My number is bigger?

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PsiCubed2
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Re: My number is bigger?

Postby PsiCubed2 » Thu May 16, 2019 11:32 am UTC

Technically phillip's number isn't well-defined, because he never defined "->->->->".

But yeah, given the most natural definition, his last number is about [w+3]1000, which is a bit bigger than mine (M3=[w+3]10) but much much smaller than 10ls's (which is something like [w^2]12=N12=[12,0]10).

And since we're already beyond the limit of [a,b]c , I'll define arbitrary-length arrays:

1. [anything]1=10
2. [0]n=10n
3. [0,...,0,a,...,n]m = [a,...,n]m
4. [a,b,c,...,n+1](m+1) = [a,b,c,...,n][a,b,c,...,n+1]m
5. [a,b,n+1,(k zeros)]m = [a,b,n,m,(k-1) zeros]m
6. Nn = [1,0,0]n ; Pn = [1,(n zeros)]10

Also, for finer tuning of my numbers, I define a binary version of P for selected rational numbers:

7. For integers n>1, 10n<=m<10n+1: (m/10n)Pn = [the digits of m seperated by commas]10

And submit this entry:

1.23P2 = (123/100)P2 = [1,2,3]10 = [w^2+w*2+3]10

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lllllllllwith10ls
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Re: My number is bigger?

Postby lllllllllwith10ls » Thu May 16, 2019 8:48 pm UTC

s(10,10,1,1,3)
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solune
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Re: My number is bigger?

Postby solune » Fri May 17, 2019 4:52 pm UTC

I've completely lost the plot for a few posts now, so I'll just say s(10,10,1,4) and hope it means anything to anyone.

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lllllllllwith10ls
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Re: My number is bigger?

Postby lllllllllwith10ls » Fri May 17, 2019 8:43 pm UTC

Too small. Also try to include fgh levels if you can
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solune
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Re: My number is bigger?

Postby solune » Fri May 17, 2019 9:45 pm UTC

oh I meant s(10,10,1,1,4)

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lllllllllwith10ls
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Re: My number is bigger?

Postby lllllllllwith10ls » Fri May 17, 2019 9:54 pm UTC

https://stepstowardinfinity.wordpress.com/array/extended/ definition

I'm putting s(10,10{2}2)
fgh level is about w^w
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SuperJedi224
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Re: My number is bigger?

Postby SuperJedi224 » Fri May 17, 2019 10:04 pm UTC

In Cascading-E Notation, my next number is E100#^#2#2, or approximately f[w^w](f[w](100)).

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lllllllllwith10ls
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Re: My number is bigger?

Postby lllllllllwith10ls » Fri May 17, 2019 10:15 pm UTC

s(10,10{3}2)

fgh level is w^w^2
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phillip1882
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Re: My number is bigger?

Postby phillip1882 » Sun May 19, 2019 9:37 pm UTC

Ackermann( Ackerman(1000,Ackerman(1000,Ackermann(1000,1000))),Ackermann(100->->100,100->->100))
good luck have fun

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lllllllllwith10ls
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Re: My number is bigger?

Postby lllllllllwith10ls » Sun May 19, 2019 9:45 pm UTC

Too small, only has a level of w^2
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phillip1882
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Re: My number is bigger?

Postby phillip1882 » Mon May 20, 2019 4:16 pm UTC

okay, not quite sure how you're determining that, but ill take your word for it.

Tree(Tree(Tree(10000000->->->->->10000000)))
good luck have fun

PsiCubed2
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Re: My number is bigger?

Postby PsiCubed2 » Mon May 20, 2019 7:20 pm UTC

The OP gave the level as w^2 because he mistakenly thought you're using Conway Chained Arrows. In that notation, a->...->a (with a arrows) is indeed on that level.

With your notation, I'm afraid the final result is much smaller.

I'll actually show you how I figured this out:

We do this by counting the levels of recursion in the notation. A "level of recursion" is every time you've taken the previous function and nested it n times:

a+b : level 1
a++b: level 2 (level 1 repeated b times)
a+++b: level 3 (level 2 repeated b times)
a++++b: level 4 (level 3 repeated b times)

and so on.

Then we have:

a->b->c = a++++...+++++b

We call this "level w", because it contains all the finite levels within itself. Think of "w" as a wildcard that can be replaced by any finite number. For a fixed value of c, a->b->c would be at level c, and since c can be as large as we want it to be, a->b->c corresponds to level w.

and then you've written

a->->b = a->a->...->a (b times)

Which is "level w" repeated b times, so we add one to the level of a->b->c and get w+1.

And that's it.

The Ackerman, which is merely an w-level function, doesn't change the final result by much. So the end result is still an w+1 function.

Back to your notation, we can continue with
a->->->b = a->->a...->->a : w+2 ("w+1"-level repeated b times)

a->->->->b = a->->->a...->->->a : w+3 ("w+2"-level repeated b times)

and so on.

This last expression is why I estimated your previous number of 100->->->->100 as roughly [w+3]100.

We could continue, adding another "->" for every level, and then we could do something like this:

a&b = a->->...->a (with b ->'s)

This is as powerful as w+b for arbitrary large b. Once again we need a "wildcard" here, so we replace that b with an w and get w+w=w*2 for a&b

Doing all of this again with the "&'s" we can get:

a&&b = a&a&a&...&a (level w*2+1)
a&&&b = a&&a&&a&&...a (level w*2+2)

and so on, until we have (say):

a@b = a&&....&&b (with b &'s)

which would be level w*2+w = w*3.

At this point, the pattern is clear, but we need to constantly come up with new symbols. So why not use numbers, instead of symbols? We can write [0] instead of "+" and [1] instead of "->" and [2] instead of "&" and [3] instead of "@", and voila! We have an a system with an infinite number of symbols!

So now:

a[1]b = a->b is level w
a[2]b = a&b is level w*2
a[3]b = a@b is level w*3
a[4]b is level w*4
and so on.

Now the function a[n]b has strength of w*n, where n is an arbitrary large number. So we replace the "n" with an "w" and we get w*w = w^2.

And this is, basically, what Conway Chained Arrows do. If you look carefully at the expansion rules for Conway Arrows, you will see that every additional arrow adds another w levels. The final number in the chain (regardless of its length) serves as a level counter within every set of w levels.

Of-course, we are way beyond this now. Which brings me to your last entry:
Tree(Tree(Tree(10000000->->->->->10000000)))

By now you should know that "10000000->->->->->10000000" is only at level w+4, so that's nothing exciting at this point.

However, "Tree" is a very fast growing-function. In term of ordinals level, TREE is way beyond anything we can describe with w's. It's roughly at a level called "The Small Veblen Ordinal" (SVO) which is way too large for me to explain exactly how it works right now.

In short, even writing Tree(3) would have been a winner at this point. On the other hand, by nesting Tree into itself, you haven't really increased the number by much. If the ordinal level of Tree(n) is X, then the ordinal level of Tree(Tree(...(Tree(n))...)) is merely X+1.
Last edited by PsiCubed2 on Tue May 21, 2019 10:27 am UTC, edited 4 times in total.

PsiCubed2
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Re: My number is bigger?

Postby PsiCubed2 » Mon May 20, 2019 7:23 pm UTC

As for my own number, I'll use a variant of pair sequences:

Rules:

(1) Expressions are of the form (a1,b1)(a2,b2)...(ak,bk)[n]
(2) If k=1 then the expression evaluates to n+1.
(3) If ak=0 then we delete the final pair and add 1 to n.
(4) If ak>0 and bk=0 then we do the following:
(4-i) We look for the rightmost pair (ai,bi) such that ai<ak. If no such pair is found, we set i=1.
(4-ii) We delete the final pair (ak,bk)
(4-iii) We append n-1 additional copies of (ai,bi)...(ak-1,bk-1) just before the [n]
(4-iv) The value of n remains unchanged.
(5) If ak>0 and bk>0 then we do the following:
(5-i) We look for the rightmost pair (ai,bi) such that:
(5-i-1) ai<ak.
(5-i-2) bi<bk.
(5-i-3) For every j>i, ai<aj
If no such pair is found, we set i=1
(5-ii) We set c=ak-ai
(5-iii) We delete the final pair (ak,bk)
(5-iv) For all numbers K from 1 to n-1 we do the following:
(5-iv-1) Insert (ai+K*c,bi)...(ak-1+K*c,bk-1) just before the [n]
(5-v) The value of n remains unchanged.


And my number:

(0,0)(1,1)(2,1)(3,1)(4,1)(0,0)(1,0)(1,0)[100] ~ [LVO]1032 >> TREE(anything reasonable)

(should be around TE32 in my letter notation, if I ever define it up to those levels)

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lllllllllwith10ls
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Re: My number is bigger?

Postby lllllllllwith10ls » Mon May 20, 2019 8:54 pm UTC

https://stepstowardinfinity.wordpress.c ... ti-expand/

My number is s(10,10{1``2}2) (around BHO level)

Also, no uncomputables
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Daggoth
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Re: My number is bigger?

Postby Daggoth » Wed May 22, 2019 7:42 pm UTC

10^50+( ( 2^^^( s(10,10{1``2}2) ) )

PsiCubed2
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Re: My number is bigger?

Postby PsiCubed2 » Wed May 22, 2019 7:47 pm UTC

(0,0)(1,1)(2,2)(1,1)(0,0)(1,0)(1,0)[100] ~ [εBHO+1](1032)

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lllllllllwith10ls
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Re: My number is bigger?

Postby lllllllllwith10ls » Wed May 22, 2019 8:48 pm UTC

s(10,10{1``1``2}2) ~ psi(W_2^2)
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Daggoth
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Re: My number is bigger?

Postby Daggoth » Wed May 22, 2019 11:22 pm UTC

X(s(10,10{1``1``2}2) ~ psi(W_2^2)) where X() is the s()th integral of s()

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lllllllllwith10ls
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Re: My number is bigger?

Postby lllllllllwith10ls » Thu May 23, 2019 12:08 am UTC

stop making naive extensions
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Daggoth
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Re: My number is bigger?

Postby Daggoth » Fri May 24, 2019 5:02 pm UTC

s(20,30{2``3``3}3

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lllllllllwith10ls
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Re: My number is bigger?

Postby lllllllllwith10ls » Fri May 24, 2019 5:22 pm UTC

>:C
s(10,10{1{2^``}2}2)
mom fapathi

solune
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Re: My number is bigger?

Postby solune » Mon May 27, 2019 2:12 pm UTC

Just to make lllllllllwith10ls angry:

s(10,10{1{2^``}2}2) + 1

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lllllllllwith10ls
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Re: My number is bigger?

Postby lllllllllwith10ls » Mon May 27, 2019 3:07 pm UTC

s(10,10{1{2^``}2}2)+4
mom fapathi

PsiCubed2
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Re: My number is bigger?

Postby PsiCubed2 » Tue May 28, 2019 12:14 pm UTC

Loader's Number.

Definition:
Spoiler:
The output of the following program in C, assuming int variables that never overflow:

#define R { return
#define P P (
#define L L (
#define T S (v, y, c,
#define C ),
#define X x)
#define F );}

int r, a;
P y, X
R y - ~y << x;
}
Z (X
R r = x % 2 ? 0 : 1 + Z (x / 2 F
L X
R x / 2 >> Z (x F
#define U = S(4,13,-4,
T t)
{
int
f = L t C
x = r;
R
f - 2 ?
f > 2 ?
f - v ? t - (f > v) * c : y :
P f, P T L X C
S (v+2, t U y C c, Z (X )))
:
A (T L X C
T Z (X ) F
}
A (y, X
R L y) - 1
? 5 << P y, X
: S (4, x, 4, Z (r) F
#define B (x /= 2) % 2 && (
D (X
{
int
f,
d,
c = 0,
t = 7,
u = 14;
while (x && D (x - 1 C B 1))
d = L L D (X ) C
f = L r C
x = L r C
c - r || (
L u) || L r) - f ||
B u = S (4, d, 4, r C
t = A (t, d) C
f / 2 & B c = P d, c C
t U t C
u U u) )
C
c && B
t = P
~u & 2 | B
u = 1 << P L c C u) C
P L c C t) C
c = r C
u / 2 & B
c = P t, c C
u U t C
t = 9 );
R a = P P t, P u, P x, c)) C
a F
}
main ()
R D (D (D (D (D (99)))) F

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lllllllllwith10ls
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Re: My number is bigger?

Postby lllllllllwith10ls » Tue May 28, 2019 3:00 pm UTC

god dammit
mom fapathi

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Meowers
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Re: My number is bigger?

Postby Meowers » Thu Jun 06, 2019 11:45 pm UTC

Loader's number+1

PsiCubed2
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Re: My number is bigger?

Postby PsiCubed2 » Thu Jun 06, 2019 11:48 pm UTC

The output of the Loader's program, when we replace the "99" with "100" "999" (just to be on the safe side)

username5243
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Re: My number is bigger?

Postby username5243 » Fri Jun 07, 2019 12:03 am UTC

The output of the Loader's number program when 99 is changed to Loader's Number


...this is going nowhere...
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PsiCubed2
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Re: My number is bigger?

Postby PsiCubed2 » Fri Jun 07, 2019 12:08 am UTC

Are you going to answer my PM?

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Meowers
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Re: My number is bigger?

Postby Meowers » Fri Jun 07, 2019 12:16 am UTC

That plus Loader's number multiplied by 10^1000

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lllllllllwith10ls
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Re: My number is bigger?

Postby lllllllllwith10ls » Fri Jun 07, 2019 12:23 am UTC

Define D(x) to be the output of the Loader function where 99 is changed to x.

My number is d(3,3{1,,,3}2) where d(#) is s(#) in strong array notation but the base rule is changed to D(a*b) instead of a^b. Definition of strong array notation: https://stepstowardinfinity.wordpress.com/array/drop/sec-drop/
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username5243
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Re: My number is bigger?

Postby username5243 » Fri Jun 07, 2019 12:23 am UTC

Salad numbers are bad... just saying.

If D is loader's function (D^5(99) is Loader's number), then I'll go:

D^Loader's Number(99)

(Yes, that's not much better than the last few. But no idea how to get out of this loop.)
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lllllllllwith10ls
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Re: My number is bigger?

Postby lllllllllwith10ls » Wed Jun 12, 2019 7:38 pm UTC

D^(D^(loaders number)(99))(99)
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