Rearrange the digits in ⟨125034⟩ to meet the rules below.
⟨5th 4th 3rd 2nd 1st 0th⟩
✅Match
Jump(2,4) = 1
⟨ ⁵ᵗʰa ⁴ᵗʰb ⟩, (ab)₁₀ ≥ 24
4th → a, 1st → b, a+b=0+5n
⛔Avoid
⟦3,4⟧ ∋ 0
5th|4th|2nd|0th → 4
⟨⋯ 3 ⋯ a ⋯⟩, a = 0|4
#125034_v2.0
WR 2023-10-22
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✅✅✅💎✅✅
✅✅✅✅✅✅
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
Step 1 │ │ │ │ │ 4 │ │
├───┼───┼───┼───┼───┼───┤
Step 2 │ │ │ 2 │ │ 4 │ │
├───┼───┼───┼───┼───┼───┤
Step 3 │ │ 1 │ 2 │ │ 4 │ │
├───┼───┼───┼───┼───┼───┤
Step 4 │ │ 1 │ 2 │ │ 4 │ 3 │
├───┼───┼───┼───┼───┼───┤
Step 5 │ 5 │ 1 │ 2 │ │ 4 │ 3 │
├───┼───┼───┼───┼───┼───┤
Step 6 │ 5 │ 1 │ 2 │ 0 │ 4 │ 3 │
└───┴───┴───┴───┴───┴───┘
Proof
═════
Notation: if nth -> a, then we write [nth] = a.
(1) By ⛔「5th|4th|2nd|0th → 4」, we have 4=[3rd] or [1st]. We claim that 4=[1st]. Suppose on the contrary 4=[3rd].
Then, by ✅【Jump(2,4) = 1】, either [5th] or [3rd] is 2:
┌───┬───┬───┬───┬───┬───┐
│*5 │4th│ 3▲│2nd│*1 │0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
(2) │ x │ │ 4 │ │ y │ │
└───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ │ 5 │ 0 │ 3 │ │
└───┴───┴───┴───┴───┴───┘
If x is 2, then by ✅【⟨ ⁵ᵗʰa ⁴ᵗʰb ⟩, (ab)₁₀ ≥ 24】, we need [4th]=5:
┌───┬───┬───┬───┬───┬───┐
│ 5▲│ 4▲│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 2 │ 5 │ 4 │ │ │ │
└───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ │ │ 0 │ 3 │ │
└───┴───┴───┴───┴───┴───┘
and then by ✅【4th → a, 1st → b, a+b=0+5n】, we have
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│ 1▲│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 2 │ 5 │ 4 │ │ 0 │ │
└───┴───┴───┴───┴───┴───┘
It is a contradiction, as either ⛔「⟦3,4⟧ ∋ 0」 or ⛔「⟨⋯ 3 ⋯ a ⋯⟩, a = 0|4」 would be matched. So, we need y=2 in (2). By ✅【4th → a, 1st → b, a+b=0+5n】, this implies
┌───┬───┬───┬───┬───┬───┐
│5th│ 4▲│3rd│2nd│ 1▲│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ 3 │ 4 │ │ 2 │ │
└───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ │ 5 │ 0 │ │ │
└───┴───┴───┴───┴───┴───┘
It then follows from ✅【⟨ ⁵ᵗʰa ⁴ᵗʰb ⟩, (ab)₁₀ ≥ 24】 that
┌───┬───┬───┬───┬───┬───┐
│ 5▲│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 5 │ 3 │ 4 │ │ 2 │ │
└───┴───┴───┴───┴───┴───┘
It is again a contradiction, as either ⛔「⟦3,4⟧ ∋ 0」 or ⛔「⟨⋯ 3 ⋯ a ⋯⟩, a = 0|4」 would be matched. This verifies our claim in (1).
Accordingly, we have
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│ 1■│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
Step 1 │ │ │ │ │ 4 │ │
└───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ 5 │ 0 │ 3 │ │
└───┴───┴───┴───┴───┴───┘
The rest are straightforward. By ✅【Jump(2,4) = 1】 and ✅【4th → a, 1st → b, a+b=0+5n】, we have
┌───┬───┬───┬───┬───┬───┐
│5th│ 4■│ 3■│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ │ │ │ 4 │ │
├───┼───┼───┼───┼───┼───┤
Step 2 │ │ │ 2 │ │ 4 │ │
├───┼───┼───┼───┼───┼───┤
Step 3 │ │ 1 │ 2 │ │ 4 │ │
└───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ 5 │ 0 │ 3 │ │
└───┴───┴───┴───┴───┴───┘
By ⛔「⟨⋯ 3 ⋯ a ⋯⟩, a = 0|4」, 3 is at the right of 4, so
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│ 0■│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ 1 │ 2 │ │ 4 │ │
├───┼───┼───┼───┼───┼───┤
Step 4 │ │ 1 │ 2 │ │ 4 │ 3 │
└───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ 5 │ 0 │ │ │
└───┴───┴───┴───┴───┴───┘
Finally, by ✅【⟨ ⁵ᵗʰa ⁴ᵗʰb ⟩, (ab)₁₀ ≥ 24】, we get [5th]=5. We reach the answer:
┌───┬───┬───┬───┬───┬───┐
│ 5■│4th│3rd│ 2■│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ 1 │ 2 │ │ 4 │ 3 │
├───┼───┼───┼───┼───┼───┤
Step 5 │ 5 │ 1 │ 2 │ │ 4 │ 3 │
├───┼───┼───┼───┼───┼───┤
Step 6 │ 5 │ 1 │ 2 │ 0 │ 4 │ 3 │
└───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.0