2024-08-27 Q1(m=6)

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

⟨6th 5th 4th 3rd 2nd 1st 0th⟩

✅Match

⟨⋯ 0 ⋯ 4 ⋯⟩

⟨⋯ Perm(0,5) ⋯⟩

⟨⋯ Perm(2,3,6) ⋯⟩

Sim⟨ ⁶ᵗʰ4 ⁵ᵗʰ3 ⁴ᵗʰ6 ³ʳᵈ5 ²ⁿᵈ1 ¹ˢᵗ2 ⁰ᵗʰ0 ⟩ = 1

⛔Avoid

max ⊢2⊣ ≥ 4

5th|4th|3rd|2nd|0th → 0

⟨⋯ 0 ⋯ ? ⋯ 1 (?−1)⟩ (?≠1,2)

4th → a, 1st → b, ab=8

#125034_v2.6

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

Proof of 2024-08-27 Q1(m=6)
═══════════════════════════

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

We begin with an observation. As ⛔「max ⊢2⊣ ≥ 4」 implies that 2 and 6 are not adjacent, so in view of ✅「⟨⋯ Perm(2,3,6) ⋯⟩」, we have the following required pattern:

(1) ⟨⋯ 2 3 6 ⋯⟩ or ⟨⋯ 6 3 2 ⋯⟩.

We consider where to place 0. By ⛔「5th|4th|3rd|2nd|0th → 0」, we have 0 = [6th] | [1st].

(2) We claim that 0 = [1st] actually.

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

For, suppose on the contrary 0 = [6th]. Combining this with ✅「⟨⋯ Perm(0,5) ⋯⟩」, we get

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

To match ✅「Sim⟨ ⁶ᵗʰ4 ⁵ᵗʰ3 ⁴ᵗʰ6 ³ʳᵈ5 ²ⁿᵈ1 ¹ˢᵗ2 ⁰ᵗʰ0 ⟩ = 1」 as well, we need to have exactly one positional digit agreed with ⟨4365120⟩. There are three choices:

(i) 6 = [4th];
(ii) 1 = [2nd];
(iii) 2 = [1st].

If case (ii) holds, then we do not have three consecutive boxes for (1):

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

which is a contradiction. So, case (i) or (iii) holds instead. Combining them with (1) respectively, we get

      ┌───┬───┬───┬───┬───┬───┬───┐
      │6th│5th│4th│3rd│2nd│1st│0th│
      ╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
(3.1) │ 0 │ 5 │ 6 │ 3 │ 2 │   │   │
      ├───┼───┼───┼───┼───┼───┼───┤
(3.2) │ 0 │ 5 │   │ 6 │ 3 │ 2 │   │
      └───┴───┴───┴───┴───┴───┴───┘

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

By ⛔「max ⊢2⊣ ≥ 4」, 2 and 4 are not adjacent, so the above become

      ┌───┬───┬───┬───┬───┬───┬───┐
      │6th│5th│4th│3rd│2nd│1st│0th│
      ╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
(3.3) │ 0 │ 5 │ 6 │ 3 │ 2 │ 1 │ 4 │
      ├───┼───┼───┼───┼───┼───┼───┤
(3.4) │ 0 │ 5 │ 4 │ 6 │ 3 │ 2 │ 1 │
      └───┴───┴───┴───┴───┴───┴───┘

This is a contradiction, however, as case (3.3) matches ⛔「⟨⋯ 0 ⋯ ? ⋯ 1 (?−1)⟩ (?≠1,2)」 while case (3.4) matches ⛔「4th → a, 1st → b, ab=8」.

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

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

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

--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ 6 │ 3 │   │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘

Then, to match ✅「⟨⋯ 0 ⋯ 4 ⋯⟩」 and ✅「⟨⋯ Perm(0,5) ⋯⟩」, we get

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

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

We consider ✅「Sim⟨ ⁶ᵗʰ4 ⁵ᵗʰ3 ⁴ᵗʰ6 ³ʳᵈ5 ²ⁿᵈ1 ¹ˢᵗ2 ⁰ᵗʰ0 ⟩ = 1」 again. To have exactly one agreed positional digit, we need

(I) 3 = [5th] and 6 != [4th]; or
(II) 3 != [5th] and 6 = [4th].

Observe that by (1), case (I) holds actually. We finish by

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

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

Q.E.D.

#125034_v2.6