Rearrange the digits in ⟨125034⟩ to meet the rules below.
⟨5th 4th 3rd 2nd 1st 0th⟩
✅Match
⟨ ⁴ᵗʰc ³ʳᵈb ¹ˢᵗa ⟩, a > b > c
⟨ ⁵ᵗʰa ²ⁿᵈb ⟩, max⟦a,b⟧ = 4
⟦0,4⟧ ∋ 1
Jump(2,3) = 3
⛔Avoid
max ⊢2⊣ = 2
#125034_v2.9
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ 0 │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ 0 │ 1 │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ │ 0 │ 1 │ │ 2 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ 3 │ 0 │ 1 │ │ 2 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ 3 │ 0 │ 1 │ 4 │ 2 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 3 │ 0 │ 1 │ 4 │ 2 │ 5 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
Proof of 2025-05-20 WR
══════════════════════
Notation: if nth -> a, then we write [nth] = a.
To match ✅「Jump(2,3) = 3」, there are two possibilities:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
(1) │ │ ▬ │ │ │ │ ▬ │
├───┼───┼───┼───┼───┼───┤
(2) │ ▬ │ │ │ │ ▬ │ │
└───┴───┴───┴───┴───┴───┘
where "▬" are occupied by 2,3.
We show that (2) holds actually.
------------------------------
Suppose on the contrary (1) holds:
┌───┬───┬───┬───┬───┬───┐
│5th│ 4▲│3rd│2nd│1st│ 0▲│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ ▬ │ │ │ │ ▬ │
└───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ │ 5 │ 0 │ │ 4 │
└───┴───┴───┴───┴───┴───┘
Then, to match ✅「⟨ ⁴ᵗʰc ³ʳᵈb ¹ˢᵗa ⟩, a > b > c」, the "b" and "a" have to be 4 and 5:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│ 3▲│2nd│ 1▲│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ ▬ │ 4 │ │ 5 │ ▬ │
└───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ │ │ 0 │ │ │
└───┴───┴───┴───┴───┴───┘
Note that we cannot match ✅「⟦0,4⟧ ∋ 1」 now. This shows that if (2) does not hold, then there is a contradiction.
------------------------------
We have verified that (2) holds. Accordingly, we get
┌───┬───┬───┬───┬───┬───┐
│ 5▲│4th│3rd│2nd│ 1▲│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ ▬ │ │ │ │ ▬ │ │
└───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ │ 5 │ 0 │ │ 4 │
└───┴───┴───┴───┴───┴───┘
Then, in view of ✅「⟨ ⁴ᵗʰc ³ʳᵈb ¹ˢᵗa ⟩, a > b > c」, we get:
┌───┬───┬───┬───┬───┬───┐
│5th│ 4■│ 3■│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ ▬ │ 0 │ │ │ ▬ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ ▬ │ 0 │ 1 │ │ ▬ │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ 5 │ │ │ 4 │
└───┴───┴───┴───┴───┴───┘
Next, let us solve the "▬". If ([5th], [1st]) = (2,3):
┌───┬───┬───┬───┬───┬───┐
│ 5▲│4th│3rd│2nd│ 1▲│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 2 │ 0 │ 1 │ │ 3 │ │
└───┴───┴───┴───┴───┴───┘
then we would fail to avoid ⛔「max ⊢2⊣ = 2」. Therefore, it has to be ([5th], [1st]) = (3,2):
┌───┬───┬───┬───┬───┬───┐
│ 5■│4th│3rd│2nd│ 1■│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ ▬ │ 0 │ 1 │ │ ▬ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ ▬ │ 0 │ 1 │ │ 2 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ 3 │ 0 │ 1 │ │ 2 │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ 5 │ │ │ 4 │
└───┴───┴───┴───┴───┴───┘
Finally, to match ✅「⟨ ⁵ᵗʰa ²ⁿᵈb ⟩, max⟦a,b⟧ = 4」, we finish by
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│ 2■│1st│ 0■│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 3 │ 0 │ 1 │ │ 2 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ 3 │ 0 │ 1 │ 4 │ 2 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 3 │ 0 │ 1 │ 4 │ 2 │ 5 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.9