Rearrange the digits in ⟨125034⟩ to meet the rules below.
⟨5th 4th 3rd 2nd 1st 0th⟩
✅Match
⟨⋯ 2 ⋯ 0 ⋯ 4 ⋯⟩
median {p5, p2, p1} = 1
Jump(1,3) ≤ 3
5th → a, 3rd → b, a+b=6
⛔Avoid
4th → a, 1st → b, ab=0+5n
#125034_v2.9
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│▒
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Step 1 │ │ │ │ 0 │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ 1 │ │ │ 0 │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ 1 │ │ 5 │ 0 │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ 1 │ 2 │ 5 │ 0 │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ 1 │ 2 │ 5 │ 0 │ 3 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 1 │ 2 │ 5 │ 0 │ 3 │ 4 │▒
└───┴───┴───┴───┴───┴───┘▒
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Proof of 2025-06-16 WR
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Notation: if nth -> a, then we write [nth] = a.
Observe that ✅「median {p5, p2, p1} = 1」 implies
(1) 0 ∈ {[5th], [2nd], [1st]}.
On the other hand, note that:
(1.1) according to ✅「⟨⋯ 2 ⋯ 0 ⋯ 4 ⋯⟩」, 0 is not in the left corner (5th);
(1.2) by ⛔「4th → a, 1st → b, ab=0+5n」, it is not at 1st either.
Therefore, (1) implies 0 = [2nd]:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│ 2■│1st│0th│▒
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Step 1 │ │ │ │ 0 │ │ │▒
└───┴───┴───┴───┴───┴───┘▒
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--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ 5 │ │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘
Then, we consider how to match ✅「5th → a, 3rd → b, a+b=6」. There are two possibilities for S := {[5th], [3rd]}:
(2) S = {2,4} or {1,5}.
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
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│ ▬ │ │ ▬ │ 0 │ │ │
└───┴───┴───┴───┴───┴───┘
If S = {2,4}, then we would fail to match ✅「⟨⋯ 2 ⋯ 0 ⋯ 4 ⋯⟩」. Therefore,
(3) S = {1,5}.
A fortiori, we have
(4) 1 ∈ S = {[5th], [3rd]}.
By ✅「median {p5, p2, p1} = 1」, 1 is in {[5th], [2nd], [1st]}. Combining this with (4), we get 1 = [5th]:
┌───┬───┬───┬───┬───┬───┐
│ 5■│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ │ │ 0 │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ 1 │ │ │ 0 │ │ │▒
└───┴───┴───┴───┴───┴───┘▒
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--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ 2 │ 5 │ │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘
It then follows from (3) that
┌───┬───┬───┬───┬───┬───┐
│5th│4th│ 3■│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 1 │ │ │ 0 │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ 1 │ │ 5 │ 0 │ │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ 2 │ │ │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘
To match ✅「⟨⋯ 2 ⋯ 0 ⋯ 4 ⋯⟩」, we need to have 2 to the left of 0. There is only one way to do so:
┌───┬───┬───┬───┬───┬───┐
│5th│ 4■│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 1 │ │ 5 │ 0 │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ 1 │ 2 │ 5 │ 0 │ │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ │ │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘
Finally, to match ✅「Jump(1,3) ≤ 3」, we finish by
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│ 1■│ 0■│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 1 │ 2 │ 5 │ 0 │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ 1 │ 2 │ 5 │ 0 │ 3 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 1 │ 2 │ 5 │ 0 │ 3 │ 4 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.9