2023-10-31 WR

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

⟨5th 4th 3rd 2nd 1st 0th⟩

✅Match

3rd → a, 0th → b, |a-b|=3

2nd → a, 0th → b, ab=3+5n

⟨⋯ a ⋯ 1 ⋯⟩, a = 0|2|5

⟨ ⁵ᵗʰ= ⁴ᵗʰ↑ ³ʳᵈ↓ ²ⁿᵈ↓ ¹ˢᵗ↑ ⁰ᵗʰ= ⟩ after 5−⟨⇌⟩

#125034_v2.1

2023-10-31 WR

✅🚧🚧🚧🚧🚧

✅🚧🚧🚧🚧✅

✅🚧✅🚧🚧✅

✅🚧✅✅🚧✅

✅🚧✅✅✅✅

✅✅✅✅✅✅

#125034_v2.1

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

Proof of 2023-10-31 WR
══════════════════════

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

Let x := [5th] and y := [0th]. By considering the "=" positions of ✅【⟨ ⁵ᵗʰ= ⁴ᵗʰ↑ ³ʳᵈ↓ ²ⁿᵈ↓ ¹ˢᵗ↑ ⁰ᵗʰ= ⟩ after 5−⟨⇌⟩】, we have three possibilities for {x,y}:

(i) {5,0}
(ii) {4,1}
(iii) {3,2}

We claim that case (iii) holds. 

------------------------------

Indeed, note that by ✅【2nd → a, 0th → b, ab=3+5n】, we have y ≠ 5 and y ≠ 0. Therefore, case (i) does not hold.

Else if case (ii) holds, then as ✅【⟨⋯ a ⋯ 1 ⋯⟩, a = 0|2|5】 implies that 1 ≠ [5th], we have (x,y) = (4,1). But then ✅【3rd → a, 0th → b, |a-b|=3】 would never be matched.

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

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

------------------------------

We have verified that case (iii) holds. Next, we determine whether (x,y) = (2,3) or (3,2). If the former holds, then it follows from ✅【3rd → a, 0th → b, |a-b|=3】 that [3rd] = 0:

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

It is a contradiction, because by ✅【⟨ ⁵ᵗʰ= ⁴ᵗʰ↑ ³ʳᵈ↓ ²ⁿᵈ↓ ¹ˢᵗ↑ ⁰ᵗʰ= ⟩ after 5−⟨⇌⟩】, [3rd] can decrease. Therefore, we have (x,y) = (3,2).

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

--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │   │ 5 │ 0 │   │ 4 │
└───┴───┴───┴───┴───┴───┘

------------------------------

The rest are straightforward. By ✅【3rd → a, 0th → b, |a-b|=3】 and ✅【2nd → a, 0th → b, ab=3+5n】, we have 

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

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

As ✅【⟨⋯ a ⋯ 1 ⋯⟩, a = 0|2|5】 implies that 1 ≠ [4th], we reach

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

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

Q.E.D.

#125034_v2.1