Rearrange the digits in ⟨125034⟩ to meet the rules below.
⟨5th 4th 3rd 2nd 1st 0th⟩
✅Match
Jump(2,3) = 1
median {p5, p3, p0} = 3
4th → 0|2|4
⛔Avoid
Jump(0,2) = 0
median {p4, p3, p1} = 2
⟨⋯ 1 ⋯ 3 ⋯⟩
⟨⋯ 2 ⋯ ? 3 ⋯ (?+1)⟩ (?≠3,2)
#125034_v2.10
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ │ 3 │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ │ 3 │ │ 2 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ │ 4 │ 3 │ │ 2 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ 0 │ 4 │ 3 │ │ 2 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ 0 │ 4 │ 3 │ │ 2 │ 5 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 0 │ 4 │ 3 │ 1 │ 2 │ 5 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
Proof of 2025-08-19 WR
══════════════════════
Notation: if nth -> a, then we write [nth] = a.
By ⛔「⟨⋯ 1 ⋯ 3 ⋯⟩」, 3 is not at 0th. Combining this with ✅「median {p5, p3, p0} = 3」, we have
(1) 3 = [5th] | [3rd].
(1.1) We show that 3 = [3rd] actually.
------------------------------
Suppose on the contrary 3 = [5th]:
┌───┬───┬───┬───┬───┬───┐
│ 5▲│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 3 │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┘
Then, it follows from ✅「Jump(2,3) = 1」 that 2 = [3rd]:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│ 3▲│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 3 │ │ 2 │ │ │ │
└───┴───┴───┴───┴───┴───┘
To match ✅「4th → 0|2|4」, we need [4th] = 0 or 4. It is not 0 in view of ⛔「Jump(0,2) = 0」. Hence [4th] = 4:
┌───┬───┬───┬───┬───┬───┐
│5th│ 4▲│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 3 │ 4 │ 2 │ │ │ │
└───┴───┴───┴───┴───┴───┘
Then, we consider where to place 0. By ⛔「Jump(0,2) = 0」, it is not at 2nd, and by ⛔「median {p4, p3, p1} = 2」, it is not at 1st either. Therefore, 0 = [0th]:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│ 0▲│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 3 │ 4 │ 2 │ │ │ 0 │
└───┴───┴───┴───┴───┴───┘
We have failed to match ✅「median {p5, p3, p0} = 3」, which is a contradiction.
------------------------------
We have verified (1.1). Accordingly, we get
┌───┬───┬───┬───┬───┬───┐
│5th│4th│ 3■│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ │ 3 │ │ │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ 5 │ 0 │ │ 4 │
└───┴───┴───┴───┴───┴───┘
Given that 3 = [3rd], to match ✅「Jump(2,3) = 1」 we need
(2) 2 = [5th] | [1st]:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ ▬ │ │ 3 │ │ ▬ │ │
└───┴───┴───┴───┴───┴───┘
(2.1) We show that 2 = [1st].
------------------------------
If on the contrary 2 = [5th]:
┌───┬───┬───┬───┬───┬───┐
│ 5▲│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 2 │ │ 3 │ │ │ │
└───┴───┴───┴───┴───┴───┘
Then, to match ✅「4th → 0|2|4」 and avoid ⛔「Jump(0,2) = 0」 at the same time, we have [4th] = 4:
┌───┬───┬───┬───┬───┬───┐
│5th│ 4▲│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 2 │ 4 │ 3 │ │ │ │
└───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ │ 5 │ 0 │ │ │
└───┴───┴───┴───┴───┴───┘
Note that there is only one way to match ✅「median {p5, p3, p0} = 3」:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│ 0▲│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 2 │ 4 │ 3 │ │ │ 5 │
└───┴───┴───┴───┴───┴───┘
But then we have matched ⛔「⟨⋯ 2 ⋯ ? 3 ⋯ (?+1)⟩ (?≠3,2)」, which is a contradiction.
------------------------------
We have verified (2.1). So we get
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│ 1■│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ │ 3 │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ │ 3 │ │ 2 │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ │ 5 │ 0 │ │ 4 │
└───┴───┴───┴───┴───┴───┘
Next, we determine what [4th] is. To match ✅「4th → 0|2|4」 and avoid ⛔「median {p4, p3, p1} = 2」 at the same time, it has to be 4:
┌───┬───┬───┬───┬───┬───┐
│5th│ 4■│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ │ 3 │ │ 2 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ │ 4 │ 3 │ │ 2 │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ │ 5 │ 0 │ │ │
└───┴───┴───┴───┴───┴───┘
Then, plainly, there is only one way to avoid ⛔「Jump(0,2) = 0」:
┌───┬───┬───┬───┬───┬───┐
│ 5■│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ 4 │ 3 │ │ 2 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ 0 │ 4 │ 3 │ │ 2 │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ │ 5 │ │ │ │
└───┴───┴───┴───┴───┴───┘
Finally, to match ✅「median {p5, p3, p0} = 3」, we finish by
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│ 2■│1st│ 0■│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 0 │ 4 │ 3 │ │ 2 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ 0 │ 4 │ 3 │ │ 2 │ 5 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 0 │ 4 │ 3 │ 1 │ 2 │ 5 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.10