2023-11-21 WR

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

⟨5th 4th 3rd 2nd 1st 0th⟩

✅Match

⟨⋯ 4 ⋯ ? ⋯ 1 (?−1)⟩ (?≠1,2)

⟨? 0 ⋯ (?−2) ⋯ 1 ⋯⟩ (?≠2)

⛔Avoid

⟨     ³ʳᵈa ²ⁿᵈb   ⁰ᵗʰc ⟩, (abc)₁₀ ≥ 341

#125034_v2.1

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

Proof of 2023-11-21 WR
══════════════════════

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

Plainly, the patterns ✅【⟨⋯ 4 ⋯ ? ⋯ 1 (?−1)⟩ (?≠1,2)】 and ✅【⟨? 0 ⋯ (?−2) ⋯ 1 ⋯⟩ (?≠2)】 require that [1st] = 1 and [4th] = 0:

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

--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│   │ 2 │ 5 │   │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘

(1) Next, let us make an observation: to avoid ⛔「⟨     ³ʳᵈa ²ⁿᵈb   ⁰ᵗʰc ⟩, (abc)₁₀ ≥ 341」, we need [3rd] = 3 or 2.

We proceed to determine the value of [0th]. Consider ✅【⟨⋯ 4 ⋯ ? ⋯ 1 (?−1)⟩ (?≠1,2)】. To have ?-1 >= 0, we need ? = 5 or 4 or 3. It is not possible that ? = 5, for otherwise we get ⟨⋯ 4 ⋯ 5 ⋯ 14⟩, which is a contradiction. Therefore, [0th] = 3 or 2. Together with (1), we have

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

--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│   │   │ 5 │   │   │ 4 │
└───┴───┴───┴───┴───┴───┘

where one of the following holds:

(i) (x,y) = (3,2);
(ii) (x,y) = (2,3).

Case (i) is impossible, for otherwise we cannot avoid ⛔「⟨     ³ʳᵈa ²ⁿᵈb   ⁰ᵗʰc ⟩, (abc)₁₀ ≥ 341」. Hence, we have

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

--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│   │   │ 5 │   │   │ 4 │
└───┴───┴───┴───┴───┴───┘

Note that since [0th] = 3, the "?" in ✅【⟨? 0 ⋯ (?−2) ⋯ 1 ⋯⟩ (?≠2)】 cannot be 5. Accordingly, we have

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

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

Q.E.D.

#125034_v2.1