Rearrange the digits in ⟨125034⟩ to meet the rules below.
⟨5th 4th 3rd 2nd 1st 0th⟩
✅Match
Jump(0,4) ≥ 2
5th → 2|5
⛔Avoid
{p5, p3, p2} = ? + {0,2,3}
⟨ ⁵ᵗʰa ³ʳᵈb ⟩, max⟦a,b⟧ = 4
⟨ ⁵ᵗʰa ³ʳᵈb ¹ˢᵗc ⟩, (abc)₁₀ ≥ 435
⟨⋯ 2 ⋯ 0 ⋯ 1 ⋯⟩
median {p2, p1, p0} = 1
#125034_v2.8
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ 2 │ │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ 2 │ │ 0 │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ 2 │ 1 │ 0 │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ 2 │ 1 │ 0 │ │ │ 4 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ 2 │ 1 │ 0 │ │ 3 │ 4 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 2 │ 1 │ 0 │ 5 │ 3 │ 4 │▒
└───┴───┴───┴───┴───┴───┘▒
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Proof of 2025-03-11 WR
══════════════════════
Notation: if Nth -> a, then we write pN = a.
Plainly, the first step follows from combining ✅「5th → 2|5」 and ⛔「⟨ ⁵ᵗʰa ³ʳᵈb ¹ˢᵗc ⟩, (abc)₁₀ ≥ 435」:
┌───┬───┬───┬───┬───┬───┐
│ 5■│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ 2 │ │ │ │ │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ │ 5 │ 0 │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘
Then, we consider where to place 0. There are five possible cases:
0 = p4 | p3 | p2 | p1 | p0.
(1) We show that 0 = p3 actually.
------------------------------
(1.1) Suppose otherwise that 0 = p1 | p0:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│ 1▲│ 0▲│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 2 │ │ │ │ ▬ │ ▬ │
└───┴───┴───┴───┴───┴───┘
Then, to match ✅「Jump(0,4) ≥ 2」, we need 4∈{p4, p3}. On the other hand, to avoid ⛔「median {p2, p1, p0} = 1」, we need 1∉{p2,p1,p0}. It follows that both 4 and 1 are in {p4, p3}. However, it would match ⛔「⟨ ⁵ᵗʰa ³ʳᵈb ⟩, max⟦a,b⟧ = 4」, which is a contradiction.
(1.2) Else if 0 = p2:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│ 2▲│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 2 │ │ │ 0 │ │ │
└───┴───┴───┴───┴───┴───┘
then it cannot match ✅「Jump(0,4) ≥ 2」.
(1.3) Else if 0 = p4:
┌───┬───┬───┬───┬───┬───┐
│5th│ 4▲│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 2 │ 0 │ │ │ │ │
└───┴───┴───┴───┴───┴───┘
then it would match ⛔「⟨⋯ 2 ⋯ 0 ⋯ 1 ⋯⟩」.
------------------------------
By (1.1), (1.2) and (1.3), we have verified the claim in (1). Accordingly, we get
┌───┬───┬───┬───┬───┬───┐
│5th│4th│ 3■│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 2 │ │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ 2 │ │ 0 │ │ │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ │ 5 │ │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘
Then, to avoid ⛔「⟨⋯ 2 ⋯ 0 ⋯ 1 ⋯⟩」, we need to have
┌───┬───┬───┬───┬───┬───┐
│5th│ 4■│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 2 │ │ 0 │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ 2 │ 1 │ 0 │ │ │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ 5 │ │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘
and to match ✅「Jump(0,4) ≥ 2」, we need
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│ 0■│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 2 │ 1 │ 0 │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ 2 │ 1 │ 0 │ │ │ 4 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ 5 │ │ 3 │ │
└───┴───┴───┴───┴───┴───┘
Finally, to avoid ⛔「{p5, p3, p2} = ? + {0,2,3}」, we finish by
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│ 2■│ 1■│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 2 │ 1 │ 0 │ │ │ 4 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ 2 │ 1 │ 0 │ │ 3 │ 4 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 2 │ 1 │ 0 │ 5 │ 3 │ 4 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.8