Rearrange the digits in ⟨125034⟩ to meet the rules below.
⟨5th 4th 3rd 2nd 1st 0th⟩
✅Match
⟨ ³ʳᵈa ⁰ᵗʰb ⟩, min⟦a,b⟧ = 1
⛔Avoid
4th|2nd|1st|0th → 0
1st → 0|1|4
5th|4th|3rd|0th → 5
⟨⋯ 4 ⋯ a ⋯⟩, a = 0|1|2|3|5
0 ∾ 1 ∾ 3 ∾ 4 ∾ 5
⟨⋯ ? ⋯ 3 ⋯ (?+3)⟩ (?≠0)
#125034_v2.9
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ │ │ │ │ 4 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ 0 │ │ │ │ │ 4 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ 0 │ │ │ 1 │ │ 4 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ 0 │ │ │ 1 │ 5 │ 4 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ 0 │ │ 3 │ 1 │ 5 │ 4 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 0 │ 2 │ 3 │ 1 │ 5 │ 4 │▒
└───┴───┴───┴───┴───┴───┘▒
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Proof of 2025-05-06 WR
══════════════════════
Notation: if nth -> a, then we write [nth] = a.
Note that to avoid ⛔「⟨⋯ 4 ⋯ a ⋯⟩, a = 0|1|2|3|5」, we need 4 is in the right corner (0th):
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│ 0■│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ │ │ │ │ 4 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ 5 │ 0 │ 3 │ │
└───┴───┴───┴───┴───┴───┘
Then, to match ✅「⟨ ³ʳᵈa ⁰ᵗʰb ⟩, min⟦a,b⟧ = 1」, we need 0 = [5th] | [4th]:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ ▬ │ ▬ │ │ │ │ 4 │
└───┴───┴───┴───┴───┴───┘
In view of ⛔「4th|2nd|1st|0th → 0」, it is 0 = [5th] actually:
┌───┬───┬───┬───┬───┬───┐
│ 5■│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ │ │ │ │ 4 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ 0 │ │ │ │ │ 4 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ 5 │ │ 3 │ │
└───┴───┴───┴───┴───┴───┘
Next, we consider where to place 1. Combining ✅「⟨ ³ʳᵈa ⁰ᵗʰb ⟩, min⟦a,b⟧ = 1」 with ⛔「1st → 0|1|4」, we see that
(1) 1 = [3rd] | [2nd].
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 0 │ │ ▬ │ ▬ │ │ 4 │
└───┴───┴───┴───┴───┴───┘
(2) We show that 1 = [2nd] actually.
------------------------------
Suppose on the contrary 1 = [3rd]:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│ 3▲│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 0 │ │ 1 │ │ │ 4 │
└───┴───┴───┴───┴───┴───┘
Then, to avoid ⛔「⟨⋯ ? ⋯ 3 ⋯ (?+3)⟩ (?≠0)」, we need
┌───┬───┬───┬───┬───┬───┐
│5th│ 4▲│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 0 │ 3 │ 1 │ │ │ 4 │
└───┴───┴───┴───┴───┴───┘
However, in view of cycle decomposition, observe that we now have
(A) 0 ∾ 4 ∾ 5, and
(B) 1 ∾ 3 ∾ 4.
Consequently, we fail to avoid ⛔「0 ∾ 1 ∾ 3 ∾ 4 ∾ 5」, which is a contradiction.
------------------------------
We have verified (2). Accordingly, we get
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│ 2■│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 0 │ │ │ │ │ 4 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ 0 │ │ │ 1 │ │ 4 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ 2 │ 5 │ │ 3 │ │
└───┴───┴───┴───┴───┴───┘
Then, it follows from ⛔「5th|4th|3rd|0th → 5」 that 5 = [1st]:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│ 1■│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 0 │ │ │ 1 │ │ 4 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ 0 │ │ │ 1 │ 5 │ 4 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ 2 │ │ │ 3 │ │
└───┴───┴───┴───┴───┴───┘
Finally, to avoid ⛔「0 ∾ 1 ∾ 3 ∾ 4 ∾ 5」, we finish by
┌───┬───┬───┬───┬───┬───┐
│5th│ 4■│ 3■│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 0 │ │ │ 1 │ 5 │ 4 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ 0 │ │ 3 │ 1 │ 5 │ 4 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 0 │ 2 │ 3 │ 1 │ 5 │ 4 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.9