Q1(m=6) 2023-10-23

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

⟨6th 5th 4th 3rd 2nd 1st 0th⟩

✅Match

⟨⋯ 3 ⋯ ? 5 ⋯ (?−2)⟩ (?≠5)

⟨⋯ 3 ⋯ ? 0 ⋯ (?+1)⟩ (?≠0)

⟨⋯ 3 ⋯ 1 ⋯⟩

6th → a, 1st → b, a+b=3+5n

⛔Avoid

⟦1,3⟧ ∋ 0

⟨⋯ a ⋯ 0 ⋯⟩, a = 2|4|6

#125034_v2.0

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

Proof
═════

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

(1) By ✅【⟨⋯ 3 ⋯ ? 5 ⋯ (?−2)⟩ (?≠5)】 and ✅【⟨⋯ 3 ⋯ ? 0 ⋯ (?+1)⟩ (?≠0)】, we see that [0th] can only be 4 or 2.

If [0th]=4, then the ? in ✅【⟨⋯ 3 ⋯ ? 0 ⋯ (?+1)⟩ (?≠0)】 is 3. In view of ✅【⟨⋯ 3 ⋯ 1 ⋯⟩】, we have 

⟨⋯ 30 ⋯ 1 ⋯⟩

which matches ⛔「⟦1,3⟧ ∋ 0」. Therefore, we cannot have [0th]=4 in (1). So, [0th]=2.

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

--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │   │ 6 │ 3 │ 0 │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘

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

Now, we consider where to place 3. By ✅【⟨⋯ 3 ⋯ ? 5 ⋯ (?−2)⟩ (?≠5)】 and ✅【⟨⋯ 3 ⋯ ? 0 ⋯ (?+1)⟩ (?≠0)】, we have

(2) ⟨⋯ 3 ⋯ 45 ⋯ 2⟩, and

(3) ⟨⋯ 3 ⋯ 10 ⋯ 2⟩,

so 3=[6th] or [5th], for otherwise there are not enough places for 4,5,1,0.

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

Claim that 3=[6th]. If on the contrary 3=[5th], then 4th, 3rd, 2nd, and 1st are occupied by 4,5,1,0. Consequently, [6th]=6:

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

It would match ⛔「⟨⋯ a ⋯ 0 ⋯⟩, a = 2|4|6」, however, which is a contradiction. 

This verifies our claim. It follows that

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

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

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

Next, by (2), (3), and ⛔「⟨⋯ a ⋯ 0 ⋯⟩, a = 2|4|6」, we have

⟨3 ⋯ 10 ⋯ 45 ⋯ 2⟩, and

⟨3 ⋯ 10 ⋯ 6 ⋯ 2⟩.

In particular, 1 is at the left of 0,4,5,6. Hence, we obtain

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

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

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

Using ✅【6th → a, 1st → b, a+b=3+5n】 and (2), we get 

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

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

Plainly, we reach ⟨3106452⟩ at the end. This completes the proof.

Q.E.D.

#125034_v2.0