2024-05-21 Q1(m=6)

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

⟨6th 5th 4th 3rd 2nd 1st 0th⟩

✅Match

3rd → a, 0th → b, a+b=6

6th|4th|2nd|1st|0th → 4

⟨⋯ Perm(0,6) ⋯⟩

⟨⋯ a ⋯ 6 ⋯⟩, a = 0|1|2|3

3rd → a, 2nd → b, |a-b|=1

⛔Avoid

⟨⋯ ? ⋯ 6 ⋯ (?+1)⟩ (?≠5)

max ⊢3⊣ = 5

⟨⋯ Perm(1,3,4) ⋯⟩

5th|4th|3rd|0th → 1

#125034_v2.5

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

Proof of 2024-05-21 Q1(m=6)
═══════════════════════════

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

By ✅「3rd → a, 0th → b, a+b=6」, we have

{[3rd],[0th]} = {0,6} | {1,5} | {2,4}.

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

Since 0,6 are adjacent by ✅「⟨⋯ Perm(0,6) ⋯⟩」, and 1 ∉ {[3rd], [0th]} by ⛔「5th|4th|3rd|0th → 1」, it is indeed

{[3rd],[0th]} = {2,4}.

Combining this with ✅「6th|4th|2nd|1st|0th → 4」, we get

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

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

Next, we decide how to place 0,6. They are adjacent by ✅「⟨⋯ Perm(0,6) ⋯⟩」. So, to avoid ⛔「⟨⋯ ? ⋯ 6 ⋯ (?+1)⟩ (?≠5)」, we need to match the following pattern:

(1) ⟨⋯ Perm(0,6) ⋯ 3 ⋯ 4⟩.

It implies one of the following holds:

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

where "▬" are occupied by 0,6.

We show that case (3) holds actually.

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

If on the contrary case (2) holds:

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

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

then in view of (1), we have

(4) S := {[2nd], [1st]} = {1,3} | {3,5}.

However, if the former holds, then we matches ⛔「⟨⋯ Perm(1,3,4) ⋯⟩」; else if the latter holds, then we matches ⛔「max ⊢3⊣ = 5」. This shows a contradiction.

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

We have verified that case (3) holds. Noting that 6 is not at the leftmost corner by ✅「⟨⋯ a ⋯ 6 ⋯⟩, a = 0|1|2|3」, we get

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

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

As the argument of (4) shows, we would get a contradiction if S = {1,3} or {3,5}. Therefore, S = {1,5}. In view of ✅「3rd → a, 2nd → b, |a-b|=1」, we get

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

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

It follows that the answer is ⟨0632154⟩.

Q.E.D.

#125034_v2.5