Rearrange the digits in ⟨1263045⟩ to meet the rules below.
⟨6th 5th 4th 3rd 2nd 1st 0th⟩
✅Match
⟨ ⁴ᵗʰa ³ʳᵈb ⁰ᵗʰc ⟩, a > b > c
⟨? ⋯ 1 (?+2) ⋯ 4 ⋯⟩ (?≠1)
⟨ ⁶ᵗʰ↓ ⁵ᵗʰ↑ ³ʳᵈ↑ ⟩ after ⟨→⟩
⛔Avoid
Jump(2,6) ≥ 1
⟨⋯ 2 ⋯ 6 ⋯⟩
6th|4th|2nd → 0
#125034_v2.2
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ 3 │ │ │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 2 │ 3 │ │ │ │ 6 │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 3 │ 3 │ │ │ │ 6 │ 2 │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 4 │ 3 │ │ │ │ 6 │ 2 │ 0 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 5 │ 3 │ 1 │ │ │ 6 │ 2 │ 0 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 6 │ 3 │ 1 │ 5 │ │ 6 │ 2 │ 0 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 7 │ 3 │ 1 │ 5 │ 4 │ 6 │ 2 │ 0 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
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Proof of 2023-12-05 Q1(m=6)
═══════════════════════════
Notation: if nth -> a, then we write [nth] = a.
We begin with determining the value of [6th]. To match ✅「⟨? ⋯ 1 (?+2) ⋯ 4 ⋯⟩ (?≠1)」, the "?" there can only be 0|2|3. It is not 0 by ⛔「6th|4th|2nd → 0」, and it is not 2 by ⛔「⟨⋯ 2 ⋯ 6 ⋯⟩」. It follows that
┌───┬───┬───┬───┬───┬───┬───┐
│ 6■│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ 3 │ │ │ │ │ │ │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ 6 │ │ 0 │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Note that the pattern ✅「⟨? ⋯ 1 (?+2) ⋯ 4 ⋯⟩ (?≠1)」 becomes
(1) ⟨3 ⋯ 15 ⋯ 4 ⋯⟩.
Also, note that to avoid ⛔「Jump(2,6) ≥ 1」 and ⛔「⟨⋯ 2 ⋯ 6 ⋯⟩」, we need to match the following pattern:
(2) ⟨⋯ 62 ⋯⟩.
We proceed to determine where to place 6,2. We cannot place 6 at a "↑" position of ✅「⟨ ⁶ᵗʰ↓ ⁵ᵗʰ↑ ³ʳᵈ↑ ⟩ after ⟨→⟩」, so there are only three possibilities:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
(2.1) │ 3 │ │ 6 │ 2 │ │ │ │
├───┼───┼───┼───┼───┼───┼───┤
(2.2) │ 3 │ │ │ │ 6 │ 2 │ │
├───┼───┼───┼───┼───┼───┼───┤
(2.3) │ 3 │ │ │ │ │ 6 │ 2 │
└───┴───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ │ │ │ 0 │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
We claim that case (2.2) holds actually.
------------------------------
Case (2.1):
If (2.1) holds, then we have only one way to match (1):
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│*4 │*3 │ 2▲│ 1▲│ 0▲│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 3 │ │ 6 │ 2 │ │ │ │
├───┼───┼───┼───┼───┼───┼───┤
│ 3 │ │ 6 │ 2 │ 1 │ 5 │ 4 │
└───┴───┴───┴───┴───┴───┴───┘
But then we would have 6th ↑ after ⟨→⟩, which contradicts ✅「⟨ ⁶ᵗʰ↓ ⁵ᵗʰ↑ ³ʳᵈ↑ ⟩ after ⟨→⟩」.
Case (2.3):
If (2.3) holds, then to match (1), we have (1,5) = ([5th], [4th]) or ([4th], [3rd]):
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│*1 │*0 │
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 3 │ 1 │ 5 │ │ │ 6 │ 2 │
├───┼───┼───┼───┼───┼───┼───┤
│ 3 │ │ 1 │ 5 │ │ 6 │ 2 │
└───┴───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ │ │ 0 │ 4 │ │
└───┴───┴───┴───┴───┴───┴───┘
In view of ✅「⟨ ⁴ᵗʰa ³ʳᵈb ⁰ᵗʰc ⟩, a > b > c」, we need [4th] > [3rd], so the second case above is not possible. The preceding pattern implies [3rd] > [0th] too. A fortiori [3rd] != 0, so we get
┌───┬───┬───┬───┬───┬───┬───┐
│6th│ 5▲│ 4▲│3rd│ 2▲│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 3 │ 1 │ 5 │ │ │ 6 │ 2 │
├───┼───┼───┼───┼───┼───┼───┤
│ 3 │ 1 │ 5 │ │ 0 │ 6 │ 2 │
└───┴───┴───┴───┴───┴───┴───┘
But then we match ⛔「6th|4th|2nd → 0」, which is a contradiction.
------------------------------
We have verified that case (2.2) holds. Consequently
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│ 2■│ 1■│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 3 │ │ │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 2 │ 3 │ │ │ │ 6 │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 3 │ 3 │ │ │ │ 6 │ 2 │ │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ │ │ │ 0 │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Next, we consider the value of [0th]. To match ✅「⟨ ⁶ᵗʰ↓ ⁵ᵗʰ↑ ³ʳᵈ↑ ⟩ after ⟨→⟩」, we need [0th] < [6th] = 3, so [0th] = 0 or 1. In view of (1), we see that [0th] != 1. Hence
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│ 0■│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 3 │ │ │ │ 6 │ 2 │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 4 │ 3 │ │ │ │ 6 │ 2 │ 0 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ │ │ │ │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Finally, there is only one way to match (1). We finish by
┌───┬───┬───┬───┬───┬───┬───┐
│6th│ 5■│ 4■│ 3■│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 3 │ │ │ │ 6 │ 2 │ 0 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 5 │ 3 │ 1 │ │ │ 6 │ 2 │ 0 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 6 │ 3 │ 1 │ 5 │ │ 6 │ 2 │ 0 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 7 │ 3 │ 1 │ 5 │ 4 │ 6 │ 2 │ 0 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.2