2024-02-27 Q1(m=6)

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

⟨6th 5th 4th 3rd 2nd 1st 0th⟩

✅Match

Sim⟨ ⁶ᵗʰ3 ⁵ᵗʰ6 ⁴ᵗʰ5 ³ʳᵈ1 ²ⁿᵈ2 ¹ˢᵗ4 ⁰ᵗʰ0 ⟩ ≥ 3

⟨⋯ 4 ⋯ ? 0 ⋯ (?+4)⟩ (?≠0)

⟨? 3 ⋯ (?+1) ⋯ 5 ⋯⟩ (?≠3,2)

#125034_v2.3

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

Proof of 2024-02-27 Q1(m=6)
═══════════════════════════

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

Plainly, it follows from ✅「⟨? 3 ⋯ (?+1) ⋯ 5 ⋯⟩ (?≠3,2)」 that [5th] = 3:

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

--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ 6 │   │ 0 │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘

Next, we consider how to match ✅「Sim⟨ ⁶ᵗʰ3 ⁵ᵗʰ6 ⁴ᵗʰ5 ³ʳᵈ1 ²ⁿᵈ2 ¹ˢᵗ4 ⁰ᵗʰ0 ⟩ ≥ 3」. By ✅「⟨⋯ 4 ⋯ ? 0 ⋯ (?+4)⟩ (?≠0)」, there are at least three digits at the right of 4, and at least one digit at the right of 0. Hence,

(1) [1st] != 4 and [0th] != 0.

On the other hand, step 1 shows that [5th] = 3. A fortiori,

(2) [6th] != 3 and [5th] != 6.

Combining ✅「Sim⟨ ⁶ᵗʰ3 ⁵ᵗʰ6 ⁴ᵗʰ5 ³ʳᵈ1 ²ⁿᵈ2 ¹ˢᵗ4 ⁰ᵗʰ0 ⟩ ≥ 3」, (1), and (2), we obtain the values of [4th], [3rd], and [2nd]:

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

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

By ✅「⟨⋯ 4 ⋯ ? 0 ⋯ (?+4)⟩ (?≠0)」, 0 is not in corners, and 4 is at the left of 0. Therefore, we finish by

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

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

Q.E.D.

#125034_v2.3