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