Your number is, in fact, not bigger!

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Vytron
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Re: Your number is, in fact, not bigger!

Postby Vytron » Tue Feb 10, 2015 8:50 am UTC

(Also, I thought about a second argument to show how 4[0-0-[0-0,0-0]] can't be 5[0-0,0-0] - At the very base of the continuation of my progression, I have 4[0-0-[0-0,0-0]],0 = n[0-0-[0-0,0-0]] n[0-0-[0-0,0-0]] n[0-0-[0-0,0-0]] n[0-0-[0-0,0-0]] 4[0-0-[0-0,0-0]] i.e. there's nesting applied to the previous function. It means 2[0-0-[0-0,0-0]],0 is (((2[0-0-[0-0,0-0]])[0-0-[0-0,0-0]])[0-0-[0-0,0-0]]) which is already much larger than 5[0-0-[0-0,0-0]])

I know how to pass that in my notation, unfortunately ordinal plugging leads to confusion, so I'll continue the previous notation with relevant parts bolded.

n[[[0-0,0-0]]] = n[[0-0,0-0]]~[[0-0,0-0]]~
n[[[[0-0,0-0]]]] = n[[[0-0,0-0]]]~[[[0-0,0-0]]]~
n[0-[0-0,0-0]] = n[~[0-0,0-0]~]~[~[0-0,0-0]~]~
n[0-[0-0,0-0],0] = n[0-[0-0,0-0]]~[0-[0-0,0-0]]~
n[0-[0-0,0-0],1] = n[0-[0-0,0-0],0,~]
n[0-[0-0,0-0],[0]] = n[0-[0-0,0-0],n]
n[0-[0-0,0-0],[0],[0]] = n[0-[0-0,0-0],[0],n]
n[0-[0-0,0-0],0,[0]] = n[0-[0-0,0-0],[0],~]
n[0-[0-0,0-0],[0]~[0]] = n[0-[0-0,0-0],n,[0]]
n[0-[0-0,0-0],[0,0]] = n[0-[0-0,0-0],[0]~[0]~]
n[0-[0-0,0-0],[1]] = n[0-[0-0,0-0],[0,~]]
n[0-[0-0,0-0],[[0]]] = n[0-[0-0,0-0],[n]]
n[0-[0-0,0-0],[0-0]] = n[0-[0-0,0-0],[~[0]~]]
n[0-[0-0,0-0],[0-0-0]] = n[0-[0-0,0-0],[~[1-0]-0~]]
n[0-[0-0,0-0],[0-0-[0-0-0]]] = n[0-[0-0,0-0],[0-0-[~[1-0]-0~]]]
n[0-[0-0,0-0],[0-0,0-0]]] = n[0-[0-0,0-0],[0-0-[~0-0-[0-0-0]~]]]
n[0-[0-0,0-0],[0-0,0-0],[0-0,0-0]]] = n[0-[0-0,0-0],[0-0,0-0],[0-0-[~0-0-[0-0-0]~]]]
n[0-0,[0-0,0-0]] = n[0-[0-0,0-0],~]
n[0-0,0,[0-0,0-0]] = n[0-0,[0-0,0-0],~~]
n[0-1,[0-0,0-0]] = n[0-0,~[0-0,0-0]]
n[0-[0],[0-0,0-0]] = n[0-n,[0-0,0-0]]
n[0-[0]~[0],[0-0,0-0]] = n[0-n,n,[0-0,0-0]]
n[0-[0,0],[0-0,0-0]] = n[0-[0]~[0]~[0-0,0-0]]
n[0-[1],[0-0,0-0]] = n[0-[0,~],[0-0,0-0]]
n[0-[[0]],[0-0,0-0]] = n[0-[n],[0-0,0-0]]
n[0-[0-0],[0-0,0-0]] = n[0-[~[0]~],[0-0,0-0]]
n[0-[0-0-0],[0-0,0-0]] = n[0-[~[1-0]-0~],[0-0,0-0]]
n[0-[0-0,0-0]~[0-0,0-0]] = n[0-[0-0-[~0-0-[0-0-0]~],[0-0,0-0]]
n[0-[0-0,0-0]~[0-0,0-0]~[0-0,0-0]] = n[0-[0-0-[~0-0-[0-0-0]~]~[0-0,0-0]]

n[0-[[0-0,0-0]]] = n[0-[0-0,0-0]~[0-0,0-0]~]
n[0-[[[0-0,0-0]]]] = n[0-[[0-0,0-0]]~[[0-0,0-0]]~]
n[0-[0-[0-0,0-0]]] = n[0-[~[0-0,0-0]~]~[~[0-0,0-0]~]~]
n[0,0-[0-0,0-0]] = n[0-[0-~[0-0,0-0]~]]
n[0,0,0-[0-0,0-0]] = n[0,0-[0,0-~[0-0,0-0]~]]
n[1-[0-0,0-0]] = n[0,~-[0,~-~[0-0,0-0]~]]
n[[0]-[0-0,0-0]] = n[n-[n-~[0-0,0-0]~]]
n[[1]-[0-0,0-0]] = n[[0,~]-[[0,~]-~[0-0,0-0]~]]
n[[[0]]-[0-0,0-0]] = n[[n]-[[n]-~[0-0,0-0]~]]
n[[0-0]-[0-0,0-0]] = n[[~[0]~]-[[~[0]~]-~[0-0,0-0]~]]
n[[0-[0-0,0-0]]-[0-0,0-0]] = n[[~[0-0,0-0]~]~[~[0-0,0-0]~]~-[[~[0-0,0-0]~]~[~[0-0,0-0]~]~-~[0-0,0-0]~]]

n[0-0-[0-0,0-0]] = n[~[~0-[0-0,0-0]~-[0-0,0-0]~]-[0-0,0-0]] = φ(1,1,1)

(Not n[0-0-[0-0,0-0]] = n[0-0-[0-0-[~0-0-[0-0-0]~]], which is from where the confusion came from - whenever some ordinal equivalent is nested it's harder to make it go up, all the previous bolded parts would now go up to this limit.)

And it begins again inside of n[0-0-X], i.e. all the stuff I've put on this post fits on X

...

n[0-0-[[0-0]-[0-0,0-0]]] = n[0-0-[~0-0[~[0]~]-[[~[0]~]-~[0-0,0-0]~]~]]
n[0-0-[[0-[0-0,0-0]]-[0-0,0-0]]] = n[0-0-[~0-0[[~[0-0,0-0]~]~[~[0-0,0-0]~]~-[[~[0-0,0-0]~]~[~[0-0,0-0]~]~-~[0-0,0-0]~]~]]

n[0-0,0-0,0] = n[0-0-~[0-0-~0-[0-0,0-0]~-[0-0,0-0]~]-[0-0,0-0]] = φ(1,1,2)
n[0-0,0-0,0,0] = n[0-0-~[0-0-~0-[0-0,0-0,0]~-[0-0,0-0,0]~]-[0-0,0-0,0]] = φ(1,1,3)
n[0-0,0-0,0,0,0] = n[0-0-~[0-0-~0-[0-0,0-0,0,0]~-[0-0,0-0,0,0]~]-[0-0,0-0,0,0]] = φ(1,1,4)

n[0-0,0-1] = n[0-0-~[0-0-~0-[0-0,0-0,~]~-[0-0,0-0,~]~]-[0-0,0-0,~]] = φ(1,1,ω)
n[0-0,0-2] = n[0-0-~[0-0-~0-[0-0,0-0,~1]~-[0-0,0-0,~1]~]-[0-0,0-0,~1]] = φ(1,1,ω^ω)
n[0-0,0-3] = n[0-0-~[0-0-~0-[0-0,0-0,~1,2]~-[0-0,0-0,~1,2]~]-[0-0,0-0,~1,2]] = φ(1,1,ω^ω^ω)

Send:

4[0-0,0-[0]]

Where: [0-0,0-[0]] = [0-0,0-n] for φ(1,1,ε_0)

----

Explanation:

I have some notation X, then I add a box, [X], and I can then put stuff inside this box up to [~[X]~], which is the limit of the boxes. However, once I reach a new place, I go back to these boxes, and can put something new in [X], which is now more powerful. And I do that with [0-X], and [0,0-X], and [0-0-X]... So that there's always a symbol that never appears in X, and once I create it, I can go back and reuse the same symbols for much stronger representations of recursions.

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Re: Your number is, in fact, not bigger!

Postby Daggoth » Tue Feb 10, 2015 7:44 pm UTC

That sort of reminds me of the boxes from illusory space

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WarDaft
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Re: Your number is, in fact, not bigger!

Postby WarDaft » Tue Feb 10, 2015 11:28 pm UTC

It actually can be done, because there is an ordinal representation for what he wanted to do but was having trouble expressing with illusory space. This may or may not be exactly what he wanted, but on the face of it, there's nothing wrong. I haven't done a precise analysis to ensure there's no errors at all, but it looks right-ish so I'm not contesting it.

[2|0[ :0[][]:0[]] for φ(1,φ(1,0),0)
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Re: Your number is, in fact, not bigger!

Postby Vytron » Wed Feb 11, 2015 2:29 am UTC

The thing of Illusory space is that I can plop it to any other notation, so, say, I could add ''s around my boxes and claim "that's Illusory Space, and 4[0] is 4'[0-0,0-[0]]', And that's Level 0, and in Level 0,0 you add n[0] 's, and in level 0,0,0 you add 4<big ordinal> 's", and then I'd say "That's Layer 0 of Illusory space, in Layer 0,0 you go to imaginary level 4[0] of Layer 0 to add that many 's", and so on.

I may have to go with it once this notation runs out, but I think it has good fuel left...

n[0-0,0-[[0]]] = n[0-0,0-[n]] = φ(1,1,ε_ε_0)
n[0-0,0-[0-0]] = n[0-0,0-[~[0]~]] = φ(1,1,φ(2,0))
n[0-0,0-[0,0-0]] = φ(1,1,φ(3,0))
n[0-0,0-[0,0,0-0]] = φ(1,1,φ(4,0))
n[0-0,0-[0-0,0-0]] = φ(1,1,φ(1,1,0))
n[0-0,0-[0-0,0-[0-0,0-0]]] = φ(1,1,φ(1,1,φ(1,1,φ(1,1,0))))
n[0-0,0,0-0] = n[0-0,0-[~0-0,0-[0-0,0-0]~]] φ(1,2,0)
n[0-0,0,0,0-0] = n[0-0,0,0-[~0-0,0,0-[0-0,0,0-0]~]] φ(1,3,0)
n[0-1-0] = n[0-0,~-0] φ(1,ω,0)
n[0-[0]-0] = n[0-n-0] φ(1,ε_0,0)
n[0-[[0]]-0] = n[0-[n]-0] φ(1,ε_ε_0,0)
n[0-[0-0]-0] = n[0-[~[0]~]-0]

Send:

4[0-[0-0]-0]

For f_φ(1,φ(2,0),0)(4)

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Re: Your number is, in fact, not bigger!

Postby WarDaft » Wed Feb 11, 2015 4:04 am UTC

Excellent! Let the rampant 1-upmanship continue unabated.

[2|0[ :[ : :0[]]:0[]]] for f{φ(1,φ(1,0,0),0)}(2)
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Re: Your number is, in fact, not bigger!

Postby Vytron » Wed Feb 11, 2015 9:19 am UTC

Okay, I've had enough.

Define:

n[X-Y]

Compute n[X] and Compute n[Y].

If X produces a larger number then return:

n[X-[X-...[X-y]] for n -s

Where y follows the rules of n,X (i.e. if y is a 1 it becomes 0,0,...,0,0 with n ,s, if it's [[0]] it becomes an n)

If Y is 0 return: n[x-[x-...[x-0]] for n -s

If Y produces a larger number then return:

n[x-[x-...[x-Y]] for n -s

Where x follows the rules of n,X (i.e. if x is a 1 it becomes 0,0,...,0,0 with n ,s, if it's [[0]] it becomes an [n])

If X is 0 return:

n[0-y]

--------

Define:

n[X-Y-Z]. Compute n[X], n[Y] and n[Z].

If n[Z] produces the largest number and n[Y] => n[X]:

Return: n[X-Y-[X-Y-...[X-Y-z]]]] for n[X-Y-z] -s (note: n*2 -s suffice here)

Where z follows the rules of n,X

If Z is a 0, return:

n[X-y-[X-y-...[X-y-0]]]] for n[X-y-0] -s

If Y is a 0 (then X is a 0 as well), return:

n[[0-Z]...-Z]-Z] for n[0-Z] -s (note: n -s suffice here)

If n[Z] produces the largest number and n[Y] < n[X]:

Return: n[X-Y-[X-Y-...[X-Y-z]]]] for n[X-Y-z] -s

If Z is a 0, return:

n[X-y-[X-y-...[X-y-0]]]] for n[X-y-0] -s

If Z and Y are 0, return:

n[x-[x-...[x-0-0]...-0]-0] for n[x-Y-0] -s

If n[Y] produces the largest number:

Return:

n[X-Y-[X-Y-...[X-Y-z]]]] for n[X-Y-z] -s

If Z is 0, return:

n[X-y-[X-y-...[X-y-0]]]] for n[X-y-0] -s

If n[X] produces the largest number and Y and Z are 0:

Return:

[x-[x-...[x-0-[x-X-0]]...-0]-0]

--------

There we go, now everything before should work all right, and:

n[0-[0-0-0]-0] = φ(1,φ(1,0,0),0)
n[0-[0-0-[0-0-0]]-0] = φ(1,φ(1,0,1),0)
n[0-[0-0,0-0]-0] = φ(1,φ(1,1,0),0)
n[0-[0-0-[0-0,0-0]]-0] = φ(1,φ(1,1,1),0)
n[0-[0-[0-0-[0-0,0-0]]-0]-0] = φ(1,φ(1,φ(1,1,1),0),0)
n[0-[0-[0-[0-0-[0-0,0-0]]-0]-0]-0] = φ(1,φ(1,φ(1,φ(1,1,1),0),0),0)
n[0,0-0-0] = φ(2,0,0)
n[0,0,0-0-0] = n[0,0-[~[0-0-[0,0-0-0]~]-0]-0] φ(3,0,0)
n[1-0-0] = n[0,~-0-0] φ(ω,0,0)
n[[0]-0-0] = n[n-0-0] φ(ε_0,0,0)
n[[0-0]-0-0] = n[[~[0]~]-0-0] φ(φ(2,0),0,0)
n[[0-0-0]-0-0] = φ(φ(1,0,0),0,0)
n[[1-0-0]-0-0] = φ(φ(ω,0,0),0,0)
n[[[0-0-[0-0-0]]-0-0]-0-0] = φ(φ(1,0,1),0,0)
n[[[[0-0-[0-0-0]]-0-0]-0-0]-0-0] = φ(φ(φ(1,0,1),0,0),0,0)
n[0-0-0-0] = n[~[0-0-[0-0-0]~]-0-0]~-0-0]

Send:

4[0-0-0-0]

For f_φ(1,0,0,0)(4)

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Re: Your number is, in fact, not bigger!

Postby WarDaft » Wed Feb 11, 2015 7:18 pm UTC

Hmm, I think I have discovered a very concise way of expressing your base rule:

(0,X)-- = X
(a,X)-- = (<a-->,a,X--)
(a)-- = (<a-->)
(X,()) = X

Though I believe this is reversed left-right.


With the first rule: 0,0,1,0 -> 0,1,0 -> 1,0
With the second rule, we have 1,0 -> <1-->,1,(0)-- = 0,0,0,0,0,1,(0)-- = 0,0,0,0,0,1,() = 0,0,0,0,0,1
(1,1,0)-- = <1-->,1,(1,0)-- = 0,0,1,(<1-->,1,(0)--) = 0,0,1,0,0,1,() = 0,0,1,0,0,1

Does that sound about right?
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Re: Your number is, in fact, not bigger!

Postby Vytron » Thu Feb 12, 2015 1:58 am UTC

Yeah, with rules reversed left-to-right you'd see if the leftmost element is a 0, and if it is, you iterate the function, otherwise, you search for the next element that is bigger than the left-most element, you "decrease" what's left of it, and make n copies of the string.

What is missing on your description is the ~ element:

Once you reach something like (left-to-right):

[0],0
[0],1,0
[0],2,1,0
[0],3,2,1,0
...

In your notation, one would advance to:

[0,0] = [0],3,2,1,0...0

In mine, you go to:

[0]~[0] = [0],3,2,1,0...0

So the ~s work as ,s, allowing for an extra layer of strings before reaching:

[0,0] = [0]~[0]~...~[0]

Because what I had wasn't enough to reach the ordinals I was wanting to reach.

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Re: Your number is, in fact, not bigger!

Postby Vytron » Thu Feb 12, 2015 2:11 am UTC

WarDaft wrote:(1,1,0)-- = <1-->,1,(1,0)-- = 0,0,1,(<1-->,1,(0)--) = 0,0,1,0,0,1,() = 0,0,1,0,0,1

Does that sound about right?


Um, it should be (left-to-right):

n[1,1,0] = n[0...0,1,0]~[0...0,1,0]~...~[0...0,1,0] with n ,s inside each box and n ~s.

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Re: Your number is, in fact, not bigger!

Postby WarDaft » Thu Feb 12, 2015 4:15 am UTC

I didn't actually want to go that far, was just trying to concisely define the base form. This helps because it has revealed an incongruity in what I thought you were doing.
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Re: Your number is, in fact, not bigger!

Postby Vytron » Thu Feb 12, 2015 5:47 am UTC

Oh, we're looking at (from right to left - i.e. my original notation):

n,0,1,1

First:

Is the rightmost element a 0? No.

Then, look for the previous element that is as high as the rightmost element:

0,1,1

This makes the string:

0,1, (1)

I.e. 0,1, is the unchanged prefix.

This means you decrease the string and make n copies.

0,0,1,1 = 0
1,0,1,1 = 1,0,1(,0) = 1,0,1,0,0
2,0,1,1 = 1,0,1(,0,0) = 2,0,1,0,0,0,0,0,0
3,0,1,1 = 3,0,1(,0,0,0) = 3,0,1,0,0,0,0,0,0,0,0,0,0,0,0
4,0,1,1 = 4,0,1(,0,0,0,0) = 4,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0

For f_{ω^2}+ω(n).

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Re: Your number is, in fact, not bigger!

Postby Vytron » Thu Feb 12, 2015 6:47 pm UTC

Okay, there we go, I thought of a way to get rid of all possible incongruities:

Define:

nXY

Where X and Y are strings:

,a,b,c,...

Of finite elements, possibly empty.

n, = n*(2^n)

The bottom case (it's this instead of n+1 so that the new base case produces numbers as high as the old one)

nXY,0 = aXY bXY ... oXY nXY with n spaces

Where o is nXY and a would be bXY, etc.

Define:

X = *,Z

(Z is the rightmost element in X)

Define:

Y = A,*,z

Where:

Z=>z

In other words, Z is the rightmost element in X that is equal or higher than z.

nXY = nXyy...yy for n+1 ys (n ys suffice but number would be lower than old notation)

n,~ = n,n,...,n,n with n ,s

Where:

Y = y
A,*,z = a,*,z

(A is decreased to a)

Decreasing rule:

A = a
0 = ()
n = 0,~1,2,3...n-2,n-1

And that's it.

Example 1:

4,0,1,2,3

z=3

There's no element equal or higher than z, so X is empty.

A=0

4,0,1,2,3 = 4(,1,2,3)

Make 4 copies:

4,0,1,2,3=4,1,2,3,1,2,3,1,2,3,1,2,3,1,2,3

Example 2:

4,3,2,1,0

z=0

4,3,2,1,0 = ((((4,3,2,1),3,2,1),3,2,1),3,2,1),3,2,1

Example 3:

4,3,2,1

z=1

2>z

A=1

X=,3,2

Y=,1

a=0,~

4,3,2,1 = 4,3,2(,0,0,0,0)

Make n copies:

4,3,2,1 = 4,3,2,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0

Example 4:

4,0,1,2,3,2,1,0

z=0

4,0,1,2,3,2,1 = ((((4,0,1,2,3,2,1),0,1,2,3,2,1),0,1,2,3,2,1),0,1,2,3,2,1),0,1,2,3,2,1

Example 5:

4,0,1,2,3,2,1

z=1

2>z

A=1

X=,0,1,2,3,2

Y=,1

a=0,~

4,0,1,2,3,2,1 = 4,0,1,2,3,2(,0,0,0,0)

Make n copies:

4,0,1,2,3,2,1 = 4,0,1,2,3,2,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0

Example 6:

4,3,2,1,0,1,2,3

z=3

3=z

A=2

X=,3

Y=,2,1,0,1,2,3

a=0,~1

4,3,2,1,0,1,2,3 = 4,3(,0,0,0,0,1,1,0,1,2,3)

Make n copies:

4,3,2,1,0,1,2,3 = 4,3,0,0,0,0,1,1,0,1,2,3,0,0,0,0,1,1,0,1,2,3,0,0,0,0,1,1,0,1,2,3
,0,0,0,0,1,1,0,1,2,3,0,0,0,0,1,1,0,1,2,3

n,n = f_{φ(1,0)}(n)

And now, for an entirely different extension:

n,{0} = n,n
n,{0},0 = a,{0} b,{0} ... o,{0} n,{0} with n spaces
n,{0},0,0 = a,{0},0 b,{0},0 ... o,{0},0 n,{0},0 with n spaces

n,{0},XY

n,0,{0} = n,{0},~

n,{0,0} = n,n+1,{0} = f_{φ(0,φ(1,0)+1)}(n)

n,{0,0,0} = n,n+1,{0},{0,0} = f_{φ(0,φ(0,φ(1,0)+1))}(n)

n,{{0}} = n,{n+1} = f_{φ(1,1)}(n)

n,{{{0}}} = n,{{n+1}} = f_{φ(1,φ(1,0))}(n)

n,{{{{0}}}} = n,{{{n+1}}} = f_{φ(1,φ(1,φ(1,0)))}(n)

n,0~0 = n,{~{0}~} = f_{φ(2,0)}(n)

n,0~0,0 = n,{n+1} = f_{φ(2,0)+1}(n)

n,{0~0,0} = f_{φ(1,φ(2,0)+1)}(n)

n,{{0~0,0}} = f_{φ(1,φ(1,φ(2,0)+1))}(n)

n,0,0~0 = n,{~{0~0,0}~} = f_{φ(2,1)}(n)

n,{0~0,0}~0 = f_{φ(2,φ(2,0)+1)}(n)

n,{{0~0,0}}~0 = f_{φ(2,φ(2,φ(2,0)+1)}(n)

n,0~0~0 = n,{~{0~0,0}~}~0 f_{φ(3,0)}(n)

n,0,, = n,0~ = f_{φ(φ(0,ω),0)}(n)

n,0,,{0,,} = n,0~ = f_{φ(φ(φ(0,ω),0),0)}(n)

n,0,,{0,,{0,,}} = n,0~ = f_{φ(φ(φ(φ(0,ω),0),0),0)}(n)

n,{0,0},, = n,0,,{0,,~{0,,}~} = f_{φ(Γ_0,0)}(n)

n,{0,0},,0 = Γ_0+1
n,{0,0},,{0,0},, = Γ_0*2
n,0,{0,0},, = Γ_0^2
n,{{0,0},,0} = f_{φ(1,Γ_0+1)}(n)
n,0,,{{0,0},,0} = f_{φ(φ(0,ω),Γ_0+1)}(n)
n,0,,{0,,{{0,0},,0}} = f_{φ(φ(φ(0,ω),0),Γ_0+1)}(n)
n,{0,0,0},, = n,0,,{~0,,{{0,0},,0}~} = φ(Γ_0,1)

n,0,,{{0,0,0},,} = f_{φ(φ(Γ_0,1),0)}(n)
n,0,,{0,,{{0,0,0},,}} = f_{φ(φ(Γ_0,1),0)}(n)
n,{0,0,0,0},, = Γ_1
n,{0,0,0,0,0},, = Γ_2
n,{0,0,0,0,0,0},, = Γ_3
n,{1},, = Γ_ω
n,{{0,0},,},, = Γ_Γ_0
n,{{{0,0},,},,},, = Γ_Γ_Γ_0

n,0,,, = n,{{~{0,0}~,,},,},,
n,0,,,, = n,{{~{0,0}~,,,},,,},,,

n,[0] = n,0,,...,, n ,s φ(1,1,0)

n,[0],XY

n,0,[0] = n,[0],~

n,[0],, = n,[0]~ φ(1,1,1)
n,[0],0,, = n,[0],,{[0],,~{[0],,}~}
n,0,[0],, = n,[0],~,,

n,[0],,, = n,{{~[0],0}~,,},,},, = φ(1,1,2)
n,[0],,,, = n,{{~[0],0}~,,,},,,},,, = φ(1,1,3)

n,[0,0] = n,[0],,...,, = φ(1,1,ω)
n,[{0}] = φ(1,1,φ(2,0))
n,[[0]] = φ(1,1,φ(1,1,0))
n,[0-0] = φ(1,2,0)
n,[0,0-0] = n,[0-~[0-0]~] = φ(1,3,0)
n,[[0],,-0] = φ(1,φ(1,1,1),0)
n,[[[0],,-0]-0] = φ(1,φ(1,φ(1,1,1),0),0)
n,[0-0-0] = φ(2,0,0)
n,[0-0-0,0] = φ(3,0,0)
n,[0-0-0,,,] = φ(φ(1,0,1),0,0)
n,[0-0-[0-0-0,,,]] = φ(φ(φ(1,0,1),0,0),0,0)
n,[0-0-[0-0-[0-0-[0-0-0,,,]]]] = φ(φ(φ(φ(1,0,1),0,0),0,0),0,0)

n,[0-0,0-0] = n,[0-0-[0-0-~[0-0-0,,,]~]]

Send:

5,[0-0,0-0]

For f_φ(1,0,0,0)(5)

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Re: Your number is, in fact, not bigger!

Postby Vytron » Sat Feb 14, 2015 3:36 pm UTC

Okay, sorry, disregard that. The ,, ,,, ,,...,, and ~ extensions are garbage because they grow like ω :roll:

So, the problem with my notation is my first extension is inherently ambiguous.

We have:

,[0] = ,n

Giving φ(1,0)

Then whatever is on that box can have:

,[X],[X] = δ*2
,[X],[X],[X] = δ*3...

Up to:

0,[X] = ,[X],~ = δ*ω

Now, to reach δ^2 we need:

,[X],[X] = n,[X] = δ^2

But ,[X],[X] already stands for something else, so my notation falls apart.

What I'm going to do is use the {} soft boxes, to make them hold δ^X.

nXY rules from this post apply.

n,[0] = n,n = f_φ(1,0)

n,[0],[0] = n,[0],n = f_φ(1,0)*2

n,0,[0] = n,[0],~ = f_φ(1,0)*ω

n,n,[0] = nXY[0] = f_φ(1,0)^2

New:

n,{0},[0] = n,n,[0] = f_φ(1,0)^2

n,{0},[0],{0},[0] = n,n,[0] = f_{φ(1,0)^2}*2

n,0,{0},[0] = n,{0},[0],~~ = f_{φ(1,0)^2}*ω

n,{0},{0},[0] = n,n,{0},[0] = f_φ(1,0)^3

n,{0,0},[0] = n,{0},~[0] = f_φ(1,0)^ω

n,{{0}},[0] = n,{n},[0] = f_ω^{φ(1,0)+1}

n,{{{0}}},[0] = n,{{n}},[0] = f_ω^ω^{φ(1,0)+1}

n,[0,0] = n,{~{0}~},[0] = f_φ(1,1)

n,[[0]] = n,{~{0}~},[0] = f_φ(1,φ(1,0))

n,[[[0]]] = n,{~{0}~},[0] = f_φ(1,φ(1,φ(1,0)))

n,[0-0] = f_φ(2,0)

n,[0-[0-0]] = f_φ(2,φ(2,0))

n,[0-[0-[0-0]]] = f_φ(2,φ(2,φ(2,0)))

n,[0,0-0] = n,[0-~[0-0]~] = f_φ(3,0)

n,[0,0,0-0] = n,[0,0-~[0,0-0]~] = f_φ(4,0)

n,[1-0] = n,[0,~-0] = f_φ(ω,0)

n,[[1-0]-0] = n,[[0,~-0]-0] = f_φ(φ(ω,0),0)

n,[0-0-0] = n,[[0,~-0]-0] = f_φ(1,0,0)

n,[0-0-0],0 = f_φ(1,0,0)+1

n,[[0-0-0],0]] = f_φ(1,φ(1,0,0)+1)

n,[0-[0-0-0],0]] = f_φ(2,φ(1,0,0)+1)

n,[[0-0-0],0]-0] = f_φ(φ(ω,0),φ(1,0,0)+1)

n,[[[0-0-0],0]-0]-0] = f_φ(φ(φ(ω,0),0),φ(1,0,0)+1)

n,[0-0-0,0] = [~[0-0-0],0]~-0] = f_φ(1,0,1)

n,[0-0-0,0,0] = [~[0-0-0,0],0]~-0] = f_φ(1,0,2)

n,[0-0-1] = [0-0-0,~] = f_φ(1,0,ω)

n,[0-0-[0]] = n,[0-0-n] = f_φ(1,0,φ(1,0))

n,[0-0-[0-0-0,0]] = n,[0-0-[~[0-0-0],0]~-0]] = f_φ(1,0,φ(1,0,1))

n,[0-0-[0-0-[0-0-0,0]]] = n,[0-0-[~[0-0-0],0]~-0]] = f_φ(1,0,f_φ(1,0,φ(1,0,1)))

n,[0-0,0-0] = n,[0-0-~[0-0-0,0]~] = f_φ(1,1,0)

n,[0-0,0-0],0 = f_φ(1,1,0)+1

n,[0-0-[0-0,0-0],0] = f_φ(1,0,φ(1,1,0)+1)

n,[0-0-[0-0-[0-0,0-0],0]] = f_φ(1,0,φ(1,0,φ(1,1,0)+1)

n,[0-0,0-0,0] = n,[0-0-~[0-0,0-0],0]~] = f_φ(1,1,1)

n,[0-0,0-0,0,0] = n,[0-0,0-~[0-0,0-0,0],0]~] = f_φ(1,1,2)

n,[0-0,0-1] = n,[0-0,0-0,~] = f_φ(1,1,ω)

n,[0-0,0-[n]] = n,[0-0,0-n] = f_φ(1,1,φ(1,0))

n,[0-0,0-[0-0,0-0]] = f_φ(1,1,φ(1,1,0))

n,[0-0,0-[0-0,0-[0-0,0-0]]] = f_φ(1,1,φ(1,1,φ(1,1,0)))

n,[0-0,0,0-0] = f_φ(1,2,0)

n,[0-1-0] = n,[0-n-0] = f_φ(1,ω,0)

n,[0-[0]-0] = n,[0-n-0] = f_φ(1,φ(1,0),0)

n,[0-[0-0]-0] = n,[0-n-0] = f_φ(1,φ(2,0),0)

n,[0-[0-0-0]-0] = n,[0-[[0,~-0]-0]-0] = f_φ(1,φ(1,0,0),0)

n,[0-[0-[0-0-0]-0]-0] = n,[0-[0-[[0,~-0]-0]-0]-0] = f_φ(1,φ(1,φ(1,0,0),0),0)

n,[0,0-0-0] = n,[0-~[0-0-0]~-0] = f_φ(2,0,0)

Send:

3,[0,0-0-0] for f_φ(2,0,0)(3)

Incongruity-free, at last.

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Re: Your number is, in fact, not bigger!

Postby Daggoth » Sun Feb 15, 2015 2:42 am UTC

Thats sort of the reason i use the ¿? to englobe the operator with all the inner separators. If i allow ¡! to touch the "argument", the notation falls apart.

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Vytron
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Re: Your number is, in fact, not bigger!

Postby Vytron » Sun Feb 15, 2015 3:56 am UTC

I've been thinking about it, probably once my notation runs out (I made some investigations, if I make n[n] = n,[n-~] it seems to have about the same growth of the TREE function...) I'll introduce separating commas as follows:

,,0,,X,,0,,

=

,X,

I.e. ,,0,, is like a comma separator.

Then I can make my entire notation fit on ,,X,,

And ,,0,,X = ,,X,,X,,X,,X,,X,,... (for an even number of elements)

Taking advantage of the ,,0,,X,,0,,s working in pairs (so they can't ever be odd, so if you find an odd number of commas it means you should make copies of X which may include any number of ,,0,,s itself)

And:

,,,0,,,X,,,0,,, = ,,X,,X

So:

n,, = n,,...,,n

Could probably reach 𝛝(𝛀^𝛀)(n) or something.

Daggoth
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Re: Your number is, in fact, not bigger!

Postby Daggoth » Sun Feb 15, 2015 5:56 am UTC

if n,, = n,,...,,n wouldnt that eventually reduce down to another n,,?

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Re: Your number is, in fact, not bigger!

Postby Vytron » Sun Feb 15, 2015 6:10 am UTC

To some pair n,,n,,X,,n,,

Basically, I could use <>s instead of ,s and do it like this:

<0<X>o>

=

,X,

I.e. <0< is like a comma separator.

Then I can make my entire notation fit on <X<

And <0>X = <X<X>X>X<X<... (for an even number of elements)

Taking advantage of the <0<X>0>s working in pairs (so they can't ever be odd, so if you find an odd number of commas it means you should make copies of X which may include any number of <0<s itself)

And:

<<0<<X>>0>> = <X>X

So:

n,, = n<<...>>n

Could probably reach 𝛝(𝛀^𝛀)(n) or something.


This would be like an extension of the base separators.

{X}
[X]
<0<X>0>
<0,0<X>0,0>
<0,0,0<X>0,0,0>
<0,0,0,0<X>0,0,0,0>
<{X}<X>{X}>
<[X]<X>[X]>
<<X><X><X>>
<<<X>><X><<X>>>
<<<<X>>><X><<<X>>>>
<<X<<X>>X>>
...

But with commas...

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Re: Your number is, in fact, not bigger!

Postby Daggoth » Sun Feb 15, 2015 4:27 pm UTC

"To some pair n,,n,,X,,n,,"

This is what i mean, the red part would break again into n,,...,,n
even if it reduces to n,,n,,X,,n
wouldnt that eventually reach an open ended n,, at some point?

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Vytron
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Re: Your number is, in fact, not bigger!

Postby Vytron » Sun Feb 15, 2015 4:51 pm UTC

No, n,, breaks into n,,...,,n only if there's an odd number of comma pairs, otherwise, you find the rightmost ,,n,,X,,n,, and decrease X, and once you hit ,,n,,0,,n,, you decrease ,,X,,0,,X,, (both X decrease simultaneously to the same value, so ,,n,,Y,,n,, can hold any value less than X because you know those commas on the inside aren't paired with the ones on the outside.)

But anyway, this will be a late expansion...

Where is WarDaft? Who am I fighting with? Do I just keep going by myself?

n,[0,0,0-0-0] = n,[0,0-~[0,0-0-0]~-0] = f_φ(3,0,0)

Send:

3,[0,0,0-0-0] for f_φ(3,0,0)(3)

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Re: Your number is, in fact, not bigger!

Postby Daggoth » Sun Feb 15, 2015 8:08 pm UTC

You stand a champion until someone challenges or beats you

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Re: Your number is, in fact, not bigger!

Postby Vytron » Sun Feb 15, 2015 8:19 pm UTC

So please challenge or beat me. How's your notation doing? I'd like to compare ordinal growth with you (to go as far as I did I had to use the principle of induction, but I should be able to represent any ordinal that you'd reach, it seems you stopped at fφ(100,100)(100).)

fφ(100,100)(100) in my notation:

100,[100-100]

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Re: Your number is, in fact, not bigger!

Postby Daggoth » Sun Feb 15, 2015 9:28 pm UTC

My notation "cascades" down to all the previous processes at each step. while also adding a layer of n self recursions. so to go side by side you would need to go to your n+1th expansion (since your notation is much more tightly close to the actual fgh)

For example

i pair these for simplicity

¿0¡1!0?4 > fω^ω^ω(4) = fω^ω^4(4)

I would get there if ¿0¡1!0?4 reduced to ¿0¡0!0¡0!0¡0!0¡0!0?4

But actually the first reduction brings me to

¿4,4,4,4¡0!4,4,4,4¡0!4,4,4,4¡0!4,4,4,4¡0!4,4,4,4?44

Increasing the left most number is multiplying by whatever ordinal the increase would have reached so

¿10¡0!0¡0!0¡0!0¡0!0?4 > f{ω^ω^4}*{10}(4)
¿4,4,4,4¡0!0¡0!0¡0!0¡0!0?4 > f{ω^ω^4}*{ω^ω}(4) = fω^ω^5(4)
Filling the other 0's would bring a few more remainders but ignoring that you can see a a closer comparison for
¿0¡1!0?n is fω^ω^n+1+1(n) which completely obliterates fω^ω^ω(n) = fω^ω^n(n)

¿0¡2!0?4 the first reduction looks like this
Spoiler:
¿4,4,4,4¡0!4,4,4,4¡0!4,4,4,4¡0!4,4,4,4¡0!4,4,4,4¡1!
4,4,4,4¡0!4,4,4,4¡0!4,4,4,4¡0!4,4,4,4¡0!4,4,4,4¡1!
4,4,4,4¡0!4,4,4,4¡0!4,4,4,4¡0!4,4,4,4¡0!4,4,4,4¡1!
4,4,4,4¡0!4,4,4,4¡0!4,4,4,4¡0!4,4,4,4¡0!4,4,4,4¡1!
4,4,4,4¡0!4,4,4,4¡0!4,4,4,4¡0!4,4,4,4¡0!4,4,4,4?
¿4,4,4,4¡0!4,4,4,4¡0!4,4,4,4¡0!4,4,4,4¡0!4,4,4,4¡1!
4,4,4,4¡0!4,4,4,4¡0!4,4,4,4¡0!4,4,4,4¡0!4,4,4,4¡1!
4,4,4,4¡0!4,4,4,4¡0!4,4,4,4¡0!4,4,4,4¡0!4,4,4,4¡1!
4,4,4,4¡0!4,4,4,4¡0!4,4,4,4¡0!4,4,4,4¡0!4,4,4,4¡1!
4,4,4,4¡0!4,4,4,4¡0!4,4,4,4¡0!4,4,4,4¡0!4,4,4,4?
¿4,4,4,4¡0!4,4,4,4¡0!4,4,4,4¡0!4,4,4,4¡0!4,4,4,4¡1!
4,4,4,4¡0!4,4,4,4¡0!4,4,4,4¡0!4,4,4,4¡0!4,4,4,4¡1!
4,4,4,4¡0!4,4,4,4¡0!4,4,4,4¡0!4,4,4,4¡0!4,4,4,4¡1!
4,4,4,4¡0!4,4,4,4¡0!4,4,4,4¡0!4,4,4,4¡0!4,4,4,4¡1!
4,4,4,4¡0!4,4,4,4¡0!4,4,4,4¡0!4,4,4,4¡0!4,4,4,4?
¿4,4,4,4¡0!4,4,4,4¡0!4,4,4,4¡0!4,4,4,4¡0!4,4,4,4¡1!
4,4,4,4¡0!4,4,4,4¡0!4,4,4,4¡0!4,4,4,4¡0!4,4,4,4¡1!
4,4,4,4¡0!4,4,4,4¡0!4,4,4,4¡0!4,4,4,4¡0!4,4,4,4¡1!
4,4,4,4¡0!4,4,4,4¡0!4,4,4,4¡0!4,4,4,4¡0!4,4,4,4¡1!
4,4,4,4¡0!4,4,4,4¡0!4,4,4,4¡0!4,4,4,4¡0!4,4,4,4?4


This cascading continues to happen in the φ(m,n) to φ(m,n+1) and the φ(m,n) to φ(m+1,n) steps, so to beat my ¿0<0>0?100 you would have to beat at least φ(102,100)100 to be certain
Last edited by Daggoth on Sun Feb 15, 2015 9:54 pm UTC, edited 1 time in total.

Daggoth
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Re: Your number is, in fact, not bigger!

Postby Daggoth » Sun Feb 15, 2015 9:52 pm UTC

The main reason for that "feat" of mine is not actually to be annoying. It's to easily escape the fixed-point traps while ascending the φ steps

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Vytron
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Re: Your number is, in fact, not bigger!

Postby Vytron » Mon Feb 16, 2015 6:04 am UTC

f{ω^ω^4}*{ω^ω} is not f{ω^ω^5}, it's f{ω^(ω^4+ω^2)}:

f{ω^ω^4}*{ω} = f{ω^(ω^4+1)}
f{ω^ω^4}*{ω2} = f{ω^(ω^4+2)}
f{ω^ω^4}*{ω^2} = f{ω^(ω^4+ω)}
f{ω^ω^4}*{ω^ω} = f{ω^(ω^4+ω^2)}

You need to reach f{ω^(ω^4+ω^4)} to reach f{ω^((ω^4)2)}, and then f{ω^((ω^4)ω)} = f{ω^ω^5}

So, all your cascading is doing pretty much doing nothing and one would need to reach φ(100,100)102 to pass it for sure.

So I'll be sending:

103,[100-100]

To make sure.

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WarDaft
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Re: Your number is, in fact, not bigger!

Postby WarDaft » Mon Feb 16, 2015 7:53 am UTC

Uh, no, sorry, I'm afraid that's not right either.

{ω^ω^4}*{ω} = {ω^(ω^4+1)} yes
{ω^ω^4}*{ω2} = {ω^(ω^4+2)} no, it's {ω^ω^4}*{ω2} = {ω^(ω^4 + 1)}*2
{ω^ω^4}*{ω^2} = {ω^(ω^4+ω)} no, it's {ω^ω^4}*{ω^2} = {ω^(ω^4 + 2)}
{ω^ω^4}*{ω^ω} = {ω^(ω^4+ω^2)} no, it's {ω^ω^4}*{ω^ω} = {ω^(ω^4+ω)}

In general, ωab = ωa + b, exactly like exponent laws in algebra.
All Shadow priest spells that deal Fire damage now appear green.
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Vytron
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Re: Your number is, in fact, not bigger!

Postby Vytron » Mon Feb 16, 2015 10:39 am UTC

Oh, thanks WarDaft :o - Are you planning an extension to your notation in the shadows?

That's why I ended getting rid of rabbit holes of my number, by the end, it seems one ends "tightly close to the actual fgh" (in Daggoth's words) no matter what one does...

For instance, I believe I could pass Daggoth's cascades with this:

Define:

[0] = {n,n}

I.e.

Old notation:

5,[0] = 5,5

New notation:

5,[0] = 5,{5,5}

The former uncollapses to:

5,0,0,0,0,0,0,1,2,3,4

The latter to:

5,0,0,... n ,s...,0,0,1,2,3...n-2,n-1,n

Where n is 5,0,0,0,0,0,0,1,2,3,4

Which produces much larger numbers initially, but ends being an offset and I end at the same ordinal I do with just 5,0,0,0,0,0,0,1,2,3,4...

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Re: Your number is, in fact, not bigger!

Postby WarDaft » Mon Feb 16, 2015 4:49 pm UTC

That's how ordinals work. Like, with my system, literally all of the size right now is coming from the colons, literally nothing else is doing anything anymore, and could be discarded.

Which is exactly what I am doing. But all shadowy like.
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Re: Your number is, in fact, not bigger!

Postby Vytron » Mon Feb 16, 2015 5:26 pm UTC

WarDaft wrote:That's how ordinals work. Like, with my system, literally all of the size right now is coming from the colons, literally nothing else is doing anything anymore, and could be discarded.

Which is exactly what I am doing. But all shadowy like.


Oh yeah? What can your colons do AGAINST THIS??

n,[0-0-0-0] = n,[~[0-0,0-0,0]~-0,0-0,0] = f_φ(1,0,0,0)(n)

n,[0-0-0-0,0] = n,[~[0-0-0-0]~-0,0-0,0] = f_φ(1,0,0,1)(n)

n,[0-0-0,0-0] = n,[0-0-0-~[0-0-0-0,0]~] = f_φ(1,0,1,0)(n)

n,[0-0,0-0-0] = n,[0-0-~[0-0-0,0-0]~-0] = f_φ(1,1,0,0)(n)

n,[0,0-0-0-0] = n,[0-~[0-0,0-0-0]~-0-0] = f_φ(2,0,0,0)(n)

n,[0-0-0-0-0] = n,[~[0-0,0-0,0-0,0]~-0,0-0,0-0,0] = f_φ(1,0,0,0,0)(n)

n,[0-0-0-0-0,0] = n,[~[0-0-0-0-0]~-0,0-0,0-0,0] = f_φ(1,0,0,0,1)(n)

n,[0-0-0-0,0-0] = n,[0-0-0-0-~[0-0-0-0,0]~] = f_φ(1,0,0,1,0)(n)

n,[0-0-0,0-0-0] = n,[0-0-0-~[0-0-0-0,0-0]~-0] = f_φ(1,0,1,0,0)(n)

n,[0-0,0-0-0-0] = n,[0-0-~[0-0-0,0-0-0]~-0-0] = f_φ(1,1,0,0,0)(n)

n,[0,0-0-0-0-0] = n,[0-~[0-0,0-0-0-0]~-0-0-0] = f_φ(2,0,0,0,0)(n)

n,[0-0-0-0-0-0] = n,[~[0-0,0-0,0-0,0-0,0]~-0,0-0,0-0,0-0,0] = f_φ(1,0,0,0,0,0)(n)

n,[0-0-0-0-0-0-0] = n,[~[0-0,0-0,0-0,0-0,0-0,0]~-0,0-0,0-0,0-0,0-0,0] = f_φ(1,0,0,0,0,0,0)(n)

n,[0-0-0-0-0-0-0-0] = n,[~[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] = f_φ(1,0,0,0,0,0,0,0)(n)

n,00 = n,[0-~]

Send:

5,00

For 𝛝(𝛀^𝛚)(5)

How's that for "1-upmanship"? :twisted:

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Re: Your number is, in fact, not bigger!

Postby Daggoth » Mon Feb 16, 2015 8:05 pm UTC

U sure about that?
If x^1 = x
and x^4 * x = x^5
and x=w^w
where am i misapplying the math here?

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Re: Your number is, in fact, not bigger!

Postby WarDaft » Mon Feb 16, 2015 8:33 pm UTC

Daggoth wrote:U sure about that?
If x^1 = x
and x^4 * x = x^5
and x=w^w
where am i misapplying the math here?


Exponentiation is not associative.

(x^x)^5 is not the same as x^(x^5)

(x^x)^5 is x^(x*5)

Also, Vytron, what ordinal is 𝛝? It's not rendering for me... I just get a box.

Typically after φ, one uses ordinal collapse to describe ordinals, where ψ((Ω^Ω^α)*β) is equivalent to φ(β,0,0...,0) where β is in the α-th place, so ψ(Ω^Ω^ω) is the supremum of φ(1), φ(1,0), φ(1,0,0), φ(1,0,0,0)....
Last edited by WarDaft on Mon Feb 16, 2015 8:45 pm UTC, edited 1 time in total.
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Re: Your number is, in fact, not bigger!

Postby Vytron » Mon Feb 16, 2015 8:45 pm UTC

You're confusing {w^w}^4 with w^{w^4} (to Daggoth).

Ninja'd.

Also, Vytron, what ordinal is 𝛝? It's not rendering for me... I just get a box.


Aw, it looked so fancy...

It's:
Image

Do you see any of these 5 symbols or do you see only boxes?

𝛝 𝜗 𝝑 𝞋 𝟅

What you actually wanted to know:

It's the Small Veblen Ordinal in Weiermann's ordinal notation.

Redundant
Spoiler:
ord.png
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Re: Your number is, in fact, not bigger!

Postby WarDaft » Mon Feb 16, 2015 8:48 pm UTC

I see only boxes.
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Re: Your number is, in fact, not bigger!

Postby Vytron » Mon Feb 16, 2015 10:47 pm UTC

Okay, but you saw this image alright?

Image

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Re: Your number is, in fact, not bigger!

Postby WarDaft » Mon Feb 16, 2015 11:07 pm UTC

Yes, that comes through. I'll be posting the revised notation later today or early tomorrow, once I have drummed up a good list of example values.
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Re: Your number is, in fact, not bigger!

Postby Vytron » Tue Feb 17, 2015 12:13 am UTC

Will be looking forward to it 8-)

Anyway, so it has come to this, I'm going to upload the image to that ordinal to the forum so every time I want to use it I just paste the image code:

Image

ordi.png
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Re: Your number is, in fact, not bigger!

Postby Daggoth » Tue Feb 17, 2015 1:10 am UTC

Ah i see where i was wrong. But still fω^{ω4+ω}+1(4) > fω^ω4(4) so my point about ¿0<0>0?100 stands.

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Re: Your number is, in fact, not bigger!

Postby mike-l » Wed Feb 18, 2015 6:38 am UTC

Pretty sure Vytrons base rules don't get anywhere near epsilon_0 when he claims. I put n,n around w^w^w.

I came back to this thread and I see a lot of posting over top of each other, but not a lot of talk about what's actually happening in each step. Is anyone actually checking any of the claims made, or are you all just claiming larger ordinals without verifying the other actually reached what was claimed?
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Re: Your number is, in fact, not bigger!

Postby WarDaft » Wed Feb 18, 2015 9:02 am UTC

Anyway, this is indeed the crossover I was waiting for to re-vamp things.

< > are the repeater operator, such that <X> means the string X repeated n times
( ) denote an array, which is to be evaluated lazily, rather than immediately.
[;] is an advanced delimiter. : and ; will not work the same, so I'm not re-using :.
Capitals represent any number of terms (including zero), lower case represent exactly one term.
{ } represent special rules for decrementing a term.

n|0 = n
n|X = n+1|X--
(0)-- = 0
(b)-- = (b--)
(X,0,0)-- where X contains no non-zero terms = <(X,>0<)>
(X,0,b,Y)-- where X contains no non-zero terms = <(X,>0<,b--,Y)>
(a,X,0,b,Y)-- where X contains no non-zero terms, and b does not contain square brackets = <(a--,X,>(a--,X,0,b,Y)<,b--,Y)>
(a,b,X)-- = <(>(a--,b,X)<,b--,X)>
Terms containing [;] are considered non-zero.
(0[;]0,X) = (<0,>X)
(a,0[;]0,X)-- where X is all zero = (<0,>(a--,0[;]0,X),X)
(a,b[;]0,X)-- = (<0,>(a--,b[;]0,X),b--[;]0,X)
(0[;]b,X)-- = <(>0<[;]b--,X)>
(a,0[;]b,X)-- = <(>(a--,0[;]b,X)<[;]b--,X)>
(a[;]b,X)-- = <(>0<[;]b--,a--[;]b,X)>
(a,b[;]c,X)-- = <(>(a--,b[;]c,X)<[;]c--,b--[;]c,X)>

Using the Hardy hierarchy.
2|0 = 2 = f{0}(1)
2|(0) = 3 = f{0}(2)
2|(((((0))))) = f{0}(6)
2|(0,0) = ((0)) = f{1}(2)
2|((0,0),0) = f{1}(f{1}(2)
2|(0,1) = f{2}(2)
2|(0,2) = f{3}(2)
2|(0,(0,0)) = f{ω}(2)
2|(0,(1,0)) = f{ω+1}(2)
2|(0,((0,0),0)) = f{ω*2}(2)
2|(0,((1,0),0)) = f{ω*2+1}(2)
2|(0,(((0,0),0),0)) = f{ω*3}(2)
2|(0,(0,1)) = f{ω^2}(2)
2|(0,((0,1),1)) = f{ω^2*2}(2)
2|(0,(0,2)) = f{ω^3}(2)
2|(0,(0,3)) = f{ω^3}(2)
2|(0,(0,(0,0))) = f{ω^ω}(2)
2|(0,(1,(0,0))) = f{ω^ω+1}(2)
2|(0,((0,(0,0)),(0,0))) = f{ω^ω*2}(2)
2|(0,(0,(1,0))) = f{ω^(ω+1)}(2)
2|(0,(0,((0,0),0))) = f{ω^(ω*2)}(2)
2|(0,(0,(((0,0),0),0))) = f{ω^(ω*3)}(2)
2|(0,(0,(0,1))) = f{ω^(ω^2)}(2)
2|(0,(1,(0,1))) = f{ω^(ω^2)+1}(2)
2|(0,(0,(1,1))) = f{ω^(ω^2+1)}(2)
2|(0,(0,((0,0),1))) = f{ω^(ω^2+ω)}(2)
2|(0,(0,((0,1),1))) = f{ω^(ω^2*2)}(2)
2|(0,(0,(0,2))) = f{ω^(ω^3)}(2)
2|(0,(0,(0,(0,0)))) = f{ω^(ω^ω)}(2)
2|(0,0,0) = f{φ(1,0)}(2) = f{ψ(0)}(2)
2|(0,(0,0,0)) = f{φ(1,0)+1}(2) = f{ψ(0)+1}(2)
2|(0,(0,(0,0,0))) = f{ω^(ψ(0)+1)}(2)
2|(0,(0,(0,(0,0,0)))) = f{ω^ω^(ψ(0)+1)}(2)
2|(1,0,0) = f{ψ(1)}(2) (ε1 already!)
2|(2,0,0) = f{ψ(2)}(2)
2|((0,0,0),0,0) = f{ψ(ψ(0))}(2)
2|(0,1,0) = f{ψ(Ω)}(2)
In general 2|(a,b,c,d) = f{φ(d,c,b,a)}(2)

As such, 2|(0[;]0) is f{ψ(Ω^Ω^ω)}(2) and 2|(0[;](0[;]0)) is f{ψ(Ω^Ω^(ψ(Ω^Ω^ω)))}(2)

mike-l wrote:Pretty sure Vytrons base rules don't get anywhere near epsilon_0 when he claims. I put n,n around w^w^w.

I came back to this thread and I see a lot of posting over top of each other, but not a lot of talk about what's actually happening in each step. Is anyone actually checking any of the claims made, or are you all just claiming larger ordinals without verifying the other actually reached what was claimed?
At some point, I realized it was just easier to beat his claims than prove them correct or incorrect.
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Re: Your number is, in fact, not bigger!

Postby Vytron » Wed Feb 18, 2015 12:25 pm UTC

mike-l wrote:Pretty sure Vytrons base rules don't get anywhere near epsilon_0 when he claims. I put n,n around w^w^w.


We have gone over this before, these are an equivalent rewrite of the rules that you yourself proved to reach w^^a in the following quote in the following spoiler:

Spoiler:
mike-l wrote:Ok, proper rules for this prefix notation.

If the final element is less or equal to the penultimate element, reduce it as follows. If it is 0, remove it and iterate the function. Otherwise, say it is m > 0, replace it with 0,0,...0,1,2,..m-1.

Otherwise, consider the longest string X preceding the final element m (necessarily > 0 since the penultimate element is less than it) consisting of symbols strictly less than m, ie we have [Y, X, m] where X is a string of elements less than m and Y is a (possibly empty) string ending in an element greater or equal to m. Reduce X using the same rules, giving Z, and we are equal to [Y, Z, m, Z, m, ... Z,m]

Every time there is a '...' between like terms you may use any number of repetitions dependent on n and 'smaller' notations, however n itself is sufficient. Also you may use 1 (ie no extra terms, [Y,Z,m]) for the last rule if X did not end in 0. Any choice within these confines will yield the same ordinal growth.

// End rules

Helpful in interpreting.

Let a be the largest symbol appearing in the notation, then separate the notation after each a (or at the end as required). Using | as a separator purely for visual aid, it has no effect and is the same as a comma. So we get something like [X, a | Y, a | ... | Z, a | V] where X, Y, ... Z, V are possibly empty strings consisting solely of symbols less than a. Only the rightmost segment is ever evaluated, so if Ord(P) represents the ordinal reached by [P], this will be equal to Ord(X,a) + Ord(Y,a) + ... + Ord(Z,a) + Ord(V).

First evaluate V using the same rules if it's nonempty (it now has a lower max element), otherwise the string ends in Z,a, which becomes Z', a | Z', a ... Z', a where Z' is the reduction of Z. By the above, we have Ord(Z,a) = Ord(Z', a)*w

//End exposition, now some proofs

I'll now show my claim that Ord(Z,a) = Ord(a)*w^Ord(Z), by induction, the base case being trivial Ord(a) = Ord(a).

Now if Z represents a successor ordinal, ie it ends in a 0, then Z' is just its predecessor, and Ord(Z') + 1 = Ord(Z), so:

Ord(Z,a) = Ord(Z', a)*w = Ord(a)*w^Ord(Z')*w = Ord(a)*w^{Ord(Z') + 1} = Ord(a)*w^Ord(Z).

If Z is a limit ordinal, ie it doesn't end in 0, then for any b < Ord(Z), there is some input n for which Ord(Z') > b.
Thus:

Ord(Z,a) = Ord(Z', a)*w > Ord(b,a)*w = Ord(a)*w^b * w = Ord(a)*w^{b+1} for any b < Ord(Z)

Thus Ord(Z,a) = Ord(a)*w^Ord(z) as claimed.

This shows that my original comment was correct, that you only need to repeat Z, m if X ended in a 0.



Finally, we can show that Ord(a) = w^^a, as [a] = [(a-1)', a-1] for a>1 (Here (a-1)' is the string obtained by reducing a-1).

Firstly Ord(0) = 1 is trivial, and Ord(1) = w as well. For the rest we can use the above definition.

Ord( (a-1)', a-1) = Ord(a-1)*w^Ord( (a-1)' ) = w^^(a-1) * w^Ord( (a-1)' ) inductively.

For any b < w^^(a-1), some input of n has Ord( (a-1)' ) > b, thus

Ord(a) = w^^(a-1) * w^Ord( (a-1)' ) > w^^(a-1) * w^b
Ord(a) = w^^(a-1) * w^(w^^(a-1)) = w^^(a-1) * w^^a.

The right term is sufficiently larger than the left that it 'eats' it (specifically we get:

w^(w^^(a-2)) * w^(w^^(a-1)) = w^(w^^(a-2) + w^^(a-1))

and the right term in the sum eats the left, to the same effect)

And so: Ord(a) = w^^a.


And and the n,YX rules can be mapped to these values that you yourself claimed went over f_w^w^w^w:

Spoiler:
mike-l wrote:w^w^w+1 - [3,0]
w^w^w+w - [3,1]
w^w^w+w2 - [3,1,1]
w^w^w+w^2 [3,0,1]
w^w^w+{w^2}2 [3,0,1,0,1]
w^w^w+w^3 [3,0,0,1]
w^w^w+w^w [3,2]
w^w^w+{w^w}2 [3,2,2]
w^w^w+w^(w+1) [3,0,2]
w^w^w+w^w2 [3,1,2]
w^w^w+w^w^2 [3,0,1,2]
w^w^w+{w^w^2}2 [3,0,1,2,0,1,2]
w^w^w+w^w^3 [3,0,0,1,2]
{w^w^w}2 [3,3]
w^({w^w}+1) [0,3]
w^({w^w}+w) [1,3]
w^({w^w}+w2) [1,1,3]
w^({w^w}+w^2) [0,1,3]
w^({w^w}+{w^2}2) [0,1,0,1,3]
w^({w^w}+w^3) [0,0,1,3]
w^({w^w}2) [2,3]
w^w^(w+1) [0,2,3]
w^w^(w2) [1,2,3]
w^w^(w^2) [0,1,2,3]
w^w^({w^2}2) [0,1,0,1,2,3]
w^w^w^3 [0,0,1,2,3]
w^w^{w^3}2 as written, the above twice.
I assume you mean w^w^{{w^3}2} [0,0,1,0,0,1,2,3]
w^w^w^w [4]

That's as far as I can go on my phone right now. There may be some errors. I'll explain the last one in more detail as it has most of what's happening.

The main operations our notation can do is add ordinals, build w^^n, and multiply w^^n by w^a where a is smaller than w^^n

So we have w^w^{{w^3}2} = w^w^w * w^(w^{{w^3}2}), so we can express this as [x,3] where x represents w^{{w^3}2}.

w^{{w^3}2} = w^w * w^({w^3}2), so [x] = [y,2] where y represents {w^3}2.

{w^3}2 = w^3 + w^3, so is just represented by [z,z] where z represents w^3

Finally, w^3 = w*w^2 which is represented by [0,0,1]. So putting it all together.
[y] = [0,0,1,0,0,1]
[x] = [0,0,1,0,0,1,2]
Original = [0,0,1,0,0,1,2,3]


Would be represented with a n,0,0,1,0,0,1,2,3.

So, prove my rewrite isn't equivalent or get in contact with your old self? :mrgreen:

Now, we're using different ordinal notations, they're equivalent at:

f{ψ(Ω^Ω^ω)}(2) = f{Image(Ω^ω)}(2)

So:

2|(0[;](0[;]0)) is f{Image(Ω^{Ω^ω})}(2)

Now, plopping my values over the old boxes is much stronger than expected but much weaker than WarDaft so I need to go over it with a finecomb. I.e. it could be claimed that:

n,01 = [00-~] = (n+1),00

Which is silly.

So I'll go over everything I do to get there, just to show (n+1),00 is beaten on the very first inner nestings.

n,00,0 = (((n),00),00...),00 (for n+1 nestings, I'm using n+1 instead of n so it matches my other n,00 n,00...n,00 with n spaces notation)

n,00,0,0 = (((n),00,0),00,0...),00,0

n,00,0,0,0 = (((n),00,0,0),00,0,0...),00,0,0

n,00,1 = n,00,0,~
n,00,1,0 = (((n),00,1),00,1...),00,1
n,00,1,0,0 = (((n),00,1,0),00,1,0...),00,1,0

n,00,1,1 = n,00,1,0,~

n,00,0,1 = n,00,1,~
n,00,0,0,1 = n,00,0,1,~~
n,00,0,0,0,1 = n,00,0,0,1,~~~

n,00,2 = n,00,0,~1

n,00,3 = n,00,0,~1,2

n,00,[0] = n,00,n
n,00,[0],[0] = n,00,[0],n
n,00,[0],[0],[0] = n,00,[0],[0],n

n,00,0,[0] = n,00,[0],~
n,00,0,0,[0] = n,00,0,[0],~~
n,00,0,0,0,[0] = n,00,0,0,[0],~~~

n,00,1,[0] = n,00,0,~[0]
n,00,2,[0] = n,00,0,~1,[0]
n,00,3,[0] = n,00,0,~1,2,[0]

n,00,{0},[0] = n,00,n,[0]

Here's where things explode, because I can fit my entire notation on the {}s

n,00,{00,{0},[0]},[0] = n,00,{n,00,n,[0]},[0] = ψ({Ω^Ω^ω}+1)
n,00,{00,{00,{0},[0]},[0]},[0] = ψ({Ω^Ω^ω}+2)
n,00,{00,{00,{00,{0},[0]},[0]},[0]},[0] = ψ({Ω^Ω^ω}+3)
n,00,[0,0] = n,00,{00,~{00,{0},[0]}~,[0]},[0] = ψ({Ω^Ω^ω}+ω)

Entire notation in n,00,[X]

n,00,[0-0] = n,00,[~[00]~] = ψ({Ω^Ω^ω}*2)
n,00,[0-0,0] = n,00,[~[0-0]~] = ψ({Ω^Ω^ω}*3)

Entire notation in n,00,[0-X]

n,00,[0,0-0] = n,00,[0-[~[0-0]~]] = ψ({Ω^Ω^ω}^2)

n,00,00 = n,00,[0-~] = Image({Ω^ω}+1)

n,00,00,00 = n,00,00,[0-~] = Image({Ω^ω}+2)

n,0,00 = n,00,00,~ = Image({Ω^ω}+ω)

n,{0},00 = n,n,00,~ = Image({Ω^ω}+φ(1,0))

Entire notation in n,{X},00

n,[00] = n,{~{00,00}~},00 = Image({Ω^ω}*2)

n,[00,0] = n,{~{00,00}~},[00] = Image({Ω^ω}*3)

n,[00,0,0] = n,{~{00,00}~},[00,0] = Image({Ω^ω}*4)

n,[00,1] = n,[00,0,~] = Image({Ω^(ω+1)})

n,[00,{0}] = n,[00,n+1] = Image({Ω^(φ(1,0))})

Entire notation in n,[00,{X}]

n,[00,{00}] = n,[00,{[0-~]}] = Image({Ω^(Ω^ω)})

^Matches WarDaft.

n,[00,{[00]}] = n,[00,{{~{00,00}~},00}] = Image({Ω^{(Ω^ω)}2})

Send:

5,[00,{[00]}]

For:

f{ψ(Ω^Ω^(ψ(Ω^{{Ω^ω}2})))}(5)

mike-l
Posts: 2758
Joined: Tue Sep 04, 2007 2:16 am UTC

Re: Your number is, in fact, not bigger!

Postby mike-l » Wed Feb 18, 2015 12:56 pm UTC

We have been over this before, I showed you that you were wrong, I gave you rules that worked, and you copied them incorrectly into rules that have the same problem you originally had.
addams wrote:This forum has some very well educated people typing away in loops with Sourmilk. He is a lucky Sourmilk.


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