Rearrange the digits in ⟨125034⟩ to meet the rules below.
⟨5th 4th 3rd 2nd 1st 0th⟩
✅Match
2nd → a, 1st → b, a+b=3+4n
⟨ ⁵ᵗʰa ²ⁿᵈb ⟩, (ab)₁₀ ≥ 31
⟨⋯ Perm(3,4) ⋯⟩
min {p5, p4} = 2
⟨⋯ Perm(0,4) ⋯⟩
⛔Avoid
⟨ − − ▧ ▧ ▢ − ⟩, Σ▧ ≤ ▢
#125034_v2.9
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ 2 │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ 5 │ 2 │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ 5 │ 2 │ │ 4 │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ 5 │ 2 │ │ 4 │ │ 1 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ 5 │ 2 │ │ 4 │ 3 │ 1 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 5 │ 2 │ 0 │ 4 │ 3 │ 1 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
Proof of 2025-05-27 WR
══════════════════════
Notation: if nth -> a, then we write [nth] = a.
By ✅「min {p5, p4} = 2」, we have 2 = [5th] | [4th]. If it is at [5th], then we will fail to match ✅「⟨ ⁵ᵗʰa ²ⁿᵈb ⟩, (ab)₁₀ ≥ 31」. Therefore, 2= [4th]:
┌───┬───┬───┬───┬───┬───┐
│5th│ 4■│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ 2 │ │ │ │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ │ 5 │ 0 │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘
Next, by combining ✅「⟨⋯ Perm(3,4) ⋯⟩」 and ✅「⟨⋯ Perm(0,4) ⋯⟩」, we see that one of the following holds:
(1) ⟨⋯ 0 4 3 ⋯⟩ or ⟨⋯ 3 4 0 ⋯⟩.
As a result, [5th] != 0,3,4. So [5th] is 1 or 5. Using ✅「⟨ ⁵ᵗʰa ²ⁿᵈb ⟩, (ab)₁₀ ≥ 31」 again, we have [5th] = 5:
┌───┬───┬───┬───┬───┬───┐
│ 5■│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ 2 │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ 5 │ 2 │ │ │ │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ │ │ 0 │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘
Then, to match (1), there are two possibilities:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│ 3▲│ 2▲│ 1▲│ 0▲│
╞═══╪═══╪═══╪═══╪═══╪═══╡
(2) │ 5 │ 2 │ 1 │ ▬ │ 4 │ ▬ │
├───┼───┼───┼───┼───┼───┤
(3) │ 5 │ 2 │ ▬ │ 4 │ ▬ │ 1 │
└───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ │ 0 │ 3 │ │
└───┴───┴───┴───┴───┴───┘
where "▬" are occupied by 0,3. Observe that we cannot avoid ⛔「⟨ − − ▧ ▧ ▢ − ⟩, Σ▧ ≤ ▢」 if (2) holds. Therefore (3) holds and we get
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│ 2■│1st│ 0■│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 5 │ 2 │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ 5 │ 2 │ │ 4 │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ 5 │ 2 │ │ 4 │ │ 1 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ │ 0 │ 3 │ │
└───┴───┴───┴───┴───┴───┘
Finally, to match ✅「2nd → a, 1st → b, a+b=3+4n」, we finish by
┌───┬───┬───┬───┬───┬───┐
│5th│4th│ 3■│2nd│ 1■│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 5 │ 2 │ │ 4 │ │ 1 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ 5 │ 2 │ │ 4 │ 3 │ 1 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 5 │ 2 │ 0 │ 4 │ 3 │ 1 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.9