## FANDOM

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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)

### 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)