CompactStar (talk  contribs) Tag: Source edit 
m (Reverting fake removements of related descriptions. Changing the section order in order to solve the fake description "all information blahblah". Removing math tags. I will check later whether there are further change, as the edits were done when I am editting.) 

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== Basic Alphabet Notation == 
== Basic Alphabet Notation == 

+  Since the current version shares one issue with the original version, we explain the original definition first. 

−  === Current definition === 

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=== Original definition === 
=== Original definition === 

−  +  Several information here only applies to the old version of the notation. 

The expressions in this notation are of this form: 
The expressions in this notation are of this form: 

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: = (a)P(1+1)<sup>2</sup>7<sup>3</sup> = (a)5<sup>2</sup>7<sup>3</sup> = ()P(len()+1)<sup>ord(a)</sup>5<sup>2</sup>7<sup>3</sup> = ()P(0+1)<sup>1</sup>5<sup>2</sup>7<sup>3</sup> 
: = (a)P(1+1)<sup>2</sup>7<sup>3</sup> = (a)5<sup>2</sup>7<sup>3</sup> = ()P(len()+1)<sup>ord(a)</sup>5<sup>2</sup>7<sup>3</sup> = ()P(0+1)<sup>1</sup>5<sup>2</sup>7<sup>3</sup> 

: = ()3<sup>1</sup>5<sup>2</sup>7<sup>3</sup> = 1×3<sup>1</sup>5<sup>2</sup>7<sup>3</sup> = 25725 
: = ()3<sup>1</sup>5<sup>2</sup>7<sup>3</sup> = 1×3<sup>1</sup>5<sup>2</sup>7<sup>3</sup> = 25725 

−  Therefore the original definition is not compatible with the intended behaviour. 
+  Therefore the original definition is not compatible with the intended behaviour. Since the creator tends to change either one of the original definition or the intended value, it is not special for a notation by the creator. For examples of the differences of actual values and intended values of notations by the creator, see [[Extensible Illion System#Example]] and [[Infra Notation#Example 3]]. 
−  
−  However, the new definition has fixed some of these issues, and is compatible with the intended value. 

−  
==== Examples ==== 
==== Examples ==== 

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(abc) = 25725 
(abc) = 25725 

−  === 
+  === Alternative definition === 
−  +  It is quite elementary to solve the issues explained above: 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 
+  Let \(\mathbb{N}\) denote the set of nonnegative 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 
We define a total computable function 

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\end{eqnarray*} 
\end{eqnarray*} 

in the following recursive way: 
in the following recursive way: 

−  # If 
+  # If \(\textrm{len}(A) = 0\), then set \((A) := 1\). 
−  # Suppose 
+  # Suppose \(\textrm{len}(A) \neq 0\). 
−  ## Denote by 
+  ## Denote by \(\alpha \in T\) the formal string of length \(1\) given as the rightmost letter of \(A\). 
−  ## Denote by 
+  ## Denote by \(B \in T\) the formal string given by removing the right most letter from \(A\). 
−  ## Set 
+  ## Set \((A) := (B)P(\textrm{len}(B))^{\textrm{ord}(\alpha)}\). 
−  The totality follows from the induction on 
+  The totality follows from the induction on \(\textrm{len}(A)\), and we have 
\begin{eqnarray*} 
\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 \\ 
(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 \\ 

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which is compatible with the creator's intention. 
which is compatible with the creator's intention. 

−  Another possible alternative definition is given by replacing 
+  Another possible alternative definition is given by replacing \(P\) by the enumeration of prime numbers with respect to onebased indexing, i.e. \(P(1) = 2\), \(P(2) = 3\), and \(P(3) = 5\), instead of changing rule 2. 
−  === 
+  === Current definition === 
−  It's hard to compare the asymptotic growth rate of this notation with the other notations like [[Fastgrowing 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 {{wFibonacci 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: 

+  After the issues in the original definition are pointed out in [[#IssuesIssues]] secition, the creator updated the definition following the second alternative definition mentioned to at the bottom of [[#Alternative definitionAlternative definition]] section. 

−  S(0)='y',S(1)='z',S(n) = S(n1)S(n2) 

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+  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 uppermostnumbered 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 issue on the weird explanation on the possibility of the application of two or more rules remains, as the creator just rephrased it as '''If there are multiple rules to apply to a single expression, the uppermostnumbered rule which is applicable and whose result is a valid expression will be applied.''' 

+  
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+  === Growth rate === 

+  The following is the analysis by the creator in the original source:<ref name=":0" /> 

+  : ''It’s hard to compare the asymptotic growth rate of this notation with the other notations like [[Fastgrowing hierarchy]], mainly because they input lists of numbers rather than letters. However, we can use [https://en.wikipedia.org/wiki/Fibonacci_word fibonacci word] function “S” to convert a number into a string of letters. After I tried some numbers, it looks like (S(N)) is likely asymptotic to f<sup>2</sup>(f<sup>2</sup>(N)) (doubleexponential growth rate) in the FGH. That’s it for this part of the notation.'' 

+  However, there is no written explicit definition of S or f<sup>2</sup> in the source, the meaning is ambiguous. 

+  
⚫  Instead, 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. 

{ class="wikitable" 
{ class="wikitable" 

! Proposition 
! Proposition 

 
 

−   For any 
+   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 
+  : 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*} 
\begin{eqnarray*} 

& & (A(n)) = (A(n1))P(\textrm{len}(A(n1)))^{\textrm{ord}(z)} = (A(n1))P(n1)^{26} \\ 
& & (A(n)) = (A(n1))P(\textrm{len}(A(n1)))^{\textrm{ord}(z)} = (A(n1))P(n1)^{26} \\ 

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\end{eqnarray*} 
\end{eqnarray*} 

−  Similarly, it is possible to obtain a lowerbound of the value for the formal string of length 
+  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. 
+  
+  Later, the creator updated the descripttion. 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(n1)S(n2). 

+  
⚫  For example, S(2)='yz'. The creator claims that the function (S(n)) has doubleexponential growth rate, because they tested the ratios (S(n))/(S(n1)) for values 114.<ref name="current" /> It is obvious that the resulting function (S(n)) is bounded by \((A(n))\), because \(\textrm{ord}(y) = 25 < 26 = \textrm{ord}(z)\). 

+  
== Sources == 
== Sources == 

<references /> 
<references /> 

+  [[Category:Numbers by Nirvana Supermind]] 

[[Category:Notations]] 
[[Category:Notations]] 

[[Category:Functions]] 
[[Category:Functions]] 
Revision as of 23:04, 13 January 2021
Alphabet notation is a notation created by Wikia user Nirvana Supermind.^{[1]}^{[2]}^{[3]}. 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:^{[2]}
 Basic Cascading Alphabet Notation
 Nested Basic Cascading Alphabet Notation
 Twolevel 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
Since the current version shares one issue with the original version, we explain the original definition first.
Original definition
Several 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:
 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.
 len(A) for the expression A is defined as is the number of letters in A.
 P(A) for the integer A is defined as the Ath prime (zerobased index). So P(0) = 2.
Note that P(A) is illdefined 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
 (#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 uppermostnumbered 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 uppermostnumbered rule which is applicable and whose result is a valid expression will be applied."^{[2]} 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.^{[2]} 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)7^{3} = (a)P(len(a)+1)^{ord(b)}7^{3}
 = (a)P(1+1)^{2}7^{3} = (a)5^{2}7^{3} = ()P(len()+1)^{ord(a)}5^{2}7^{3} = ()P(0+1)^{1}5^{2}7^{3}
 = ()3^{1}5^{2}7^{3} = 1×3^{1}5^{2}7^{3} = 25725
Therefore the original definition is not compatible with the intended behaviour. Since the creator tends to change either one of the original definition or the intended value, it is not special for a notation by the creator. For examples of the differences of actual values and intended values of notations by the creator, see Extensible Illion System#Example and Infra Notation#Example 3.
Examples
Intended Value
(abc) = 2250
Actual Value
(abc) = 25725
Alternative definition
It is quite elementary to solve the issues explained above: 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 nonnegative 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:
 If \(\textrm{len}(A) = 0\), then set \((A) := 1\).
 Suppose \(\textrm{len}(A) \neq 0\).
 Denote by \(\alpha \in T\) the formal string of length \(1\) given as the rightmost letter of \(A\).
 Denote by \(B \in T\) the formal string given by removing the right most letter from \(A\).
 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 onebased indexing, i.e. \(P(1) = 2\), \(P(2) = 3\), and \(P(3) = 5\), instead of changing rule 2.
Current definition
After the issues in the original definition are pointed out in Issues secition, the creator updated the definition following the second alternative definition mentioned to at the bottom of Alternative definition section.
Several 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:
 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.
 len(A) for the expression A is defined as is the number of letters in A.
 P(A) for the integer A is defined as the Ath prime (zerobased index). So P(0) = 2.
Note that P(A) is illdefined when A = 1. Every expression output a large number. To solve a (possibly empty) expression, we need some rules as follows:
 () = 1
 (#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 uppermostnumbered 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 issue on the weird explanation on the possibility of the application of two or more rules remains, as the creator just rephrased it as If there are multiple rules to apply to a single expression, the uppermostnumbered rule which is applicable and whose result is a valid expression will be applied.
Examples
(abc) = 2250
Growth rate
The following is the analysis by the creator in the original source:^{[2]}
 It’s hard to compare the asymptotic growth rate of this notation with the other notations like Fastgrowing hierarchy, mainly because they input lists of numbers rather than letters. However, we can use fibonacci word function “S” to convert a number into a string of letters. After I tried some numbers, it looks like (S(N)) is likely asymptotic to f^{2}(f^{2}(N)) (doubleexponential growth rate) in the FGH. That’s it for this part of the notation.
However, there is no written explicit definition of S or f^{2} in the source, the meaning is ambiguous.
Instead, 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(n1))P(\textrm{len}(A(n1)))^{\textrm{ord}(z)} = (A(n1))P(n1)^{26} \\ & \leq & (A(n1)) (2^n)^{26} < ((A(n1)) 10^{8n} < 10^{4(n1)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.
Later, the creator updated the descripttion. 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(n1)S(n2).
For example, S(2)='yz'. The creator claims that the function (S(n)) has doubleexponential growth rate, because they tested the ratios (S(n))/(S(n1)) for values 114.^{[3]} It is obvious that the resulting function (S(n)) is bounded by \((A(n))\), because \(\textrm{ord}(y) = 25 < 26 = \textrm{ord}(z)\).
Sources
 ↑ Nirvana Supermind. Alphabet Notation. (Retrieved at UTC 12:00 13/01/2020)
 ↑ ^{2.0} ^{2.1} ^{2.2} ^{2.3} ^{2.4} Nirvana Supermind. Basic Alphabet Notation. (Retrieved at UTC 12:00 13/01/2020)
 ↑ ^{3.0} ^{3.1} Nirvana Supermind. Basic Alphabet Notation. (Retrieved at Wed, 13 Jan 2021 22:44:39 GMT)