Rearrange the digits in ⟨125034⟩ to meet the rules below.
⟨5th 4th 3rd 2nd 1st 0th⟩
✅Match
{p5, p3, p0} = ? + {0,1,2}
1st → a, 0th → b, |a-b|=1
⟨⋯ 0 ⋯ ? 3 ⋯ (?−3)⟩ (?≠3)
5th → a, 1st → b, |a-b|=3
#125034_v2.8
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ │ │ │ 3 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ 0 │ │ │ │ 3 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ 0 │ │ │ │ 3 │ 2 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ 0 │ │ │ 5 │ 3 │ 2 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ 0 │ │ 1 │ 5 │ 3 │ 2 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 0 │ 4 │ 1 │ 5 │ 3 │ 2 │▒
└───┴───┴───┴───┴───┴───┘▒
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Proof of 2025-02-18 WR
══════════════════════
Notation: if nth -> a, then we write [nth] = a.
By ✅「⟨⋯ 0 ⋯ ? 3 ⋯ (?−3)⟩ (?≠3)」, one of the following holds:
(1) ⟨⋯ 0 ⋯ 43 ⋯ 1⟩ or ⟨⋯ 0 ⋯ 53 ⋯ 2⟩.
A fortiori, there are at least two digits to the left of 3, and at least one digit to the right of it. So the possible positions of 3 are:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
(2) │ │ │ ▬ │ │ │ │
├───┼───┼───┼───┼───┼───┤
(3) │ │ │ │ ▬ │ │ │
├───┼───┼───┼───┼───┼───┤
(4) │ │ │ │ │ ▬ │ │
└───┴───┴───┴───┴───┴───┘
(5) We show that case (4) holds actually.
------------------------------
To begin with, note that by ✅「{p5, p3, p0} = ? + {0,1,2}」, we have
(6) [5th], [3rd], [0th] are consecutive integers.
Now, if case (2) holds, then by (1), we have
┌───┬───┬───┬───┬───┬───┐
│ 5▲│4th│ 3▲│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 0 │ │ 3 │ │ │ │
└───┴───┴───┴───┴───┴───┘
It is a contradiction as it would not match (6).
Else, if case (3) holds, then by (1), we have two possibilities:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│ 3▲│ 2▲│1st│ 0▲│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ │ 4 │ 3 │ │ 1 │
├───┼───┼───┼───┼───┼───┤
│ │ │ 5 │ 3 │ │ 2 │
└───┴───┴───┴───┴───┴───┘
But again, none of them would match (6).
------------------------------
We have verified (5). Accordingly, we get
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│ 1■│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ │ │ │ 3 │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ 5 │ 0 │ │ 4 │
└───┴───┴───┴───┴───┴───┘
As a result, using ✅「5th → a, 1st → b, |a-b|=3」, we get:
┌───┬───┬───┬───┬───┬───┐
│ 5■│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ │ │ │ 3 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ 0 │ │ │ │ 3 │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ 5 │ │ │ 4 │
└───┴───┴───┴───┴───┴───┘
We then determine what [0th] is. By (1), it is 1 or 2. Combining this with ✅「1st → a, 0th → b, |a-b|=1」, it is 2 indeed:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│ 0■│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 0 │ │ │ │ 3 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ 0 │ │ │ │ 3 │ 2 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ │ 5 │ │ │ 4 │
└───┴───┴───┴───┴───┴───┘
Then (1) implies [2nd] = 5:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│ 2■│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 0 │ │ │ │ 3 │ 2 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ 0 │ │ │ 5 │ 3 │ 2 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ │ │ │ │ 4 │
└───┴───┴───┴───┴───┴───┘
Finally, to match ✅「{p5, p3, p0} = ? + {0,1,2}」, we finish by
┌───┬───┬───┬───┬───┬───┐
│5th│ 4■│ 3■│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 0 │ │ │ 5 │ 3 │ 2 │▒
├───┼───┼───┼───┼───┼───┤▒
│ 0 │ │ 1 │ 5 │ 3 │ 2 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 0 │ 4 │ 1 │ 5 │ 3 │ 2 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.8