Rearrange the digits in ⟨125034⟩ to meet the rules below.
⟨ ⁵ᵗʰ▨ ⁴ᵗʰ▨ ³ʳᵈ▨ ²ⁿᵈ▨ ¹ˢᵗ▨ ⁰ᵗʰ▨ ⟩
✅Match
4th → a, 2nd → b, a+b=2
{p4, p3, p1} = ? + {0,1,2}
⛔Avoid
Jump(0,1) ≥ 2
⟨⋯ ᵃb ⋯⟩, |a-b|=1
⟨ ⁵ᵗʰ↓ ⁴ᵗʰ↓ ³ʳᵈ↓ ²ⁿᵈ↑ ¹ˢᵗ↑ ⁰ᵗʰ↑ ⟩ after ⟨⇌⟩
#125034_v2.12
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ 2 │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ 2 │ │ 0 │ │ │▒
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Step 3 │ │ 2 │ │ 0 │ │ 4 │▒
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Step 4 │ 5 │ 2 │ │ 0 │ │ 4 │▒
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Step 5 │ 5 │ 2 │ │ 0 │ 3 │ 4 │▒
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Step 6 │ 5 │ 2 │ 1 │ 0 │ 3 │ 4 │▒
└───┴───┴───┴───┴───┴───┘▒
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Proof of 2026-03-24 WR
══════════════════════
To match ✅「{p4, p3, p1} = ? + {0,1,2}」, we need
(1) [4th], [3rd], [1st] are consecutive integers.
┌───┬───┬───┬───┬───┬───┐
│5th│*4 │*3 │2nd│*1 │0th│
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│ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┘
On the other hand, ✅「4th → a, 2nd → b, a+b=2」 implies
(2) ([4th], [2nd]) = (0,2) or (2,0).
┌───┬───┬───┬───┬───┬───┐
│5th│*4 │*3 │2nd│*1 │0th│
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│ │0 2│ │0 2│ │ │
└───┴───┴───┴───┴───┴───┘
Note that if ([4th], [2nd]) = (0,2), then there is no way to match (1):
┌───┬───┬───┬───┬───┬───┐
│5th│*4 │*3 │2nd│*1 │0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ 0 │ │ 2 │ │ │
└───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ │ 5 │ │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘
Therefore, we have ([4th], [2nd]) = (2,0) instead.
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│5th│ 4■│3rd│ 2■│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ 2 │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ 2 │ │ 0 │ │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ │ 5 │ │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘
Then, to avoid ⛔「Jump(0,1) ≥ 2」, we need
(3) 1 = [3rd] | [1st] | [0th].
┌───┬───┬───┬───┬───┬───┐
│5th│*4 │*3 │2nd│*1 │0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ 2 │ - │ 0 │ - │ - │
└───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ │ 5 │ │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘
If 1 = [0th] then we fail to avoid ⛔「⟨⋯ ᵃb ⋯⟩, |a-b|=1」. Therefore, 1 = [3rd] | [1st]. Combining this with (1), we have
(4) {[3rd], [1st]} = {1,3}.
This implies {[5th], [0th]} = {4,5}:
┌───┬───┬───┬───┬───┬───┐
│5th│*4 │*3 │2nd│*1 │0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│4 5│ 2 │1 3│ 0 │1 3│4 5│
└───┴───┴───┴───┴───┴───┘
To avoid ⛔「⟨⋯ ᵃb ⋯⟩, |a-b|=1」, we cannot have 4 = [5th]. It follows that 4=[0th] and 5=[5th]:
┌───┬───┬───┬───┬───┬───┐
│ 5■│4th│3rd│2nd│1st│ 0■│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ 2 │ │ 0 │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ │ 2 │ │ 0 │ │ 4 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ 5 │ 2 │ │ 0 │ │ 4 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ │ │ │ 3 │ │
└───┴───┴───┴───┴───┴───┘
Finally, to avoid ⛔「⟨ ⁵ᵗʰ↓ ⁴ᵗʰ↓ ³ʳᵈ↓ ²ⁿᵈ↑ ¹ˢᵗ↑ ⁰ᵗʰ↑ ⟩ after ⟨⇌⟩」, we finish by
┌───┬───┬───┬───┬───┬───┐
│5th│4th│ 3■│2nd│ 1■│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 5 │ 2 │ │ 0 │ │ 4 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ 5 │ 2 │ │ 0 │ 3 │ 4 │▒
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Step 6 │ 5 │ 2 │ 1 │ 0 │ 3 │ 4 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.12