2024-05-14 Q1(m=6)

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

⟨6th 5th 4th 3rd 2nd 1st 0th⟩

✅Match

6th → a, 1st → b, |a-b|=3

2nd → 0|3

⟨? ⋯ 6 ⋯ (?+3) ⋯⟩ (?≠3)

max ⊢0⊣ ≤ 4

max ⊢3⊣ = 3

#125034_v2.5

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

Proof of 2024-05-14 Q1(m=6)
═══════════════════════════

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

By ✅「⟨? ⋯ 6 ⋯ (?+3) ⋯⟩ (?≠3)」, we have [6th] = 0|1|2. Actually [6th] != 0, for otherwise it follows from ✅「6th → a, 1st → b, |a-b|=3」 that

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

But then we cannot match ✅「2nd → 0|3」, which is a contradiction.

Hence, we have

(1) [6th] = 1|2.

Combining this with ✅「6th → a, 1st → b, |a-b|=3」, we see that one of the following holds:

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

No matter which case happens, if we place 3 at 2nd, then we cannot match ✅「max ⊢3⊣ = 3」. So, in view of ✅「2nd → 0|3」, we get [2nd] = 0 as our first step:

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

--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ 6 │ 3 │   │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘

It then follows from ✅「max ⊢0⊣ ≤ 4」 that case (3) does not hold. Therefore, case (2) holds, and we get

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

--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│   │ 2 │ 6 │ 3 │   │   │ 5 │
└───┴───┴───┴───┴───┴───┴───┘

Next, we consider how to place 6. Plainly, there are four possibilities:

(4) 6 = [5th] | [4th] | [3rd] | [0th].

We proceed to show that 6 is at 5th.

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

(4.1) By ✅「⟨? ⋯ 6 ⋯ (?+3) ⋯⟩ (?≠3)」, we need 6 is between 1 and 4, so 6 != [0th].

(4.2) On the other hand, to match ✅「max ⊢0⊣ ≤ 4」, we cannot have 6 = [3rd].

(4.3) If 6 = [4th]:

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

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

then we have no way to match ✅「max ⊢3⊣ = 3」. So, we have 6 != [4th] as well.

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

Combining (4), (4.1), (4.2), and (4.3), we conclude that 6 = [5th]:

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

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

Then, observe that to match ✅「max ⊢3⊣ = 3」, there is only one way to place 3:

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

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

Finally, applying ✅「max ⊢3⊣ = 3」 once more, we finish by

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

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

Q.E.D.

#125034_v2.5