Primary dropper-dropping notation

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Primary dropper-dropping notation (pDDN) is the 12th part of my array notation. In this part, you probably don’t know what “dropper dropping” means, but it’ll be explained in next several parts.

Symbol + is called plus sign, with ASCII = 43.

Here I introduce a very high-leveled separator – the plus sign. To avoid confusing between this separator and addition, we also write it as {1⁺}. Here’s how WDmEN can be embedded into pDDN. The dropper (1)² in WDmEN is (1 {1⁺} 2), (1)³ is (1 {1⁺} 3), and so on. Generally, dropper (A)ⁿ is (A {1⁺} n).

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:

• {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 it’s not ended with a plus sign, and {#} is a separator.
• {#⁺} is a separator, where # is a string such that it’s not ended with a plus sign, and {#} is a separator.
• AB is a separator, where A is a dropper and B is a separator.

Definition of droppers:

• (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:

1. The comma is a shorthand for {1}, the plus sign is a shorthand for {1⁺}.
2. The dropper for an n-separator is (n), the dropper for the colon is (1 {1⁺} 2), the dropper for the double colon is (1 {1⁺} 3), and so on.

For separator A, define F(A,n) to be the n-th sections of A, with (#) changed into {#}. Generally, if (A) is a dropper and B is a separator, then F((A)B,1) = {A} and F((A)B,n) = F(B,n-1) for n > 1.

Rules and process

• Rule 1: (base rule – only 2 entries) s(a,b) = a^b
• Rule 2: (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, B3, B5 and B6 are terminal but case A and B4 are not. b, n, M, B_t′s and K are parts of the original array but t, u, i, S_i, R_i, A_t′s, P and Q are not. Before we start, let A₀ and F(A₀,1) 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(F(A_t,1)) < lv(D(F(M,1))), 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(F(A_t,1)) ≥ lv(D(F(M,1))), 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 the plus sign (i.e. {1⁺}) is immediately before you, then
    1. Let t be such that the plus sign 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 plus sign is inside it.
       3. If t = 1, then break the repeating, or else continue repeating.
    3. Find the maximum of u such that lv(A_u) < lv(+).
    4. Find the minimum of t such that lv(F(A_u,t)) < lv(F(A_u,1)). If there’s no such t, jump to step 5a.
    5. If lv(F(A_u,t)) < lv(D(F(A_u,1))), then
       1. Let string P and Q be such that P is the first t-1 sections of A_u, and A_u = “P Q”.
       2. Change the B_u into “P X(F(A_u,1)) A_u′ Y(F(A_u,1)) Q”, where A_u′ is A_u with the first entry increased by 1.
       3. For sections of the “P X(F(A_u,1)) A_u′ Y(F(A_u,1)) Q”, if it’s not the last one, convert “{#}” into “(#)”; if it’s the last one, convert “(#)” into “{#}”.
       4. The process ends.
    6. If lv(F(A_u,t)) ≥ lv(D(F(A_u,1))), then
       1. Let string Q be such that A_u = “P Q”, where P is the first section of A_u.
       2. Change the B_u into “S_b Q”, where b is the iterator, S₁ is “(1)” and S_i+1 = “S_i R_t“, where R₁ is empty string and R_i+1 = “R_i F(A_u,i+1)”.
       3. For sections of the “S_b Q”, if it’s not the last one, convert “{#}” into “(#)”; if it’s the last one, convert “(#)” into “{#}”.
       4. The process ends.
  • Case B4: If a separator K is immediately before you, and it doesn’t fit case B1, B2 or B3, 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 B5: 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 B6: 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(F(A_t,1)) < lv(D(F(M,1))), 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(F(A_t,1)) ≥ lv(D(F(M,1))), 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. And note that the same separators have the same level. To compare levels of other separators A and B, we follow these steps.

1. 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 –
2. 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 –
3. 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.
4. 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).

Normal rules:

1. Apply rule 2 to A and B until rule 2 cannot apply any more.
2. If A has a plus sign at the left-superscript position of rbrace and B hasn’t, then lv(A) > lv(B); if B has a plus sign at the left-superscript position of rbrace and A hasn’t, then lv(A) < lv(B); or else –
3. Let A = {a₁A₁a₂A₂…a_k-1A_k-1a_k} and B = {b₁B₁b₂B₂…b_l-1B_l-1b_l} (or A = {a₁A₁a₂A₂…a_k-1A_k-1a_k⁺} and B = {b₁B₁b₂B₂…b_l-1B_l-1b_l⁺})
4. 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 5; if k > 1 and l > 1, follow step 6 ~ 11
5. If a₁ < b₁, then lv(A) < lv(B); if a₁ > b₁, then lv(A) > lv(B); if a₁ = b₁, then lv(A) = lv(B)
6. Let , and .
7. 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 –
8. If |M(A)| < |M(B)|, then lv(A) < lv(B); if |M(A)| > |M(B)|, then lv(A) > lv(B); or else –
9. Let A = {#₁ A_maxM(A) #₂} and B = {#₃ B_maxM(B) #₄}
10. If lv({#₂}) < lv({#₄}), then lv(A) < lv(B); if lv({#₂}) > lv({#₄}), then lv(A) > lv(B); or else –
11. 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

For a separator or dropper A, U(A), D(A), X(A) and Y(A) are determined as follows. Start the subprocess shown below. Process start from the first entry of A. Note that case B1 and B3 give definitions but case A and B2 don’t.

• 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 plus sign (i.e. {1⁺}) is immediately before you, then
    1. Let string P and Q be such that A = “P {1⁺} n Q”.
    2. Define  D(A) = “P {1⁺} n-1 Q”, X(A) = “P” and Y(A) = “2 {1⁺} n-1 Q”.
  • Case B2: If a separator K not the plus sign is immediately before you, then
    1. Change the “K n” into “K 2 K n-1”.
    2. Jump to the first entry of the former K.
  • Case B3: If the “(” is immediately before you, then
    1. Let string P and Q be such that A = “P (n #)M 2 Q”, where (n #)M is the separator.
    2. Define U(A) = “(n #)M”, D(A) = “P Q”, X(A) = “P” and Y(A) = “2 Q”.

Explanation

In the main process, when we meet a plus sign, first it drops down once, then we search for the leftmost section with lower level than the first section (that’s similar to WDmEN). The adding rule (step 5 of case B3) is also similar to WDmEN.