2024-08-13 Q1(m=6)

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

⟨6th 5th 4th 3rd 2nd 1st 0th⟩

✅Match

{p6, p3, p1, p0} = ? + {0,1,3,4}

Jump(3,6) ≤ 1

2nd → a, 0th → b, ab=5+6n

⟨           ¹ˢᵗ↓   ⟩ after 3×⟨→⟩

⛔Avoid

⟨⋯ 6 ⋯ 2 ⋯ 0 ⋯ 4 ⋯⟩

6th|4th|3rd|2nd → 1

#125034_v2.6

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

Proof of 2024-08-13 Q1(m=6)
═══════════════════════════

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

To begin with, observe that ✅「2nd → a, 0th → b, ab=5+6n」 implies

(1) {[2nd], [0th]} = {1,5}.

So, in view of ⛔「6th|4th|3rd|2nd → 1」, we get

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

--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│   │ 2 │ 6 │ 3 │ 0 │ 4 │   │
└───┴───┴───┴───┴───┴───┴───┘

Let S := {p6, p3, p1, p0}. Given that 1 = [0th] ∈ S and 5 = [2nd] ∉ S, to match ✅「{p6, p3, p1, p0} = ? + {0,1,3,4}」 we need

(2) {p6, p3, p1, p0} = {0,1,3,4}.

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

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

where "▬" are occupied by 0,3,4.

It follows that {[5th], [4th]} = {2,6}. Since we need a "↓" at 1st to match ✅「⟨           ¹ˢᵗ↓   ⟩ after 3×⟨→⟩」, we have [4th] != 6. Consequently, we get

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

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

We check the value of [1st]. By ✅「⟨           ¹ˢᵗ↓   ⟩ after 3×⟨→⟩」, it is not 0, and by ✅「Jump(3,6) ≤ 1」, it is not 3. So, [1st] = 4:

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

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

Finally, to avoid ⛔「⟨⋯ 6 ⋯ 2 ⋯ 0 ⋯ 4 ⋯⟩」, we finish by

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

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

Q.E.D.

#125034_v2.6