2023-11-07 WR

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

⟨5th 4th 3rd 2nd 1st 0th⟩

✅Match

⟨⋯ 0 ⋯ ? 2 ⋯ (?−1)⟩ (?≠2,3)

⟦0,1⟧ ∋ 2,3

⛔Avoid

Sim⟨ ⁵ᵗʰ2 ⁴ᵗʰ0 ³ʳᵈ4 ²ⁿᵈ3 ¹ˢᵗ5 ⁰ᵗʰ1 ⟩ ≥ 1

#125034_v2.1

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

Proof of 2023-11-07 WR
══════════════════════

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

We first determine the value of "?" in ✅【⟨⋯ 0 ⋯ ? 2 ⋯ (?−1)⟩ (?≠2,3)】.  We need ?>=1, so ? = 5|4|1. It is not 1 since this pattern implies 0 != [0th], and it is not 4, for otherwise 3 = ?-1 is in corners, which contradicts ✅【⟦0,1⟧ ∋ 2,3】. Therefore, ? = 5, and this pattern becomes

(1) ⟨⋯ 0 ⋯ 5 2 ⋯ 4⟩

Accordingly, our first step is

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

--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ 5 │ 0 │ 3 │   │
└───┴───┴───┴───┴───┴───┘

Next, note that (1) and ✅【⟦0,1⟧ ∋ 2,3】 imply 

(2) ⟨⋯ 0 ⋯ 5 2 ⋯ 1 ⋯⟩, and

(3) ⟨⋯ 0 ⋯ 3 ⋯ 1 ⋯⟩.

To match (2) and (3), we need [5th] = 0 and [1st] = 1:

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

--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│   │ 2 │ 5 │   │ 3 │   │
└───┴───┴───┴───┴───┴───┘

There are two ways to arrange 5,2 so that (2) is matched:

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

If it is (4), then [2nd] = 3, which contradicts ⛔「Sim⟨ ⁵ᵗʰ2 ⁴ᵗʰ0 ³ʳᵈ4 ²ⁿᵈ3 ¹ˢᵗ5 ⁰ᵗʰ1 ⟩ ≥ 1」. Therefore, it is (5) and we obtain ⟨035214⟩.

Q.E.D.

#125034_v2.1