Rearrange the digits in ⟨125034⟩ to meet the rules below.
⟨5th 4th 3rd 2nd 1st 0th⟩
✅Match
5th → 3|4|5
⟨⋯ 0 ⋯ 1 ⋯ 2 ⋯⟩
⟨ ⁵ᵗʰa ³ʳᵈb ⟩, max⟦a,b⟧ = 5
⛔Avoid
⟨ ³ʳᵈa ⁰ᵗʰb ⟩, (ab)₁₀ ≥ 21
⟨⋯ 5 ⋯ ? ⋯ 2 (?+3)⟩ (?≠2)
⟨⋯ ? ⋯ 3 ⋯ (?−2)⟩ (?≠3,5)
Jump(2,3) ≥ 1
#125034_v2.8
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ 0 │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ 5 │ 0 │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ 5 │ 0 │ 1 │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ 5 │ 0 │ 1 │ │ │ 4 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ 5 │ 0 │ 1 │ │ 3 │ 4 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 5 │ 0 │ 1 │ 2 │ 3 │ 4 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
Proof of 2025-03-04 WR
══════════════════════
Notation: if nth -> a, then we write [nth] = a.
Combining ✅「⟨⋯ 0 ⋯ 1 ⋯ 2 ⋯⟩」 with ⛔「Jump(2,3) ≥ 1」, we find the required pattern
(1) ⟨⋯ 0 ⋯ 1 ⋯ Perm(2,3) ⋯⟩.
A fortiori, there are at least three digits to the right of 0. Therefore, 0 = [5th] | [4th] | [3rd].
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ ? │ ? │ ? │ │ │ │
└───┴───┴───┴───┴───┴───┘
By ✅「5th → 3|4|5」, we have 0 != [5th].
(2) We claim that 0 = [4th] actually.
For, suppose on the contrary 0 = [3rd]. Then, it follows from (1) that
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ │ 0 │ 1 │ ▬ │ ▬ │
└───┴───┴───┴───┴───┴───┘
where "▬" are occupied by 2 and 3.
However, if it is ⟨⋯ 0123⟩, then we would match ⛔「⟨⋯ 5 ⋯ ? ⋯ 2 (?+3)⟩ (?≠2)」, while if it is ⟨⋯ 0132⟩, then we would match ⛔「⟨⋯ ? ⋯ 3 ⋯ (?−2)⟩ (?≠3,5)」. So, the assumption 0 = [3rd] is incorrect.
We have verified our claim in (2). Accordingly, we get
┌───┬───┬───┬───┬───┬───┐
│5th│ 4■│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ 0 │ │ │ │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ 5 │ │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘
Then, we consider where to place 5. By ✅「⟨ ⁵ᵗʰa ³ʳᵈb ⟩, max⟦a,b⟧ = 5」, we have 5 = [5th] or [3rd]. It is not [3rd] in view of ⛔「⟨ ³ʳᵈa ⁰ᵗʰb ⟩, (ab)₁₀ ≥ 21」. So, we get
┌───┬───┬───┬───┬───┬───┐
│ 5■│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ 0 │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ 5 │ 0 │ │ │ │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ │ │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘
Then, we consider what [3rd] is. By ⛔「⟨ ³ʳᵈa ⁰ᵗʰb ⟩, (ab)₁₀ ≥ 21」, it is 1 or 2. Combining this with (1), we get [3rd] = 1:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│ 3■│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 5 │ 0 │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ 5 │ 0 │ 1 │ │ │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ 2 │ │ │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘
Now, by (1) there are two possible cases:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
(3) │ 5 │ 0 │ 1 │ 4 │ ▬ │ ▬ │
├───┼───┼───┼───┼───┼───┤
(4) │ 5 │ 0 │ 1 │ ▬ │ ▬ │ 4 │
└───┴───┴───┴───┴───┴───┘
where "▬" are occupied by 2 and 3. If case (3) holds, then we cannot avoid ⛔「⟨⋯ 5 ⋯ ? ⋯ 2 (?+3)⟩ (?≠2)」 and ⛔「⟨⋯ ? ⋯ 3 ⋯ (?−2)⟩ (?≠3,5)」 at the same time. Therefore, case (3) does not hold and case (4) holds instead:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│ 0■│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 5 │ 0 │ 1 │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ 5 │ 0 │ 1 │ │ │ 4 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ 2 │ │ │ 3 │ │
└───┴───┴───┴───┴───┴───┘
Finally, to avoid ⛔「⟨⋯ 5 ⋯ ? ⋯ 2 (?+3)⟩ (?≠2)」, we finish by
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│ 2■│ 1■│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 5 │ 0 │ 1 │ │ │ 4 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ 5 │ 0 │ 1 │ │ 3 │ 4 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 5 │ 0 │ 1 │ 2 │ 3 │ 4 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.8