Rearrange the digits in ⟨125034⟩ to meet the rules below.
⟨ ⁵ᵗʰ▨ ⁴ᵗʰ▨ ³ʳᵈ▨ ²ⁿᵈ▨ ¹ˢᵗ▨ ⁰ᵗʰ▨ ⟩
✅Match
⟨ ¹ˢᵗa ⁰ᵗʰb ⟩, (ab)₁₀ ≥ 30
⟨⋯ ᵃb ⋯⟩, |a-b|=2
4th|3rd|1st|0th → 2
⛔Avoid
1 ∾ 5
⟨⋯ Perm(0,2) ⋯⟩
5th|3rd|2nd|1st → 3
Jump(1,2) ≥ 1
#125034_v2.12
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ │ │ │ 4 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ │ │ │ 4 │ 3 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ 5 │ │ │ │ 4 │ 3 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ 5 │ │ │ 0 │ 4 │ 3 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ 5 │ 2 │ │ 0 │ 4 │ 3 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 5 │ 2 │ 1 │ 0 │ 4 │ 3 │▒
└───┴───┴───┴───┴───┴───┘▒
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Proof of 2026-02-03 WR
══════════════════════
Notation: if nth -> a, then we write [nth] = a.
By ✅「⟨ ¹ˢᵗa ⁰ᵗʰb ⟩, (ab)₁₀ ≥ 30」, we have
(1) [1st] = 5 | 4 | 3.
If it is 3 then we match ⛔「5th|3rd|2nd|1st → 3」, while if it is 5 then we match ⛔「1 ∾ 5」. Therefore, [1st] = 4:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│ 1■│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ │ │ │ 4 │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ 5 │ 0 │ 3 │ │
└───┴───┴───┴───┴───┴───┘
Then, in view of ⛔「5th|3rd|2nd|1st → 3」, we have
(2) 3 = [4th] | [0th].
(2.1) We show that 3 = [0th] actually.
------------------------------
Suppose on the contrary 3 = [4th]:
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│5th│ 4▲│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ 3 │ │ │ 4 │ │
└───┴───┴───┴───┴───┴───┘
In order to avoid ⛔「Jump(1,2) ≥ 1」, we need 1 and 2 to be adjacent. Therefore, {1,2} = {[3rd], [2nd]}:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ 3 │1 2│1 2│ 4 │ │
└───┴───┴───┴───┴───┴───┘
It then follows from ✅「4th|3rd|1st|0th → 2」 that
┌───┬───┬───┬───┬───┬───┐
│5th│4th│ 3▲│ 2▲│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ 3 │ 2 │ 1 │ 4 │ │
└───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ 5 │ 0 │ │ │
└───┴───┴───┴───┴───┴───┘
Note that we will fail to match ✅「⟨⋯ ᵃb ⋯⟩, |a-b|=2」. This shows a contradiction.
------------------------------
We have verified (2.1). Accordingly, we get
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│ 0■│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ │ │ │ 4 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ │ │ │ 4 │ 3 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ 5 │ 0 │ │ │
└───┴───┴───┴───┴───┴───┘
Then, as ⛔「Jump(1,2) ≥ 1」 implies that 1,2 are adjacent, there are three ways to place them:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
(3.1) │ ▬ │ ▬ │ │ │ 4 │ 3 │
├───┼───┼───┼───┼───┼───┤
(3.2) │ │ ▬ │ ▬ │ │ 4 │ 3 │
├───┼───┼───┼───┼───┼───┤
(3.3) │ │ │ ▬ │ ▬ │ 4 │ 3 │
└───┴───┴───┴───┴───┴───┘
We show that case (3.2) holds actually.
------------------------------
If on the contrary (3.1) or (3.3) hold, then using ✅「4th|3rd|1st|0th → 2」, we have
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
(4) │ 1 │ 2 │ │ │ 4 │ 3 │
├───┼───┼───┼───┼───┼───┤
(5) │ │ │ 2 │ 1 │ 4 │ 3 │
└───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ 5 │ 0 │ │ │
└───┴───┴───┴───┴───┴───┘
Note that we match ⛔「1 ∾ 5」 in case (4), while we cannot match ✅「⟨⋯ ᵃb ⋯⟩, |a-b|=2」 in case (5).
------------------------------
We have shown that (3.2) holds. Therefore, we have
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│0 5│1 2│1 2│0 5│ 4 │ 3 │
└───┴───┴───┴───┴───┴───┘
Then, in view of ⛔「⟨⋯ Perm(0,2) ⋯⟩」, there are two cases:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
(6) │ 0 │ 1 │ 2 │ 5 │ 4 │ 3 │
├───┼───┼───┼───┼───┼───┤
(7) │ 5 │ 2 │ 1 │ 0 │ 4 │ 3 │
└───┴───┴───┴───┴───┴───┘
We need to match ✅「⟨⋯ ᵃb ⋯⟩, |a-b|=2」, so (9) is the answer.
Q.E.D.
#125034_v2.12