Rearrange the digits in ⟨125034⟩ to meet the rules below.
⟨5th 4th 3rd 2nd 1st 0th⟩
✅Match
{p4, p2, p1} = ? + {0,1,2}
⟨⋯ Perm(0,1,4) ⋯⟩
3rd → a, 2nd → b, a+b=3
⛔Avoid
4th → a, 2nd → b, ab=0+4n
#125034_v2.8
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ │ │ 3 │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ │ 0 │ 3 │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ │ 1 │ 0 │ 3 │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ │ 1 │ 0 │ 3 │ 2 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ 4 │ 1 │ 0 │ 3 │ 2 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 4 │ 1 │ 0 │ 3 │ 2 │ 5 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
Proof of 2025-04-15 WR
══════════════════════
Notation: if Nth -> a, then we write pN = a.
By ✅「{p4, p2, p1} = ? + {0,1,2}」, we see that
(1) S := {p4, p2, p1} consists of consecutive integers.
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ ▬ │ │ ▬ │ ▬ │ │
└───┴───┴───┴───┴───┴───┘
(2) We claim that 4 ∉ S.
------------------------------
For, suppose on the contrary 4 ∈ S. To avoid ⛔「4th → a, 2nd → b, ab=0+4n」, 4 is not at 4th; and to match ✅「3rd → a, 2nd → b, a+b=3」, it is not at 2nd. Therefore, 4 = p1:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│ 1▲│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ ▬ │ │ ▬ │ 4 │ │
└───┴───┴───┴───┴───┴───┘
As S consists of consecutive integers, there are two possibilities:
S = {2,3,4} or {3,4,5}.
However, regardless of which case happens, we would fail to match ✅「⟨⋯ Perm(0,1,4) ⋯⟩」. This shows a contradiction.
------------------------------
We have verified (2). It follows that
(3) S = {0,1,2} or {1,2,3}.
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ ▬ │ │ ▬ │ ▬ │ │
└───┴───┴───┴───┴───┴───┘
As a result, we have 1,2 ∈ S. A fortiori,
(4) p3 != 1 and p3 != 2.
Observe that to match ✅「3rd → a, 2nd → b, a+b=3」, we need {p3,p2} = {0,3} or {1,2}. In view of (4), indeed
(5) {p3,p2} = {0,3}.
To avoid ⛔「4th → a, 2nd → b, ab=0+4n」, we need p2 != 0. Therefore, it follows from (5) that (p3,p2) = (0,3):
┌───┬───┬───┬───┬───┬───┐
│5th│4th│ 3■│ 2■│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ │ │ 3 │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ │ 0 │ 3 │ │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ 5 │ │ │ 4 │
└───┴───┴───┴───┴───┴───┘
Then, using (3), we get S = {1,2,3}. It follows two cases:
┌───┬───┬───┬───┬───┬───┐
│5th│ 4▲│3rd│2nd│ 1▲│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
(6) │ │ 2 │ 0 │ 3 │ 1 │ │
├───┼───┼───┼───┼───┼───┤
(7) │ │ 1 │ 0 │ 3 │ 2 │ │
└───┴───┴───┴───┴───┴───┘
In view of ✅「⟨⋯ Perm(0,1,4) ⋯⟩」, case (7) holds and we get
┌───┬───┬───┬───┬───┬───┐
│5th│ 4■│3rd│2nd│ 1■│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ │ 0 │ 3 │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ │ 1 │ 0 │ 3 │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ │ 1 │ 0 │ 3 │ 2 │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ 5 │ │ │ 4 │
└───┴───┴───┴───┴───┴───┘
Finally, to match ✅「⟨⋯ Perm(0,1,4) ⋯⟩」, we finish by
┌───┬───┬───┬───┬───┬───┐
│ 5■│4th│3rd│2nd│1st│ 0■│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ 1 │ 0 │ 3 │ 2 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ 4 │ 1 │ 0 │ 3 │ 2 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 4 │ 1 │ 0 │ 3 │ 2 │ 5 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.8