My number is bigger!

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Blatm
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Postby Blatm » Tue Jul 10, 2007 3:34 am UTC

I have officially lost my bearings. I'll go look up chained arrow notation and get back to this.

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Postby zomgmouse » Tue Jul 10, 2007 4:37 am UTC

999999999999999999999999999999999999999999999999999999999999
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999999999999993
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Blatm
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Postby Blatm » Tue Jul 10, 2007 5:26 am UTC

gmalivuk's number is still much much much bigger. The previous few have been, actually.

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Postby gmalivuk » Tue Jul 10, 2007 3:52 pm UTC

Yeah, they've actually been much, much bigger than that annoying block of 9's (even with the exponent) since Saturday evening.

Your wimpy little number is scarcely more than 10^(1.836x10^3876). Which pales in comparison to Rodan's 10^(10^10000000000).
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Postby ijmaxwell » Tue Jul 10, 2007 7:36 pm UTC

I'll have to go with

2→2→2→2→2
= 2→2→2→(2→2→2→(2→2→2→(2→2→2→(2→2→2)→1)→1)→1)→1
= 2→2→2→(2→2→2→(2→2→2→(2→2→2→(2→2→2))))
= 2→2→(2→2)→[2→2→2→(2→2→2→(2→2→2→(2→2→2))) - 1]
= 2→2→4→[2→2→2→(2→2→2→(2→2→2→(2→2→2))) - 1]
= 2→2→(2→2→(2→2→(2→2)→[2→2→2→(2→2→2→(2→2→2→(2→2→2))) - 1])→[2→2→2→(2→2→2→(2→2→2→(2→2→2))) - 1])→[2→2→2→(2→2→2→(2→2→2→(2→2→2))) - 1]
etc.....

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Postby gmalivuk » Tue Jul 10, 2007 8:22 pm UTC

Ümläüt wrote:I'll have to go with

2→2→2→2→2
= 2→2→2→(2→2→2→(2→2→2→(2→2→2→(2→2→2)→1)→1)→1)→1
= 2→2→2→(2→2→2→(2→2→2→(2→2→2→(2→2→2))))
= 2→2→(2→2)→[2→2→2→(2→2→2→(2→2→2→(2→2→2))) - 1]
= 2→2→4→[2→2→2→(2→2→2→(2→2→2→(2→2→2))) - 1]
= 2→2→(2→2→(2→2→(2→2)→[2→2→2→(2→2→2→(2→2→2→(2→2→2))) - 1])→[2→2→2→(2→2→2→(2→2→2→(2→2→2))) - 1])→[2→2→2→(2→2→2→(2→2→2→(2→2→2))) - 1]
etc.....


Terribly sorry to burst your bubble here, but 4 is not bigger than any number that's been posted so far. :-)

(2→2→anything = 4)
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Postby Blatm » Wed Jul 11, 2007 12:28 am UTC

That's almost as funny as the nine thousand bit at the beginning of the thread. I still don't have a grasp on how chained arrow notation works beyond 3 numbers, so I'll refrain from posting something that I don't understand.

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Postby gmalivuk » Wed Jul 11, 2007 1:43 am UTC

I was sorely tempted to just go with 5 and see who noticed. :-)

The Wikipedia article technically explains everything you need to know about chained arrows, but it didn't really click for me that well until I realized what is basically an intermediate step in the recursive relation.

It is, for any sequence X of arrows, with m and n integers,

X→m→n = X→(X→(m-1)→n)→n-1

In this way, you can eventually step down any tailing arrows. Which means that a 1 anywhere in the sequence kills itself and everything after it, and that a pair of 2's at the beginning mean it's going to eventually reduce to 2→2→N, where N is generally some rediculously fucking huge number. But a simple argument from up-arrow notation shows that 2→2→N = 2→2→(N-1) = ... = 2→2 = 4
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Postby Mouffles » Wed Jul 11, 2007 10:18 am UTC

I'll go one up on gmalivuk's number.
Define a function f:
f(1)=1
f(2)=2→2
f(3)=3→3→3
etc.

Then the number is f(f(...f(g_64)...)), where the function is applied g_64 times. Beat that!

(I'm just waiting for someone to introduce Busy Beaver numbers, which would pwn that number.)
In the spirit of taking things too far - the 5x5x5x5x5 Rubik's Cube.

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Postby gmalivuk » Wed Jul 11, 2007 2:54 pm UTC

I disallowed busy beavers by saying everything has to be computable in the first post. In particular, which busy beaver number would pwn yours? Sure, you could pick some really huge argument, but could you be sure it'd work?

But yeah, your number is pretty fuckin' huge, and I may just have to concede this match to you, good sir. When something's too big to be described by chained arrow notation, even if we could write a character on every particle in the universe, it's effectively really goddamn big.
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Postby LE4dGOLEM » Wed Jul 11, 2007 4:27 pm UTC

Mouffles wrote:I'll go one up on gmalivuk's number.
Define a function f:
f(1)=1
f(2)=2→2
f(3)=3→3→3
etc.

Then the number is f(f(...f(g_64)...)), where the function is applied g_64 times. Beat that!

(I'm just waiting for someone to introduce Busy Beaver numbers, which would pwn that number.)


I suppose calling that number, say, pwnt, and then calling ackerman's function on it is against the rules?
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Postby Patashu » Wed Jul 11, 2007 4:32 pm UTC

Man, I was going to use busy beaver numbers but apparantly those are forbidden since they're not computable.

In case you were interested, it was Σ(A(64,64),A(64,64))

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Postby gmalivuk » Thu Jul 12, 2007 1:21 am UTC

LE4dGOLEM wrote:
Mouffles wrote:I'll go one up on gmalivuk's number.
Define a function f:
f(1)=1
f(2)=2→2
f(3)=3→3→3
etc.

Then the number is f(f(...f(g_64)...)), where the function is applied g_64 times. Beat that!

(I'm just waiting for someone to introduce Busy Beaver numbers, which would pwn that number.)


I suppose calling that number, say, pwnt, and then calling ackerman's function on it is against the rules?


Yeah, since that would be explicitly using the previous entry to define your number.
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Postby Cheese » Sat Jul 14, 2007 8:17 pm UTC

Hmm, had the great idea of not using the last posted number, but every number before it and adding them up or multiplying them or raising them to powers of each other or whatever the hell you want...

Then realised I couldn't tell what the last acceptable number was, this thread's getting a little messed up.
Anyway, to be half-useful, I recommend you edit your first post again to disallow such inaproppriate use of math (especially as all those huge strings of digits at the start would be a real page-killer =P)
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Postby warriorness » Sat Jul 14, 2007 10:28 pm UTC

Mouffles wrote:I'll go one up on gmalivuk's number.
Define a function f:
f(1)=1
f(2)=2→2
f(3)=3→3→3
etc.

Then the number is f(f(...f(g_64)...)), where the function is applied g_64 times. Beat that!

(I'm just waiting for someone to introduce Busy Beaver numbers, which would pwn that number.)


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)
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Postby ARVash » Sat Jul 14, 2007 11:46 pm UTC

(truncated for the sake of not annoying the crap out of everyone in here XD)

while(1)
{
double i=4;
i++;
}

beh take that.
Last edited by ARVash on Sat Jul 14, 2007 11:59 pm UTC, edited 2 times in total.
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Postby crazyjimbo » Sat Jul 14, 2007 11:51 pm UTC

Not even close. That's 2^ something, and if I understand the chained arrow notation correctly, then that 'something' is pretty damn huge. A lot bigger than you could ever write out.

EDIT: Also an obnoxiously long post :(

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Postby Rodan » Sat Jul 14, 2007 11:53 pm UTC

a googolplex still eligible?

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Postby gmalivuk » Sun Jul 15, 2007 12:44 am UTC

A googolplex hasn't been eligible for ages. The number of digits in the number of digits in 3^^^3 is trillions upon trillions of times more than a googolplexian (1 with a googolplex zeros after it).
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Postby Rodan » Sun Jul 15, 2007 12:52 am UTC

damn. That's a big number... I shall shuffle out again...

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Postby warriorness » Sun Jul 15, 2007 3:31 am UTC

ARVash wrote:(truncated for the sake of not annoying the crap out of everyone in here XD)

while(1)
{
double i=4;
i++;
}

beh take that.


6. My number beats yours!
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Postby The Sleeping Tyrant » Sun Jul 15, 2007 4:00 am UTC

warriorness wrote:
ARVash wrote:(truncated for the sake of not annoying the crap out of everyone in here XD)

while(1)
{
double i=4;
i++;
}

beh take that.


6. My number beats yours!


How so? His is a loop that eventually takes i up to the highest value of a double floating point variable on the system that's compiled on. This value is, invariably, higher than 6.

Also, some of the numbers in this thread are amazingly large. I mean, awe-inspiring.

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Postby Mouffles » Sun Jul 15, 2007 4:42 am UTC

Let's take this one step further.

Let f1(x) = x→x→...→x, with x x's, as in my previous number.
Now let f2(x) = f1(f1(...f1(x)...)), where f1 is applied f1(x) times.
Then let f3(x) = f2(f2(...f2(x)...)), where f2 is applied f2(x) times.
etc.

My number is f g_64(g_64), that is, the g_64th iteration of the process above.

(btw, the factorial in warriorness' number makes very little difference: applying Q one more time would produce a hugely vaster number.)
In the spirit of taking things too far - the 5x5x5x5x5 Rubik's Cube.

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Postby warriorness » Sun Jul 15, 2007 6:25 am UTC

The Sleeping Tyrant wrote:
warriorness wrote:
ARVash wrote:(truncated for the sake of not annoying the crap out of everyone in here XD)

while(1)
{
double i=4;
i++;
}

beh take that.


6. My number beats yours!


How so? His is a loop that eventually takes i up to the highest value of a double floating point variable on the system that's compiled on. This value is, invariably, higher than 6.

Also, some of the numbers in this thread are amazingly large. I mean, awe-inspiring.


Neh, he sets i equal to 4 at the beginning of each loop iteration!

Also, re: factorial, I know. It was just for humor. It's the same as setting the chain on a door that has fifty padlocks and deadbolts already on it.
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Postby The Sleeping Tyrant » Sun Jul 15, 2007 5:45 pm UTC

warriorness wrote:Neh, he sets i equal to 4 at the beginning of each loop iteration!


Hoshit. Didn't notice that.
*egg on face*

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Postby ijmaxwell » Sun Jul 22, 2007 10:57 pm UTC

eesh, lemme try again.

I define a sequence of sequences:

a(0)(n) = n→.....→n (with n→n occurrences of n)
a(m)(n) = a(m-1)(a(m-1)(....(a(m-1)(n))....)) (with a(m-1)(n) nested occurrences of a(m-1))

Define b(0) = a(Ack(G,G))(Ack(G,G)), where Ack is the Ackermann function and G is Graham's number.
Define b(n) = a(b(n-1))(b(n-1)).

My number is b(Ack(G,G))

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Mine is bigger I believe.

Postby Sasha » Wed Jul 16, 2008 7:46 pm UTC

![!(!10)]
or ten factorial factorial factorial.
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Re: My number is bigger!

Postby Robin S » Wed Jul 16, 2008 7:50 pm UTC

If I may: are people still, at this stage, checking that their numbers are, in fact, larger than the previous ones, or merely trying to post large numbers in general? If the former, perhaps we should start adding brief explanations of why each number beats its predecessor? Make things a bit more interesting, perhaps.

Edit: just saw how old the antepenultimate post was. I thought I hadn't seen this thread in a while.
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Re: My number is bigger!

Postby AvalonXQ » Wed Jul 16, 2008 11:24 pm UTC

My number is the product of all valid submissions made before this one.

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Re: My number is bigger!

Postby Robin S » Thu Jul 17, 2008 12:06 am UTC

(Except for the previous valid submission.) How do you know that that number is larger than the previous valid submission?
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Re: My number is bigger!

Postby AvalonXQ » Thu Jul 17, 2008 1:53 am UTC

Robin S wrote:(Except for the previous valid submission.) How do you know that that number is larger than the previous valid submission?


I know mine is. The first valid submission was 9000, which is greater than 1. Each subsequent valid submission, by definition, must be larger than the previous. Multiplying numbers all of which are greater than 1 always results in a product greater than any of them. QED.

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Re: My number is bigger!

Postby Sasha » Thu Jul 17, 2008 3:36 am UTC

Yours plus Pi.
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Re: My number is bigger!

Postby drconcon » Thu Jul 17, 2008 3:56 am UTC

Now, maths is not my strong point, so I'm going to have to do my best to beat that load of gibberish.

99999 x 99999.99999 + 3

Eat that.
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Re: My number is bigger!

Postby AvalonXQ » Thu Jul 17, 2008 3:57 am UTC

That's not even NEAR the size of the previous numbers. Try again.

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Re: My number is bigger!

Postby Blatm » Thu Jul 17, 2008 8:20 am UTC

Of course, as stated in the original post, you're not allowed to define your number by using previous ones, so no "my number is your number plus 1!". I doubt anyone is going to best ijmaxwell's number unless they define a bunch of recursive functions, but that wouldn't be very elegant.

And yes, that was a pretty spectacular necro. Around the time of this thread, there were a lot of good games that I liked a lot.

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Re: My number is bigger!

Postby Robin S » Thu Jul 17, 2008 4:01 pm UTC

Just to clarify, AvalonXQ, the reason I pointed out "except for the previous valid submission" and then asked how you could prove your number to be larger than its predecessor was that the first post of this thread specified that you couldn't define your number in terms of the previous one. Therefore, your number would obviously be larger than all of the previous numbers except for its immediate predecessor, but you would still need to show that it was larger than that too.
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Re: My number is bigger!

Postby Tigerlion » Thu Jul 17, 2008 4:31 pm UTC

One, two, many, lots!
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Re: My number is bigger!

Postby Robin S » Thu Jul 17, 2008 5:22 pm UTC

I understand there are, or have been in the past, actual languages with counting systems similar to that.
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Re: My number is bigger!

Postby The Hyphenator » Fri Jul 18, 2008 3:56 pm UTC

Like the Pirahã people. They also have no religion, no mythology, little artwork, and a crazy language. :D
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Re: My number is bigger!

Postby felltir » Fri Jul 18, 2008 4:02 pm UTC

Can I just say my number mod 1 is one, and leave? It's bigger because I say it was.
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