2025-12-09 WR

Rearrange the digits in ⟨125034⟩ to meet the rules below.

⟨ ⁵ᵗʰ▨ ⁴ᵗʰ▨ ³ʳᵈ▨ ²ⁿᵈ▨ ¹ˢᵗ▨ ⁰ᵗʰ▨ ⟩

✅Match

3rd|1st|0th → 1

3rd → a, 0th → b, |a-b|=1

⟨? ⋯ 5 (?−1) ⋯ 4 ⋯⟩ (?≠5)

5th → a, 0th → b, |a-b|=1

#125034_v2.11

       ┌───┬───┬───┬───┬───┬───┐
       │5th│4th│3rd│2nd│1st│0th│▒
       ╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ 3 │   │   │   │   │   │▒
       ├───┼───┼───┼───┼───┼───┤▒
Step 2 │ 3 │   │   │   │   │ 4 │▒
       ├───┼───┼───┼───┼───┼───┤▒
Step 3 │ 3 │   │ 5 │   │   │ 4 │▒
       ├───┼───┼───┼───┼───┼───┤▒
Step 4 │ 3 │   │ 5 │ 2 │   │ 4 │▒
       ├───┼───┼───┼───┼───┼───┤▒
Step 5 │ 3 │   │ 5 │ 2 │ 1 │ 4 │▒
       ├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 3 │ 0 │ 5 │ 2 │ 1 │ 4 │▒
       └───┴───┴───┴───┴───┴───┘▒
        ▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒

Proof of 2025-12-09 WR
══════════════════════

Notation: if nth -> a, then we write [nth] = a.

Combining ✅「3rd → a, 0th → b, |a-b|=1」 with ✅「5th → a, 0th → b, |a-b|=1」, we see that

(1) [5th], [3rd], [0th] are consecutive integers, and [0th] is the median.

On the other hand, combining ✅「⟨? ⋯ 5 (?−1) ⋯ 4 ⋯⟩ (?≠5)」 with ✅「3rd|1st|0th → 1」, we have

(2) [5th] = 2 or 3.

There are four ways to match (1) and (2) simultaneously:

    ┌───┬───┬───┬───┬───┬───┐
    │ 5▲│4th│ 3▲│2nd│1st│ 0▲│
    ╞═══╪═══╪═══╪═══╪═══╪═══╡
(3) │ 2 │   │ 0 │   │   │ 1 │
    ├───┼───┼───┼───┼───┼───┤
(4) │ 2 │   │ 4 │   │   │ 3 │
    ├───┼───┼───┼───┼───┼───┤
(5) │ 3 │   │ 1 │   │   │ 2 │
    ├───┼───┼───┼───┼───┼───┤
(6) │ 3 │   │ 5 │   │   │ 4 │
    └───┴───┴───┴───┴───┴───┘

Observe that only case (6) is possible to satisfy ✅「⟨? ⋯ 5 (?−1) ⋯ 4 ⋯⟩ (?≠5)」. Therefore, we get

       ┌───┬───┬───┬───┬───┬───┐
       │ 5■│4th│ 3■│2nd│1st│ 0■│▒
       ╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ 3 │   │   │   │   │   │▒
       ├───┼───┼───┼───┼───┼───┤▒
Step 2 │ 3 │   │   │   │   │ 4 │▒
       ├───┼───┼───┼───┼───┼───┤▒
Step 3 │ 3 │   │ 5 │   │   │ 4 │▒
       └───┴───┴───┴───┴───┴───┘▒
        ▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒

--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │   │ 0 │   │   │
└───┴───┴───┴───┴───┴───┘

It then follows from ✅「⟨? ⋯ 5 (?−1) ⋯ 4 ⋯⟩ (?≠5)」 that [2nd] = 2:

       ┌───┬───┬───┬───┬───┬───┐
       │5th│4th│3rd│ 2■│1st│0th│▒
       ╞═══╪═══╪═══╪═══╪═══╪═══╡▒
       │ 3 │   │ 5 │   │   │ 4 │▒
       ├───┼───┼───┼───┼───┼───┤▒
Step 4 │ 3 │   │ 5 │ 2 │   │ 4 │▒
       └───┴───┴───┴───┴───┴───┘▒
        ▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒

--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │   │   │ 0 │   │   │
└───┴───┴───┴───┴───┴───┘

Finally, in view of ✅「3rd|1st|0th → 1」, we finish by

       ┌───┬───┬───┬───┬───┬───┐
       │5th│ 4■│3rd│2nd│ 1■│0th│▒
       ╞═══╪═══╪═══╪═══╪═══╪═══╡▒
       │ 3 │   │ 5 │ 2 │   │ 4 │▒
       ├───┼───┼───┼───┼───┼───┤▒
Step 5 │ 3 │   │ 5 │ 2 │ 1 │ 4 │▒
       ├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 3 │ 0 │ 5 │ 2 │ 1 │ 4 │▒
       └───┴───┴───┴───┴───┴───┘▒
        ▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒

--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│   │   │   │   │   │   │
└───┴───┴───┴───┴───┴───┘

Q.E.D.

#125034_v2.11