Rearrange the digits in ⟨1263045⟩ to meet the rules below.
⟨6th 5th 4th 3rd 2nd 1st 0th⟩
✅Match
6th → a, 2nd → b, a+b=2
5th → a, 4th → b, a+b=7
1st → 3
⛔Avoid
Sim⟨ ⁶ᵗʰ6 ⁵ᵗʰ0 ⁴ᵗʰ2 ³ʳᵈ1 ²ⁿᵈ3 ¹ˢᵗ5 ⁰ᵗʰ4 ⟩ ≥ 1
Jump(2,5) = 1
6th|5th|2nd|1st|0th → 6
#125034_v2.2
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ │ │ │ │ 3 │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ │ 6 │ │ │ 3 │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 3 │ │ 1 │ 6 │ │ │ 3 │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 4 │ │ 1 │ 6 │ 4 │ │ 3 │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 5 │ │ 1 │ 6 │ 4 │ │ 3 │ 5 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 6 │ 2 │ 1 │ 6 │ 4 │ │ 3 │ 5 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 7 │ 2 │ 1 │ 6 │ 4 │ 0 │ 3 │ 5 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
Proof of 2023-12-13 Q1(m=6)
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Notation: if nth -> a, then we write [nth] = a.
Plainly, our first step follows from ✅「1st → 3」.
In view of ✅「6th → a, 2nd → b, a+b=2」, we have
┌───┬───┬───┬───┬───┬───┬───┐
│*6 │5th│4th│3rd│*2 │1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
(1) │ X │ │ │ │ Y │ 3 │ │
└───┴───┴───┴───┴───┴───┴───┘
where {X,Y} = {0,2}. Now the idle numbers are
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ │ 6 │ │ │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
so to match ✅「5th → a, 4th → b, a+b=7」, we need
┌───┬───┬───┬───┬───┬───┬───┐
│6th│*5 │*4 │3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ X │ E │ F │ │ Y │ 3 │ │
└───┴───┴───┴───┴───┴───┴───┘
where {E,F} = {1,6}. By ⛔「6th|5th|2nd|1st|0th → 6」, we have E != 6. Accordingly, we get
┌───┬───┬───┬───┬───┬───┬───┐
│6th│ 5■│ 4■│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ X │ 1 │ 6 │ │ Y │ 3 │ │
└───┴───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
4,5 are placed at 3rd and 0th. To avoid ⛔「Sim⟨ ⁶ᵗʰ6 ⁵ᵗʰ0 ⁴ᵗʰ2 ³ʳᵈ1 ²ⁿᵈ3 ¹ˢᵗ5 ⁰ᵗʰ4 ⟩ ≥ 1」, we cannot have 4 = [0th]. So 4 = [3rd], and 5 = [0th].
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│ 3■│2nd│1st│ 0■│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ X │ 1 │ 6 │ 4 │ Y │ 3 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Finally, to avoid ⛔「Jump(2,5) = 1」, we have Y != 2. As {X,Y} = {0,2} by (1), we see that the answer is
┌───┬───┬───┬───┬───┬───┬───┐
│ 6■│5th│4th│3rd│ 2■│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 2 │ 1 │ 6 │ 4 │ 0 │ 3 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.2