FANDOM


Note: first solve non-superscript arrays. If you stuck, go to the next level superscript. \(\triangle\) is an row of zeroes \(\bullet\) is the remainder of the array and \(\circ\) contains only brackets here contains \(\diamond\) only brackets with level bigger than b and zeroes

What to solve first?

In order of importance:

• brackets with the lowest legion level (Ultimate Array Notation V)

• brackets with least nested angle bracket (Ultimate Array Notation IV)

• brackets with the lowest & level (Ultimate Array Notation III)

• brackets with least & symbols (Ultimate Array Notation II)

• many brackets solve on right to left & order (Ultimate Array Notation I)

• brackets with the lowest arrow level (Extended Array Notation)

• brackets with the least number of entries (Array Notation)

• brackets with the lowest level (Extended Bracket Notation)

• least nested bracket with the smallest brackets in it (Bracket Notation)

All brackets have the same a$.

Bracket Notation

1. a$b \(\bullet\) = (a+b)$ \(\bullet\) 2. a\(\circ\)[0 \(\bullet\)]\(\circ\) = a$\(\circ\)a[ \(\bullet\)]\(\circ\) 3. a$\(\circ\)[b+1\(\bullet\)]\(\circ\) = a$\(\circ\)[b\(\bullet\)][b\(\bullet\)]...[b\(\bullet\)][b\(\bullet\)]\(\circ\) a [b\(\bullet\)]'s Where 0 and b are the less nested numbers \(\bullet\) is the remainder of the array and \(\circ\) contains only brackets

Extended Bracket Notation

4. a$[\(\diamond\)]_(b+1)\(\bullet\)  = a$[[...[[\(\diamond\)]_b ]...]_b ]_b\(\bullet\)  where there are a b-brackets and the b-bracket is the bracket with the lowest level.

5. a$\(\circ\)[\(\diamond\)]\(\circ\) = a$\(\circ\)\(\diamond\)\(\diamond\)...\(\diamond\)\(\diamond\)\(\circ\) where there are a \(\diamond\)'s  ( here is 'b' 1 )

6. a$[b\(\bullet\)]_(c+1) = a$[[[...[[[b-1\(\bullet\)]_(c+1)\(\bullet\)]_(c)\(\bullet\)...]_(c)\(\bullet\)]_(c)\(\bullet\) triangle may or may not exist.

Linear Array Notation

7. \(\bullet\)0 = \(\bullet\) applies everywhere, for example [a \(\rightarrow\)0] = a if there is a ] bracket after the 0.

8. a$[\(\triangle\),0,b\(\bullet\)] Call the whole array ( including & structures etc. ) with the b with one desecared without the a$ X, the new array is a$[\(\triangle\),X_X...X_X,b\(\bullet\)]

9. a$[\(\triangle\),0,c\(\bullet\)]_(b+1)\(\bullet\) = a$[\(\triangle\),[...[\(\triangle\),[\(\triangle\),0,c\(\bullet\)]_b,c\(\bullet\)]_b...]_b,c\(\bullet\)]_b\(\bullet\) \(\triangle\) is an row of zeroes

Extended Array Notation

10. This rule applies if and only if there is something in the form of a$[0\(\rightarrow_b\)c], and the 'c' is the first non-zero entry in the array. If you got somthing like 'a$[0\(\rightarrow_b\)0,c]' you have to solve 0,c first. Rule 10: a$[0...0\(\rightarrow_b\)c\(\bullet\)] = a$[0...a\(\rightarrow_{b-1}\)a...a\(\rightarrow_{b-1}\)a\(\rightarrow_b\)c-1\(\bullet\)] where there are a new a's a comma is a 0-arrow, and a non-indexed a 1-arrow

Ultimate Array Notation I

This is the point where the notations gets its power. Each bracket can have a & suffix, witch is part of the bracket so have to be copied etc.

Rule 11: a$[0]&b = a$[0\(\rightarrow_{[0\rightarrow_{...[0\rightarrow_{[0]\&b-1)}1]\&b-1)...}1]\&b-1)}\)1]&(b-1) with a nests/ For &0 applies rule 7 so &0 can be removed.

Rule 12. a$[b]&c = a$[[b]&c\(\rightarrow_{[[b]&c\(\rightarrow_{..._{[[b]&c\(\rightarrow_{[b]&c}1}...}1}1]

Continue until a$[0]&([0]&([0]&(...([0]&[0])...))) Here comes a level 2 array:

Rule 13. a$[0]&[0,{2}1]  = a$[0]&([0]&([0]&(...([0]&[0])...)))

rest of the ,{2} and even \(\rightarrow\){2} works normal. 


Ultimate Array Notation II

Of course we can also have a$[0]&[0\(\rightarrow\){2}1], a$[0]&[0\(\rightarrow_2\){2}1] and these arrays can also be nested. a$[0]&[0]&b = a$[0]&[0\(\rightarrow_{[0\rightarrow_{...[0\rightarrow_{[0]}\{2\}\&\(b-1)}\{2\}1]\&\(b-1)...}1]\&\(b-1)}\)\{2\}1]&(b-1) so can we also have level 3 arrays after the second &.

Rules: The bold italic & is the nth &

Rule 11. a$\(\square\)[0]&(b)\(\bullet\) = a$\(\square\)[0\(\rightarrow_{\square[0\rightarrow_{...{\square[0\rightarrow_{\square[0]\&(b-1)\bullet}1]\&(b-1)\{b-1\}\bullet}...}1]\&(b-1)\{b-1\}\bullet}1]\&(b-1){b-1}\(\bullet\)

Rule 12. a$\(\square\)[c]&(b)\(\bullet\) = a$\(\square\)[0\(\rightarrow_{\square[0\rightarrow_{...{\square[0\rightarrow_{\square[c-1]\&(b)\bullet}1]\&(b)\{b-1\}\bullet}...}1]\&(b)\{b-1\}\bullet}1]\&(b){b-1}\(\bullet\)

Rule 13. a$ \(\square\) &[0,{n-1}1]  = a$(\(\square\)&[(\(\square\)&(...([\(\square\)&[0])...))) \(\square\) means more &'s

Ultimate Array Notation III

Rule 14. a$[0]\(\square\)(&_c)b = a$[0](&_(c-1))[0]...[0](&_(c-1))[0](&_c)(b-1) with a 's.

The c can also be an array.

Rule 15. a$[0]\(\square\)&$b = a$[0]\(\square\)&_([0]\(\square\)&_..._([0]\(\square\)&_[0]$b-1)$b-1)$b-1

Rule 16.\(\bullet\)&_0\(\bullet\)= \(\bullet\)\(\bullet\)

structures like &$&_1 = &$& have to be solved with rule 14, remove &_0's by rule 16.

Ultimate Array Notation IV

Rule 17. a$[0]# = a$[0]&$&$...&$&$# b &$'s

Rule 18. a$[0]<<...<>...>># = a$[0]<<...<[0]<<......>>#>...>>#

Of course are arrays based on <> allowed, <<...<<>>...>> = <>_2 etc.

if the normal array has level theta(X), then the <> array has level lambda(X).

a$[0](<0>&2)# is possible, even a$[0](<0><0>[0])# or nested a$[0](<0>(<0>(<0>(......)[0])[0])[0])[0]

Ultimate array notation V

a$[0]/b-1 = a$[0](<0>(<0>(...(<0><0>/b-1[0])...)[0]/b-1)[0]/b-1)[0]/b-1

More coming soon!

FGH

a$[0,0...0,0,1] ~ BHO so the limit of linear array notation is still the BHO

Limit of extended arrays: theta(Omega_2)

Limit of Ultimate Array Notation I: theta(Omega_Omega_2)

Limit of Ultimate Array Notation II: theta(I)

Limit of Ultimate Array Notation III: theta(I*w)

Limit of Ultimate Array Notation IV: theta(I^2)

Limit of Ultimate Array Notation V >> theta(I^Omega)

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