User blog:Ikosarakt1/Formal definition of tetrational arrays

A few notes firstly:


 * \(\#\) represents the remainder of an array.


 * Batch of unfilled separators means that it is either first of separators or the separators immediately after 1.


 * In my notation, separators are written in their natural polynomial form (without commas and nested separators), i.e. (0,1) should be written as \((X^X)\), and ((1) 2) = \((X^{(X^X)2})\).


 * The expression \(A \&\ b[p]\) means A array of a's using the prime p. If the prime is unspecified, it means that \(A \&\ b\) = \(A \&\ b[b]\).


 * For the Rule M3, rank of (A) and (B) can be determined when we put that separators on my algorithm of ranking separators (notice the similarity with Chris Bird's nested arrays).


 * The main set of rules cannot be applied until expression contains the \(\&\\) symbol.


 * Array of rules must be considered starting from the bottom exponent to the topmost.


 * A and B represents the remainder of the power tower and commonly used in the "array of" rules.


 * Like Chris Bird does, I've used inverted commas to represent the repeatedly expanding string (in the Rule A4).

Rule M1
Condition: Only 2 entries, all in the main row.

\(\{a,b\} = a^b\)

Rule M2
Condition: 2nd entry is 1.

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

Rule M3
Condition: (A) < (B) or 1 ends the array.

\(\{\# (A) 1 (B) \#\} = \{\# (B) \#\}\)

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

Rule M4
Condition: 2 entries in the main row followed by the batch of unfilled separators and non-1 entry immediately after that.

\(\{a,b (A) \cdots 1 (B) c \#\} = \{A \&\ a[b] (A) \cdots B \&\ a[b] (B) c-1 \#\}\)

Rule M5
Condition: 2 entries in the main row followed by the batch of unfilled separators, string of 1's and non-1 entry.

\(\{a,b (A) \cdots 1 (B) 1,1,...,1,1,c \#\} = \{A \&\ a[b] (A) \cdots B \&\ a[b] (B) a,a,...,a,\{a,b-1 (A) \cdots 1 (B) 1,1,...,1,1,c \#\},c-1 \#\}\)

Rule M6
Condition: string of 1's in the main row, from 3rd entry to nth followed by non-1 entry.

\(\{a,b,1,1,...,1,1,c \#\} = \{a,a,a,a,...,a,\{a,b-1,1,1,...,1,1,c \#\},c-1 \#\}

Rule M7
Condition: Rules M1-M6 don't apply.

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

Rule A1
Condition: X tops the expression at the current exponent level.

\(A^X \&\ b[p] = A^p \&\ b[p]\)

\(A*X \&\ b[p] = A*p \&\ b[p]\)

\(A+X \&\ b[p] = A+p \&\ b[p]\)

Rule A2
Condition: X is not at the first exponent level.

\(A^(X+1) \&\ b[p] = A^X*X \&\ b[p]\)

Rule A3
Condition: n tops the expression at the current exponent level.

\(A^B*n \&\ b[p] = A^B*(n-1)+A^B \&\ b[p]\)

Rule A4
Condition: (A+1) at the first exponent level.

\(X^(A+1) \&\ b[p] = 'X^A \&\ b[p] (X^A) X^A \&\ b[p] (X^A) \cdots (X^A) X^A \&\ b[p] (X^A) X^A \&\ b[p]' (with p \(X^A \&\ b[p]\)'s)\)

Step 1
There are two cases:

1) Separator in the form \(n_1*B^{A_1}+P_1\), where \(A_1\) is not a number and P is the remainder of the polynomial. Transform \(n_1*B^{A_1}+P_1 = n_1*B^{n_2*X^{A_2}+P_2}+P_1\) and repeat the Step 1.

2) Separator in the form \(n_1*B^m+P_1\), where m is a number and \(P_1\) is a remainder of the polynomial. Take \(Pol(A)=n_1*X^m+P_1\) and replace \(Pol(A)\) in the power tower by the number 1.

Step 2
Now, \(Pol(A)\) is in the form \(n_1*X^m+n_2*X^{m-1}+n_3*X^{m-2}+ \cdots +n_{m+1}\) and \(Pol(B)\) is in the form \(a_1*X^b+a_2*X^{b-1}+a_3*X^{b-2}+ \cdots +n_{b+1}\).

There are three cases:

If \(mb\), then \((A)>(B)\)

If \(m=b\), then go to the Step 3.

Step 3
Take \(x=1\) and repeat until \(xa_x\), then \((A)>(B)\)

If \(n_x<a_x\), then \((A)<(B)\)

If \(n_x=a_x\), then \(x:=x+1\)

After that, if \(x=m+1\), \(n_x=a_x\) and the number of ^'s in the power tower = 2, then (A)=(B).

Otherwise, return to the step 1.