2023-12-05 Q1(m=6)

Rearrange the digits in ⟨1263045⟩ to meet the rules below.

⟨6th 5th 4th 3rd 2nd 1st 0th⟩

✅Match

⟨     ⁴ᵗʰa ³ʳᵈb     ⁰ᵗʰc ⟩, a > b > c

⟨? ⋯ 1 (?+2) ⋯ 4 ⋯⟩ (?≠1)

⟨ ⁶ᵗʰ↓ ⁵ᵗʰ↑   ³ʳᵈ↑       ⟩ after ⟨→⟩

⛔Avoid

Jump(2,6) ≥ 1

⟨⋯ 2 ⋯ 6 ⋯⟩

6th|4th|2nd → 0

#125034_v2.2

       ┌───┬───┬───┬───┬───┬───┬───┐
       │6th│5th│4th│3rd│2nd│1st│0th│▒
       ╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ 3 │   │   │   │   │   │   │▒
       ├───┼───┼───┼───┼───┼───┼───┤▒
Step 2 │ 3 │   │   │   │ 6 │   │   │▒
       ├───┼───┼───┼───┼───┼───┼───┤▒
Step 3 │ 3 │   │   │   │ 6 │ 2 │   │▒
       ├───┼───┼───┼───┼───┼───┼───┤▒
Step 4 │ 3 │   │   │   │ 6 │ 2 │ 0 │▒
       ├───┼───┼───┼───┼───┼───┼───┤▒
Step 5 │ 3 │ 1 │   │   │ 6 │ 2 │ 0 │▒
       ├───┼───┼───┼───┼───┼───┼───┤▒
Step 6 │ 3 │ 1 │ 5 │   │ 6 │ 2 │ 0 │▒
       ├───┼───┼───┼───┼───┼───┼───┤▒
Step 7 │ 3 │ 1 │ 5 │ 4 │ 6 │ 2 │ 0 │▒
       └───┴───┴───┴───┴───┴───┴───┘▒
        ▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒

Proof of 2023-12-05 Q1(m=6)
═══════════════════════════

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

We begin with determining the value of [6th]. To match ✅「⟨? ⋯ 1 (?+2) ⋯ 4 ⋯⟩ (?≠1)」, the "?" there can only be 0|2|3. It is not 0 by ⛔「6th|4th|2nd → 0」, and it is not 2 by ⛔「⟨⋯ 2 ⋯ 6 ⋯⟩」. It follows that

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

--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ 6 │   │ 0 │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘

Note that the pattern ✅「⟨? ⋯ 1 (?+2) ⋯ 4 ⋯⟩ (?≠1)」 becomes

(1) ⟨3 ⋯ 15 ⋯ 4 ⋯⟩.

Also, note that to avoid ⛔「Jump(2,6) ≥ 1」 and ⛔「⟨⋯ 2 ⋯ 6 ⋯⟩」, we need to match the following pattern:

(2) ⟨⋯ 62 ⋯⟩.

We proceed to determine where to place 6,2. We cannot place 6 at a "↑" position of ✅「⟨ ⁶ᵗʰ↓ ⁵ᵗʰ↑   ³ʳᵈ↑       ⟩ after ⟨→⟩」, so there are only three possibilities:

      ┌───┬───┬───┬───┬───┬───┬───┐
      │6th│5th│4th│3rd│2nd│1st│0th│
      ╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
(2.1) │ 3 │   │ 6 │ 2 │   │   │   │
      ├───┼───┼───┼───┼───┼───┼───┤
(2.2) │ 3 │   │   │   │ 6 │ 2 │   │
      ├───┼───┼───┼───┼───┼───┼───┤
(2.3) │ 3 │   │   │   │   │ 6 │ 2 │
      └───┴───┴───┴───┴───┴───┴───┘

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

We claim that case (2.2) holds actually.

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

Case (2.1):

If (2.1) holds, then we have only one way to match (1):

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

But then we would have 6th ↑ after ⟨→⟩, which contradicts ✅「⟨ ⁶ᵗʰ↓ ⁵ᵗʰ↑   ³ʳᵈ↑       ⟩ after ⟨→⟩」.

Case (2.3):

If (2.3) holds, then to match (1), we have (1,5) = ([5th], [4th]) or ([4th], [3rd]):

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

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

In view of ✅「⟨     ⁴ᵗʰa ³ʳᵈb     ⁰ᵗʰc ⟩, a > b > c」, we need [4th] > [3rd], so the second case above is not possible. The preceding pattern implies [3rd] > [0th] too. A fortiori [3rd] != 0, so we get

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

But then we match ⛔「6th|4th|2nd → 0」, which is a contradiction.

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

We have verified that case (2.2) holds. Consequently

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

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

Next, we consider the value of [0th]. To match ✅「⟨ ⁶ᵗʰ↓ ⁵ᵗʰ↑   ³ʳᵈ↑       ⟩ after ⟨→⟩」, we need [0th] < [6th] = 3, so [0th] = 0 or 1. In view of (1), we see that [0th] != 1. Hence

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

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

Finally, there is only one way to match (1). We finish by

       ┌───┬───┬───┬───┬───┬───┬───┐
       │6th│ 5■│ 4■│ 3■│2nd│1st│0th│▒
       ╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
       │ 3 │   │   │   │ 6 │ 2 │ 0 │▒
       ├───┼───┼───┼───┼───┼───┼───┤▒
Step 5 │ 3 │ 1 │   │   │ 6 │ 2 │ 0 │▒
       ├───┼───┼───┼───┼───┼───┼───┤▒
Step 6 │ 3 │ 1 │ 5 │   │ 6 │ 2 │ 0 │▒
       ├───┼───┼───┼───┼───┼───┼───┤▒
Step 7 │ 3 │ 1 │ 5 │ 4 │ 6 │ 2 │ 0 │▒
       └───┴───┴───┴───┴───┴───┴───┘▒
        ▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒

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

Q.E.D.

#125034_v2.2