Alphabet notation

Alphabet notation is a notation created by Wikia user Nirvana Supermind.. It is a based on recursion, and inputs a string of English letters. It has only this part currently:

• Basic Alphabet Notation

The creator clarifies that he or she intends to create at least six other parts:

• Basic Cascading Alphabet Notation
• Nested Basic Cascading Alphabet Notation
• Two-level Cascading Alphabet Notation
• Cascading Alphabet notation
• Tetrational Alphabet Notation
• Arrow Alphabet Notation

Note that the creator is using (single) capital letters to represent variables, and single small letters to mean the actual letters. For example, "A" in this article means a variable instead of the capital 'A', while "a" in this article means the actual letter 'a'.

Basic Alphabet Notation

Current definition

All information here only applies to the current version of the notation, which is given after the issues on the original definition were pointed out.

The expressions in this notation are of this form:

(ABCDEFGHIJKLMN…)

Here the “ABCDEFGHIJKLMN…” are a sequence of small letters in the Latin alphabet. The wrapping braces are simply to distinguish the expressions from actual words. () is also a valid expression. An example of a valid expression is (abc).

Terminology because they make the definition easier to write:

1. ord(A) for the letter A is defined as 1 if A = "a" 2 if A = "b" 3 if A = "c" 4 if A = "d" etc.
2. len(A) for the expression A is defined as is the number of letters in A.
3. P(A) for the integer A is defined as the Ath prime (zero-based index). So P(0) = 2.

Note that P(A) is ill-defined when A = -1. Every expression output a large number. To solve a (possibly empty) expression, we need some rules as follows:

1. () = 1
2. (#A) = (#)P(len(#))^ord(A)

Here # denotes a substring of the current expression. It can also be empty. If there are multiple rules to apply to a single expression, the uppermost-numbered rule which is applicable and whose result is a valid expression will be applied.

Examples

(abc) = 2250

Original definition

All information here only applies to the old version of the notation.

The expressions in this notation are of this form:

(ABCDEFGHIJKLMN…)

Here the “ABCDEFGHIJKLMN…” are a sequence of small letters in the Latin alphabet. The wrapping braces are simply to distinguish the expressions from actual words. () is also a valid expression. An example of a valid expression is (abc).

Terminology because they make the definition easier to write:

1. ord(A) for the letter A is defined as 1 if A = "a" 2 if A = "b" 3 if A = "c" 4 if A = "d" etc.
2. len(A) for the expression A is defined as is the number of letters in A.
3. P(A) for the integer A is defined as the Ath prime (zero-based index). So P(0) = 2.

Note that P(A) is ill-defined when A = -1. Every expression is intended to output a large number. To solve a (possibly empty) expression, we need some rules as follows:

1. () = 1
2. (#A) = (#)P(len(#)+1)^ord(A)

Here # denotes a substring of the current expression. It can also be empty. If there are two or more distinct rules to apply to a single expression, the uppermost-numbered rule which is applicable and whose result is a valid expression will be applied. Although it is not clarified, "A" in rule 2 is a variable which means a single small letter rather than a valid expression, because the creator considers ord(A). Readers should be careful that the creator uses "A" also as variables for a valid expression and an integer, as the defitions of len and P show.

Issues

The description "If there are two or more distinct rules to apply to a single expression, the uppermost-numbered rule which is applicable and whose result is a valid expression will be applied." is weird, because there is no valid expression to which two rules are applicable. Moreover, there are no more rules, while the creator expresses "two or more rules".

Moreover, (abc) is intended to coincide with 2250, according to the creator. However, the actual value should be computed in the following way:

(abc) = (ab)P(len(ab)+1)^ord(c) = (ab)P(2+1)³ = (ab)7³ = (a)P(len(a)+1)^ord(b)7³

= (a)P(1+1)²7³ = (a)5²7³ = ()P(len()+1)^ord(a)5²7³ = ()P(0+1)¹5²7³

= ()3¹5²7³ = 1×3¹5²7³ = 25725

Therefore the original definition is not compatible with the intended behaviour. However, the new definition has fixed some of these issues, and is compatible with the intended value.

Examples

Intended Value

(abc) = 2250

Actual Value

(abc) = 25725

P進大好きbot's definition

User:P進大好きbot created a fixed definition of the notation, before the creator updated their definition. It is quite elementary to solve the issues of the old notation: We have only to remove the weird description of the application of two or more rules and change rule 2. In order to make the solution clearer, we explain the precise alternative formulation.

Let \(\mathbb{N}\) denote the set of non-negative integers, and \(T\) the set of formal strings consisting of small letters in the Latin alphabet. For an \(A \in T\), we denote by \(\textrm{len}(A)\) the length of \(A\). For an \(\alpha \in T\) of length \(1\), we denote by \(\textrm{ord}(\alpha)\) the positive integer corresponding to the ordinal numeral of α with respect to the usual ordering of small letters in the Latin alphabet. For example, we have \(\textrm{ord}(a) = 1\), \(\textrm{ord}(b) = 2\), and \(\textrm{ord}(c) = 3\). For an \(n \in \mathbb{N}\), we denote by \(P(n)\) the \((1+n)\)-th prime number. For example, we have \(P(0) = 2\), \(P(1) = 3\), and \(P(2) = 5\).

We define a total computable function \begin{eqnarray*} () \colon T & \to & \mathbb{N} \\ A & \mapsto & (A) \end{eqnarray*} in the following recursive way:

1. If \(\textrm{len}(A) = 0\), then set \((A) := 1\).
2. Suppose \(\textrm{len}(A) \neq 0\).
   1. Denote by \(\alpha \in T\) the formal string of length \(1\) given as the rightmost letter of \(A\).
   2. Denote by \(B \in T\) the formal string given by removing the right most letter from \(A\).
   3. Set \((A) := (B)P(\textrm{len}(B))^{\textrm{ord}(\alpha)}\).

The totality follows from the induction on \(\textrm{len}(A)\), and we have \begin{eqnarray*} (abc) & = & (ab)P(\textrm{len}(ab))^{\textrm{ord}(c)} = (ab)P(2)^3 = (ab)5^3 = (a)P(\textrm{len}(a))^{\textrm{ord}(b)}5^3 \\ & = & (a)P(1)^2 5^3 = (a)3^2 5^3 = ()P(\textrm{len}())^{\textrm{ord}(a)}3^2 5^3 = ()P(0)^1 3^2 5^3 \\ & = & ()2^1 3^2 5^3 = 1 \cdot 2^1 3^2 5^3 = 2250, \end{eqnarray*} which is compatible with the creator's intention.

Another possible alternative definition is given by replacing \(P\) by the enumeration of prime numbers with respect to one-based indexing, i.e. \(P(1) = 2\), \(P(2) = 3\), and \(P(3) = 5\), instead of changing rule 2.

Growth rate

It's hard to compare the asymptotic growth rate of this notation with the other notations like Fast-growing hierarchy, mainly because the limit function of them can be easily computed, while this one cannot as it inputs strings rather than numbers. The creator used the of Fibonacci word to convert numbers to strings. Using the highest possible letters, the Fibonacci word function S is given by the following recursive definition according to the creator:

S(0)='y',S(1)='z',S(n) = S(n-1)S(n-2)

For example, S(2)='yz'. The creator claims that the function (S(n)) has double-exponential growth rate, because they tested the ratios (S(n))/(S(n-1)) for values 1-14.

An alternative limit function, also provided by User:P進大好きbot, is given by the map assigning to each \(n \in \mathbb{N}\) the value \((A(n))\) of the formal string \(A(n)\) of length \(n\) consisting of \(z\). It is quite elementary to show an upperbound, thanks to Bertrand's postulate.

Proposition
For any \(n \in \mathbb{N}\), \((A(n)) \leq 10^{4n(n+1)}\) holds with respect to the alternative definition.

Proof

We show the assertion by the induction on \(n\). If \(n = 0\), then we have \((A(n)) = () = 1 = 10^{4n(n+1)}\). Suppose \(n > 0\). We have

\begin{eqnarray*} & & (A(n)) = (A(n-1))P(\textrm{len}(A(n-1)))^{\textrm{ord}(z)} = (A(n-1))P(n-1)^{26} \\ & \leq & (A(n-1)) (2^n)^{26} < ((A(n-1)) 10^{8n} < 10^{4(n-1)n} \times 10^{8n} \\ & = & 10^{4n(n+1)}. \end{eqnarray*}

Similarly, it is possible to obtain a lowerbound of the value for the formal string of length \(n\) consisting of \(a\). In this sense, the asymptotic growth rate of the limit function can be easily estimated.