2024-10-01 Q1(m=6)

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

⟨6th 5th 4th 3rd 2nd 1st 0th⟩

✅Match

Jump(2,3) = 2

6th → a, 5th → b, |a-b|=3

⟨ ⁶ᵗʰ↓ ⁵ᵗʰ↓ ⁴ᵗʰ↑ ³ʳᵈ↑ ²ⁿᵈ↑ ¹ˢᵗ↓ ⁰ᵗʰ↓ ⟩ after 6−⟨⇌⟩

5th → a, 0th → b, a+b=0+5n

5th → a, 2nd → b, a+b=7

#125034_v2.6

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

Proof of 2024-10-01 Q1(m=6)
═══════════════════════════

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

By ✅「5th → a, 2nd → b, a+b=7」, there are six possibilities for ([5th], [2nd]):

(A) (1,6)
(B) (2,5)
(C) (3,4)
(D) (4,3)
(E) (5,2)
(F) (6,1)

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

Observing that in order to match ✅「Jump(2,3) = 2」, case (B)-(E) do not hold. Case (A) also does not hold, for otherwise we have

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

and then we cannot have a "↓" at 5th after 6−⟨⇌⟩, contradicting ✅「⟨ ⁶ᵗʰ↓ ⁵ᵗʰ↓ ⁴ᵗʰ↑ ³ʳᵈ↑ ²ⁿᵈ↑ ¹ˢᵗ↓ ⁰ᵗʰ↓ ⟩ after 6−⟨⇌⟩」.

Therefore, case (F) holds. We get

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

--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│   │ 2 │   │ 3 │ 0 │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘

Then, it follows from ✅「6th → a, 5th → b, |a-b|=3」 that [6th] = 3:

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

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

Using ✅「5th → a, 0th → b, a+b=0+5n」 and ✅「Jump(2,3) = 2」, we get [0th] = 4 and [3rd] = 2 as well:

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

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

Finally, since 0 cannot be at a "↓" position of ✅「⟨ ⁶ᵗʰ↓ ⁵ᵗʰ↓ ⁴ᵗʰ↑ ³ʳᵈ↑ ²ⁿᵈ↑ ¹ˢᵗ↓ ⁰ᵗʰ↓ ⟩ after 6−⟨⇌⟩」, we finish by

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

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

Q.E.D.

#125034_v2.6