2024-04-16 Q1(m=6)

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

⟨6th 5th 4th 3rd 2nd 1st 0th⟩

✅Match

3rd → a, 0th → b, ab=4+6n

⟨⋯ Perm(0,3,5) ⋯⟩

⟨     ⁴ᵗʰc ³ʳᵈa ²ⁿᵈb     ⟩, a > b > c

Jump(2,5) ≥ 3

#125034_v2.4

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

Proof of 2024-04-16 Q1(m=6)
═══════════════════════════

Notation: if Nth -> a, then we write pN = a.

To begin with, observe that ✅「3rd → a, 0th → b, ab=4+6n」 implies (p3)(p0) = 4|10. Therefore, we have two cases:

i) {p3,p0} = {2,5};
ii) {p3,p0} = {1,4}.

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

Note that if the former case holds, then we cannot match ✅「Jump(2,5) ≥ 3」. Therefore, the latter case holds actually.

Also, in view of ✅「⟨     ⁴ᵗʰc ³ʳᵈa ²ⁿᵈb     ⟩, a > b > c」, we cannot place 1 at 3rd. Consequently, we get (p3,p0) = (4,1):

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

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

Next, note that we need three consecutive boxes to match ✅「⟨⋯ Perm(0,3,5) ⋯⟩」. Accordingly

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

where "▬" ∈ {0,3,5} and "●" ∈ {2,6}. As 6 is the maximum digit, by ✅「⟨     ⁴ᵗʰc ³ʳᵈa ²ⁿᵈb     ⟩, a > b > c」 we cannot place it at 2nd. Therefore, 6 = p1, and so 2 = p2:

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

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

Now there is only one way to match ✅「Jump(2,5) ≥ 3」:

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

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

Finally, to match ✅「⟨     ⁴ᵗʰc ³ʳᵈa ²ⁿᵈb     ⟩, a > b > c」, we finish by

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

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

Q.E.D.

#125034_v2.4