WR 2023-10-22

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

⟨5th 4th 3rd 2nd 1st 0th⟩

✅Match

Jump(2,4) = 1

⟨ ⁵ᵗʰa ⁴ᵗʰb         ⟩, (ab)₁₀ ≥ 24

4th → a, 1st → b, a+b=0+5n

⛔Avoid

⟦3,4⟧ ∋ 0

5th|4th|2nd|0th → 4

⟨⋯ 3 ⋯ a ⋯⟩, a = 0|4

#125034_v2.0

WR 2023-10-22

💎💎💎💎✅💎

💎💎✅💎✅💎

💎✅✅💎✅💎

💎✅✅💎✅✅

✅✅✅💎✅✅

✅✅✅✅✅✅

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

Proof
═════

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

(1) By ⛔「5th|4th|2nd|0th → 4」, we have 4=[3rd] or [1st]. We claim that 4=[1st]. Suppose on the contrary 4=[3rd].

Then, by ✅【Jump(2,4) = 1】, either [5th] or [3rd] is 2:

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

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

If x is 2, then by ✅【⟨ ⁵ᵗʰa ⁴ᵗʰb         ⟩, (ab)₁₀ ≥ 24】, we need [4th]=5:

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

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

and then by ✅【4th → a, 1st → b, a+b=0+5n】, we have 

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

It is a contradiction, as either ⛔「⟦3,4⟧ ∋ 0」 or ⛔「⟨⋯ 3 ⋯ a ⋯⟩, a = 0|4」 would be matched. So, we need y=2 in (2). By ✅【4th → a, 1st → b, a+b=0+5n】, this implies

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

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

It then follows from ✅【⟨ ⁵ᵗʰa ⁴ᵗʰb         ⟩, (ab)₁₀ ≥ 24】 that

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

It is again a contradiction, as either ⛔「⟦3,4⟧ ∋ 0」 or ⛔「⟨⋯ 3 ⋯ a ⋯⟩, a = 0|4」 would be matched. This verifies our claim in (1).

Accordingly, we have 

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

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

The rest are straightforward. By ✅【Jump(2,4) = 1】 and ✅【4th → a, 1st → b, a+b=0+5n】, we have

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

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

By ⛔「⟨⋯ 3 ⋯ a ⋯⟩, a = 0|4」, 3 is at the right of 4, so

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

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

Finally, by ✅【⟨ ⁵ᵗʰa ⁴ᵗʰb         ⟩, (ab)₁₀ ≥ 24】, we get [5th]=5. We reach the answer:

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

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

Q.E.D.

#125034_v2.0