2023-12-13 Q1(m=6)

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

⟨6th 5th 4th 3rd 2nd 1st 0th⟩

✅Match

6th → a, 2nd → b, a+b=2

5th → a, 4th → b, a+b=7

1st → 3

⛔Avoid

Sim⟨ ⁶ᵗʰ6 ⁵ᵗʰ0 ⁴ᵗʰ2 ³ʳᵈ1 ²ⁿᵈ3 ¹ˢᵗ5 ⁰ᵗʰ4 ⟩ ≥ 1

Jump(2,5) = 1

6th|5th|2nd|1st|0th → 6

#125034_v2.2

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

Proof of 2023-12-13 Q1(m=6)
═══════════════════════════

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

Plainly, our first step follows from ✅「1st → 3」.

In view of ✅「6th → a, 2nd → b, a+b=2」, we have 

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

where {X,Y} = {0,2}. Now the idle numbers are

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

so to match ✅「5th → a, 4th → b, a+b=7」, we need

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

where {E,F} = {1,6}. By ⛔「6th|5th|2nd|1st|0th → 6」, we have E != 6. Accordingly, we get

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

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

4,5 are placed at 3rd and 0th. To avoid ⛔「Sim⟨ ⁶ᵗʰ6 ⁵ᵗʰ0 ⁴ᵗʰ2 ³ʳᵈ1 ²ⁿᵈ3 ¹ˢᵗ5 ⁰ᵗʰ4 ⟩ ≥ 1」, we cannot have 4 = [0th]. So 4 = [3rd], and 5 = [0th].

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

Finally, to avoid ⛔「Jump(2,5) = 1」, we have Y != 2. As {X,Y} = {0,2} by (1), we see that the answer is

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

Q.E.D.

#125034_v2.2