2024-10-15 Q1(m=6)

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

⟨6th 5th 4th 3rd 2nd 1st 0th⟩

✅Match

⟨⋯ 2 ⋯ 3 ⋯⟩

⟦4,6⟧ ∋ 2,5

2nd → a, 0th → b, a+b=11

Jump(0,6) = 3

⟨⋯ 3 ⋯ 1 ⋯⟩

#125034_v2.7

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

Proof of 2024-10-15 Q1(m=6)
═══════════════════════════

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

By ✅「2nd → a, 0th → b, a+b=11」, we have

(1) { [2nd], [0th] } = {5,6}.

By ✅「⟦4,6⟧ ∋ 2,5」, the right corner (0th) cannot be 5. Therefore, we get

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

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

Then, it follows from ✅「Jump(0,6) = 3」 that [4th] = 0:

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

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

Next, we consider how to place 2. Combining ✅「⟨⋯ 2 ⋯ 3 ⋯⟩」 with ✅「⟨⋯ 3 ⋯ 1 ⋯⟩」, we have the following required pattern:

(2) ⟨⋯ 2 ⋯ 3 ⋯ 1 ⋯⟩.

In particular, we need to reserve two empty boxes at the right of 2 for placing 3,1. Noting that ✅「⟦4,6⟧ ∋ 2,5」 implies 2 is not at the left corner, we finish by

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

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

Q.E.D.

#125034_v2.7