2024-08-06 Q1(m=6)

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

⟨6th 5th 4th 3rd 2nd 1st 0th⟩

✅Match

⟨ ⁶ᵗʰ↑       ²ⁿᵈ↓ ¹ˢᵗ↓   ⟩ after 2×⟨←⟩

3rd → a, 0th → b, ab=3+4n

⟦3,4⟧ ∋ 0,5

#125034_v2.6

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

Proof of 2024-08-06 Q1(m=6)
═══════════════════════════

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

We first consider where to place 0. We cannot place 0 at a "↓" position of ✅「⟨ ⁶ᵗʰ↑       ²ⁿᵈ↓ ¹ˢᵗ↓   ⟩ after 2×⟨←⟩」, so 0 != [2nd] and [1st]. To have a "↑" at 6th, we need 0 != [4th] as well.

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

Moreover, by ✅「3rd → a, 0th → b, ab=3+4n」, we have 0 != [3rd] and [0th]:

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

and by ✅「⟦3,4⟧ ∋ 0,5」, 0 is not at the left corner:

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

It follows that 0 = [5th]:

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

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

Then, to match ✅「⟦3,4⟧ ∋ 0,5」, we need 

(1) [6th] = 3|4.

Observe that ✅「3rd → a, 0th → b, ab=3+4n」 implies

(2) {[3rd], [0th]} = {1,3} | {3,5}.

A fortiori, 3 ∈ {[3rd], [0th]}. Combining this with (1), we get

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

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

Next, we use ✅「⟨ ⁶ᵗʰ↑       ²ⁿᵈ↓ ¹ˢᵗ↓   ⟩ after 2×⟨←⟩」. Given that [6th] = 4, to have a "↑" at 6th and a "↓" at 1st, we need [4th] > 4 and [1st] > 4. This means

(3) {[4th], [1st]} = {5,6}.

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

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

(where "▬" are occupied by 5,6)

And combining (3) with (2), we get

(4) {[3rd], [0th]} = {1,3}.

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

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

(where "#" are occupied by 1,3)

Plainly, it follows that [2nd] = 2:

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

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

Now, in view of (4), there are two possibilities:

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

If case (5) holds, then we cannot have a "↓" at 2nd, contradicting ✅「⟨ ⁶ᵗʰ↑       ²ⁿᵈ↓ ¹ˢᵗ↓   ⟩ after 2×⟨←⟩」. Therefore, case (6) holds:

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

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

Finally, in view of ✅「⟦3,4⟧ ∋ 0,5」, we finish by

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

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

Q.E.D.

#125034_v2.6