Q1(m=6) 2023-10-18

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

⟨6th 5th 4th 3rd 2nd 1st 0th⟩

✅Match

5th → a, 0th → b, ab=4+6n

⟨ ⁶ᵗʰd ⁵ᵗʰc     ²ⁿᵈa ¹ˢᵗb   ⟩, a > b > c > d

Jump(2,5) = 4

⟨⋯ 0 ⋯ ? 6 ⋯ (?+2)⟩ (?≠4)

⛔Avoid

4th → 1|2|4|5

#125034_v2.0

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

Proof of 2023-10-18 Q1(m=6)
═══════════════════════════

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

We start with ✅【Jump(2,5) = 4】. It implies two possibilities:

    ┌───┬───┬───┬───┬───┬───┬───┐
    │ 6 │ 5 │ 4 │ 3 │ 2 │ 1 │ 0 │
    ╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
(1) │ x │   │   │   │   │ y │   │
    ├───┼───┼───┼───┼───┼───┼───┤
(2) │   │ x │   │   │   │   │ y │
    └───┴───┴───┴───┴───┴───┴───┘

(3) where {x,y}={2,5}. By ✅【⟨ ⁶ᵗʰd ⁵ᵗʰc     ²ⁿᵈa ¹ˢᵗb   ⟩, a > b > c > d】 we have x=2 and y=5. We claim that (2) holds actually.

Suppose (1) holds on the contrary. ✅【⟨ ⁶ᵗʰd ⁵ᵗʰc     ²ⁿᵈa ¹ˢᵗb   ⟩, a > b > c > d】 would then imply [2nd]=6:

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

--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │   │   │ 3 │ 0 │ 4 │   │
└───┴───┴───┴───┴───┴───┴───┘

and then ✅【5th → a, 0th → b, ab=4+6n】 implies

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

with {E,F}={1,4}. As [0th]≥2 by ✅【⟨⋯ 0 ⋯ ? 6 ⋯ (?+2)⟩ (?≠4)】, we have

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

But then [5th]<[6th], contradicting ✅【⟨ ⁶ᵗʰd ⁵ᵗʰc     ²ⁿᵈa ¹ˢᵗb   ⟩, a > b > c > d】. This verifies our claim in (3). Accordingly we reach

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

--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │   │ 6 │ 3 │ 0 │ 4 │   │
└───┴───┴───┴───┴───┴───┴───┘

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

(4) Next we utilize ⛔「4th → 1|2|4|5」, which is equivalent to [4th]=0|3|6. Note that ✅【⟨⋯ 0 ⋯ ? 6 ⋯ (?+2)⟩ (?≠4)】 rejects [4th]=6. We claim that it is 0 actually.

If not, then [4th]=3 and we have

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

where by ✅【⟨ ⁶ᵗʰd ⁵ᵗʰc     ²ⁿᵈa ¹ˢᵗb   ⟩, a > b > c > d】 we have A>B>2. Since the idle are

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

we have

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

This would never match ✅【⟨⋯ 0 ⋯ ? 6 ⋯ (?+2)⟩ (?≠4)】, which is a contradiction. Accordingly, back to (4), we have

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

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

Now, as [6th]<[5th] by ✅【⟨ ⁶ᵗʰd ⁵ᵗʰc     ²ⁿᵈa ¹ˢᵗb   ⟩, a > b > c > d】, we have 

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

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

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

The rest are straightforward. By ✅【⟨⋯ 0 ⋯ ? 6 ⋯ (?+2)⟩ (?≠4)】, we see that ?=3, thus 3,6 come together in this order. As [1st]≠6 by ✅【⟨ ⁶ᵗʰd ⁵ᵗʰc     ²ⁿᵈa ¹ˢᵗb   ⟩, a > b > c > d】, we conclude that

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

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

Q.E.D.

#125034_v2.0