2023-11-21 Q1(m=6)

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

⟨6th 5th 4th 3rd 2nd 1st 0th⟩

✅Match

⟨? ⋯ 3 ⋯ (?+2) ⋯⟩ (?≠3,1)

4th → a, 1st → b, a+b=4+5n

6th → a, 2nd → b, a+b=8

4th → a, 0th → b, |a-b|=2

#125034_v2.1

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

Proof of 2023-11-21 Q1(m=6)
═══════════════════════════

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

We begin with the left corner. In view of ✅【⟨? ⋯ 3 ⋯ (?+2) ⋯⟩ (?≠3,1)】, we need [6th] = 0 or 2 or 4 in order that (?+2) <= 6. We need to match ✅【6th → a, 2nd → b, a+b=8】 too. When [6th] = 0 or 4, there is no solution to "[6th] + [2nd] = 8". Therefore, we have [6th] = 2, and thus [2nd] = 6.

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

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

(1) Next, we determine the value of [4th]. Claim that indeed [4th] = 3.

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

(2) If [4th] = 5 or 1, then by ✅【4th → a, 0th → b, |a-b|=2】 we have [0th] = 3:

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

Then 3 is in corners, which contradictis ✅【⟨? ⋯ 3 ⋯ (?+2) ⋯⟩ (?≠3,1)】.

(3) Else, if [4th] = 4, then ✅【4th → a, 0th → b, |a-b|=2】 cannot be matched:

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

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

(4) Else, if [4th] = 0:

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

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

then we cannot match ✅【4th → a, 0th → b, |a-b|=2】.

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

Combining (1), (2), (3), (4), we have verified that [4th] = 3.

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

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

Now there is only one way to match ✅【4th → a, 1st → b, a+b=4+5n】:

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

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

and only one way to match ✅【4th → a, 0th → b, |a-b|=2】:

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

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

Finally, as ✅【⟨? ⋯ 3 ⋯ (?+2) ⋯⟩ (?≠3,1)】 implies that 3 is at the left of 4, we finish by

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

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

Q.E.D.

#125034_v2.1