2024-03-05 Q1(m=6)

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

⟨6th 5th 4th 3rd 2nd 1st 0th⟩

✅Match

⟦1,3⟧ ∋ 0,2,4,6

⟨? ⋯ 5 (?+3) ⋯ 6 ⋯⟩ (?≠2)

⟨⋯ ? ⋯ 5 ⋯ (?+3)⟩ (?≠2)

⛔Avoid

⟨   ⁵ᵗʰc     ²ⁿᵈa   ⁰ᵗʰb ⟩, a > b > c

#125034_v2.3

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

Proof of 2024-03-05 Q1(m=6)
═══════════════════════════

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

By ✅「⟦1,3⟧ ∋ 0,2,4,6」, we see that 4 and 6 are not in the right corner. Hence, the "?" in ✅「⟨⋯ ? ⋯ 5 ⋯ (?+3)⟩ (?≠2)」 is 0, and we have

(1) ⟨⋯ 0 ⋯ 5 ⋯ 3⟩.

Combining (1) with ✅「⟦1,3⟧ ∋ 0,2,4,6」, we have

(2) ⟨⋯ 1 ⋯ 0 ⋯ 5 ⋯ 3⟩,

whence ⟦1,3⟧ ∋ 0,2,4,5,6. As ⟦1,3⟧ contains five digits, we see that 1,3 are in the corners:

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

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

Then, using ✅「⟨? ⋯ 5 (?+3) ⋯ 6 ⋯⟩ (?≠2)」, we get

(3) ⟨1 ⋯ 5 4 ⋯ 6 ⋯⟩.

Combining (2) and (3), we get

(4) ⟨1 ⋯ 0 ⋯ 5 4 ⋯ 6 ⋯ 3⟩.

A fortiori, 0 = [5th] | [4th]:

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

Actually, we have 0 != [4th], for otherwise we get from (4) that

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

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

and we would match ⛔「⟨   ⁵ᵗʰc     ²ⁿᵈa   ⁰ᵗʰb ⟩, a > b > c」. Therefore, 0 = [5th]:

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

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

Now, to avoid ⛔「⟨   ⁵ᵗʰc     ²ⁿᵈa   ⁰ᵗʰb ⟩, a > b > c」, we need [2nd] < 3. It follows that [2nd] = 2:

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

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

Finally, using (4), we finish by ⟨1054263⟩.

Q.E.D.

#125034_v2.3