2023-12-17 Q1(m=6)

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

⟨6th 5th 4th 3rd 2nd 1st 0th⟩

✅Match

⟨⋯ 0 ⋯ 2 ⋯ 3 ⋯⟩

⟦4,5⟧ ∋ 1

⟨⋯ 4 ⋯ 2 ⋯ 1 ⋯⟩

⟨⋯ ? ⋯ 6 ⋯ (?−1)⟩ (?≠6)

⛔Avoid

⟨ ⁶ᵗʰa ⁵ᵗʰb ⁴ᵗʰc         ⟩, (abc)₁₀ ≤ 163

1st → a, 0th → b, a+b=8

#125034_v2.2

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

Proof of 2023-12-17 Q1(m=6)
═══════════════════════════

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

(1) We begin with determining the value of [0th]. We have [0th] = 3, because:

(1.1) By ✅「⟨⋯ ? ⋯ 6 ⋯ (?−1)⟩ (?≠6)」, we have [0th] != 6,5;

(1.2) By ✅「⟨⋯ 4 ⋯ 2 ⋯ 1 ⋯⟩」, we have [0th] != 4,2;

(1.3) By ✅「⟦4,5⟧ ∋ 1」, we have [0th] != 1;

(1.4) By ✅「⟨⋯ 0 ⋯ 2 ⋯ 3 ⋯⟩」, we have [0th] != 0.

Accordingly, our first step is:

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

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

Observe that ✅「⟨⋯ 0 ⋯ 2 ⋯ 3 ⋯⟩」, ✅「⟦4,5⟧ ∋ 1」, and ✅「⟨⋯ 4 ⋯ 2 ⋯ 1 ⋯⟩」 then imply the following patterns:

(2) 

⟨⋯ 0 ⋯ 2 ⋯ 1 ⋯ 3⟩, and
⟨⋯ 4 ⋯ 2 ⋯ 1 ⋯ 5 ⋯ 3⟩.

As a consequence, 5 is at the right of 0,1,2,4, so 5 can only placed at the 2nd or 1st position. Since ⛔「1st → a, 0th → b, a+b=8」 forbids 5 = [1st], we have 5 = [2nd].

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

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

Now, the places at the left of 5 are occupied by 0,1,2,4, so [1st] = 6.

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

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

Back to (2), we need to match:

⟨⋯ 0 ⋯ 2 ⋯ 1 ⋯⟩, and
⟨⋯ 4 ⋯ 2 ⋯ 1 ⋯⟩,

therefore ([4th], [3rd]) = (2,1):

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

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

Finally, to avoid ⛔「⟨ ⁶ᵗʰa ⁵ᵗʰb ⁴ᵗʰc         ⟩, (abc)₁₀ ≤ 163」, we have [6th] != 0. As a result, we have

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

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

Q.E.D.

#125034_v2.2