Rearrange the digits in ⟨125034⟩ to meet the rules below.
⟨ ⁵ᵗʰ▨ ⁴ᵗʰ▨ ³ʳᵈ▨ ²ⁿᵈ▨ ¹ˢᵗ▨ ⁰ᵗʰ▨ ⟩
✅Match
⟨⋯ ᵃb ⋯⟩, |a-b|=0
⟨ ⁵ᵗʰa ²ⁿᵈb ⟩, max⟦a,b⟧ = 4
⟨⋯ Perm(1,2,4) ⋯⟩
⟨⋯ ᵃb ⋯⟩, |a-b|=2
⟦2,5⟧ ∋ 4
⛔Avoid
⟨? ⋯ 5 ⋯ (?−1) ⋯⟩ (?≠5)
⟨⋯ ? ⋯ 1 ⋯ (?+2)⟩ (?≠1)
⟨ − − ▧ ▧ − ▢ ⟩, Σ▧ ≤ ▢
{p4, p3, p2} = ? + {0,1,3}
#125034_v2.11
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ 2 │ │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ 2 │ 4 │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ 2 │ 4 │ 1 │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ 2 │ 4 │ 1 │ 3 │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ 2 │ 4 │ 1 │ 3 │ 5 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 2 │ 4 │ 1 │ 3 │ 5 │ 0 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
Proof of 2026-01-06 WR
══════════════════════
Notation: if nth -> a, then we write [nth] = a.
By ✅「⟨⋯ Perm(1,2,4) ⋯⟩」, there are four possibilities of placing 1,2,4:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
(1) │ │ │ │ ▬ │ ▬ │ ▬ │
├───┼───┼───┼───┼───┼───┤
(2) │ │ │ ▬ │ ▬ │ ▬ │ │
├───┼───┼───┼───┼───┼───┤
(3) │ │ ▬ │ ▬ │ ▬ │ │ │
├───┼───┼───┼───┼───┼───┤
(4) │ ▬ │ ▬ │ ▬ │ │ │ │
└───┴───┴───┴───┴───┴───┘
(5) We claim the case (4) holds actually.
------------------------------
(5.1) Note that if case (1) holds, then 5 is one of [5th], [4th] or [3rd], so we shall fail to match ✅「⟨ ⁵ᵗʰa ²ⁿᵈb ⟩, max⟦a,b⟧ = 4」.
(5.2) Else if case (2) holds, then to match ✅「⟨ ⁵ᵗʰa ²ⁿᵈb ⟩, max⟦a,b⟧ = 4」, we need
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ │ ▬ │ ▬ │ ▬ │ 5 │
└───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ │ 0 │ 3 │ │
└───┴───┴───┴───┴───┴───┘
However, we shall match ⛔「⟨⋯ ? ⋯ 1 ⋯ (?+2)⟩ (?≠1)」, which is a contradiction.
(5.3) Else if case (3) holds, then we have matched ⛔「{p4, p3, p2} = ? + {0,1,3}」, which is also a contradiction.
(5.4) Therefore, case (4) indeed holds.
------------------------------
Since ✅「⟦2,5⟧ ∋ 4」 implies 4 is not in the left corner, so (4) implies
(6) [5th] = 1 or 2.
(6.1) We show that [5th] = 2 actually.
For, if it is 1 instead, then to match ✅「⟦2,5⟧ ∋ 4」, we need
┌───┬───┬───┬───┬───┬───┐
│ 5▲│ 4▲│ 3▲│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 1 │ 2 │ 4 │ │ │ │
└───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ 5 │ 0 │ 3 │ │
└───┴───┴───┴───┴───┴───┘
Then, there is only one way to match ✅「⟨⋯ ᵃb ⋯⟩, |a-b|=0」:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│ 0▲│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 1 │ 2 │ 4 │ │ │ 0 │
└───┴───┴───┴───┴───┴───┘
We have matched ⛔「⟨? ⋯ 5 ⋯ (?−1) ⋯⟩ (?≠5)」, which is a contradiction.
We have verified (6.1). Therefore, we get
┌───┬───┬───┬───┬───┬───┐
│ 5■│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ 2 │ │ │ │ │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ │ 5 │ 0 │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘
Then, to match (4), one of the following holds:
┌───┬───┬───┬───┬───┬───┐
│5th│ 4▲│ 3▲│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
(7) │ 2 │ 1 │ 4 │ │ │ │
├───┼───┼───┼───┼───┼───┤
(8) │ 2 │ 4 │ 1 │ │ │ │
└───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ 5 │ 0 │ 3 │ │
└───┴───┴───┴───┴───┴───┘
Actually case (8) holds. For, if case (7) holds instead, then to match ✅「⟨⋯ ᵃb ⋯⟩, |a-b|=0」, we need
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│ 0▲│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 2 │ 1 │ 4 │ │ │ 0 │
└───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ 5 │ │ 3 │ │
└───┴───┴───┴───┴───┴───┘
and in view of ✅「⟨ ⁵ᵗʰa ²ⁿᵈb ⟩, max⟦a,b⟧ = 4」, we reach
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│ 2▲│ 1▲│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 2 │ 1 │ 4 │ 3 │ 5 │ 0 │
└───┴───┴───┴───┴───┴───┘
We have failed to match ✅「⟨⋯ ᵃb ⋯⟩, |a-b|=2」, which is a contradiction.
------------------------------
As case (8) holds, we get
┌───┬───┬───┬───┬───┬───┐
│5th│ 4■│ 3■│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 2 │ │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ 2 │ 4 │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ 2 │ 4 │ 1 │ │ │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ 5 │ 0 │ 3 │ │
└───┴───┴───┴───┴───┴───┘
Next, we consider [2nd]. If it is 5, then we cannot match ✅「⟨ ⁵ᵗʰa ²ⁿᵈb ⟩, max⟦a,b⟧ = 4」; else if it is 0, then we cannot avoid ⛔「⟨ − − ▧ ▧ − ▢ ⟩, Σ▧ ≤ ▢」. Consequently, [2nd] = 3:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│ 2■│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 2 │ 4 │ 1 │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ 2 │ 4 │ 1 │ 3 │ │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ 5 │ 0 │ │ │
└───┴───┴───┴───┴───┴───┘
Finally, to avoid ⛔「⟨ − − ▧ ▧ − ▢ ⟩, Σ▧ ≤ ▢」, we finish by
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│ 1■│ 0■│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 2 │ 4 │ 1 │ 3 │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ 2 │ 4 │ 1 │ 3 │ 5 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 2 │ 4 │ 1 │ 3 │ 5 │ 0 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.11