Rearrange the digits in ⟨1263045⟩ to meet the rules below.
⟨6th 5th 4th 3rd 2nd 1st 0th⟩
✅Match
3rd → a, 0th → b, a+b=6
6th|4th|2nd|1st|0th → 4
⟨⋯ Perm(0,6) ⋯⟩
⟨⋯ a ⋯ 6 ⋯⟩, a = 0|1|2|3
3rd → a, 2nd → b, |a-b|=1
⛔Avoid
⟨⋯ ? ⋯ 6 ⋯ (?+1)⟩ (?≠5)
max ⊢3⊣ = 5
⟨⋯ Perm(1,3,4) ⋯⟩
5th|4th|3rd|0th → 1
#125034_v2.5
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ │ │ │ │ │ 4 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ │ │ 2 │ │ │ 4 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 3 │ 0 │ │ │ 2 │ │ │ 4 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 4 │ 0 │ 6 │ │ 2 │ │ │ 4 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 5 │ 0 │ 6 │ │ 2 │ │ 5 │ 4 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 6 │ 0 │ 6 │ │ 2 │ 1 │ 5 │ 4 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 7 │ 0 │ 6 │ 3 │ 2 │ 1 │ 5 │ 4 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
Proof of 2024-05-21 Q1(m=6)
═══════════════════════════
Notation: if nth -> a, then we write [nth] = a.
By ✅「3rd → a, 0th → b, a+b=6」, we have
{[3rd],[0th]} = {0,6} | {1,5} | {2,4}.
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│ 3▲│2nd│1st│ 0▲│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ │ │ │ ? │ │ │ ? │
└───┴───┴───┴───┴───┴───┴───┘
Since 0,6 are adjacent by ✅「⟨⋯ Perm(0,6) ⋯⟩」, and 1 ∉ {[3rd], [0th]} by ⛔「5th|4th|3rd|0th → 1」, it is indeed
{[3rd],[0th]} = {2,4}.
Combining this with ✅「6th|4th|2nd|1st|0th → 4」, we get
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│ 3■│2nd│1st│ 0■│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ │ │ │ │ │ 4 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ │ │ 2 │ │ │ 4 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ │ 6 │ 3 │ 0 │ │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Next, we decide how to place 0,6. They are adjacent by ✅「⟨⋯ Perm(0,6) ⋯⟩」. So, to avoid ⛔「⟨⋯ ? ⋯ 6 ⋯ (?+1)⟩ (?≠5)」, we need to match the following pattern:
(1) ⟨⋯ Perm(0,6) ⋯ 3 ⋯ 4⟩.
It implies one of the following holds:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
(2) │ │ ▬ │ ▬ │ 2 │ │ │ 4 │
├───┼───┼───┼───┼───┼───┼───┤
(3) │ ▬ │ ▬ │ │ 2 │ │ │ 4 │
└───┴───┴───┴───┴───┴───┴───┘
where "▬" are occupied by 0,6.
We show that case (3) holds actually.
------------------------------
If on the contrary case (2) holds:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ │ ▬ │ ▬ │ 2 │ │ │ 4 │
└───┴───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ │ │ 3 │ │ │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
then in view of (1), we have
(4) S := {[2nd], [1st]} = {1,3} | {3,5}.
However, if the former holds, then we matches ⛔「⟨⋯ Perm(1,3,4) ⋯⟩」; else if the latter holds, then we matches ⛔「max ⊢3⊣ = 5」. This shows a contradiction.
------------------------------
We have verified that case (3) holds. Noting that 6 is not at the leftmost corner by ✅「⟨⋯ a ⋯ 6 ⋯⟩, a = 0|1|2|3」, we get
┌───┬───┬───┬───┬───┬───┬───┐
│ 6■│ 5■│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ │ │ 2 │ │ │ 4 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 3 │ 0 │ │ │ 2 │ │ │ 4 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 4 │ 0 │ 6 │ │ 2 │ │ │ 4 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ │ │ 3 │ │ │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
As the argument of (4) shows, we would get a contradiction if S = {1,3} or {3,5}. Therefore, S = {1,5}. In view of ✅「3rd → a, 2nd → b, |a-b|=1」, we get
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│ 2■│ 1■│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 0 │ 6 │ │ 2 │ │ │ 4 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 5 │ 0 │ 6 │ │ 2 │ │ 5 │ 4 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 6 │ 0 │ 6 │ │ 2 │ 1 │ 5 │ 4 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ │ 3 │ │ │ │
└───┴───┴───┴───┴───┴───┴───┘
It follows that the answer is ⟨0632154⟩.
Q.E.D.
#125034_v2.5