Rearrange the digits in ⟨125034⟩ to meet the rules below.
⟨5th 4th 3rd 2nd 1st 0th⟩
✅Match
4th → a, 1st → b, |a-b|=1
4th|1st → 0
⟨ ³ʳᵈa ²ⁿᵈb ⟩, (ab)₁₀ ≥ 43
⛔Avoid
2nd → a, 1st → b, ab=0
5th → a, 0th → b, |a-b|=1
⟨⋯ 4 ⋯ ? 1 ⋯ (?+2)⟩ (?≠1)
⟨⋯ ? ⋯ 5 ⋯ (?+2)⟩ (?≠3)
#125034_v2.10
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ 0 │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ 0 │ │ │ 1 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ │ 0 │ │ │ 1 │ 2 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ 5 │ 0 │ │ │ 1 │ 2 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ 5 │ 0 │ 4 │ │ 1 │ 2 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 5 │ 0 │ 4 │ 3 │ 1 │ 2 │▒
└───┴───┴───┴───┴───┴───┘▒
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Proof of 2025-09-02 WR
══════════════════════
Notation: if nth -> a, then we write [nth] = a.
Combining ✅「4th|1st → 0」 with ✅「4th → a, 1st → b, |a-b|=1」, we have:
(1) ([4th], [1st]) = (1,0) | (0,1).
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ ▬ │ │ │ ▬ │ │
└───┴───┴───┴───┴───┴───┘
If it is (1,0), then we would match ⛔「2nd → a, 1st → b, ab=0」. Therefore, it has to be (0,1).
┌───┬───┬───┬───┬───┬───┐
│5th│ 4■│3rd│2nd│ 1■│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ 0 │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ 0 │ │ │ 1 │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ 2 │ 5 │ │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘
It follows that [0th] = 5|4|3|2.
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ 0 │ │ │ 1 │ ▬ │
└───┴───┴───┴───┴───┴───┘
(2) We show that [0th] = 2 actually.
------------------------------
(2.1) If [0th] = 5:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│ 0▲│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ 0 │ │ │ 1 │ 5 │
└───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ 2 │ │ │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘
then to match ✅「⟨ ³ʳᵈa ²ⁿᵈb ⟩, (ab)₁₀ ≥ 43」, we need ([3rd], [2nd]) = (4,3):
┌───┬───┬───┬───┬───┬───┐
│5th│4th│ 3▲│ 2▲│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ 0 │ 4 │ 3 │ 1 │ 5 │
└───┴───┴───┴───┴───┴───┘
We have matched ⛔「⟨⋯ 4 ⋯ ? 1 ⋯ (?+2)⟩ (?≠1)」, which is a contradiction.
(2.2) Else if [0th] = 4:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│ 0▲│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ 0 │ │ │ 1 │ 4 │
└───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ 2 │ 5 │ │ 3 │ │
└───┴───┴───┴───┴───┴───┘
then to match ✅「⟨ ³ʳᵈa ²ⁿᵈb ⟩, (ab)₁₀ ≥ 43」, we need [3rd] = 5:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│ 3▲│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ 0 │ 5 │ │ 1 │ 4 │
└───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ 2 │ │ │ 3 │ │
└───┴───┴───┴───┴───┴───┘
To avoid ⛔「⟨⋯ ? ⋯ 5 ⋯ (?+2)⟩ (?≠3)」, we reach
┌───┬───┬───┬───┬───┬───┐
│ 5▲│4th│3rd│ 2▲│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 3 │ 0 │ 5 │ 2 │ 1 │ 4 │
└───┴───┴───┴───┴───┴───┘
However, we have matched ⛔「5th → a, 0th → b, |a-b|=1」. This shows a contradiction.
(2.3) Else if [0th] = 3:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│ 0▲│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ 0 │ │ │ 1 │ 3 │
└───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ 2 │ 5 │ │ │ 4 │
└───┴───┴───┴───┴───┴───┘
then to avoid ⛔「5th → a, 0th → b, |a-b|=1」, we need [5th] = 5:
┌───┬───┬───┬───┬───┬───┐
│ 5▲│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 5 │ 0 │ │ │ 1 │ 3 │
└───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ 2 │ │ │ │ 4 │
└───┴───┴───┴───┴───┴───┘
We cannot match ✅「⟨ ³ʳᵈa ²ⁿᵈb ⟩, (ab)₁₀ ≥ 43」 now, which is a contradiction.
------------------------------
We have verified (2). Accordingly, we get
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│ 0■│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ 0 │ │ │ 1 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ │ 0 │ │ │ 1 │ 2 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ 5 │ │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘
Then, note that there is only one way to avoid ⛔「⟨⋯ ? ⋯ 5 ⋯ (?+2)⟩ (?≠3)」:
┌───┬───┬───┬───┬───┬───┐
│ 5■│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ 0 │ │ │ 1 │ 2 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ 5 │ 0 │ │ │ 1 │ 2 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ │ │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘
Finally, to match ✅「⟨ ³ʳᵈa ²ⁿᵈb ⟩, (ab)₁₀ ≥ 43」, we finish by
┌───┬───┬───┬───┬───┬───┐
│5th│4th│ 3■│ 2■│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 5 │ 0 │ │ │ 1 │ 2 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ 5 │ 0 │ 4 │ │ 1 │ 2 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 5 │ 0 │ 4 │ 3 │ 1 │ 2 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.10