Talk:Bird's array notation

Plugging array-like notations to functions
I have the plan how to adopt Bird's array notation (or probably any recursive notation) to any other function. Just change some main rules:

Rule M1: (no separators)

\(\{a\} = \Sigma(a)\)

Rule M2: (first entry is 1)

\(\{1,\#\} = 1\)

Rule M6: (string of 1's from 2rd entry to nth)

\(\{a,1,1,\cdots,1,1,c \#\} = \{a,a,a,\cdots,a,\{a-1,1,1,\cdots,1,1,c \#\},c-1 \#\}\)

Rule M7: (otherwise)

\(\{a,b \#\} = \{\{a-1,b \#\},b-1 \#\}\)

All other rules remain unchanged. The limit ordinal of this notation is \(\omega_1^{CK}+\theta(\varepsilon_{\Omega+1})\), in other words, Bachmann-Howard ordinal-typed recursion around busy beaver function. There will be also more powerful variant of H(n) function (that grows on par with \(f_{\omega_1^{CK}+\theta(\varepsilon_{\Omega+1})}(n)\)). In this notation, H(1) = \(\Sigma(3) = 6\) and H(2) will be certainly larger than anything we can write down in the observable universe, using recursive definitions.

In general, we can plug any so-far-defined recursive notation around any so-far-defined function, if we replace the terminating rule of the notation to computing function and probably change some other rules. If the function has ordinal level \(\alpha\) and the notation \(\beta\), then super-notation will has the ordinal level \(\alpha+\beta\). Ikosarakt1 (talk ^ contribs) 12:28, May 25, 2013 (UTC)

New Expansion Pack - Ready for downloading!
The new Expansion Pack for Battle of Array Nations, dubbed "Newer Subscripting Adventure Network", is now available for download! This Expansion Pack adds new quests, with a new super-strong final boss, called \(\theta(\theta_1(\Omega))\).

So yeah, download it on here, now! Have fun! -- &#9729; I want more clouds! &#9925; 10:31, May 28, 2013 (UTC)

Define CB(t) as function time->Ord defining upper bound of current Chris Bird's work. I state CB(t) eventually outgrows every recursive function. Proof of this would make Church-Turing thesis false. So waiting for results. LittlePeng9 (talk) 17:49, May 28, 2013 (UTC)
 * The problem is that Bird can eventually exhaust all symbols invented by mankind and all possible well-defined recursive combinations of them. I wonder whether happen earlier: this point or his numbers will be bigger than \(\Sigma(1000000000)\). Ikosarakt1 (talk ^ contribs) 16:02, June 10, 2013 (UTC)
 * If Bird uses all symbols mankind invented, he will invent new ones! And this will happen waaay before we reach even like S(10000). LittlePeng9 (talk) 20:51, June 10, 2013 (UTC)
 * Nonetheless, as the number of symbols becomes too large, say, million of symbols, it is hard even to distinguish some of them, let alone parse how to solve arrays using them. For example, normal backslash and backslash rotated to the right by 1 degree. Currently, he use backslash as a main symbol allowing so big level of recursion, but I feel that soon he will use a new symbol since all combinations on \ will be used. Ikosarakt1 (talk ^ contribs) 12:06, June 11, 2013 (UTC)
 * So how about using some arbitrary function c(n) to denote characters? Just like Bird did to replace negation sign, diamond and sun with \(\backslash_n\)? But, as I guess, for any such numbering he will further diagonalize through it to reach further, and now I ran out of ideas. LittlePeng9 (talk) 15:33, June 11, 2013 (UTC)
 * Eagerly awaiting the release of the Ordinals server mod. FB100Z &bull; talk &bull; contribs 20:20, June 10, 2013 (UTC)

Nested separators
Where I was wrong?

[A] = [1 [1 [2] 2] 3]

[B] = [1 [1 [2] 2] 2]

Step 1:

[A_1] = [1 [1 [2] 2] 3]

[B_1] = [1 [1 [2] 2] 2]

Step 2:

[A_2] = [1 [1 [2] 2] 3]

[B_2] = [1 [1 [2] 2] 2]

Step 3:

Lev(A_2) = 2

Lev(B_2) = 2

Lev(A_2)=Lev(B_2)>0

Step 4:

[A*] = [1 [2] 2]

[B*] = [1 [2] 2]

[A*]=[B*]=[H]

Step 5:

Num(H,A_2)=1

Num(H,B_2)=1

Num(H,A_2)=Num(H,B_2)

Step 6:

[A_2] = [1]

[B_2] = [1]

Step 3:

Lev(A_2)=0

Lev(B_2)=0

Lev(A_2)=Lev(B_2)=0

Step 7:

[1]=[1]

Step 8:

[A_1] = [1]

[B_1] = [1]

Step 2:

[A_2] = [1]

[B_2] = [1]

Step 3:

Lev(A_2) = [1]

Lev(B_2) = [1]

Step 4:

No separators? Ikosarakt1 (talk ^ contribs) 14:17, June 10, 2013 (UTC)

I think the problem came from the vaguity of Step 6: what means "remove all entries up to and including [H] separator"? There are two ways to interpret it, and given something like [1,1,1,2 [2] 2], after Step 6 it can be either [2] or [1,1,1,2]. Ikosarakt1 (talk ^ contribs) 15:58, June 10, 2013 (UTC)