2023-12-10 Q1(m=6)

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

⟨6th 5th 4th 3rd 2nd 1st 0th⟩

✅Match

⟨? ⋯ 1 (?+2) ⋯ 2 ⋯⟩ (?≠1)

Jump(4,5) = 2

⟨ ⁶ᵗʰa         ¹ˢᵗb   ⟩, (ab)₁₀ ≥ 13

6th|5th|4th|2nd|0th → 6

⛔Avoid

⟦2,4⟧ ∋ 1

5th → a, 2nd → b, a+b=2+4n

2nd → a, 1st → b, a+b=10

⟨? ⋯ 2 ⋯ (?−3) ⋯⟩ (?≠5)

#125034_v2.2

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

Proof of 2023-12-10 Q1(m=6)
═══════════════════════════

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

We begin with determining the value of [6th]. By ✅「⟨? ⋯ 1 (?+2) ⋯ 2 ⋯⟩ (?≠1)」, we have [6th] = 0|3|4. If [6th] = 4, then the preceding pattern becomes:

    ⟨4 ⋯ 16 ⋯ 2 ⋯⟩

Matching it would be a contradiction, however, as we need to avoid ⛔「⟦2,4⟧ ∋ 1」. Hence, [6th] = 0|3. To match ✅「⟨ ⁶ᵗʰa         ¹ˢᵗb   ⟩, (ab)₁₀ ≥ 13」, we nedd [6th] >= 1, thus [6th] = 3.

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

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

Now, note that to avoid ⛔「⟨? ⋯ 2 ⋯ (?−3) ⋯⟩ (?≠5)」, we need to match:

(1)     ⟨3 ⋯ 0 ⋯ 2 ⋯⟩

Also, note that ✅「⟨? ⋯ 1 (?+2) ⋯ 2 ⋯⟩ (?≠1)」 becomes:

(2)     ⟨3 ⋯ 15 ⋯ 2 ⋯⟩

In particular, 1,5 are adjacent to each other in this order. To match (2), there are four possibilities:

      ┌───┬───┬───┬───┬───┬───┬───┐
      │6th│5th│4th│3rd│2nd│1st│0th│
      ╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
(2.1) │ 3 │ 1 │ 5 │   │   │   │   │
      ├───┼───┼───┼───┼───┼───┼───┤
(2.2) │ 3 │   │ 1 │ 5 │   │   │   │
      ├───┼───┼───┼───┼───┼───┼───┤
(2.3) │ 3 │   │   │ 1 │ 5 │   │   │
      ├───┼───┼───┼───┼───┼───┼───┤
(2.4) │ 3 │   │   │   │ 1 │ 5 │ 2 │
      └───┴───┴───┴───┴───┴───┴───┘

We proceed to show that (2.1) holds actually.

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

Case (2.2):

If (2.2) holds, then by ✅「Jump(4,5) = 2」, we have

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

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

If 0 is at 5th or 2nd, then we would match ⛔「5th → a, 2nd → b, a+b=2+4n」. Hence, 0 = [1st].

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

--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│   │ 2 │ 6 │   │   │   │   │
└───┴───┴───┴───┴───┴───┴───┘

This results in a contradiction, however, as we cannot match (1) now.

Case (2.3):

If (2.3) holds, then by ✅「Jump(4,5) = 2」, we have

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

But then to match (2) we would match ⛔「⟦2,4⟧ ∋ 1」 too, which is a contradiction.

Case (2.4):

If (2.4) holds:

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

then unavoidably we would match ⛔「⟦2,4⟧ ∋ 1」.

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

We have verified that (2.1) holds indeed. Combining it with ✅「Jump(4,5) = 2」, we obtain

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

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

Now we consider where to place 6. By ✅「6th|5th|4th|2nd|0th → 6」, we see that 6 = [2nd] or [0th]. To avoid ⛔「2nd → a, 1st → b, a+b=10」, we need 6 = [0th].

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

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

Finally, recall that we need to match (1). We finish by

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

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

Q.E.D.

#125034_v2.2