2024-09-24 Q1(m=6)

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

⟨6th 5th 4th 3rd 2nd 1st 0th⟩

✅Match

6th → a, 4th → b, ab=2+6n

⟨⋯ ? ⋯ 0 ⋯ (?+3)⟩ (?≠0)

⟨⋯ 1 ⋯ 4 ⋯ 2 ⋯ 3 ⋯⟩

Sim⟨ ⁶ᵗʰ5 ⁵ᵗʰ4 ⁴ᵗʰ0 ³ʳᵈ2 ²ⁿᵈ1 ¹ˢᵗ6 ⁰ᵗʰ3 ⟩ = 2

⛔Avoid

⟨ ⁶ᵗʰa ⁵ᵗʰb           ⟩, max⟦a,b⟧ = 5

⟨⋯ 2 ⋯ ? 6 ⋯ (?+2)⟩ (?≠4)

#125034_v2.6

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

Proof of 2024-09-24 Q1(m=6)
═══════════════════════════

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

By ✅「⟨⋯ 1 ⋯ 4 ⋯ 2 ⋯ 3 ⋯⟩」, 4 is not at the right corner. Therefore, ✅「⟨⋯ ? ⋯ 0 ⋯ (?+3)⟩ (?≠0)」 implies one of the following holds:

(1) 

(a) ⟨⋯ 2 ⋯ 0 ⋯ 5⟩; or
(b) ⟨⋯ 3 ⋯ 0 ⋯ 6⟩.

Combining this with ✅「⟨⋯ 1 ⋯ 4 ⋯ 2 ⋯ 3 ⋯⟩」, we see that no matter which case happens, there are at least four digits at the right of 4, and at least one digit at the left of 4. So, 4 = [5th] or [4th]:

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

(2) We show that 4 = [5th] actually.

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

Suppose on the contrary 4 = [4th]:

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

Then, by ✅「6th → a, 4th → b, ab=2+6n」, we have [6th] = 2|5. This is a contradiction however, because

(2.1) if [6th] = 2, then we cannot match ✅「⟨⋯ 1 ⋯ 4 ⋯ 2 ⋯ 3 ⋯⟩」:

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

(2.2) Else if [6th] = 5, then to match ✅「⟨⋯ 1 ⋯ 4 ⋯ 2 ⋯ 3 ⋯⟩」 we need [5th] = 1:

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

but then we match ⛔「⟨ ⁶ᵗʰa ⁵ᵗʰb           ⟩, max⟦a,b⟧ = 5」.

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

We have verified our claim in (2). Accordingly, we get

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

--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ 6 │ 3 │ 0 │   │ 5 │
└───┴───┴───┴───┴───┴───┴───┘

Then, using ✅「⟨⋯ 1 ⋯ 4 ⋯ 2 ⋯ 3 ⋯⟩」, we get [6th] = 1. By ✅「6th → a, 4th → b, ab=2+6n」, we get [4th] = 2 as well:

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

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

To continue, we consider ✅「Sim⟨ ⁶ᵗʰ5 ⁵ᵗʰ4 ⁴ᵗʰ0 ³ʳᵈ2 ²ⁿᵈ1 ¹ˢᵗ6 ⁰ᵗʰ3 ⟩ = 2」. To match it, we need exactly one of the following:

(3) 

(a) [1st] = 6; or 
(b) [0th] = 3.

Observe that (3) cannot happen if case (1)(b) holds. Therefore, case (1)(a) holds, and we get [0th] = 5:

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

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

Case (3)(b) becomes impossible. So, case (3)(a) holds:

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

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

Finally, to avoid ⛔「⟨⋯ 2 ⋯ ? 6 ⋯ (?+2)⟩ (?≠4)」, we finish by

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

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

Q.E.D.

#125034_v2.6