Rearrange the digits in ⟨1263045⟩ to meet the rules below.
⟨6th 5th 4th 3rd 2nd 1st 0th⟩
✅Match
6th → a, 4th → b, |a-b|=1
⟨⋯ 1 ⋯ 2 ⋯ 5 ⋯ 6 ⋯⟩
⛔Avoid
⟨⋯ 5 ⋯ a ⋯⟩, a = 2|3|4
⟨⋯ 1 ⋯ ? 5 ⋯ (?+4)⟩ (?≠1)
⟨⋯ 2 ⋯ a ⋯⟩, a = 1|4
⟨? 4 ⋯ (?−1) ⋯ 6 ⋯⟩ (?≠4,5)
3rd → 0|2|4|6
#125034_v2.3
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ │ │ 3 │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ │ 2 │ 3 │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 3 │ 1 │ │ 2 │ 3 │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 4 │ 1 │ 4 │ 2 │ 3 │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 5 │ 1 │ 4 │ 2 │ 3 │ 5 │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 6 │ 1 │ 4 │ 2 │ 3 │ 5 │ 6 │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 7 │ 1 │ 4 │ 2 │ 3 │ 5 │ 6 │ 0 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
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Proof of 2024-01-30 Q1(m=6)
═══════════════════════════
Notation: if nth -> a, then we write [nth] = a.
We begin with the 3rd position. To avoid ⛔「3rd → 0|2|4|6」, we need
(1) [3rd] = 1|3|5.
We show that [3rd] = 3 actually.
------------------------------
(1.1) If [3rd] = 1, then by ✅「⟨⋯ 1 ⋯ 2 ⋯ 5 ⋯ 6 ⋯⟩」, we have
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│ 3▲│ 2▲│ 1▲│ 0▲│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ │ │ │ 1 │ 2 │ 5 │ 6 │
└───┴───┴───┴───┴───┴───┴───┘
which contradicts ⛔「⟨⋯ 1 ⋯ ? 5 ⋯ (?+4)⟩ (?≠1)」.
(1.2) Else if [3rd] = 5, then to avoid ⛔「⟨⋯ 5 ⋯ a ⋯⟩, a = 2|3|4」, we have
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ * │ * │ * │ 5 │ │ │ │
└───┴───┴───┴───┴───┴───┴───┘
where the "*" are occupied by 2,3,4. But to match ✅「⟨⋯ 1 ⋯ 2 ⋯ 5 ⋯ 6 ⋯⟩」, we also need 1 at the left of 5. This shows a contradiction.
------------------------------
Therefore, it follows from (1), (1.1), and (1.2) that [3rd] = 3.
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│ 3■│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ │ │ 3 │ │ │ │▒
└───┴───┴───┴───┴───┴───┴───┘▒
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--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ 6 │ │ 0 │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Next, we consider where to place 2. To match ✅「⟨⋯ 1 ⋯ 2 ⋯ 5 ⋯ 6 ⋯⟩」, we need 2 = [5th] | [4th] | [2nd]:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│*5 │*4 │3rd│*2 │1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ │ ● │ ● │ 3 │ ● │ │ │
└───┴───┴───┴───┴───┴───┴───┘
(2) We show that 2 = [4th].
------------------------------
(2.1) If 2 = [5th], then by ✅「⟨⋯ 1 ⋯ 2 ⋯ 5 ⋯ 6 ⋯⟩」, we have [6th] = 1:
┌───┬───┬───┬───┬───┬───┬───┐
│ 6▲│ 5▲│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 1 │ 2 │ │ 3 │ │ │ │
└───┴───┴───┴───┴───┴───┴───┘
but then we would match ⛔「⟨⋯ 2 ⋯ a ⋯⟩, a = 1|4」.
(2.2) Else if 2 = [2nd], then by ✅「⟨⋯ 1 ⋯ 2 ⋯ 5 ⋯ 6 ⋯⟩」, we have
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│ 2▲│ 1▲│ 0▲│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ │ │ │ 3 │ 2 │ 5 │ 6 │
└───┴───┴───┴───┴───┴───┴───┘
and we would match ⛔「⟨⋯ 1 ⋯ ? 5 ⋯ (?+4)⟩ (?≠1)」.
------------------------------
By (2.1) and (2.2), we have verified our claim in (2). Accordingly, we have
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│ 4■│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ │ │ 3 │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ │ 2 │ 3 │ │ │ │▒
└───┴───┴───┴───┴───┴───┴───┘▒
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--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ │ 6 │ │ 0 │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Then, by ✅「6th → a, 4th → b, |a-b|=1」, plainly we have [6th] = 1:
┌───┬───┬───┬───┬───┬───┬───┐
│ 6■│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ │ 2 │ 3 │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 3 │ 1 │ │ 2 │ 3 │ │ │ │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ 6 │ │ 0 │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Next, to avoid ⛔「⟨⋯ 2 ⋯ a ⋯⟩, a = 1|4」, we have to place 4 at 5th:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│ 5■│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 1 │ │ 2 │ 3 │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 4 │ 1 │ 4 │ 2 │ 3 │ │ │ │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ 6 │ │ 0 │ │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Let us consider where to place 5. By ✅「⟨⋯ 1 ⋯ 2 ⋯ 5 ⋯ 6 ⋯⟩」, it is at [2nd] or [1st]. We have 5 != [1st], for otherwise we get
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│ 2▲│ 1▲│ 0▲│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 1 │ 4 │ 2 │ 3 │ │ 5 │ │
├───┼───┼───┼───┼───┼───┼───┤
│ 1 │ 4 │ 2 │ 3 │ │ 5 │ 6 │
├───┼───┼───┼───┼───┼───┼───┤
│ 1 │ 4 │ 2 │ 3 │ 0 │ 5 │ 6 │
└───┴───┴───┴───┴───┴───┴───┘
which matches ⛔「⟨? 4 ⋯ (?−1) ⋯ 6 ⋯⟩ (?≠4,5)」. Therefore, 5 = [2nd]:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│ 2■│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 1 │ 4 │ 2 │ 3 │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 5 │ 1 │ 4 │ 2 │ 3 │ 5 │ │ │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ 6 │ │ 0 │ │ │
└───┴───┴───┴───┴───┴───┴───┘
Finally, noting that there is only one way to avoid ⛔「⟨? 4 ⋯ (?−1) ⋯ 6 ⋯⟩ (?≠4,5)」, we finish by
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│ 1■│ 0■│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 1 │ 4 │ 2 │ 3 │ 5 │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 6 │ 1 │ 4 │ 2 │ 3 │ 5 │ 6 │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 7 │ 1 │ 4 │ 2 │ 3 │ 5 │ 6 │ 0 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.3