# Weak dropper-expanding notation

Weak dropper-expanding notation (WDEN) is the ninth part of my array notation.

Symbol : is called colon, with ASCII = 58.

In NDAN we say that, the colon has the highest level, and even higher than the triangle. Here I explain what’s the “:”. The colon is used as a separator, {1 : 1} = {1} by rule 2, and {1 : 2} is the triangle. In NDAN, the triangle always appears as a section, a part of a separator. But in WDEN, we can have a single {1 : 2}. e.g. s(a,b {1 : 2} 2) = s(a,b (1)(1 : 2)(1 : 2)…(1 : 2){1 : 2} 2) with b plus signs, and the (1 : 2) is exactly the circle.

The colon works as follows. When we meet a colon in the process, we just simply change {1 : 1 : … 1 : 1 : n #} into (1)(1 : 1 : … 1 : 1 : n #)(1 : 1 : … 1 : 1 : n #)……(1 : 1 : … 1 : 1 : n #){1 : 1 : … 1 : 1 : n #} with b-1 (1 : 1 : … 1 : 1 : n #)′s.

Something about levels. The colon has the highest level, so the triangle (= {1 : 2}) has lower level than it. But non-triangle separators in NDAN have no colons inside them, so they have lower level than {1 : 2}.

Finally, let {1:} be the colon. So we can go further. e.g. {1 : 2:}, {1 {2:} 2:}, {1 {1 : 2:} 2:}, etc.

# 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 colon, and {#} is a separator.
• {#:} is a separator, where # is a string such that it’s not ended with a colon, 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 colon is a shorthand for {1:}, and the comma is a shorthand for {1}.
2. The grave accent in DAN corresponds to (1){1 : 2}, the double comma corresponds to (2){1 : 2}, the triple comma corresponds to (3){1 : 2}, and so on.

For separator A, define F(A) to be the first section of A, with (#) changed into {#}.

## 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} = {#}, {# 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, Bt′s and K are parts of the original array but t, i, Si, At′s, P and Q are not. Before we start, let A0 and F(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 Bt 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(At) < lv((1)M).
4. If lv(F(At)) < lv(D(F(M))), then
1. Find the minimum of v such that v > t and lv(Av) < lv(U(M)).
2. Let string P and Q be such that Bt = “P Bv Q”.
3. Change Bt into “P X(M) Av Y(M) Q”.
4. The process ends.
5. If lv(F(At)) ≥ lv(D(F(M))), then
1. Let string P and Q be such that Bt = “P (1)M n Q”.
2. Change Bt into Sb, where b is the iterator, S1 is comma, and Si+1 = “P Si 2 (1)M n-1 Q”.
3. The process ends.
• Case B3: If a colon is immediately before you, then
1. Let t be such that the colon is are layer t. And let B0 be the whole array now.
2. Repeat this:
1. Subtract t by 1.
2. Let separator Bt be such that it’s at layer t, and the colon is inside it.
3. If t = 1, then break the repeating, or else continue repeating.
3. Find the maximum of t such that lv(At) < lv(:).
4. Let string P be such that At = {P} or At = (P)#, where # is the remaining sections.
5. Change Bt into “SbAt“, where b is the iterator, S1 = “(1)”, and Si+1 = “Si(P)”.
6. 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 At = K, here K is at layer t.
• Case B5: If an lbrace is immediately before you, then
1. Change separator {n #} into string Sb, where b is the iterator, S1 = “{n-1 #}” and Si+1 = “Si 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 Bt 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(At) < lv((n #)M).
4. If lv(F(At)) < lv(D(F(M))), then
1. Find the minimum of v such that v > t and lv(Av) < lv(U(M)).
2. Let string P and Q be such that Bt = “P Bv Q”.
3. Change Bt into “P X(M) Av Y(M) Q”.
4. The process ends.
5. If lv(F(At)) ≥ lv(D(F(M))), then
1. Change Bt into “(n-1 #)At“.
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 colon at the left-superscript position of rbrace and B hasn’t, then lv(A) > lv(B); if B has a colon at the left-superscript position of rbrace and A hasn’t, then lv(A) < lv(B); or else –
3. Let A = {a1A1a2A2…ak-1Ak-1ak} and B = {b1B1b2B2…bl-1Bl-1bl} (or A = {a1A1a2A2…ak-1Ak-1ak:} and B = {b1B1b2B2…bl-1Bl-1bl:})
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 ${M(A)=\{i\in\{1,2,\cdots,k-1\}|\forall j\in\{1,2,\cdots,k-1\}(lv(A_i)\ge lv(A_j))\}}$, and ${M(B)=\{i\in\{1,2,\cdots,l-1\}|\forall j\in\{1,2,\cdots,l-1\}(lv(B_i)\ge lv(B_j))\}}$.
7. If lv(AmaxM(A)) < lv(BmaxM(B)), then lv(A) < lv(B); if lv(AmaxM(A)) > lv(BmaxM(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 AmaxM(A) #2} and B = {#3 BmaxM(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 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 colon is immediately before you, then
1. Let string P and Q be such that A = “P : n Q”.
2. Define U(A) = “:”, D(A) = “P : n-1 Q”, X(A) = “P” and Y(A) = “2 : n-1 Q”.
• Case B2: If a separator K not the colon is immediately before you, then
1. Change the “K n” into “K 2 K n-1”.
• 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

All the dropping and expanding rules in WDEN are the same as NDAN, but case B3 in the process make it stronger.

Above {1 : 2}. There’s (1){1 : 3} – simply the 1-separator of { ____ : 2}. And then (2){1 : 3} is the 2-separator of { ____ : 2}. And further, (1,2){1 : 3}, (1 (1){1 : 2} 2){1 : 3}, (1 (1)(1 : 2){1 : 2} 2){1 : 3}, (1 (1)(1 : 2)(1 : 2){1 : 2} 2){1 : 3}, etc. these lead to (1 {1 : 2} 2){1 : 3}.

Then (1 (1){1 : 3} 2){1 : 3}, (1 (2){1 : 3} 2){1 : 3}, (1 (1,2){1 : 3} 2){1 : 3}, (1 (1 {1 : 2} 2){1 : 3} 2){1 : 3}, and so on, the limit of (1 (1,2){1 : 3} 2){1 : 3}, (1 (1 (1,2){1 : 3} 2){1 : 3} 2){1 : 3}, (1 (1 (1 (1,2){1 : 3} 2){1 : 3} 2){1 : 3} 2){1 : 3}, etc. is (1 (1)(1 : 2){1 : 3} 2){1 : 3}, where (1)(1 : 2){1 : 3} (not (1)(1 : 3){1 : 3}) is the 1-separator of ( ____ ){1 : 3}.

The limit of (1)(1 : 2){1 : 3}, (1)(1 : 2)(1 : 2){1 : 3}, (1)(1 : 2)(1 : 2)(1 : 2){1 : 3}, etc. is (1 : 2){1 : 3}. The limit of (1,2 : 2){1 : 3}, (1 (1,2 : 2){1 : 3} 2 : 2){1 : 3}, (1 (1 (1,2 : 2){1 : 3} 2 : 2){1 : 3} 2 : 2){1 : 3}, etc. is (1 (1)(1 : 3){1 : 3} 2 : 2){1 : 3}, where (1)(1 : 3){1 : 3} is the 1-separator of ( ____ : 2){1 : 3}.

And the limit of (1)(1 : 3){1 : 3}, (1)(1 : 3)(1 : 3){1 : 3}, (1)(1 : 3)(1 : 3)(1 : 3){1 : 3}, etc. is {1 : 3}. That’s what we need to get {1 : 3}.

Separator (#)(1 : 2)(1 : 2)…(1 : 2){1 : 2} always searches out for no-plus separators.
Separator (#)(1 : 3)(1 : 3)…(1 : 3){1 : 3} searches out for { ____ : 2} separators, and
separator (#)(1 : n)(1 : n)…(1 : n){1 : n} searches out for { ____ : n-1} separators.

In case B3 in the process, a separator {1 : 1 : … 1 : 1 : n #} expands to nested droppers, and it works on lower-leveled separators, so this notation is called “dropper-expanding” notation. However, that’s just a weak type of “dropper expanding”, and a strong type will come several parts later, so we call it weak dropper-expanding notation.

## 11 thoughts on “Weak dropper-expanding notation”

1. Aarex Tiaokhiao says:

(B C)(D 1:E F)G = (B C)(D 1 (B+1 C)(D 1:E F)G 2:E-1 F)G
(D 1:E F)G = (1)(D 1:E F)(D 1:E F)(D 1:E F)…G
(D 1 (D 1 (…)G 2:E F)G 2:E F)G 2:E F)G = (D 1 (1)(D 1:E+1 F)G 2:E F)G

Where G is the rest of the sections; C, D, and F are the rest of the array; and B and E are the numbers.

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2. Aarex Tiaokhiao says:

What array A, which equals into 1(1:2){1:3}B, expands into?

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• The (1)(1:2){1:3}B search out for a separator lower-leveled than it. Then (1 (1)(1:2){1:3}B 2){1:3}B expands to (1 (1 …(1 (1,2){1:3}B 2){1:3}B… 2){1:3}B 2){1:3}B.

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• Aarex Tiaokhiao says:

and what separator (1:2){1:3}?

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3. Aarex Tiaokhiao says:

Is separator {2:2} using different case than case 5B?

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• Only one of the cases applies for one separator. So for {2:2} only case B5 applies.

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• Aarex Tiaokhiao says:

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• It’s case B2. It drops down once and then expands.

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4. Aarex Tiaokhiao says:

So DEN will splits to 2 major extensions?

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• Wait for the strong one. It’ll come much later.

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• Aarex Tiaokhiao says:

Let me see….

WDEN
mWDEN
WDmEN
pDDN
sDDN
DDN

SDEN

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