2023-12-04 Q1(m=6)

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

⟨6th 5th 4th 3rd 2nd 1st 0th⟩

✅Match

⟨⋯ ? ⋯ 2 ⋯ (?+2)⟩ (?≠2,0)

⟨⋯ Perm(1,5) ⋯⟩

4th → a, 0th → b, a+b=1+5n

⛔Avoid

⟨⋯ Perm(2,6) ⋯⟩

5th → a, 3rd → b, ab=0+5n

3rd → 1|2|4|5

⟨⋯ Perm(2,3,5) ⋯⟩

4th → a, 1st → b, a+b=1+6n

#125034_v2.2

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

Proof of 2023-12-04 Q1(m=6)
═══════════════════════════

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

To begin with, we consider the value of [0th]. By ✅「⟨⋯ ? ⋯ 2 ⋯ (?+2)⟩ (?≠2,0)」, we have [0th] = 3 or 5 or 6. We need to match ✅「4th → a, 0th → b, a+b=1+5n」 too, so [0th] != 3. Accordingly, we have:

(1) [0th] = 5 or 6.

Next, we consider the 3rd position. To avoid ⛔「3rd → 1|2|4|5」, we have [3rd] = 0 or 3 or 6, and to avoid ⛔「5th → a, 3rd → b, ab=0+5n」, we need [3rd] != 0.

(2) Hence [3rd] = 3 or 6. We proceed to show that [3rd] = 3.

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

Suppose on the contrary [3rd] = 6. Using (1) and ✅「⟨⋯ Perm(1,5) ⋯⟩」, we get

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

However, we cannot match ✅「4th → a, 0th → b, a+b=1+5n」 now, which is a contradiction.

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

We have verified our claim in (2). So, our first step is:

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

--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ 6 │   │ 0 │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘

(3) We proceed to show that [0th] = 6.

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

If on the contrary [0th] != 6, then by (1), we have [0th] = 5. Using ✅「⟨⋯ Perm(1,5) ⋯⟩」 and ✅「4th → a, 0th → b, a+b=1+5n」, we have

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

It matches ⛔「4th → a, 1st → b, a+b=1+6n」, however, which is a contradiction.

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

Our claim in (3) is verified. Accordingly, we have:

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

--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │   │   │ 0 │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘

Now, by ✅「4th → a, 0th → b, a+b=1+5n」, we have [4th] = 0 or 5. If [4th] = 5, then to avoid ⛔「⟨⋯ Perm(2,6) ⋯⟩」 and ⛔「⟨⋯ Perm(2,3,5) ⋯⟩」, 2 can only be placed at 6th:

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

But to match ✅「⟨⋯ ? ⋯ 2 ⋯ (?+2)⟩ (?≠2,0)」, 2 cannot be in the left corner. Therefore, we need [4th] = 0.

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

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

In view of ✅「⟨⋯ ? ⋯ 2 ⋯ (?+2)⟩ (?≠2,0)」 and ✅「⟨⋯ Perm(1,5) ⋯⟩」, the remaining two possibilities are:

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

where the "-" are occupied by 1,5. We need to avoid ⛔「⟨⋯ Perm(2,6) ⋯⟩」 too, so (4) holds indeed.

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

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

Finally, to avoid ⛔「4th → a, 1st → b, a+b=1+6n」, we reach

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

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

Q.E.D.

#125034_v2.2