Rearrange the digits in ⟨125034⟩ to meet the rules below.
⟨ ⁵ᵗʰ▨ ⁴ᵗʰ▨ ³ʳᵈ▨ ²ⁿᵈ▨ ¹ˢᵗ▨ ⁰ᵗʰ▨ ⟩
✅Match
{p4, p3, p0} = ? + {0,1,2}
⟨⋯ a ⋯ 4 ⋯⟩, a = 1|5
5th → a, 1st → b, |a-b|=2
⛔Avoid
Jump(0,1) = 0
⟨ ³ʳᵈb ¹ˢᵗa ⟩, a > b
⟨ ³ʳᵈa ⁰ᵗʰb ⟩, a > b
#125034_v2.12
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ │ │ 1 │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ 0 │ │ │ 1 │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ 0 │ │ │ 1 │ 2 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ 0 │ │ │ 1 │ 2 │ 4 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ 0 │ 5 │ │ 1 │ 2 │ 4 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 0 │ 5 │ 3 │ 1 │ 2 │ 4 │▒
└───┴───┴───┴───┴───┴───┘▒
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Proof of 2026-04-07 WR
══════════════════════
Notation: if Nth -> a, then we write pN = a.
To match ✅「{p4, p3, p0} = ? + {0,1,2}」, we need p4, p3, p0 to be consecutive integers.
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ ▬ │ ▬ │ │ │ ▬ │
└───┴───┴───┴───┴───┴───┘
There are four possibilities for {p4, p3, p0}:
(A) {0,1,2} (idle: 3,4,5)
(B) {1,2,3} (idle: 0,4,5)
(C) {2,3,4} (idle: 0,1,5)
(D) {3,4,5} (idle: 0,1,2)
Observe that if it is (B) or (C), then we will fail to match ✅「5th → a, 1st → b, |a-b|=2」:
┌───┬───┬───┬───┬───┬───┐
│ 5▲│4th│3rd│2nd│ 1▲│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ ▬ │ ▬ │ │ │ ▬ │
└───┴───┴───┴───┴───┴───┘
Therefore, it is (A) or (D). It is (D) indeed, for otherwise we will fail to avoid ⛔「⟨ ³ʳᵈb ¹ˢᵗa ⟩, a > b」.
As {p4, p3, p0} = {3,4,5}, we have
(1) {p5, p2, p1} = {0,1,2}.
To match ✅「5th → a, 1st → b, |a-b|=2」, we need
(2) {p5, p1} = {0,2}.
Combining (1) with (2), we get p2 = 1:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│ 2■│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ │ │ 1 │ │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ 2 │ 5 │ 0 │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘
It then follows from (1) and ⛔「Jump(0,1) = 0」 that
┌───┬───┬───┬───┬───┬───┐
│ 5■│4th│3rd│2nd│ 1■│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ │ │ 1 │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ 0 │ │ │ 1 │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ 0 │ │ │ 1 │ 2 │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ 5 │ │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘
Next, we consider where to place 4. If it is p4, then we cannot match ✅「⟨⋯ a ⋯ 4 ⋯⟩, a = 1|5」. Else if it is p3, then the preceding pattern implies
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 0 │ 5 │ 4 │ 1 │ 2 │ │
└───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ │ │ 3 │ │
└───┴───┴───┴───┴───┴───┘
and we will fail to avoid ⛔「⟨ ³ʳᵈa ⁰ᵗʰb ⟩, a > b」. Consequently, 4 = p0:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│ 0■│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 0 │ │ │ 1 │ 2 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ 0 │ │ │ 1 │ 2 │ 4 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ 5 │ │ 3 │ │
└───┴───┴───┴───┴───┴───┘
Finally, to avoid ⛔「⟨ ³ʳᵈa ⁰ᵗʰb ⟩, a > b」, we finish by
┌───┬───┬───┬───┬───┬───┐
│5th│ 4■│ 3■│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 0 │ │ │ 1 │ 2 │ 4 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ 0 │ 5 │ │ 1 │ 2 │ 4 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 0 │ 5 │ 3 │ 1 │ 2 │ 4 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.12