pepAAN

Primitive Expanding Aarex Array Notation is an third part of Aarex's Array Notation. This part adds {A}^B definition.

Separators

Separators is the object between the entries. Let define row of Ax as separators and define the separator:

• , is the 4th-ranking separator.
• {n} is the 3rd-ranking separator.
• {a A1 b A2 c A3 ...} is the 2nd-ranking separator.
• {}^n is the 1st-ranking separator.

There are finite layers of the separators, so we need to define the separator layer too:

• a() array entries are the separator layer 0.
• Separators on the array A in the are the separator layer, 1 more than A.

Now here is the definition.

Definition

The primitive expanding array notation look like this:

a(a,b,c A1 d A2 e...)x[y]

The integers a, b, c, etc.; are the entries, the Ax group are the separators where is between the entries, and the outermost "a" (like this can be place outside of the array entries) meaning this expression can be Aarex's Array Notation.

The first entry and second entry of AAN are called base and iterator.

The string x[y] is optional and can be removable. x and y means nester and multiplier. Outermost a and x are always be 'a' character and 'x' character.

Note: The comma symbol ',' is shorthand for {0}^1. Also, {n}^m is shorthand of {{{...{{{n}}}...}}} with m {}s surrounding.

Rules and process

Every array needs to results in a number. To solve it, we need some rules and processes as follows: (Thanks to Nayuto Ito for helping me defining english version.)

• Rule 1: (base rule - if it have 3 entries and the 3rd entry is 0) a(a,b,0) = a*b
  • English: If it have 3 entries and the 3rd entry is 0, the whole expression will resulted as the multiplication product with a and b.
  • Rule 1a: (if it have 2 entries) a(a,b) = a(a,b,0)
    • English: If it have 2 entries, the whole expression will resulted as the multiplication product with a and b.
  • Rule 1b: (if it have 1 entry) a(a) = a(a,1,0)
    • English: If it have 2 entries, this will be resulted as the entry.
• Rule 2: (tailing rule - if the last entry is 0) a(# 0) = a(#)
  • English: Remove the tailing zeroes in the last entries.
  • Rule 2a: (in separators) {# 0}^n = {#}^n
    • English: Remove the tailing zeroes in the last entries on the separators.
  • Rule 2b (if B > A): (different spaces) {# A 0 B %}^n = {# B %}^n
• Rule 3: (recursion rule - if the 3rd entry isn't 0, but there is no nester) a(a,b,c #) = a(a,b,c-1 #)x[b]
  • English: If the third entry is not 0, subtract the entry by one and add 'x[b]' in the end of the expression.
• Rule 4: (nesting-end rule - if nester and multiplier are exists and multiplier is 1) a(#)x[1] = a(#)
  • English: If the nester are exists and multiplier is 1, remove 'x[1]' in the end.
• Rule 5: (nesting rule - if nester and multiplier are exists and multiplier isn't 1) a(a,b #)x[y] = a(a,a(a,b #)x[y-1] #)
  • English: If the nester are exists and multiplier isn't 1, replace the second entry with the original array, but subtract y by one; then in the original array, remove 'x[y]' in the end.

where # and % is the part of array (a string of separators between the entries or empty string.)

If neither of 5 rules applied, start the process shown below. First, start from the third entry.

• Case A: If the entry is 0, move on to the next entry.
  • English: Else, find the third non-zero entry.
• Case B: If the entry is not 0:
  • Decrease the entry by 1.
  • Case B1: If there is the comma in the left:
    • Change the previous entry to the iterator.
    • This process ends.
  • Case B2: If there is the {0}^y in the left:
    • Define S0 and Sn as '0' and '0{Sn-1}^(y-1)1' (with y-1 {} surrounding 'Sn-1')
    • Replace the previous entry with Sb.
    • This process ends.
  • Case B3: If there is the {x #}^y in the left:
    • Define S0 and Sn as '1' and '0{x-1 #}^y Sn-1' (with y {} surrounding 'x-1 #')
    • Replace the previous entry with Sb.
    • This process ends.
  • Case B4: If there is the separator in the left and cases B2 and B3 did not applied:
    • Copy the separator to R.
    • Replace the entry before the separator with '0 R 1'.
    • Continue the process within R separator, beside the process with a(x,y,z).

Separator Levels

To compare, we need 2 functions as lv(A) and lv(B) and start at case 1. Set the comma into {0}.

• Case 1: Find the number from {}^n and call it A' and B'.
  • If A' < B', then lv(A) < lv(B) and finish.
  • If A' > B', then lv(A) > lv(B) and finish.
  • If A' = B', then go into case 2.
• Case 2: If A and B were all numbers, then use this case; else go to Case 3.
  • If A < B, then lv(A) < lv(B) and finish.
  • If A > B, then lv(A) > lv(B) and finish.
  • If A = B, then lv(A) = lv(B) and finish.
• Case 3: Find the highest separator possible and call it H. Then find the amounts of H and call it A' and B'.
  • If A' < B', then lv(A) < lv(B) and finish.
  • If A' > B', then lv(A) > lv(B) and finish.
  • If A' = B', then go into case 4.
• Case 4: Find the array preceded by the first H separator and call it A' and B'.
  • If lv({A'}) < lv({B'}), then lv(A) < lv(B) and finish.
  • If lv({A'}) > lv({B'}), then lv(A) > lv(B) and finish.
  • If lv({A'}) = lv({B'}), then remove A' and B' and the first H separator that isn't removed, then:
    • If the separator was {}, then lv(A) = lv(B) and finish.
    • Else, repeat case 4.