User blog:Wythagoras/Dollar function formal definition

'''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

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

Linear Array Notation
7. \(\bullet\)0 = \(\bullet\)

8. a$[\(\triangle\),0,b\(\bullet\)] = a$[\(\triangle\),[\(\triangle\),0,b-1\(\bullet\)]_[\(\triangle\),0,b-1\(\bullet\)]...[\(\triangle\),0,b-1\(\bullet\)]_[\(\triangle\),0,b-1\(\bullet\)],b-1\(\bullet\)]

9. a$[\(\triangle\),c]_(b+1)\(\bullet\) = a$[\(\triangle\)[...[\(\triangle\)[\(\triangle\),c-1]_b ]...]_b]_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

Ultimate Array Notation
Rule 11. a$(&_0)0 = a$[0\(\rightarrow_{[0\rightarrow_{..._[0\rightarrow_{[0]}1]...}1]}\)1] with a nests

Rule 12. a$(&_b)0 = a$(&_(b-1))[0\(\rightarrow_{[0\rightarrow_{..._[0\rightarrow_{[0]}1]...}1]}\)1] with a nests

