Rearrange the digits in ⟨125034⟩ to meet the rules below.
⟨ ⁵ᵗʰ▨ ⁴ᵗʰ▨ ³ʳᵈ▨ ²ⁿᵈ▨ ¹ˢᵗ▨ ⁰ᵗʰ▨ ⟩
✅Match
⟨ ²ⁿᵈa ⁰ᵗʰb ⟩, max⟦a,b⟧ = 4
⟨⋯ 2 ⋯ 5 ⋯⟩
⛔Avoid
{p5, p2, p1} = ? + {0,2,3}
⟨⋯ a ⋯ 3 ⋯⟩, a = 0|2|4
4th → a, 2nd → b, ab=0+4n
#125034_v2.12
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ 3 │ │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ 3 │ 2 │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ 3 │ 2 │ 5 │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ 3 │ 2 │ 5 │ │ 0 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ 3 │ 2 │ 5 │ 1 │ 0 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 3 │ 2 │ 5 │ 1 │ 0 │ 4 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
Proof of 2026-03-17 WR
══════════════════════
Notation: if nth -> a, then we write [nth] = a.
To avoid ⛔「⟨⋯ a ⋯ 3 ⋯⟩, a = 0|2|4」, 3 is to the left of 0,2,4. On the other hand, by ✅「⟨⋯ 2 ⋯ 5 ⋯⟩」, 2 is to the left of 5. Combining these, we have
(1) 3 is to the left of 0,2,4,5
and
(2) ⟨⋯ 3 ⋯ 2 ⋯ 5 ⋯⟩.
By (1), we have 3 = [5th] | [4th]:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ ▬ │ ▬ │ │ │ │ │
└───┴───┴───┴───┴───┴───┘
If 3 = [4th], then to match (2), 5 will be at the 2nd, 1st, or 0th position. But then we will fail to match ✅「⟨ ²ⁿᵈa ⁰ᵗʰb ⟩, max⟦a,b⟧ = 4」, which is a contradiction. Therefore, we have 3 = [5th] instead, which is our first step:
┌───┬───┬───┬───┬───┬───┐
│ 5■│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ 3 │ │ │ │ │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ 5 │ 0 │ │ 4 │
└───┴───┴───┴───┴───┴───┘
Having this, there is only one way to match (2) and ✅「⟨ ²ⁿᵈa ⁰ᵗʰb ⟩, max⟦a,b⟧ = 4」 at the same time:
┌───┬───┬───┬───┬───┬───┐
│5th│ 4■│ 3■│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 3 │ │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ 3 │ 2 │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ 3 │ 2 │ 5 │ │ │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ │ │ 0 │ │ 4 │
└───┴───┴───┴───┴───┴───┘
Then, to avoid ⛔「{p5, p2, p1} = ? + {0,2,3}」, we cannot have {[2nd], [1st]} = {1,4}. It implies
(3) 0 = [2nd] | [1st].
If 0 = [2nd], then we fail to avoid ⛔「4th → a, 2nd → b, ab=0+4n」. Therefore, 0 = [1st]:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│ 1■│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 3 │ 2 │ 5 │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ 3 │ 2 │ 5 │ │ 0 │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ │ │ │ │ 4 │
└───┴───┴───┴───┴───┴───┘
To avoid ⛔「4th → a, 2nd → b, ab=0+4n」, we cannot have [2nd] = 4 as well. Therefore, we finish by
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│ 2■│1st│ 0■│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 3 │ 2 │ 5 │ │ 0 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ 3 │ 2 │ 5 │ 1 │ 0 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 3 │ 2 │ 5 │ 1 │ 0 │ 4 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.12