Rearrange the digits in ⟨125034⟩ to meet the rules below.
⟨5th 4th 3rd 2nd 1st 0th⟩
✅Match
Jump(2,5) = 2
⟦0,2⟧ ∋ 3,4
⛔Avoid
min ⊢3⊣ ≤ 2
5th → 0|1|3
⟨? ⋯ 1 ⋯ (?−2) ⋯⟩ (?≠3)
#125034_v2.8
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ 2 │ │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ 2 │ │ │ 5 │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ 2 │ │ 3 │ 5 │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ 2 │ 4 │ 3 │ 5 │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ 2 │ 4 │ 3 │ 5 │ │ 1 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 2 │ 4 │ 3 │ 5 │ 0 │ 1 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
Proof of 2025-02-25 WR
══════════════════════
Notation: if nth -> a, then we write [nth] = a.
From ⛔「min ⊢3⊣ ≤ 2」, we see that 3 can only be adjacent to 4 or 5. As 3 is not in corners (5th or 0th) by ✅「⟦0,2⟧ ∋ 3,4」, we get the folloing required pattern:
(1) ⟨⋯ 435 ⋯⟩ or ⟨⋯ 534 ⋯⟩.
Let us consider the value of [5th]. By ⛔「5th → 0|1|3」, it is 2|4|5. It has to be 2, for otherwise (1) implies
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ ▬ │ 3 │ ▬ │ │ │ │
└───┴───┴───┴───┴───┴───┘
where "▬" are occupied by 4 and 5. But then we cannot match ✅「⟦0,2⟧ ∋ 3,4」, which is a contradiction.
As a result, we have
┌───┬───┬───┬───┬───┬───┐
│ 5■│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ 2 │ │ │ │ │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ │ 5 │ 0 │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘
Then, by ✅「Jump(2,5) = 2」, we get
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│ 2■│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 2 │ │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ 2 │ │ │ 5 │ │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ │ │ 0 │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘
Given [2nd] = 5, it follows from (1) that one of the following holds:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
(2) │ 2 │ │ │ 5 │ 3 │ 4 │
├───┼───┼───┼───┼───┼───┤
(3) │ 2 │ 4 │ 3 │ 5 │ │ │
└───┴───┴───┴───┴───┴───┘
If (2) holds then we cannot match ✅「⟦0,2⟧ ∋ 3,4」. Therefore, (3) holds instead and we get
┌───┬───┬───┬───┬───┬───┐
│5th│ 4■│ 3■│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 2 │ │ │ 5 │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ 2 │ │ 3 │ 5 │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ 2 │ 4 │ 3 │ 5 │ │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ │ │ 0 │ │ │
└───┴───┴───┴───┴───┴───┘
Finally, to avoid ⛔「⟨? ⋯ 1 ⋯ (?−2) ⋯⟩ (?≠3)」, we finish by
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│ 1■│ 0■│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 2 │ 4 │ 3 │ 5 │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ 2 │ 4 │ 3 │ 5 │ │ 1 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 2 │ 4 │ 3 │ 5 │ 0 │ 1 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.8