Rearrange the digits in ⟨1263045⟩ to meet the rules below.
⟨6th 5th 4th 3rd 2nd 1st 0th⟩
✅Match
⟨⋯ 3 ⋯ ? 5 ⋯ (?−2)⟩ (?≠5)
⟨⋯ 3 ⋯ ? 0 ⋯ (?+1)⟩ (?≠0)
⟨⋯ 3 ⋯ 1 ⋯⟩
6th → a, 1st → b, a+b=3+5n
⛔Avoid
⟦1,3⟧ ∋ 0
⟨⋯ a ⋯ 0 ⋯⟩, a = 2|4|6
#125034_v2.0
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
Step 1 │ │ │ │ │ │ │ 2 │
├───┼───┼───┼───┼───┼───┼───┤
Step 2 │ 3 │ │ │ │ │ │ 2 │
├───┼───┼───┼───┼───┼───┼───┤
Step 3 │ 3 │ 1 │ │ │ │ │ 2 │
├───┼───┼───┼───┼───┼───┼───┤
Step 4 │ 3 │ 1 │ 0 │ │ │ │ 2 │
├───┼───┼───┼───┼───┼───┼───┤
Step 5 │ 3 │ 1 │ 0 │ │ │ 5 │ 2 │
├───┼───┼───┼───┼───┼───┼───┤
Step 6 │ 3 │ 1 │ 0 │ │ 4 │ 5 │ 2 │
├───┼───┼───┼───┼───┼───┼───┤
Step 7 │ 3 │ 1 │ 0 │ 6 │ 4 │ 5 │ 2 │
└───┴───┴───┴───┴───┴───┴───┘
Proof
═════
Notation: if nth -> a, then we write [nth] = a.
(1) By ✅【⟨⋯ 3 ⋯ ? 5 ⋯ (?−2)⟩ (?≠5)】 and ✅【⟨⋯ 3 ⋯ ? 0 ⋯ (?+1)⟩ (?≠0)】, we see that [0th] can only be 4 or 2.
If [0th]=4, then the ? in ✅【⟨⋯ 3 ⋯ ? 0 ⋯ (?+1)⟩ (?≠0)】 is 3. In view of ✅【⟨⋯ 3 ⋯ 1 ⋯⟩】, we have
⟨⋯ 30 ⋯ 1 ⋯⟩
which matches ⛔「⟦1,3⟧ ∋ 0」. Therefore, we cannot have [0th]=4 in (1). So, [0th]=2.
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│ 0■│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
Step 1 │ │ │ │ │ │ │ 2 │
└───┴───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ │ 6 │ 3 │ 0 │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
------------------------------
Now, we consider where to place 3. By ✅【⟨⋯ 3 ⋯ ? 5 ⋯ (?−2)⟩ (?≠5)】 and ✅【⟨⋯ 3 ⋯ ? 0 ⋯ (?+1)⟩ (?≠0)】, we have
(2) ⟨⋯ 3 ⋯ 45 ⋯ 2⟩, and
(3) ⟨⋯ 3 ⋯ 10 ⋯ 2⟩,
so 3=[6th] or [5th], for otherwise there are not enough places for 4,5,1,0.
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ │ │ ▬ │ ▬ │ ▬ │ ▬ │ 2 │
└───┴───┴───┴───┴───┴───┴───┘
Claim that 3=[6th]. If on the contrary 3=[5th], then 4th, 3rd, 2nd, and 1st are occupied by 4,5,1,0. Consequently, [6th]=6:
┌───┬───┬───┬───┬───┬───┬───┐
│ 6▲│ 5▲│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 6 │ 3 │ ▬ │ ▬ │ ▬ │ ▬ │ 2 │
└───┴───┴───┴───┴───┴───┴───┘
It would match ⛔「⟨⋯ a ⋯ 0 ⋯⟩, a = 2|4|6」, however, which is a contradiction.
This verifies our claim. It follows that
┌───┬───┬───┬───┬───┬───┬───┐
│ 6■│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ │ │ │ │ │ │ 2 │
├───┼───┼───┼───┼───┼───┼───┤
Step 2 │ 3 │ │ │ │ │ │ 2 │
└───┴───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ │ 6 │ │ 0 │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
------------------------------
Next, by (2), (3), and ⛔「⟨⋯ a ⋯ 0 ⋯⟩, a = 2|4|6」, we have
⟨3 ⋯ 10 ⋯ 45 ⋯ 2⟩, and
⟨3 ⋯ 10 ⋯ 6 ⋯ 2⟩.
In particular, 1 is at the left of 0,4,5,6. Hence, we obtain
┌───┬───┬───┬───┬───┬───┬───┐
│6th│ 5■│ 4■│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 3 │ │ │ │ │ │ 2 │
├───┼───┼───┼───┼───┼───┼───┤
Step 3 │ 3 │ 1 │ │ │ │ │ 2 │
├───┼───┼───┼───┼───┼───┼───┤
Step 4 │ 3 │ 1 │ 0 │ │ │ │ 2 │
└───┴───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ 6 │ │ │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
------------------------------
Using ✅【6th → a, 1st → b, a+b=3+5n】 and (2), we get
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│ 2■│ 1■│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 3 │ 1 │ 0 │ │ │ │ 2 │
├───┼───┼───┼───┼───┼───┼───┤
Step 5 │ 3 │ 1 │ 0 │ │ │ 5 │ 2 │
├───┼───┼───┼───┼───┼───┼───┤
Step 6 │ 3 │ 1 │ 0 │ │ 4 │ 5 │ 2 │
└───┴───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ 6 │ │ │ │ │
└───┴───┴───┴───┴───┴───┴───┘
Plainly, we reach ⟨3106452⟩ at the end. This completes the proof.
Q.E.D.
#125034_v2.0