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Note: this version of the notation doesn’t work.

Nested dropper array notation (NDAN) is the eighth part of my array notation. In this part, the definition will change a lot.

Symbol △ is called triangle, with unicode = 9651 and LaTeX = \triangle
Symbol ○ is called circle, with unicode = 9675 and LaTeX = \circ

Let the triangle be a high-leveled separator. And we use a pair of parentheses to be a prefix of separators. If A is a separator, then we can have (m)A for some A, where m is a positive integer.

(1)A is a 1-separator, (2)A is a 2-separator, (3)A is a 3-separator, and so on. When this prefix is applied on the triangle, then (1)△ is the grave accent – the least 1-separator, (2)△ is the double comma – the least 2-separator, (3)△ is the triple comma – the least 3-separator, and so on.

In DAN, the (m)-separator (m)△ solves in this way: Search out for a separator A_u(m) that has lower level than it, then B_u(m) becomes (m-1)B_u(m). As a (m-1)-separator, the (m-1)B_u(m) search out for a separator A_u(m-1) that has lower level than it. If it has lower level than “X Y” (where (m-1)B_u(m) = “X (m)B_u(m+1) 2 Y” = “X (m)△ 2 Y”), then add “X Y” somewhere inside B_u(m-1). Or B_u(m-1) becomes (m-2)B_u(m-1). Continue downward, until (1)B_u(2). Finally B_u(1) = “P (1)B_u(2) Q”, then it expands.

So, an (m)A drops down to (m-1)B, and {P (1)A Q} expands. So the ()-prefix is called dropper. But in DAN, all the dropping process are inside one rule, and it can just drop finite times, meaning that we cannot have (1,2) dropper, (1`2) dropper, etc. In NDAN, we can because the “case B2” is separated into several parts. The “1,2” in (1,2)A reduces in normal comma way, and the “1`2” in (1`2)A reduces in normal way (just search out, it doesn’t matter if you’re inside a dropper).

Next, let the circle be a high-leveled dropper. We can have (1(1)△2)△ = (1`2)△, (1(2)△2)△ = (1,,2)△, (1(3)△2)△ = (1,,,2)△, (1(1,2)△2)△, (1(1(1,2)△2)△2)△, and so on. ○△ is a high-leveled separator, and (2)○△ is the 1-separator of ( ____ )△ – it’s the least 1-separator in a dropper! If A is a dropper and B is a no-prefixed separator, then (1)AB is a 1-separator of dropper, (2)AB is a 2-separator of dropper, etc. If A and B are droppers, C is a no-prefixed separator, then (1)ABC is a 1-separator of dropper of dropper, (2)ABC is a 2-separator of dropper of dropper. And (1)○○○△, (1)○○○○△, etc. – all these lead to the full nested dropper array notation.

Definition

Syntax:

s(a,b) is an array where a and b are positive integers.
s(#Am) is an array, where m is a positive integer, A is a separator, and # is a string such that s(#) is an array.

Definition of separators:

The m-ple comma (,,…,) with m ≥ 2 are separators. (basic separators)
{m} is a separator, where m is a positive integer.
{#Am} is a separator, where m is a positive integer, A is a separator, and # is a string such that {#} is a separator.
AB is a separator, where A is a dropper and B is a separator.

Definition of droppers:

The ○ is a dropper.
(m) is a dropper, where m is a positive integer.
(#Am) is a dropper, where m is a positive integer, A is a separator, and # is a string such that (#) is a dropper.

Note:

The comma (,) is a shorthand for {1}.
The grave accent ` = (1)△, the double comma ,, = (2)△, the triple comma ,,, = (3)△, and so on, the m-ple comma ,,…, = (m)△ where m ≥ 2.
When a separator A = A₁A₂…A_k-1A_k, where A₁, A₂, …, A_k-1 are dropper and A_k is a single separator (not jointed by a dropper and a separator), we say A₁, A₂, …, A_k-1, A_k are sections in separator A.

Rules and process

Rule 1: (base rule – only 2 entries) s(a,b) = a^b
Rule 2a: (tailing rule – the last entry is 1) s(# A 1) = s(#), {# A 1} = {#} and (# A 1) = (#)
Rule 3: (recursion rule – neither the 2nd nor 3rd entry is 1) s(a,b,c #) = s(a, s(a,b-1,c #) ,c-1 #)

where # is a string of entries and separators, it can also be empty. A is a separator.

If none of the 3 rules above applies, start the process shown below. Note that case B1, B2, B4 and B5 are terminal but case A and B3 are not. b, n, M, B_t′s and K are parts of the original array but t, i, S_i, A_t′s, P and Q are not. Before we start, let A₀ be the whole array. First start from the 3rd entry.

Case A: If the entry is 1, then you jump to the next entry.
Case B: If the entry n is not 1, look to your left:
- Case B1: If the comma is immediately before you, then
  1. Change the “1,n” into “b,n-1” where n is this non-1 entry and the b is the iterator.
  2. Change all the entries at base layer before them into the base.
  3. The process ends.
- Case B2: If a separator (1)M (where M is the remaining sections) is immediately before you, then
  1. Let t be such that the (1)M is are layer t. And let B₀ be the whole array now.
  2. Repeat this:
    1. Subtract t by 1.
    2. Let separator B_t be such that it’s at layer t, and the (1)M is inside it.
    3. If t = 1, then break the repeating, or else continue repeating.
  3. Find the maximum of t such that lv(A_t) < lv((1)M).
  4. If lv(A_t) < lv(D(M)), then
    1. Find the minimum of v such that v > t and lv(A_v) < lv(U(M)).
    2. Let string P and Q be such that B_t = “P B_v Q”.
    3. Change B_t into “P X(M) A_v Y(M) Q”.
    4. The process ends.
  5. If lv(A_t) ≥ lv(D(M)), then
    1. Let string P and Q be such that B_t = “P (1)M n Q”.
    2. Change B_t into S_b, where b is the iterator, S₁ is comma, and S_i+1 = “P S_i 2 (1)M n-1 Q”.
    3. The process ends.
- Case B3: If a separator K is immediately before you, and it doesn’t fit case B1 or B2, then
  1. Change the “K n” into “K 2 K n-1”.
  2. Set separator A_t = K, here K is at layer t.
  3. Jump to the first entry of the former K.
- Case B4: If an lbrace is immediately before you, then
  1. Change separator {n #} into string S_b, where b is the iterator, S₁ = “{n-1 #}” and S_i+1 = “S_i 1 {n-1 #}”.
  2. The process ends.
- Case B5: If a “(” is immediately before you, then
  1. Let t be such that the separator (n #)M is are layer t, here (n #) is the first section and M is the remaining sections. And let B₀ be the whole array now.
  2. Repeat this:
    1. Subtract t by 1.
    2. Let separator B_t be such that it’s at layer t, and the (n #)M is inside it.
    3. If t = 1, then break the repeating, or else continue repeating.
  3. Find the maximum of t such that lv(A_t) < lv((n #)M).
  4. If lv(A_t) < lv(D(M)), then
    1. Find the minimum of v such that v > t and lv(A_v) < lv(U(M)).
    2. Let string P and Q be such that B_t = “P B_v Q”.
    3. Change B_t into “P X(M) A_v Y(M) Q”.
    4. The process ends.
  5. If lv(A_t) ≥ lv(D(M)), then
    1. Change B_t into “(n-1 #)A_t“.
    2. The process ends.

Level comparison

All arrays have the lowest and the same level. Then the “+” has the highest level, the triangle has the second highest level. And note that the same separators have the same level. To compare levels of other separators A and B, we follow these steps.

Compare levels of the last section of A and B using normal rules. If lv(last section of A) < lv(last section of B), then lv(A) < lv(B); if lv(last section of A) > lv(last section of B), then lv(A) > lv(B); or else –
If A has only 1 section but B has more than 1 sections, then lv(A) > lv(B); if A has more than 1 sections but B has only 1 section, then lv(A) < lv(B); if both A and B have only 1 section, then lv(A) = lv(B); or else –
Let A = “#₁ (P) M” where M is the last section and (P) is the second last section, B = “#₂ (Q) N” where N is the last section and (Q) is the second last section.
If lv(“#₁ {P}”) < lv(“#₂ {Q}”), then lv(A) < lv(B); if lv(“#₁ {P}”) > lv(“#₂ {Q}”), then lv(A) > lv(B); if lv(“#₁ {P}”) = lv(“#₂ {Q}”), then lv(A) = lv(B). Note that if (P) (or (Q)) is the circle, then {P} (or {Q}) is the triangle.

Normal rules:

Apply rule 2 to A and B until rule 2 cannot apply any more.
Let A = {a₁A₁a₂A₂…a_k-1A_k-1a_k} and B = {b₁B₁b₂B₂…b_l-1B_l-1b_l}
If k = 1 and l > 1, then lv(A) < lv(B); if k > 1 and l = 1, then lv(A) > lv(B); if k = l = 1, follow step 4; if k > 1 and l > 1, follow step 5 ~ 10
If a₁ < b₁, then lv(A) < lv(B); if a₁ > b₁, then lv(A) > lv(B); if a₁ = b₁, then lv(A) = lv(B)
Let ${M(A)=\{i\in\{1,2,\cdots,k-1\}|\forall j\in\{1,2,\cdots,k-1\}(lv(A_i)\ge lv(A_j))\}}$ , and ${M(B)=\{i\in\{1,2,\cdots,l-1\}|\forall j\in\{1,2,\cdots,l-1\}(lv(B_i)\ge lv(B_j))\}}$ .
If lv(A_maxM(A)) < lv(B_maxM(B)), then lv(A) < lv(B); if lv(A_maxM(A)) > lv(B_maxM(B)), then lv(A) > lv(B); or else –
If |M(A)| < |M(B)|, then lv(A) < lv(B); if |M(A)| > |M(B)|, then lv(A) > lv(B); or else –
Let A = {#₁ A_maxM(A) #₂} and B = {#₃ B_maxM(B) #₄}
If lv({#₂}) < lv({#₄}), then lv(A) < lv(B); if lv({#₂}) > lv({#₄}), then lv(A) > lv(B); or else –
If lv({#₁}) < lv({#₃}), then lv(A) < lv(B); if lv({#₁}) > lv({#₃}), then lv(A) > lv(B); if lv({#₁}) = lv({#₃}), then lv(A) = lv(B)

Subfunctions

4 subfunctions appear in the process: U, D, X and Y function. The inputs of them are separators, the outputs of U and D are separators, and the outputs of X and Y are strings.

The primitive definitions of U, D, X and Y are as follows. If A = ○○…○△ (there also can be no circles), then

U(A) = “+”
D(A) = “,”
X(A) = “{1”
Y(A) = “2}”

For other A such that the first section of A is neither △ nor ○, U(A), D(A), X(A) and Y(A) are determined in those steps:

Start the process, and start from the first entry of B. But we use case A, B3 and B5 only. Until we meet case B5.
When we meet case B5, don’t go into it. Let string P and Q be such that A = “P (n #)M 2 Q”.
Define U(A) = “(n #)M”, D(A) = “P Q”, X(A) = “P” and Y(A) = “2 Q”.

Explanation

In the process, the case B2 is the dropping and expanding of 1-separators. The case B5 is the dropping of higher-separators. The four functions, U, D, X and Y are the linking between different “drop down process”.

Since U(○○…○△) = “+”, when it drops in case B5, and we jump into step 4, “lv(A_v) < lv(U(M))” is always true, so v = t+1.

Pay attention to the step 2 of the 4 new rules in level comparisons. It means that “(P)M” always have lower level than M. In step 5 of case B5 in the process, the B_t becomes “(n-1 #)A_t“, so the level of it reduces.

Since we separate the dropping process into several parts, we’ll meet case B3 more often. Every time we meet case B3, it pulls the K out, so NDAN is not exactly the same as DAN even below (1,2)△. The simplest example, s(a,b{1{1(2)△2}3}2) = s(a,b{1(1){1(2)△2}2{1(2)△2}2}2) = s(a,b{1…{1{1,2(1){1(2)△2}1{1(2)△2}2}2(1){1(2)△2}1{1(2)△2}2}…2(1){1(2)△2}1{1(2)△2}2}2) with b-1 (1){1(2)△2}′s. After many steps, s(a,b{1(1){1(2)△2}1{1(2)△2}2}2) = s(a,b{1(1){1(2)△2}1(1){1(2)△2}2{1(2)△2}1}2) = s(a,b{1(1){1(2)△2}1(1){1(2)△2}2}2), and it has growth rate ${\varphi(2,0)}$ , so s(n,n{1{1(2)△2}3}2) has growth rate ${\varepsilon_{\varphi(2,0)+1}}$ , while s(n,n{1{1,,2}3}2) has growth rate ${\varepsilon_1}$ in DAN. But both s(n,n{1{1,2(2)△2}2}2) and s(n,n{1{1,2,,2}3}2) have growth rate ${\varphi(\omega^\omega,0)}$ , so this difference doesn’t matter the final strength of DAN and NDAN.