Rearrange the digits in ⟨125034⟩ to meet the rules below.
⟨ ⁵ᵗʰ▨ ⁴ᵗʰ▨ ³ʳᵈ▨ ²ⁿᵈ▨ ¹ˢᵗ▨ ⁰ᵗʰ▨ ⟩
✅Match
Jump(1,2) = 1
2nd → a, 1st → b, |a-b|=4
⟨⋯ 1 ⋯ 5 ⋯ 4 ⋯⟩
⛔Avoid
4th|2nd|1st|0th → 2
⟨ ⁵ᵗʰc ³ʳᵈb ¹ˢᵗa ⟩, a > b > c
#125034_v2.11
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ │ 2 │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ 1 │ │ 2 │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ 1 │ 5 │ 2 │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ 1 │ 5 │ 2 │ │ │ 3 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ 1 │ 5 │ 2 │ 4 │ │ 3 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 1 │ 5 │ 2 │ 4 │ 0 │ 3 │▒
└───┴───┴───┴───┴───┴───┘▒
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Proof of 2025-12-02 WR
══════════════════════
Notation: if nth -> a, then we write [nth] = a.
By ⛔「4th|2nd|1st|0th → 2」, we have
(1) 2 = [5th] or [3rd].
(1.1) We show that 2 = [3rd].
------------------------------
For, suppose on the contrary 2 = [5th]:
┌───┬───┬───┬───┬───┬───┐
│ 5▲│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 2 │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┘
Then by ✅「Jump(1,2) = 1」, we have
┌───┬───┬───┬───┬───┬───┐
│5th│4th│ 3▲│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 2 │ │ 1 │ │ │ │
└───┴───┴───┴───┴───┴───┘
To match ✅「2nd → a, 1st → b, |a-b|=4」, we need
(2) { [2nd], [1st] } = {0,4}.
But having (2), we will fail to match ✅「⟨⋯ 1 ⋯ 5 ⋯ 4 ⋯⟩」, which is a contradiction.
------------------------------
We have verified (1.1). Accordingly, we get 2 = [3rd]:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│ 3■│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ │ 2 │ │ │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ │ 5 │ 0 │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘
Then, to match ✅「Jump(1,2) = 1」, there are two possibilities:
┌───┬───┬───┬───┬───┬───┐
│ 5▲│4th│3rd│2nd│ 1▲│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 1 │ │ 2 │ │ 1 │ │
└───┴───┴───┴───┴───┴───┘
If 1 = [1st], then we cannot match ✅「⟨⋯ 1 ⋯ 5 ⋯ 4 ⋯⟩」. Therefore, 1 = [5th]:
┌───┬───┬───┬───┬───┬───┐
│ 5■│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ │ 2 │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ 1 │ │ 2 │ │ │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ 5 │ 0 │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘
Then, to match ✅「2nd → a, 1st → b, |a-b|=4」, we need
(3) { [2nd], [1st] } = {0,4}.
Since 5 is to the left of 4 by ✅「⟨⋯ 1 ⋯ 5 ⋯ 4 ⋯⟩」, we get 5 = [4th]:
┌───┬───┬───┬───┬───┬───┐
│5th│ 4■│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 1 │ │ 2 │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ 1 │ 5 │ 2 │ │ │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ │ 0 │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘
Note that (3) now implies [0th] = 3:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│ 0■│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 1 │ 5 │ 2 │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ 1 │ 5 │ 2 │ │ │ 3 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ │ 0 │ │ 4 │
└───┴───┴───┴───┴───┴───┘
Finally, to avoid ⛔「⟨ ⁵ᵗʰc ³ʳᵈb ¹ˢᵗa ⟩, a > b > c」, we finish by
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│ 2■│ 1■│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 1 │ 5 │ 2 │ │ │ 3 │▒
├───┼───┼───┼───┼───┼───┤▒
│ 1 │ 5 │ 2 │ 4 │ │ 3 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 1 │ 5 │ 2 │ 4 │ 0 │ 3 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.11