Rearrange the digits in ⟨125034⟩ to meet the rules below.
⟨5th 4th 3rd 2nd 1st 0th⟩
✅Match
⟦4,5⟧ ∋ 0,3
⟨? ⋯ 3 ⋯ (?−2) ⋯⟩ (?≠3,5)
4th → a, 0th → b, |a-b|=2
3rd → a, 0th → b, |a-b|=1
⛔Avoid
⟨⋯ Perm(0,3,4) ⋯⟩
3rd → a, 2nd → b, a+b=2+5n
#125034_v2.0
WR 2023-10-17
✅⚪⚪⚪⚪⚪
✅⚪⚪⚪⚪✅
✅✅⚪⚪⚪✅
✅✅✅⚪⚪✅
✅✅✅✅⚪✅
✅✅✅✅✅✅
#125034_v2.0
┌───┬───┬───┬───┬───┬───┐
│ 5 │ 4 │ 3 │ 2 │ 1 │ 0 │
╞═══╪═══╪═══╪═══╪═══╪═══╡
Step 1 │ 4 │ │ │ │ │ │
├───┼───┼───┼───┼───┼───┤
Step 2 │ 4 │ │ │ │ │ 2 │
├───┼───┼───┼───┼───┼───┤
Step 3 │ 4 │ 0 │ │ │ │ 2 │
├───┼───┼───┼───┼───┼───┤
Step 4 │ 4 │ 0 │ 1 │ │ │ 2 │
├───┼───┼───┼───┼───┼───┤
Step 5 │ 4 │ 0 │ 1 │ 3 │ │ 2 │
├───┼───┼───┼───┼───┼───┤
Step 6 │ 4 │ 0 │ 1 │ 3 │ 5 │ 2 │
└───┴───┴───┴───┴───┴───┘
Proof of 2023-10-17 WR
══════════════════════
Notation: if nth -> a, then we write [nth] = a.
We start with ✅【⟨? ⋯ 3 ⋯ (?−2) ⋯⟩ (?≠3,5)】. We need ?-2≥0, so ?=[5th] can only be 4 or 2.
We claim that [5th] = 4. Suppose on the contrary that [5th]=2. Then to match ✅【⟦4,5⟧ ∋ 0,3】, there are three possibilities:
┌───┬───┬───┬───┬───┬───┐
│*5 │ 4 │ 3 │ 2 │ 1 │ 0 │
╞═══╪═══╪═══╪═══╪═══╪═══╡
(1) │ 2 │ [ │ ● │ ● │ ] │ │
├───┼───┼───┼───┼───┼───┤
(2) │ 2 │ │ [ │ ● │ ● │ ] │
├───┼───┼───┼───┼───┼───┤
(3) │ 2 │ [ │ │ │ │ ] │
└───┴───┴───┴───┴───┴───┘
where the brackets indicate 4,5 and the dots are 0,3. But case (1) and (2) would match ⛔「⟨⋯ Perm(0,3,4) ⋯⟩」, while case (3) would not match ✅【4th → a, 0th → b, |a-b|=2】. This shows contradictions and verifies our claim. So, we have
┌───┬───┬───┬───┬───┬───┐
│ 5■│ 4 │ 3 │ 2 │ 1 │ 0 │
╞═══╪═══╪═══╪═══╪═══╪═══╡
Step 1 │ 4 │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ 5 │ 0 │ 3 │ │
└───┴───┴───┴───┴───┴───┘
---------------------------------------
Next, by ✅【⟦4,5⟧ ∋ 0,3】, we see that 0,3 cannot be at the boundary, so [0th]=1|2|5. Note that by ✅【3rd → a, 0th → b, |a-b|=1】, [0th] cannot be 5.
(4) We show by contradiction that [0th]=2 . If on the contrary [0th]=1, then ✅【4th → a, 0th → b, |a-b|=2】 gives
┌───┬───┬───┬───┬───┬───┐
│ 5 │ 4▲│*3 │ 2 │ 1 │ 0▲│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 4 │ 3 │ x │ │ │ 1 │
└───┴───┴───┴───┴───┴───┘
with x=0|2 by ✅【3rd → a, 0th → b, |a-b|=1】. As x≠0 by ⛔「⟨⋯ Perm(0,3,4) ⋯⟩」, we reach
┌───┬───┬───┬───┬───┬───┐
│ 5 │ 4 │ 3▲│ 2 │ 1 │ 0 │
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 4 │ 3 │ 2 │ │ │ 1 │
└───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ 5 │ 0 │ │ │
└───┴───┴───┴───┴───┴───┘
It then unavoidably matches ⛔「3rd → a, 2nd → b, a+b=2+5n」, which is a contradiction. So, back to (4), we have
┌───┬───┬───┬───┬───┬───┐
│ 5 │ 4 │ 3 │ 2 │ 1 │ 0■│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 4 │ │ │ │ │ │
├───┼───┼───┼───┼───┼───┤
Step 2 │ 4 │ │ │ │ │ 2 │
└───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ │ 5 │ 0 │ 3 │ │
└───┴───┴───┴───┴───┴───┘
---------------------------------------
Now, using ✅【4th → a, 0th → b, |a-b|=2】, we get [4th]=0, and then using ✅【3rd → a, 0th → b, |a-b|=1】 with ⛔「⟨⋯ Perm(0,3,4) ⋯⟩」 in mind, we have [3rd]=1:
┌───┬───┬───┬───┬───┬───┐
│ 5 │ 4■│ 3■│ 2 │ 1 │ 0 │
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 4 │ │ │ │ │ 2 │
├───┼───┼───┼───┼───┼───┤
Step 3 │ 4 │ 0 │ │ │ │ 2 │
├───┼───┼───┼───┼───┼───┤
Step 4 │ 4 │ 0 │ 1 │ │ │ 2 │
└───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ 5 │ │ 3 │ │
└───┴───┴───┴───┴───┴───┘
---------------------------------------
Finally, to match ✅【⟦4,5⟧ ∋ 0,3】, we finish by
┌───┬───┬───┬───┬───┬───┐
│ 5 │ 4 │ 3 │ 2■│ 1■│ 0 │
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 4 │ 0 │ 1 │ │ │ 2 │
├───┼───┼───┼───┼───┼───┤
Step 5 │ 4 │ 0 │ 1 │ 3 │ │ 2 │
├───┼───┼───┼───┼───┼───┤
Step 6 │ 4 │ 0 │ 1 │ 3 │ 5 │ 2 │
└───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.0