Rearrange the digits in ⟨125034⟩ to meet the rules below.
⟨ ⁵ᵗʰ▨ ⁴ᵗʰ▨ ³ʳᵈ▨ ²ⁿᵈ▨ ¹ˢᵗ▨ ⁰ᵗʰ▨ ⟩
✅Match
⟨ − ▧ ▧ − ▢ ▢ ⟩, Σ▧ = Σ▢
max {p5, p4, p1} = 5
4th → a, 3rd → b, a+b=5
4th|0th → 3
Jump(2,4) ≤ 1
⛔Avoid
⟨⋯ ᵃb ⋯⟩, |a-b|=2
#125034_v2.11
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ 4 │ │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ 4 │ │ │ 1 │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ 4 │ 3 │ │ 1 │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ 4 │ 3 │ 2 │ 1 │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ 4 │ 3 │ 2 │ 1 │ 5 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 4 │ 3 │ 2 │ 1 │ 5 │ 0 │▒
└───┴───┴───┴───┴───┴───┘▒
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Proof of 2025-11-18 WR
══════════════════════
Notation: if nth -> a, then we write [nth] = a.
Combining ✅「4th → a, 3rd → b, a+b=5」 with ✅「⟨ − ▧ ▧ − ▢ ▢ ⟩, Σ▧ = Σ▢」, we have
(1) [4th] + [3rd] = 5 = [1st] + [0th].
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ ● │ ● │ │ ▬ │ ▬ │
└───┴───┴───┴───┴───┴───┘
Observe that this implies [5th] + [2nd] = 5 as well.
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ # │ ● │ ● │ # │ ▬ │ ▬ │
└───┴───┴───┴───┴───┴───┘
Therefore,
(2)
Each of the following sets:
• {[5th], [2nd]}
• {[4th], [3rd]}
• {[1st], [0th]}
is equal to one of the following:
{0,5}, {1,4}, {2,3}.
We consider what {[5th], [2nd]} is. If it is {0,5}, then in view of ✅「max {p5, p4, p1} = 5」, we have
┌───┬───┬───┬───┬───┬───┐
│ 5▲│4th│3rd│ 2▲│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 5 │ │ │ 0 │ │ │
└───┴───┴───┴───┴───┴───┘
This matches ⛔「⟨⋯ ᵃb ⋯⟩, |a-b|=2」, which is a contradiction. Else if it is {2,3}, then we cannot match ✅「4th|0th → 3」. Consequently, it is {1,4}. To avoid ⛔「⟨⋯ ᵃb ⋯⟩, |a-b|=2」, we need 4 != [2nd]. So we have
┌───┬───┬───┬───┬───┬───┐
│ 5■│4th│3rd│ 2■│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ 4 │ │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ 4 │ │ │ 1 │ │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ 2 │ 5 │ 0 │ 3 │ │
└───┴───┴───┴───┴───┴───┘
Next, by (2), we have {[4th], [3rd]} = {0,5} or {2,3}. If it is {0,5} then we cannot match ✅「Jump(2,4) ≤ 1」. Therefore, it is {2,3}. Combining this with ✅「4th|0th → 3」, we get
┌───┬───┬───┬───┬───┬───┐
│5th│ 4■│ 3■│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 4 │ │ │ 1 │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ 4 │ 3 │ │ 1 │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ 4 │ 3 │ 2 │ 1 │ │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ 5 │ 0 │ │ │
└───┴───┴───┴───┴───┴───┘
Finally, to match ✅「max {p5, p4, p1} = 5」, we finish by
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│ 1■│ 0■│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 4 │ 3 │ 2 │ 1 │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ 4 │ 3 │ 2 │ 1 │ 5 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 4 │ 3 │ 2 │ 1 │ 5 │ 0 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.11