2024-01-30 Q1(m=6)

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

⟨6th 5th 4th 3rd 2nd 1st 0th⟩

✅Match

6th → a, 4th → b, |a-b|=1

⟨⋯ 1 ⋯ 2 ⋯ 5 ⋯ 6 ⋯⟩

⛔Avoid

⟨⋯ 5 ⋯ a ⋯⟩, a = 2|3|4

⟨⋯ 1 ⋯ ? 5 ⋯ (?+4)⟩ (?≠1)

⟨⋯ 2 ⋯ a ⋯⟩, a = 1|4

⟨? 4 ⋯ (?−1) ⋯ 6 ⋯⟩ (?≠4,5)

3rd → 0|2|4|6

#125034_v2.3

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

Proof of 2024-01-30 Q1(m=6)
═══════════════════════════

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

We begin with the 3rd position. To avoid ⛔「3rd → 0|2|4|6」, we need

(1) [3rd] = 1|3|5.

We show that [3rd] = 3 actually.

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

(1.1) If [3rd] = 1, then by ✅「⟨⋯ 1 ⋯ 2 ⋯ 5 ⋯ 6 ⋯⟩」, we have

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

which contradicts ⛔「⟨⋯ 1 ⋯ ? 5 ⋯ (?+4)⟩ (?≠1)」.

(1.2) Else if [3rd] = 5, then to avoid ⛔「⟨⋯ 5 ⋯ a ⋯⟩, a = 2|3|4」, we have

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

where the "*" are occupied by 2,3,4. But to match ✅「⟨⋯ 1 ⋯ 2 ⋯ 5 ⋯ 6 ⋯⟩」, we also need 1 at the left of 5. This shows a contradiction.

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

Therefore, it follows from (1), (1.1), and (1.2) that [3rd] = 3.

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

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

Next, we consider where to place 2. To match ✅「⟨⋯ 1 ⋯ 2 ⋯ 5 ⋯ 6 ⋯⟩」, we need 2 = [5th] | [4th] | [2nd]:

┌───┬───┬───┬───┬───┬───┬───┐
│6th│*5 │*4 │3rd│*2 │1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│   │ ● │ ● │ 3 │ ● │   │   │
└───┴───┴───┴───┴───┴───┴───┘

(2) We show that 2 = [4th].

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

(2.1) If 2 = [5th], then by ✅「⟨⋯ 1 ⋯ 2 ⋯ 5 ⋯ 6 ⋯⟩」, we have [6th] = 1:

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

but then we would match ⛔「⟨⋯ 2 ⋯ a ⋯⟩, a = 1|4」.

(2.2) Else if 2 = [2nd], then by ✅「⟨⋯ 1 ⋯ 2 ⋯ 5 ⋯ 6 ⋯⟩」, we have

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

and we would match ⛔「⟨⋯ 1 ⋯ ? 5 ⋯ (?+4)⟩ (?≠1)」.

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

By (2.1) and (2.2), we have verified our claim in (2). Accordingly, we have

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

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

Then, by ✅「6th → a, 4th → b, |a-b|=1」, plainly we have [6th] = 1:

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

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

Next, to avoid ⛔「⟨⋯ 2 ⋯ a ⋯⟩, a = 1|4」, we have to place 4 at 5th:

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

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

Let us consider where to place 5. By ✅「⟨⋯ 1 ⋯ 2 ⋯ 5 ⋯ 6 ⋯⟩」, it is at [2nd] or [1st]. We have 5 != [1st], for otherwise we get

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

which matches ⛔「⟨? 4 ⋯ (?−1) ⋯ 6 ⋯⟩ (?≠4,5)」. Therefore, 5 = [2nd]:

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

--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│   │   │ 6 │   │ 0 │   │   │
└───┴───┴───┴───┴───┴───┴───┘

Finally, noting that there is only one way to avoid ⛔「⟨? 4 ⋯ (?−1) ⋯ 6 ⋯⟩ (?≠4,5)」, we finish by

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

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

Q.E.D.

#125034_v2.3