2024-07-16 Q1(m=6)

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

⟨6th 5th 4th 3rd 2nd 1st 0th⟩

✅Match

⟨⋯ 2 ⋯ ? 3 ⋯ (?−1)⟩ (?≠3,4)

⟦0,2⟧ ∋ 1,3,4,5

⛔Avoid

⟨⋯ Perm(4,5) ⋯⟩

⟦2,6⟧ ∋ 3,4

⟨⋯ ? ⋯ 1 ⋯ (?−6)⟩

⟨ ⁶ᵗʰc   ⁴ᵗʰd ³ʳᵈb ²ⁿᵈa     ⟩, a > b > c > d

#125034_v2.5

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

Proof of 2024-07-16 Q1(m=6)
═══════════════════════════

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

By ✅「⟨⋯ 2 ⋯ ? 3 ⋯ (?−1)⟩ (?≠3,4)」, there are at least two digits at the right of 2. Combining this with ✅「⟦0,2⟧ ∋ 1,3,4,5」, we have 2 = [6th] | [5th]:

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

(1) We claim that 2 = [6th].

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

For, if on the contrary 2 = [5th], then to match ✅「⟦0,2⟧ ∋ 1,3,4,5」, we need

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

where "/" are occupied by 1,3,4,5. It follows that [6th] = 6:

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

So we would match ⛔「⟨⋯ ? ⋯ 1 ⋯ (?−6)⟩」, which is a contradiction.

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

We have verified our claim in (1). Accordingly, we get

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

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

Then, there are two ways to match ✅「⟦0,2⟧ ∋ 1,3,4,5」:

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

Since ✅「⟨⋯ 2 ⋯ ? 3 ⋯ (?−1)⟩ (?≠3,4)」 requires that [0th] < 6, case (2) does not hold. Therefore, case (3) holds, and we get

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

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

Now, to match ✅「⟨⋯ 2 ⋯ ? 3 ⋯ (?−1)⟩ (?≠3,4)」 and avoid ⛔「⟨⋯ ? ⋯ 1 ⋯ (?−6)⟩」 at the same time, we need to match the following pattern:

(4) ⟨2 ⋯ 1 3 ⋯ 6 ⋯ 0⟩.

We consider how to place 6. By (4), the digits 1,3 are at the left of 6, so 6 = [3rd] | [2nd] | [1st]:

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

(5) We proceed to show that 6 = [2nd].

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

If 6 = [3rd], then (4) gives

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

and we would match ⛔「⟨⋯ Perm(4,5) ⋯⟩」. Else if 6 = [1st]:

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

then we would match ⛔「⟦2,6⟧ ∋ 3,4」.

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

We have verified (5). As a result, we get

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

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

In view of (4), 3 is at the left of 6. So, in order to avoid ⛔「⟦2,6⟧ ∋ 3,4」, we need 4 to be at the right of 6:

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

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

Now, there is only one way to match (4) and avoid ⛔「⟨ ⁶ᵗʰc   ⁴ᵗʰd ³ʳᵈb ²ⁿᵈa     ⟩, a > b > c > d」 at the same time. We finish by

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

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

Q.E.D.

#125034_v2.5