Rearrange the digits in ⟨1263045⟩ to meet the rules below.
⟨6th 5th 4th 3rd 2nd 1st 0th⟩
✅Match
⟨ ⁵ᵗʰb ⁴ᵗʰc ⁰ᵗʰa ⟩, a > b > c
⟨? ⋯ 2 ⋯ (?−2) ⋯⟩ (?≠2,4)
Jump(0,5) = 0
⟨⋯ 3 ⋯ ? 6 ⋯ (?+4)⟩ (?≠2)
⛔Avoid
⟨⋯ Perm(0,4,6) ⋯⟩
#125034_v2.1
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
Step 1 │ 3 │ │ │ │ │ │ │
├───┼───┼───┼───┼───┼───┼───┤
Step 2 │ 3 │ │ │ │ │ │ 5 │
├───┼───┼───┼───┼───┼───┼───┤
Step 3 │ 3 │ │ │ │ │ 0 │ 5 │
├───┼───┼───┼───┼───┼───┼───┤
Step 4 │ 3 │ │ │ 1 │ │ 0 │ 5 │
├───┼───┼───┼───┼───┼───┼───┤
Step 5 │ 3 │ │ │ 1 │ 6 │ 0 │ 5 │
├───┼───┼───┼───┼───┼───┼───┤
Step 6 │ 3 │ 4 │ │ 1 │ 6 │ 0 │ 5 │
├───┼───┼───┼───┼───┼───┼───┤
Step 7 │ 3 │ 4 │ 2 │ 1 │ 6 │ 0 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Proof of 2023-10-27 Q1(m=6)
═══════════════════════════
Notation: if nth -> a, then we write [nth] = a.
(1) By ✅【⟨⋯ 3 ⋯ ? 6 ⋯ (?+4)⟩ (?≠2)】, we have [0th] = 4 or 5. It also implies that 6 is not in the corners. A fortiori, 6 != [6th].
(2) Therefore, the ? in ✅【⟨? ⋯ 2 ⋯ (?−2) ⋯⟩ (?≠2,4)】 can only can 5,3. It cannot be 5, for otherwise by ✅【Jump(0,5) = 0】 and (1), we have
┌───┬───┬───┬───┬───┬───┬───┐
│ 6▲│ 5▲│4th│3rd│2nd│1st│ 0▲│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 5 │ 0 │ │ │ │ │ 4 │
└───┴───┴───┴───┴───┴───┴───┘
and ✅【⟨⋯ 3 ⋯ ? 6 ⋯ (?+4)⟩ (?≠2)】 cannot be matched. Hence, back to (2), we have [6th] = 3 as our first step:
┌───┬───┬───┬───┬───┬───┬───┐
│ 6■│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
Step 1 │ 3 │ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ 6 │ │ 0 │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
------------------------------
Next, we claim that [0th] = 5.
(3) Suppose on the contrary [0th] != 5. Then by (1), we have
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│ 0▲│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 3 │ │ │ │ │ │ 4 │
└───┴───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ 6 │ │ 0 │ │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Furthermore, combining ✅【Jump(0,5) = 0】 with ✅【⟨⋯ 3 ⋯ ? 6 ⋯ (?+4)⟩ (?≠2)】, we have
<3 ⋯ 506 ⋯ 4>.
By ✅【⟨ ⁵ᵗʰb ⁴ᵗʰc ⁰ᵗʰa ⟩, a > b > c】, 5 cannot be at 5th or 4th. Hence, 5,0,6 are placed as follows:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│ 3▲│ 2▲│ 1▲│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 3 │ │ │ 5 │ 0 │ 6 │ 4 │
└───┴───┴───┴───┴───┴───┴───┘
This is a contradiction as ⛔「⟨⋯ Perm(0,4,6) ⋯⟩」 is matched. Accordingly, back to (3) and using ✅【Jump(0,5) = 0】, we have
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│ 1■│ 0■│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 3 │ │ │ │ │ │ │
├───┼───┼───┼───┼───┼───┼───┤
Step 2 │ 3 │ │ │ │ │ │ 5 │
├───┼───┼───┼───┼───┼───┼───┤
Step 3 │ 3 │ │ │ │ │ 0 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ 6 │ │ │ 4 │ │
└───┴───┴───┴───┴───┴───┴───┘
------------------------------
Now, we see that the ? in ✅【⟨⋯ 3 ⋯ ? 6 ⋯ (?+4)⟩ (?≠2)】 is 1, so we need 1,6 adjacent to each other in this order. There are three possibilies:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
(4) │ 3 │ 1 │ 6 │ │ │ 0 │ 5 │
├───┼───┼───┼───┼───┼───┼───┤
(5) │ 3 │ │ 1 │ 6 │ │ 0 │ 5 │
├───┼───┼───┼───┼───┼───┼───┤
(6) │ 3 │ │ │ 1 │ 6 │ 0 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
By ✅【⟨ ⁵ᵗʰb ⁴ᵗʰc ⁰ᵗʰa ⟩, a > b > c】, (4) is impossible. (5) is also impossible, for otherwise ✅【⟨? ⋯ 2 ⋯ (?−2) ⋯⟩ (?≠2,4)】 implies that the answer is <3216405>, which matches ⛔「⟨⋯ Perm(0,4,6) ⋯⟩」 however. Therefore, we have (6) as our next steps.
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│ 3■│ 2■│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 3 │ │ │ │ │ 0 │ 5 │
├───┼───┼───┼───┼───┼───┼───┤
Step 4 │ 3 │ │ │ 1 │ │ 0 │ 5 │
├───┼───┼───┼───┼───┼───┼───┤
Step 5 │ 3 │ │ │ 1 │ 6 │ 0 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ 2 │ │ │ │ 4 │ │
└───┴───┴───┴───┴───┴───┴───┘
Finally, to match ✅【⟨ ⁵ᵗʰb ⁴ᵗʰc ⁰ᵗʰa ⟩, a > b > c】, we reach <3421605>. This completes the proof.
Q.E.D.
#125034_v2.1