2023-12-09 Q1(m=6)

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

⟨6th 5th 4th 3rd 2nd 1st 0th⟩

✅Match

6th → a, 4th → b, a+b=5

⟨? ⋯ 3 ⋯ (?+2) ⋯⟩ (?≠3,1)

⟨ ⁶ᵗʰa       ²ⁿᵈb ¹ˢᵗc   ⟩, (abc)₁₀ ≥ 210

⛔Avoid

Jump(0,3) = 0

⟨⋯ 1 ⋯ a ⋯⟩, a = 3|4|6

0th → 0|1|4|6

#125034_v2.2

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

Proof of 2023-12-09 Q1(m=6)
═══════════════════════════

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

Firstly, observe that to avoid ⛔「⟨⋯ 1 ⋯ a ⋯⟩, a = 3|4|6」, we need to match the following patterns:

(1) 

⟨⋯ 3 ⋯ 1 ⋯⟩
⟨⋯ 4 ⋯ 1 ⋯⟩
⟨⋯ 6 ⋯ 1 ⋯⟩

We proceed to determine the value of [6th]. By ✅「⟨? ⋯ 3 ⋯ (?+2) ⋯⟩ (?≠3,1)」, we have [6th] = 0|2|4. In order to match ✅「⟨ ⁶ᵗʰa       ²ⁿᵈb ¹ˢᵗc   ⟩, (abc)₁₀ ≥ 210」, we cannot have [6th] = 0. Hence [6th] = 2 or 4.

If [6th] = 4, then ✅「6th → a, 4th → b, a+b=5」 implies [4th] = 1:

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

But then we cannot match all of the patterns in (1). This is a contradiction. As a result, we have [6th] = 2.

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

--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │   │ 6 │ 3 │ 0 │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘

Next, we check the value of [0th]. To avoid ⛔「0th → 0|1|4|6」, we need [0th] = 2,3,5. It cannot be 2 as we have used it, and it cannot be 3 as we need to match (1). Consequently, we get [0th] = 5. On the other hand, it follows from ✅「6th → a, 4th → b, a+b=5」 that [4th] = 3.

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

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

Now, we consider where to place 0. By ⛔「Jump(0,3) = 0」, 0 is not adjacent to 3. Also, to match ✅「⟨ ⁶ᵗʰa       ²ⁿᵈb ¹ˢᵗc   ⟩, (abc)₁₀ ≥ 210」, it cannot be placed at the 2nd position. Hence, we have

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

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

Recall that we need to match (1). It can be done only when 1 is placed at the 2nd position.

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

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

Finally, noting that ✅「⟨? ⋯ 3 ⋯ (?+2) ⋯⟩ (?≠3,1)」 becomes:

⟨2 ⋯ 3 ⋯ 4 ⋯⟩,

we finish by

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

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

Q.E.D.

#125034_v2.2