Rearrange the digits in ⟨125034⟩ to meet the rules below.
⟨5th 4th 3rd 2nd 1st 0th⟩
✅Match
⟦1,5⟧ ∋ 2,3
⟨⋯ Perm(3,4) ⋯⟩
{p4, p2, p0} = ? + {0,1,2}
⛔Avoid
⟨ ⁴ᵗʰa ²ⁿᵈb ⟩, min⟦a,b⟧ = 2
#125034_v2.10
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ │ │ │ │ 1 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ 5 │ │ │ │ │ 1 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ 5 │ │ 0 │ │ │ 1 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ 5 │ │ 0 │ │ 4 │ 1 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ 5 │ │ 0 │ 3 │ 4 │ 1 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 5 │ 2 │ 0 │ 3 │ 4 │ 1 │▒
└───┴───┴───┴───┴───┴───┘▒
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Proof of 2025-09-23 WR
══════════════════════
Notation: if Nth -> a, then we write pN = a.
Combining ✅「⟦1,5⟧ ∋ 2,3」 with ✅「⟨⋯ Perm(3,4) ⋯⟩」, we see that
(1) digits 2,3,4 are positioned between digits 1 and 5.
It implies three possibliities:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
(2) │1/5│ │ │ │1/5│ 0 │
├───┼───┼───┼───┼───┼───┤
(3) │ 0 │1/5│ │ │ │1/5│
├───┼───┼───┼───┼───┼───┤
(4) │1/5│ │ │ │ │1/5│
└───┴───┴───┴───┴───┴───┘
(4.1) We show that case (4) holds actually.
------------------------------
(5.1) Suppose case (2) holds:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│1/5│ │ │ │1/5│ 0 │
└───┴───┴───┴───┴───┴───┘
Then, to match ✅「{p4, p2, p0} = ? + {0,1,2}」, we need {p4, p2, p0} = {0,1,2}:
┌───┬───┬───┬───┬───┬───┐
│5th│ 4▲│3rd│ 2▲│1st│ 0▲│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│1/5│ ▬ │ │ ▬ │1/5│ 0 │
└───┴───┴───┴───┴───┴───┘
As 1 ∈ {p5, p1}, this shows a contradiction.
(5.2) Else, suppose case (3) holds:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 0 │1/5│ │ │ │1/5│
└───┴───┴───┴───┴───┴───┘
Note that there is no way to make {p4, p2, p0} consist of consecutive integer. Therefore, we would fail to match ✅「{p4, p2, p0} = ? + {0,1,2}」 again.
------------------------------
We have verify (4.1). Accordingly, (4) holds and we have:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│1/5│ │ │ │ │1/5│
└───┴───┴───┴───┴───┴───┘
Next, we claim that (p5, p0) = (5, 1). For, if on the contrary (p5, p0) = (1, 5):
┌───┬───┬───┬───┬───┬───┐
│ 5▲│4th│3rd│2nd│1st│ 0▲│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 1 │ │ │ │ │ 5 │
└───┴───┴───┴───┴───┴───┘
then to match ✅「{p4, p2, p0} = ? + {0,1,2}」, we need {p4, p2} = {3,4}. It follows that 3,4 are not adjacent and we would fail to match ✅「⟨⋯ Perm(3,4) ⋯⟩」.
We have thus shown that (p5, p0) = (5, 1):
┌───┬───┬───┬───┬───┬───┐
│ 5■│4th│3rd│2nd│1st│ 0■│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ │ │ │ │ 1 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ 5 │ │ │ │ │ 1 │▒
└───┴───┴───┴───┴───┴───┘▒
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--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ 2 │ │ 0 │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘
Next, we consider ✅「{p4, p2, p0} = ? + {0,1,2}」 again. To match it, we need:
(6) {p4, p2, p0} = {0,1,2} or {1,2,3}.
┌───┬───┬───┬───┬───┬───┐
│5th│ 4▲│3rd│ 2▲│1st│ 0▲│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 5 │ ▬ │ │ ▬ │ │ 1 │
└───┴───┴───┴───┴───┴───┘
Note that 3,4 will not be adjacent if {p4, p2, p0} = {0,1,2}:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│ 3▲│2nd│ 1▲│ 0 │
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 5 │0/2│3/4│0/2│3/4│ 1 │
└───┴───┴───┴───┴───┴───┘
So, to match ✅「⟨⋯ Perm(3,4) ⋯⟩」, we need {p4, p2, p0} = {1,2,3} instead, and {p3, p1} = {0,4} follows:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 5 │2/3│0/4│2/3│0/4│ 1 │
└───┴───┴───┴───┴───┴───┘
If 0 = p1, then we would match ⛔「⟨ ⁴ᵗʰa ²ⁿᵈb ⟩, min⟦a,b⟧ = 2」. Hence, we get 0 = p3, and 4 = p1:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│ 3■│2nd│ 1■│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 5 │ │ │ │ │ 1 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ 5 │ │ 0 │ │ │ 1 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ 5 │ │ 0 │ │ 4 │ 1 │▒
└───┴───┴───┴───┴───┴───┘▒
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--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ 2 │ │ │ 3 │ │
└───┴───┴───┴───┴───┴───┘
Finally, to match ✅「⟨⋯ Perm(3,4) ⋯⟩」, we finish by
┌───┬───┬───┬───┬───┬───┐
│5th│ 4■│3rd│ 2■│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 5 │ │ 0 │ │ 4 │ 1 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ 5 │ │ 0 │ 3 │ 4 │ 1 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 5 │ 2 │ 0 │ 3 │ 4 │ 1 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.10