Rearrange the digits in ⟨125034⟩ to meet the rules below.
⟨5th 4th 3rd 2nd 1st 0th⟩
✅Match
5th|2nd → 0
Jump(2,4) = 2
⛔Avoid
⟨ ▧ ▧ ▧ ▢ ▢ ▢ ⟩, Σ▧ ≤ Σ▢
⟨⋯ a ⋯ 1 ⋯⟩, a = 2|4|5
⟨ ⁵ᵗʰa ⁰ᵗʰb ⟩, (ab)₁₀ ≤ 30
#125034_v2.9
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ 1 │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ 1 │ │ 0 │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ 3 │ 1 │ │ 0 │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ 3 │ 1 │ │ 0 │ 5 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ 3 │ 1 │ 4 │ 0 │ 5 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 3 │ 1 │ 4 │ 0 │ 5 │ 2 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
Proof of 2025-07-29 WR
══════════════════════
Notation: if nth -> a, then we write [nth] = a.
To avoid ⛔「⟨⋯ a ⋯ 1 ⋯⟩, a = 2|4|5」, we need
(1) 1 is to the left of 2,4,5.
This implies
(2) 1 = [5th] | [4th] | [3rd].
If 1 = [5th], then we cannot avoid ⛔「⟨ ⁵ᵗʰa ⁰ᵗʰb ⟩, (ab)₁₀ ≤ 30」. Else if 1 = [3rd]:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│ 3▲│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ │ 1 │ │ │ │
└───┴───┴───┴───┴───┴───┘
then we cannot match (1) and ✅「Jump(2,4) = 2」 at the same time. Therefore, 1 = [4th]:
┌───┬───┬───┬───┬───┬───┐
│5th│ 4■│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ 1 │ │ │ │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ 2 │ 5 │ 0 │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘
Then we consider how to match ✅「5th|2nd → 0」. If [5th] = 0:
┌───┬───┬───┬───┬───┬───┐
│ 5▲│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 0 │ 1 │ │ │ │ │
└───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ 2 │ 5 │ │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘
then we would fail to avoid ⛔「⟨ ▧ ▧ ▧ ▢ ▢ ▢ ⟩, Σ▧ ≤ Σ▢」. So, we have 0 = [2nd] instead:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│ 2■│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ 1 │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ 1 │ │ 0 │ │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ 2 │ 5 │ │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘
By (1), the boxes to the right of 1 are occupied by 2,4,5. It follows that 3 = [5th]:
┌───┬───┬───┬───┬───┬───┐
│ 5■│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ 1 │ │ 0 │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ 3 │ 1 │ │ 0 │ │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ 2 │ 5 │ │ │ 4 │
└───┴───┴───┴───┴───┴───┘
By ✅「Jump(2,4) = 2」, 2 and 4 are separated by 2 boxes. There is only one way to do so:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│ 3▲│2nd│1st│ 0▲│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 3 │ 1 │ ▬ │ 0 │ │ ▬ │
└───┴───┴───┴───┴───┴───┘
whence 5 = [1st]:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│ 1■│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 3 │ 1 │ │ 0 │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ 3 │ 1 │ │ 0 │ 5 │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ 2 │ │ │ │ 4 │
└───┴───┴───┴───┴───┴───┘
Finally, to avoid ⛔「⟨ ▧ ▧ ▧ ▢ ▢ ▢ ⟩, Σ▧ ≤ Σ▢」, we finish by
┌───┬───┬───┬───┬───┬───┐
│5th│4th│ 3■│2nd│1st│ 0■│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 3 │ 1 │ │ 0 │ 5 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ 3 │ 1 │ 4 │ 0 │ 5 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 3 │ 1 │ 4 │ 0 │ 5 │ 2 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.9