Dropping array notation

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Dropping array notation (DAN) is the seventh part of my array notation.

Let’s have an overview of EAN, pDAN and sDAN. In EAN, the grave accent works as the least 1-separator, and it drops down once until the expansion. In pDAN, the double comma works as the least 2-separator, and it drops down twice until the expansion. In sDAN, the triple comma works as the least 3-separator, and it drops down 3 times until the expansion.

Continue upward, we have the least 4-separator – the quadruple comma (,,,,), the least 5-separator – the quintuple comma (,,,,,), and so on. Generally, an m-separator drops down m times until the expansion. Here comes the definition of dropping array notation.



  • 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.


  1. The comma (,) is a shorthand for {1}.
  2. The dot . = {1{1,,2}2}, the grave accent ` = {1,,2}, the double comma ,, = {1,,,2}, the triple comma ,,, = {1,,,,2}, and so on, the m-ple comma ,,…, = {1,,…,,2} with an (m+1)-ple comma inside.

Rules and process

  • Rule 1: (base rule – only 2 entries) s(a,b) = a^b
  • Rule 2:(recursion rule – neither the 2nd nor 3rd entry is 1) s(a,b,c #) = s(a, s(a,b-1,c #) ,c-1 #)
  • Rule 3: (tailing rule – the last entry is 1) s(# A 1) = s(#) and {# A 1} = {#}
  • Rule 4: (if lv(A) < lv(B)) {# A 1 B #′} = {# B #′}

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

If none of the 4 rules above applies, start the process shown below. Note that case B1, B2 and B4 are terminal but case A and B3 are not. b, n, m, M, Bu′s and K are parts of the original array but j, t, u(j), v(j), i, Si, Au′s, X, Y, 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 the m-ple comma (m ≥ 2) M is immediately before you, then
      1. Change “M n” into “M 2 M n-1”, and let Bu(m+1) be the former M.
      2. Let t be such that the M is at layer t. And let B0 be the whole array now.
      3. Repeat this:
        1. Subtract t by 1.
        2. Let separator Bt be such that it’s at layer t, and the M is inside it.
        3. If t = 1, then break the repeating, or else continue repeating.
      4. Find the maximum of u(m) such that lv(Au(m)) < lv(M).
      5. If lv(Au(m)) < lv(,), then
        1. Let string P and Q be such that Bu(m) = “P Bu(m)+1 Q”.
        2. Change Bu(m) into “P {1 Au(m)+1 2} Q”.
        3. The process ends.
      6. Set j = m. Set Au(m+1) = M.
      7. Repeat this:
        1. Subtract j by 1.
        2. Find the maximum of u(j) such that lv(Au(j)) < lv(Au(j+1)).
        3. Let string X and Y be such that Bu(j+1) = “X Bu(j+2) 2 Y”.
        4. If lv(Au(j)) < lv(“X Au(j+1) 2 Y”), then
          1. Find the minimum of v(j) such that v(j) > u(j) and lv(Av(j)) < lv(Au(j+2)).
          2. Let string P and Q be such that Bu(j) = “P Bv(j) Q”.
          3. Change Bu(j) into “P X Av(j) 2 Y Q”.
          4. The process ends.
        5. If j = 1, then break the repeating, or else continue repeating.
      8. Let string P and Q be such that Bu(1) = “P Bu(2) Q”.
      9. Change Bu(1) into Sb, where b is the iterator, S1 is comma, and Si+1 = “P Si Q”.
      10. 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 At = 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 Sb, where b is the iterator, S1 = “{n-1 #}” and Si+1 = “Si 1 {n-1 #}”.
      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.

First some preparation.

  1. Find the maximum of m that there’re some m-ple commas in the expressions of A or B.
  2. Convert all the multiple comma of A and B into expressions of m-ple comma, using ,,…, (i commas) = {1,,…,,2} (i+1 commas)

Then the m-ple comma has the highest level, and other separators have level lower than it.

  1. Apply rule 2 to A and B until rule 2 cannot apply any more.
  2. Let A = {a1A1a2A2…ak-1Ak-1ak} and B = {b1B1b2B2…bl-1Bl-1bl}
  3. 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
  4. If a1 < b1, then lv(A) < lv(B); if a1 > b1, then lv(A) > lv(B); if a1 = b1, then lv(A) = lv(B)
  5. 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))\}}.
  6. 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 –
  7. If |M(A)| < |M(B)|, then lv(A) < lv(B); if |M(A)| > |M(B)|, then lv(A) > lv(B); or else –
  8. Let A = {#1 AmaxM(A) #2} and B = {#3 BmaxM(B) #4}
  9. If lv({#2}) < lv({#4}), then lv(A) < lv(B); if lv({#2}) > lv({#4}), then lv(A) > lv(B); or else –
  10. 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)


The At in case B3 and the Bt in step 1 ~ 3 of case B2 are preparation. Step 4 and 5 are where the m-ple comma (the least m-separator) drops down to an (m-1)-separator. Step 7 is where the (m-1)-separator drops down m-1 times, to a 0-separator, and step 6 is a preparation for it. And step 8 ~ 10 are the expansion. Here, the Au(m+1) = M is an m-separator, and the Au(j) is a (j-1)-separator.

The step 5 and step 7d are where the adding rules apply. Once an adding rule apply, the expansion doesn’t apply; only when all the adding rules don’t apply, the expansion applies.

The level comparison rules also change. The preparation step change A and B into such that the m-ple comma is the only “basic separator”.


One thought on “Dropping array notation

  1. Aarex Tiaokhiao says:

    ,, corresponds to M.
    ,,, corresponds to K.
    ,,,, corresponds to Pi^1_2.
    So the limit ordinal is psi(Pi^1_w).



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