2023-12-08 Q1(m=6)

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

⟨6th 5th 4th 3rd 2nd 1st 0th⟩

✅Match

2nd → a, 0th → b, a+b=3

⟨ ⁶ᵗʰa ⁵ᵗʰd ⁴ᵗʰb ³ʳᵈc       ⟩, a > b > c > d

⟨     ⁴ᵗʰa   ²ⁿᵈb ¹ˢᵗc   ⟩, (abc)₁₀ ≥ 531

⟨⋯ Perm(4,5) ⋯⟩

⛔Avoid

1 ∾ 2 ∾ 5

#125034_v2.2

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

Proof of 2023-12-08 Q1(m=6)
═══════════════════════════

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

To begin with, note that to match ✅「⟨     ⁴ᵗʰa   ²ⁿᵈb ¹ˢᵗc   ⟩, (abc)₁₀ ≥ 531」, we need [4th] = 5 or 6. It cannot be 6, because by ✅「⟨ ⁶ᵗʰa ⁵ᵗʰd ⁴ᵗʰb ³ʳᵈc       ⟩, a > b > c > d」, [4th] cannot be the maximum number. Therefore, [4th] = 5.

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

--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ 6 │ 3 │ 0 │ 4 │   │
└───┴───┴───┴───┴───┴───┴───┘

By ✅「⟨⋯ Perm(4,5) ⋯⟩」, 4 is adjacent to 5, so 4 = [5th] or [3rd]. The former is not possible, for otherwise by ✅「⟨ ⁶ᵗʰa ⁵ᵗʰd ⁴ᵗʰb ³ʳᵈc       ⟩, a > b > c > d」, we have 5 > [3rd] > 4, which is a contradiction. It follows that 4 = [3rd]. The preceding pattern also implies [6th] > 5, so [6th] = 6.

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

--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │   │ 3 │ 0 │   │   │
└───┴───┴───┴───┴───┴───┴───┘

Next, we determine the value of [2nd]. To match ✅「⟨     ⁴ᵗʰa   ²ⁿᵈb ¹ˢᵗc   ⟩, (abc)₁₀ ≥ 531」, we can only take [2nd] = 3. Then, to match ✅「2nd → a, 0th → b, a+b=3」, we need [0th] = 0.

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

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

Finally, to avoid ⛔「1 ∾ 2 ∾ 5」, we simply place 1 at the 1st position so that 1 is itself a permutation circle. We finish by

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

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

Q.E.D.

#125034_v2.2