2024-04-30 Q1(m=6)

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

⟨6th 5th 4th 3rd 2nd 1st 0th⟩

✅Match

⟨ ⁶ᵗʰb ⁵ᵗʰa ⁴ᵗʰc         ⟩, a > b > c

min ⊢4⊣ = 1

2nd → a, 1st → b, ab=2

⟨   ⁵ᵗʰb ⁴ᵗʰd ³ʳᵈa ²ⁿᵈc     ⟩, a > b > c > d

#125034_v2.4

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

Proof of 2024-04-30 Q1(m=6)
═══════════════════════════

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

By ✅「2nd → a, 1st → b, ab=2」, we have

(1) {[2nd],[1st]} = {1,2}.

Combining ✅「⟨   ⁵ᵗʰb ⁴ᵗʰd ³ʳᵈa ²ⁿᵈc     ⟩, a > b > c > d」 with (1), we have

(2) [4th] < [2nd] <= max{[2nd],[1st]} = 2; and

(3) [4th] != 1.

It follows that [4th] = 0:

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

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

Next, observe that ✅「min ⊢4⊣ = 1」 implies 4 is adjacent to 1. Combining this with (1), we have two possibilities:

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

Since case (4) contradicts ✅「min ⊢4⊣ = 1」, we see that case (5) holds actually. Accordingly, we have

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

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

We proceed to determine where to place 6. It is the maximum digit, so cannot be the "b" in ✅「⟨ ⁶ᵗʰb ⁵ᵗʰa ⁴ᵗʰc         ⟩, a > b > c」 and ✅「⟨   ⁵ᵗʰb ⁴ᵗʰd ³ʳᵈa ²ⁿᵈc     ⟩, a > b > c > d」. That is,

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

It follows that 6 = [3rd]:

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

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

Finally, to match ✅「⟨ ⁶ᵗʰb ⁵ᵗʰa ⁴ᵗʰc         ⟩, a > b > c」, we finish by

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

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

Q.E.D.

#125034_v2.4