{2} = l(B_{2}), etc.)
s(X) = X_{1}^(X_{2}^(X_{3}^(...^X_{m}))) where m is the number of elements in sequence X
So here is my number:
l(l(l(l(l(l(l(l(l(l(s(p(p(p(p(p(p(p(p(p(p(A)))))))))))))))))))))
A is the sequence of all welldefined, finite numbers greater than 2 posted in this thread so far (whether or not they were the largest at the time of posting), excluding the previous winner, each rounded to the nearest integer. m is the number of numbers in A. The elements in A are arranged by size; A_{1} is the smallest number in A and A_{m} is the largest.
f(x) converts the input number x into a string equal to x written out in base 2. Example: f(3) = “11”
g(x) converts the input numeric string x into a number by interpreting it as a number written out in base one googolplex. Example: g(“11”) = 10^(10^100) + 1
h(x) = g(f(x)) Example: h(3) = 10^(10^100) + 1
j(x) = h(h(h(h(h(h(h(h(h(h(x))))))))))
k(x) is the function defined in the Matlab code below (where j(x) is the function defined above)
function y = k(x)
y = x
for a=1:(x^x)
y = j(y)
end
Example : k(3) = j(j(j(j(j(j(j(j(j(j(j(j(j(j(j(j(j(j(j(j(j(j(j(j(j(j(j(3))))))))))))))))))))))))))) = a really big number
l(x) = k(k(k(k(k(k(k(k(k(k(x))))))))))
p(X) applies l(x) to each element in sequence X (So if C = p(B), then C_{1} = l(B_{1}), C_{2} = l(B_{2}), etc.)
s(X) = X_{1}^(X_{2}^(X_{3}^(...^X_{m}))) where m is the number of elements in sequence X
So here is my number:
l(l(l(l(l(l(l(l(l(l(s(p(p(p(p(p(p(p(p(p(p(A)))))))))))))))))))))
 Yakk
 Poster with most posts but no title.
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Re: My number is bigger!
Probably not a winner. Everything you did is rounding error at the scales of the top numbers in this thread.
So the top two numbers are going to differ by much more than that, even if they are so close we cannot tell them apart.
They are big.
You never built a generalization of recursion. You never used something like godel numbering. Ya got no growth kid.
So the top two numbers are going to differ by much more than that, even if they are so close we cannot tell them apart.
They are big.
You never built a generalization of recursion. You never used something like godel numbering. Ya got no growth kid.
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.
Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.
 Vytron
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 Joined: Mon Oct 19, 2009 10:11 am UTC
 Location: The Outside. I use She/He/Her/His/Him as gender neutral pronouns :P
Re: My number is bigger!
I do think that SirGabriel's number could be bigger if it was valid, as the second biggest and third biggest number are possibly bigger already than the biggest one (a tie was agreed between the contestants), but they haven't been proven to be bigger. While SirGabriel hasn't proven that his is the biggest, he could be.
Something like:
SirGabriel = f_w+3(second biggest number+third biggest number) (fourth and later don't really make a difference.)
However, it's referring directly a previous number, and not even in a specific way, but a fuzzy one ("second biggest, whatever it is"), and referencing many, which I don't think is actually valid, though gmalivuk can correct me on that.
Something like:
SirGabriel = f_w+3(second biggest number+third biggest number) (fourth and later don't really make a difference.)
However, it's referring directly a previous number, and not even in a specific way, but a fuzzy one ("second biggest, whatever it is"), and referencing many, which I don't think is actually valid, though gmalivuk can correct me on that.
 SirGabriel
 Posts: 40
 Joined: Wed Jul 16, 2014 11:54 pm UTC
Re: My number is bigger!
The rules only say "you can't refer directly to the previous number", and I didn't. I referred directly to every number posted so far except the previous one.
 Vytron
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 Joined: Mon Oct 19, 2009 10:11 am UTC
 Location: The Outside. I use She/He/Her/His/Him as gender neutral pronouns :P
Re: My number is bigger!
Now you have to prove that that's enough to beat the highest number, because as it is, the second biggest number on the thread has a recursive function much stronger than what you're doing, so that recursing one more time would be bigger than your number, but it wouldn't be bigger than the biggest number of the thread, so your number would be smaller still.
 Yakk
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Re: My number is bigger!
Vytron wrote:I do think that SirGabriel's number could be bigger if it was valid, as the second biggest and third biggest number are possibly bigger already than the biggest one (a tie was agreed between the contestants), but they haven't been proven to be bigger. While SirGabriel hasn't proven that his is the biggest, he could be.
If they where already bigger, they not the second and third biggest numbers.
So even if we generously give SirGabriel the freedom to refer to the biggest number (by exclusion), he still won't be the biggest.
The problem is, as you noted later, that SirGabriel's recursive system is extremely weak. It doesn't increase the size of the input numbers significantly, at the scale that the big numbers in this thread are working with. Everything is rounding error.
It is extremely unlikely that two numbers in this thread in the "big" range are close enough to each other that SirGabriel's recursive system would be sufficient to make one bigger than the other, even if we did not know which was larger.
Imagine if someone was playing a more conventional game of "bigger number".
SirGabriel's would be like saying "Pretend the second largest number is mass of the universe. I add the mass of a neutrino. I must be bigger than the largest number!" ... except that idiom is still way way too weak.
SirGabriel's would be like saying "Pretend the second largest number is the TREE3 in kg. I add a gram to it. I must be bigger than the largest number!" ... except that idiom is still way way too weak.
SirGabriel's would be like saying "Pretend the second largest number 1. I add zero to it. I must be bigger than the largest number!" ... that is closer.
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.
Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.
 Vytron
 Posts: 429
 Joined: Mon Oct 19, 2009 10:11 am UTC
 Location: The Outside. I use She/He/Her/His/Him as gender neutral pronouns :P
Re: My number is bigger!
Yeah, i get it.
My head hurts when I try to understand it, though, I guess the numbers are way too big for it.
Let's say that these are true facts:
The biggest number of the thread is using the most powerful recursion on the thread, and it would beat the second biggest even if it didn't recurse as many times.
The second biggest number of the thread is using the second most powerful recursion, much more powerful than any other entry below, but would require an extremely big number of recursions to pass the biggest one.
The third biggest number of the thread is using the third most powerful recursion, much more powerful than any other entry below, but would require an extremely big number of recursions to pass the second one.
Okay, that'd make sense.
Now, enter SirGabriel's function, which doesn't actually use the recursions of those big numbers, but uses them directly. Since it's using them directly, then, by force, his number should be between the second and biggest one.
And here's my problem:
The third biggest number's function is much stronger than SirGabriel, which means, that if the third biggest number recursed twice, it would eat SirGabriel's number for breakfast.
However, it has been stated that the third biggest number's function would need to recurse an extremely high number of times to pass the second biggest number, yet it can pass a number between the second and biggest number by recursing twice, leading to a contradiction.
So by the principle of explosion SirGabriel's number may actually be the biggest one, or the biggest one may be below 0, etc.
My head hurts when I try to understand it, though, I guess the numbers are way too big for it.
Let's say that these are true facts:
The biggest number of the thread is using the most powerful recursion on the thread, and it would beat the second biggest even if it didn't recurse as many times.
The second biggest number of the thread is using the second most powerful recursion, much more powerful than any other entry below, but would require an extremely big number of recursions to pass the biggest one.
The third biggest number of the thread is using the third most powerful recursion, much more powerful than any other entry below, but would require an extremely big number of recursions to pass the second one.
Okay, that'd make sense.
Now, enter SirGabriel's function, which doesn't actually use the recursions of those big numbers, but uses them directly. Since it's using them directly, then, by force, his number should be between the second and biggest one.
And here's my problem:
The third biggest number's function is much stronger than SirGabriel, which means, that if the third biggest number recursed twice, it would eat SirGabriel's number for breakfast.
However, it has been stated that the third biggest number's function would need to recurse an extremely high number of times to pass the second biggest number, yet it can pass a number between the second and biggest number by recursing twice, leading to a contradiction.
So by the principle of explosion SirGabriel's number may actually be the biggest one, or the biggest one may be below 0, etc.
 SirGabriel
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 Joined: Wed Jul 16, 2014 11:54 pm UTC
Re: My number is bigger!
I haven't been following this thread very closely. At the moment, what are the largest few numbers (other than mine)?
 Yakk
 Poster with most posts but no title.
 Posts: 11058
 Joined: Sat Jan 27, 2007 7:27 pm UTC
 Location: E pur si muove
Re: My number is bigger!
Vytron wrote:Now, enter SirGabriel's function, which doesn't actually use the recursions of those big numbers, but uses them directly. Since it's using them directly, then, by force, his number should be between the second and biggest one.
Yes. His number is at least as big as "second biggest", as a^b > a*b when a and b are at least 2, and the end of his mess has ^C where C the the result of taking the 2nd biggest number and passing it to a function z such that z(x)>x for x>1.
The recursive strength of his mess is about 4 or so (where + is 0, * is 1, ^ is 2). Maybe 5, I may have miscounted. You could work through each step, or you could just note that each step is merely a recursion of a previous step, and starts with ^, and count.
The mediumbig answers in this thread all use large cardinals to describe the recursion they are using (on similar scales).
The third biggest number's function is much stronger than SirGabriel, which means, that if the third biggest number recursed twice, it would eat SirGabriel's number for breakfast.
Not quite. SirGabriel's uses as input a really big number (well a bunch of them) which was stolen from other people's work.
All of the work SirGabriel did did not change it in any measurable sense, but it remains that really big number.
Had SirGabriel posted "the 2nd largest number, +1", SirGabriel's number would still be larger than the 2nd largest number by about the same margin.
Suppose the second largest number was x_y(3), where _y is some large cardinal. They could say x_{y+w} (3) or x_y(4) and beat SirGabriel's number.
The third largest number is going to have a massively weaker recursive strength. That if they are a_b(3), then they won't be able to beat x_y(3) by saying a_b(4) or a_b(1000) or a_b(1 googol).
At the same time, they could beat x_y(3) with a_b( x_y(3) ) because a_b doesn't decrease its input. But the "work" being done to make the result bigger is all in x_y(3). What is worse is that a_b( x_y(3) ) would probably be smaller than x_y(4), as passing the value through a_b **does not significantly make it bigger** at the scales that x_y works in.
That is why talking about the "second largest value" and using it to beat the largest doesn't work, unless the largest is basically "I copied what the second largest value was, then I added the logically equivalent of 1", and such addition occurring at an inefficient spot. Especially if you use it as an input to your function.
Now, I believe the largest numbers in this thread no longer use ordinal based recursion, but I am not sure: certain kinds of ordinal based recursion are probably equivalent to proof theories. But my naive guess would be that the proof theoretic ones may exceed them  the ones that talk about finding "short" proofs of "short" Turing machines that halt, then finding the longest running one.
The ordinal based recursion problems start out with an operation like +, then recursively call it a large number of times (with each layer of recursion increasing the number of times the previous layer is called, and then the number of layers, then the number of layers of layers, ... (and then the number of such layer of layers of ... stacks (and then the number of nested brackets like this talking about nesting,... (and then ...))))): they map such increasingly complex recursive structures over to the ordinals, which lets them invoke already well known large ordinals with known properties to obscenely complex recursive structures to generate rather fast growing functions.
The proof theoretic ones start with a faster growing function  basically a pseudobusy beaver. And it may be enough to beat out the above.
My problem is that really powerful and large ordinals have so much structure inside of them that using them as a description of "mere recursion" might end up producing a machine as powerful as a naive proof checker "implicitly". And I wouldn't know where to begin to determine if one was larger than the other.
In short, my complexity theory isn't up to determining which one of those is bigger.
But I do know enough that throwing a few layers of recursion on top of any of these beasts isn't going to make one pass another. Which is why the bigger numbers start with a really powerful function, and then evaluated it at 3 or somesuch "small" number (or 9 if they are feeling cheeky: same number of characters).
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.
Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.
Re: My number is bigger!
This thread is alive again! :O
 Vytron
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 Location: The Outside. I use She/He/Her/His/Him as gender neutral pronouns :P
Re: My number is bigger!
Thanks Yakk for the explanation, now I understand better why SirGabriel's approach can't be used to beat the biggest number of the thread.
This is the biggest number of the thread:
EliezerYudkowsky's
It's so big that it could use a much smaller number as input, but it does use 10 as input.
Second biggest is here. tomtom2357 claims it's even bigger than Yudkowsky's, but doesn't prove it.
(tomtom2357's was challenged and changed his number to "F(n) as the largest number such that there exists a statement that uniquely defines it, where that is provable in a proof of at most f(n) symbols (where f(n) is your favorite function  I guess tomtom2357 would be using Deedlit's function below here.).)
And here's the third biggest one:
Deedlit's
The one to beat before Yudkowsky arrived, and, it may actually be the current champion since Yudkowsky never proved that his was bigger (people just went like, "oh, yeah, Yudkowsky's should be bigger", because you can't invent a notation that represents Yudkowsky number by definition, or something.)
SirGabriel wrote:I haven't been following this thread very closely. At the moment, what are the largest few numbers (other than mine)?
This is the biggest number of the thread:
EliezerYudkowsky's
It's so big that it could use a much smaller number as input, but it does use 10 as input.
Second biggest is here. tomtom2357 claims it's even bigger than Yudkowsky's, but doesn't prove it.
(tomtom2357's was challenged and changed his number to "F(n) as the largest number such that there exists a statement that uniquely defines it, where that is provable in a proof of at most f(n) symbols (where f(n) is your favorite function  I guess tomtom2357 would be using Deedlit's function below here.).)
And here's the third biggest one:
Deedlit's
The one to beat before Yudkowsky arrived, and, it may actually be the current champion since Yudkowsky never proved that his was bigger (people just went like, "oh, yeah, Yudkowsky's should be bigger", because you can't invent a notation that represents Yudkowsky number by definition, or something.)
Re: My number is bigger!
Define:
Rayo_{a}(n) = Rayo(Rayo(Rayo(Rayo(Rayo(...(n)...))))) where the number of nested brackets is a.
Define:
Rayo_{a}[c](n) = (Rayo_a(n))_{(Rayo}_{_a(n))}..._{Rayo}_{_a(n)}(n) where the amount of subscripts after one another is equal to [c]. (Apologies; I couldn't do multiple subscripts in this forum so I used the underscore _ as subscript too).
Rayo_{a}(n) = Rayo(Rayo(Rayo(Rayo(Rayo(...(n)...))))) where the number of nested brackets is a.
Define:
Rayo_{a}[c](n) = (Rayo_a(n))_{(Rayo}_{_a(n))}..._{Rayo}_{_a(n)}(n) where the amount of subscripts after one another is equal to [c]. (Apologies; I couldn't do multiple subscripts in this forum so I used the underscore _ as subscript too).
 Vytron
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Re: My number is bigger!
You forgot to define Rayo(n).
Re: My number is bigger!
Even if you had defined rayo(n)
The amount of power you gained on rayo(n) vs Rayo_{a}[c](n) is truly insignificant.
Its like going from 4^^^^4 to 4>4>4>4
(edit: ill go all the way and explain my comment)
lets asume there is an ordinal ? for which f_{?}(n) has growth matching with Rayo(n), what you did is like f_{?*2}(n) at most.
The amount of power you gained on rayo(n) vs Rayo_{a}[c](n) is truly insignificant.
Its like going from 4^^^^4 to 4>4>4>4
(edit: ill go all the way and explain my comment)
lets asume there is an ordinal ? for which f_{?}(n) has growth matching with Rayo(n), what you did is like f_{?*2}(n) at most.
Re: My number is bigger!
Dang, knew I was outclassed. Ah well; it's not even computable.
Re: My number is bigger!
I know it can be confusing at first (it was for me) but once you dip into understanding the Fast Growing Hierarchy and how it works with the basic ordinals ω and ε things clear up alot.
basically fgh is represented with an f and a three other...elements: a superscript which is a sort of "nesting indicator" and a subscript which is_{} known as an "Ordinal" and finanlly the argument, written inside parenthesis
So to speak, the "f" is not actually a function, its a list of functions, and the subscript indicates which function on this list we are referring to.
There are two types of ordinals:
Successors and Limits
Successors are ordinals that succeed an ordinal. Much like 5 succeeds 4.
The first function in FGH is
f_{0}(n) = n+1
so for example f_{0}(100)=101
The nesting indicator, when it appears, just indicates that a function is nested on itself that many times.
for example
f^{2}_{0}(100)=f_{0}(f_{0}(100))=f_{0}(101)=102
The second function in FGH is
f_{1}(n) = f^{n}_{0}(n).
Remember these are all singleargument functions. However this one argument continues to have more and more power as we go further up the list.
Here the argument "unfolds" as nestings of f_{0}(n). So basically you are doing +1 n times to n itself. This is also known as n+n or n*2. f_{1}(n) is the doubling function.
f_{2}(n)=is thus the repeated doubling function. so like n*2*2*2.... or n*2^{n} (so its exponentiationlevel)
f_{3}(n)=is repeated exponentiation. like if you drop the n*2 in the above example you get something like 2^2^2... or around 2^^n
So basically when you create a function G that nests a previous function onto itself x amount of times, (much like you did) its like going +1 on the subscript of fgh and setting the argument to be the X. When you do this again with a new function H that nests G on itself its like doing +2 on that subscript etc..
basically fgh is represented with an f and a three other...elements: a superscript which is a sort of "nesting indicator" and a subscript which is_{} known as an "Ordinal" and finanlly the argument, written inside parenthesis
So to speak, the "f" is not actually a function, its a list of functions, and the subscript indicates which function on this list we are referring to.
There are two types of ordinals:
Successors and Limits
Successors are ordinals that succeed an ordinal. Much like 5 succeeds 4.
The first function in FGH is
f_{0}(n) = n+1
so for example f_{0}(100)=101
The nesting indicator, when it appears, just indicates that a function is nested on itself that many times.
for example
f^{2}_{0}(100)=f_{0}(f_{0}(100))=f_{0}(101)=102
The second function in FGH is
f_{1}(n) = f^{n}_{0}(n).
Remember these are all singleargument functions. However this one argument continues to have more and more power as we go further up the list.
Here the argument "unfolds" as nestings of f_{0}(n). So basically you are doing +1 n times to n itself. This is also known as n+n or n*2. f_{1}(n) is the doubling function.
f_{2}(n)=is thus the repeated doubling function. so like n*2*2*2.... or n*2^{n} (so its exponentiationlevel)
f_{3}(n)=is repeated exponentiation. like if you drop the n*2 in the above example you get something like 2^2^2... or around 2^^n
So basically when you create a function G that nests a previous function onto itself x amount of times, (much like you did) its like going +1 on the subscript of fgh and setting the argument to be the X. When you do this again with a new function H that nests G on itself its like doing +2 on that subscript etc..
 Vytron
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Re: My number is bigger!
Gingeas wrote:Dang, knew I was outclassed. Ah well; it's not even computable.
It is computable.
Re: My number is bigger!
Ah okay, thanks for the clarification.
 Yakk
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Re: My number is bigger!
Daggoth, f_{0}^{n}(n) = f_{1}(n) rather than f^{n}_{0}(n) to line up with more traditional notation. f_{0} is "just some function". For a function g, g^{n}(x) = g(g(g(...(g(x))...))) with n nests. So f_{0}^{n} doesn't need "special notation" to be defined as what it is.
f_{succ(x)}(n) := f_{x}^{n}(n)
You have handled successor. Next, cover what happens with limit ordinals in the subscript?
If phi is a limit ordinal, then it is "after" some infinite "fundamental sequence" of ordinals.
Then we denote phi[n] to be the nth element of that fundamental sequence, and define
f_{phi}(n) := f_{phi[n]}(n)
In the case of the first infinite ordinal  ω  the fundamental sequence is the natural numbers.
So this means that f_{ω}(x) := f_{x}(x)
If we plug in 3, we get
f_{3}(3)
= f_{2}^{3}(3)
= f_{2} f_{2} f_{2} (3)
= f_{2} f_{2} ( f_{1} f_{1} f_{1} (3) )
As noted above, f_{1}(x) = 2x
= f_{2} f_{2} ( f_{1} f_{1} 6 )
= f_{2} f_{2} ( f_{1} 12 )
= f_{2} f_{2} ( 24 )
As noted above, f_{2}(x) = x 2^x
= f_{2} 24 * (2^24)
= 24 * (2^24) * 2^(24 * (2^24))
now, 24 2^24 =~ 384 million
= 384 million * 2^(384 million)
2^384 million =~ 10^115.2
384 million =~ 10^8.5
so f_{ω}(3) = f_{3}(3) =~ 10^123.7, or about a trillion trillion googol.
Now, f_{ω+1}(3) is..
Let g := f_{ω}
Then g^{g(3)}(3)
Ie, we take the process above that generated 10^123 from 3, and we repeat it 10^123 times with its last output as its input. This includes replacing the _{3} with _{10^123} on the second iteration. And we do this a trillion trillion google times.
f_{ω+2} then repeats the above trick the number of times the first call to the above trick results in.
(!)
Etc, using the succ(ω+n) = ω+n+1 rule.
f_{2ω}(x) is f_{2ω[x]}(x). The limit sequence for 2ω is {ω, ω+1, ω+2, ...}, so here
so f_{2ω}(x) = f_{ω+x}(x)
f_{2ω+1}(x) takes the output of f_{2ω}(x), and uses it to repeat f_{2ω} that many times on the initial input. Ditto for f_{2ω+n}.
This continues.
f_{ω^2}(x) is f_{xω}(x), etc
f_{ω^ω}(x) is f_{ω^x}(x), etc
f_{ω^ω^ω}(x) is f_{ω^ω^x}(x), etc
ε_{0} is defined as the limit of {ω, ω^ω, ω^ω^ω, ω^ω^ω^ω, ...}.
So we can feed ε_{0} as a subscript, and get a tower of ω^ω^...^ω based off x, then evaluate that, etc.
Now, the thing is, these ordinals keep on going up. The ordinals above are relatively simple: as this is a studied bit of mathematics, you can get even more ridiculous kinds of recursion simply by naming a higher ordinal.
Long before we reach this point, even describing a sketch of the kind of recursion required to reach the base case becomes far beyond the power of the human mind. What we can do is reduce each step to a known halting step, and convince ourselves that it does "eventually" stop.
This is called the "fast growing hierarchy". Note I am far from an expert at this, and I could easily have made errors above.
Now, what we can use this for is to describe the nature of your recursion. It is relatively easy to show that a given description of recursion cannot "pass some ordinal in the fast growing hierarchy"  or, more precisely, you can first show that the function you start with is weaker than some f_{phi_0} in the hierarchy, and that the recursive system you called it with adds less than some transformation on the ordinal, and thus bound the result below some (usually modestly near) other f_{phi_1}.
In short, in order to generate a decent value, you first need to describe not just recursion, but a method to generate recursion and a way to name extremely complex recursive structures. The fast growing hierarchy is a famous one that is hard to naively beat. In this thread, if you aren't doing something at least that powerful, your submission could be described as small. If you have at least used the fast growing hierarchy, I would denote your answer as at least medium.
Now for large, I think the question is how well extremely large ordinals compete with the prooftheoretic attacks (basically pseudobusybeaver). I don't know enough complexity theory, but the idea would be that a sufficiently complex ordinal could have more structure than the proof theory that the prooftheoretic system is using, and it could exploit that structure to effectively beat out the growth of the proof theoretic ones.
So, the competition for large would be (in my inexperienced eye) be between the robust prooftheoretic ones, and the obscenely large ordinal ones. But I could easily be wrong: the reason both are large is I personally lack the skills to determine their relative size.
f_{succ(x)}(n) := f_{x}^{n}(n)
You have handled successor. Next, cover what happens with limit ordinals in the subscript?
If phi is a limit ordinal, then it is "after" some infinite "fundamental sequence" of ordinals.
Then we denote phi[n] to be the nth element of that fundamental sequence, and define
f_{phi}(n) := f_{phi[n]}(n)
In the case of the first infinite ordinal  ω  the fundamental sequence is the natural numbers.
So this means that f_{ω}(x) := f_{x}(x)
If we plug in 3, we get
f_{3}(3)
= f_{2}^{3}(3)
= f_{2} f_{2} f_{2} (3)
= f_{2} f_{2} ( f_{1} f_{1} f_{1} (3) )
As noted above, f_{1}(x) = 2x
= f_{2} f_{2} ( f_{1} f_{1} 6 )
= f_{2} f_{2} ( f_{1} 12 )
= f_{2} f_{2} ( 24 )
As noted above, f_{2}(x) = x 2^x
= f_{2} 24 * (2^24)
= 24 * (2^24) * 2^(24 * (2^24))
now, 24 2^24 =~ 384 million
= 384 million * 2^(384 million)
2^384 million =~ 10^115.2
384 million =~ 10^8.5
so f_{ω}(3) = f_{3}(3) =~ 10^123.7, or about a trillion trillion googol.
Now, f_{ω+1}(3) is..
Let g := f_{ω}
Then g^{g(3)}(3)
Ie, we take the process above that generated 10^123 from 3, and we repeat it 10^123 times with its last output as its input. This includes replacing the _{3} with _{10^123} on the second iteration. And we do this a trillion trillion google times.
f_{ω+2} then repeats the above trick the number of times the first call to the above trick results in.
(!)
Etc, using the succ(ω+n) = ω+n+1 rule.
f_{2ω}(x) is f_{2ω[x]}(x). The limit sequence for 2ω is {ω, ω+1, ω+2, ...}, so here
so f_{2ω}(x) = f_{ω+x}(x)
f_{2ω+1}(x) takes the output of f_{2ω}(x), and uses it to repeat f_{2ω} that many times on the initial input. Ditto for f_{2ω+n}.
This continues.
f_{ω^2}(x) is f_{xω}(x), etc
f_{ω^ω}(x) is f_{ω^x}(x), etc
f_{ω^ω^ω}(x) is f_{ω^ω^x}(x), etc
ε_{0} is defined as the limit of {ω, ω^ω, ω^ω^ω, ω^ω^ω^ω, ...}.
So we can feed ε_{0} as a subscript, and get a tower of ω^ω^...^ω based off x, then evaluate that, etc.
Now, the thing is, these ordinals keep on going up. The ordinals above are relatively simple: as this is a studied bit of mathematics, you can get even more ridiculous kinds of recursion simply by naming a higher ordinal.
Long before we reach this point, even describing a sketch of the kind of recursion required to reach the base case becomes far beyond the power of the human mind. What we can do is reduce each step to a known halting step, and convince ourselves that it does "eventually" stop.
This is called the "fast growing hierarchy". Note I am far from an expert at this, and I could easily have made errors above.
Now, what we can use this for is to describe the nature of your recursion. It is relatively easy to show that a given description of recursion cannot "pass some ordinal in the fast growing hierarchy"  or, more precisely, you can first show that the function you start with is weaker than some f_{phi_0} in the hierarchy, and that the recursive system you called it with adds less than some transformation on the ordinal, and thus bound the result below some (usually modestly near) other f_{phi_1}.
In short, in order to generate a decent value, you first need to describe not just recursion, but a method to generate recursion and a way to name extremely complex recursive structures. The fast growing hierarchy is a famous one that is hard to naively beat. In this thread, if you aren't doing something at least that powerful, your submission could be described as small. If you have at least used the fast growing hierarchy, I would denote your answer as at least medium.
Now for large, I think the question is how well extremely large ordinals compete with the prooftheoretic attacks (basically pseudobusybeaver). I don't know enough complexity theory, but the idea would be that a sufficiently complex ordinal could have more structure than the proof theory that the prooftheoretic system is using, and it could exploit that structure to effectively beat out the growth of the proof theoretic ones.
So, the competition for large would be (in my inexperienced eye) be between the robust prooftheoretic ones, and the obscenely large ordinal ones. But I could easily be wrong: the reason both are large is I personally lack the skills to determine their relative size.
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.
Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.
Re: My number is bigger!
i prefer to use f^{a}_{b}(c) so as to not allow a to be confused for an exponent in b (but this only because of the limitations here, the superscript should be directly above the indexer)
 Yakk
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 Posts: 11058
 Joined: Sat Jan 27, 2007 7:27 pm UTC
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Re: My number is bigger!
Daggoth wrote:i prefer to use f^{a}_{b}(c) so as to not allow a to be confused for an exponent in b (but this only because of the limitations here, the superscript should be directly above the indexer)
It should be (f_{b})^{a}, but for the annoyance of brackets  the order of operations is pretty clear, there is no need to combine them as one concept. We have this family of functions (subscript), and powers of said functions.
Using the other way around involves inventing new notation that is redundant. Which I dislike.
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.
Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.
Re: My number is bigger!
I'm sorry timmy, but i don't dip that way.
Re: My number is bigger!
Yakk wrote:So, the competition for large would be (in my inexperienced eye) be between the robust prooftheoretic ones, and the obscenely large ordinal ones. But I could easily be wrong: the reason both are large is I personally lack the skills to determine their relative size.
I'm fairly sure it's the proof ones that are stronger, because presumably one can actually prove that the ordinal collapse functions are finite (else we wouldn't be able to submit them as entries) and then the length of that proof is the point where the proof based numbers exceed that ordinal entry.
The problem with the big proof numbers is we don't really know which large cardinal axioms are actually ultimately consistent right now, so we can't really do... anything with them, really.
And when people mentioned Rayo's number... Rayo's function is so not computable. It's not even definable in first order set theory, and there are lots of things easy to define in FOST yet incomputable.
All Shadow priest spells that deal Fire damage now appear green.
Big freaky cereal boxes of death.
 Yakk
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 Posts: 11058
 Joined: Sat Jan 27, 2007 7:27 pm UTC
 Location: E pur si muove
Re: My number is bigger!
WarDaft, I am not so sure. See large ordinals are exceedingly complex: in some cases you need strong proof theories to show they are ordinals! And I think I can sketch an argument that shows a neat little bijection between large ordinal recursive functions and large cardinal proof theory based recursive functions:
So, assuming axiom of choice, each cardinal is indexed by an ordinal.
It existing (the large cardinal axiom) proves that the ordinal that indexes exists.
The recursive definitions terminate if and only if the ordinal recursion terminates.
Let us suppose we have f as roughly defined in my post above (a recursive function whose structure is generated by an ordinal).
I define P( foo ) to be the proof theory of ZFC + Large Cardinal Axiom א_{foo}.
So if you have a proof theory that is P(foo) that is "just barely" able to prove that foo is an ordinal. f_{foo} being shown to terminate is equivalent to the Large Cardinal Axiom _{foo}!
Suppose we have a larger cardinal, א_{foo} (sufficiently larger, such that the LCA of foo does not imply the LCA of bar). Then f_{bar} cannot be shown to terminate in the above theory. So if f_{bar} is proposed by poster A, and poster B proposes some recursive structure involving the fastest growing function that can be shown to terminate in "ZFC + Large Cardinal Axiom א_{foo}" (let alone "that has a short proof it can terminate"), then f_{bar} may out perform it. Despite this, the proof that f_{bar} terminates is short and simple (namely, "f_{ordinal} terminates, and bar is an ordinal."), but this only works in a proof theory strong enough to prove that bar is an ordinal, and if bar is an ordinal then א_{bar} is a cardinal, which means your proof theory is as strong as P(bar).
This is because the structure of bar is so complex, it requires a strong proof theory to prove it is a ordinal.
The thing is, no matter how fancy or long you make your proof in P(foo), it cannot show that f_{bar} terminates: so lots of recursion in the proof theory isn't enough.
In short, I think the termination of large ordinal f_{ordinal} for bar ordinal may be as strong as using proof searching techniques. The proof searchers definitely cannot prove themselves faster growing than larger ordinal f_{ordinal}s.
Now if we throw out the axiom of choice, there are cardinals that are not of the form א_{foo} for some ordinal foo. I haven't looked into that yet.
Another issue is that while P(foo) may not be able to prove that the function we are talking about exceeds the growth rate of f_{bar}, it may still out grow it?
Another problem is that I assumed that the larger cardinal axiom was independent. That was probably sloppy, because if it was we'd basically be outputting nominally infinite numbers (sort of). I could instead talk about the lack of existence of a "short" proof that the larger one is consistent in the smaller one, where naming the length of the proof in the weaker axiom system is impractical (even using crazy techniques like this thread does).
Any other flaws with the above argument?
Hmm. I guess the lack of "short" proof makes it difficult for someone to believe that the systems are consistent, no? And if there is a short proof in the weaker system, then the proof based one can find it and fold the larger cardinal proof theory into itself.
So, assuming axiom of choice, each cardinal is indexed by an ordinal.
It existing (the large cardinal axiom) proves that the ordinal that indexes exists.
The recursive definitions terminate if and only if the ordinal recursion terminates.
Let us suppose we have f as roughly defined in my post above (a recursive function whose structure is generated by an ordinal).
I define P( foo ) to be the proof theory of ZFC + Large Cardinal Axiom א_{foo}.
So if you have a proof theory that is P(foo) that is "just barely" able to prove that foo is an ordinal. f_{foo} being shown to terminate is equivalent to the Large Cardinal Axiom _{foo}!
Suppose we have a larger cardinal, א_{foo} (sufficiently larger, such that the LCA of foo does not imply the LCA of bar). Then f_{bar} cannot be shown to terminate in the above theory. So if f_{bar} is proposed by poster A, and poster B proposes some recursive structure involving the fastest growing function that can be shown to terminate in "ZFC + Large Cardinal Axiom א_{foo}" (let alone "that has a short proof it can terminate"), then f_{bar} may out perform it. Despite this, the proof that f_{bar} terminates is short and simple (namely, "f_{ordinal} terminates, and bar is an ordinal."), but this only works in a proof theory strong enough to prove that bar is an ordinal, and if bar is an ordinal then א_{bar} is a cardinal, which means your proof theory is as strong as P(bar).
This is because the structure of bar is so complex, it requires a strong proof theory to prove it is a ordinal.
The thing is, no matter how fancy or long you make your proof in P(foo), it cannot show that f_{bar} terminates: so lots of recursion in the proof theory isn't enough.
In short, I think the termination of large ordinal f_{ordinal} for bar ordinal may be as strong as using proof searching techniques. The proof searchers definitely cannot prove themselves faster growing than larger ordinal f_{ordinal}s.
Now if we throw out the axiom of choice, there are cardinals that are not of the form א_{foo} for some ordinal foo. I haven't looked into that yet.
Another issue is that while P(foo) may not be able to prove that the function we are talking about exceeds the growth rate of f_{bar}, it may still out grow it?
Another problem is that I assumed that the larger cardinal axiom was independent. That was probably sloppy, because if it was we'd basically be outputting nominally infinite numbers (sort of). I could instead talk about the lack of existence of a "short" proof that the larger one is consistent in the smaller one, where naming the length of the proof in the weaker axiom system is impractical (even using crazy techniques like this thread does).
Any other flaws with the above argument?
Hmm. I guess the lack of "short" proof makes it difficult for someone to believe that the systems are consistent, no? And if there is a short proof in the weaker system, then the proof based one can find it and fold the larger cardinal proof theory into itself.
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.
Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.
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