2024-08-20 Q1(m=6)

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

⟨6th 5th 4th 3rd 2nd 1st 0th⟩

✅Match

4th → a, 2nd → b, ab=2+4n

Jump(0,2) ≤ 1

5th → a, 2nd → b, a+b=1+6n

⟨⋯ 6 ⋯ 4 ⋯⟩

⛔Avoid

6th|5th|4th|3rd|1st → 0

6th|5th|2nd|0th → 3

{p4, p3, p1} = ? + {0,2,3}

#125034_v2.6

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

Proof of 2024-08-20 Q1(m=6)
═══════════════════════════

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

By ⛔「6th|5th|4th|3rd|1st → 0」, we have 0 = [2nd] | [0th]. We need [2nd] != 0 in order to match ✅「4th → a, 2nd → b, ab=2+4n」, so it is 0 = [0th].

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

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

Next, by ✅「Jump(0,2) ≤ 1」, we have 2 = [2nd] | [1st].

(1) We claim that 2 = [2nd].

------------------------------

For, suppose on the contrary 2 = [1st]:

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

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

Then, by ✅「5th → a, 2nd → b, a+b=1+6n」, one of the following holds:

(1.1) {[5th], [2nd]} = {3,4};

(1.2) {[5th], [2nd]} = {1,6}.

Observe that in order to avoid ⛔「6th|5th|2nd|0th → 3」, case (1.1) does not hold. Therefore, case (1.2) holds. Two possibilities follow:

      ┌───┬───┬───┬───┬───┬───┬───┐
      │6th│ 5▲│4th│3rd│ 2▲│1st│0th│
      ╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
(1.3) │   │ 1 │   │   │ 6 │ 2 │ 0 │
      ├───┼───┼───┼───┼───┼───┼───┤
(1.4) │   │ 6 │   │   │ 1 │ 2 │ 0 │
      └───┴───┴───┴───┴───┴───┴───┘

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

However, this is a contradiction because if case (1.3) holds, then we cannot match ✅「⟨⋯ 6 ⋯ 4 ⋯⟩」, while if case (1.4) holds, then we cannot match ✅「4th → a, 2nd → b, ab=2+4n」.

------------------------------

We have verified our claim in (1). Accordingly, we get 

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

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

Then, it follows from ✅「5th → a, 2nd → b, a+b=1+6n」 that [5th] = 5:

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

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

Let S := {p4, p3, p1}. Observe that by ⛔「6th|5th|2nd|0th → 3」 and ✅「⟨⋯ 6 ⋯ 4 ⋯⟩」, we have 3 ∈ S and 4 ∈ S. So, to avoid ⛔「{p4, p3, p1} = ? + {0,2,3}」, we need 1 ∉ S. It follows that 1 = [6th]:

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

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

Then, note that there is only one way to match ✅「4th → a, 2nd → b, ab=2+4n」:

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

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

Finally, in view of ✅「⟨⋯ 6 ⋯ 4 ⋯⟩」, we finish by

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

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

Q.E.D.

#125034_v2.6