Rearrange the digits in ⟨125034⟩ to meet the rules below.
⟨ ⁵ᵗʰ▨ ⁴ᵗʰ▨ ³ʳᵈ▨ ²ⁿᵈ▨ ¹ˢᵗ▨ ⁰ᵗʰ▨ ⟩
✅Match
3rd|2nd → 3
Jump(2,4) = 1
⟦1,5⟧ ∋ 0,2
⛔Avoid
⟨ ⁵ᵗʰa ³ʳᵈb ⟩, a > b
0th → 0|2|5
#125034_v2.11
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ 1 │ │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ 1 │ │ │ │ │ 4 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ 1 │ │ │ 2 │ │ 4 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ 1 │ │ 3 │ 2 │ │ 4 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ 1 │ │ 3 │ 2 │ 5 │ 4 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 1 │ 0 │ 3 │ 2 │ 5 │ 4 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
Proof of 2025-12-30 WR
══════════════════════
Notation: if nth -> a, then we write [nth] = a.
Firstly, we consider what [5th] is.
• By ✅「3rd|2nd → 3」, it is not 3.
• By ✅「⟦1,5⟧ ∋ 0,2」, it is not 0 or 2.
• By ⛔「⟨ ⁵ᵗʰa ³ʳᵈb ⟩, a > b」, it is not 5.
Therefore,
(1) [5th] = 1 or 4.
------------------------------
If [5th] = 4, then to avoid ⛔「⟨ ⁵ᵗʰa ³ʳᵈb ⟩, a > b」, we need
┌───┬───┬───┬───┬───┬───┐
│ 5▲│4th│ 3▲│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 4 │ │ 5 │ │ │ │
└───┴───┴───┴───┴───┴───┘
Then, in view of ✅「⟦1,5⟧ ∋ 0,2」, we have
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│ 2▲│ 1▲│ 0▲│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 4 │ │ 5 │0 2│0 2│ 1 │
└───┴───┴───┴───┴───┴───┘
We just failed to match ✅「3rd|2nd → 3」, which is a contradiction.
------------------------------
Consequently, it follows from (1) that [5th] = 1:
┌───┬───┬───┬───┬───┬───┐
│ 5■│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ 1 │ │ │ │ │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ 2 │ 5 │ 0 │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘
Next, we determine the value of [0th]. Combining ⛔「0th → 0|2|5」 with ✅「3rd|2nd → 3」, we see that it is 4:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│ 0■│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 1 │ │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ 1 │ │ │ │ │ 4 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ 2 │ 5 │ 0 │ 3 │ │
└───┴───┴───┴───┴───┴───┘
Then, using ✅「Jump(2,4) = 1」 and ✅「3rd|2nd → 3」 respectively, we get
┌───┬───┬───┬───┬───┬───┐
│5th│4th│ 3■│ 2■│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 1 │ │ │ │ │ 4 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ 1 │ │ │ 2 │ │ 4 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ 1 │ │ 3 │ 2 │ │ 4 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ 5 │ 0 │ │ │
└───┴───┴───┴───┴───┴───┘
Finally, to match ✅「⟦1,5⟧ ∋ 0,2」, we finish by
┌───┬───┬───┬───┬───┬───┐
│5th│ 4■│3rd│2nd│ 1■│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 1 │ │ 3 │ 2 │ │ 4 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ 1 │ │ 3 │ 2 │ 5 │ 4 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 1 │ 0 │ 3 │ 2 │ 5 │ 4 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.11