Rearrange the digits in ⟨1263045⟩ to meet the rules below.
⟨6th 5th 4th 3rd 2nd 1st 0th⟩
✅Match
⟨ ⁶ᵗʰa ²ⁿᵈb ⟩, min⟦a,b⟧ = 1
⟨? 6 ⋯ (?+4) ⋯ 5 ⋯⟩ (?≠2)
Jump(2,4) = 0
⟦1,2⟧ ∋ 3,4,6
⛔Avoid
3rd → 3|4
#125034_v2.3
┌───┬───┬───┬───┬───┬───┬───┐ │6th│5th│4th│3rd│2nd│1st│0th│▒ ╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒ Step 1 │ │ 6 │ │ │ │ │ │▒ ├───┼───┼───┼───┼───┼───┼───┤▒ Step 2 │ 1 │ 6 │ │ │ │ │ │▒ ├───┼───┼───┼───┼───┼───┼───┤▒ Step 3 │ 1 │ 6 │ │ │ 4 │ │ │▒ ├───┼───┼───┼───┼───┼───┼───┤▒ Step 4 │ 1 │ 6 │ │ │ 4 │ 2 │ │▒ ├───┼───┼───┼───┼───┼───┼───┤▒ Step 5 │ 1 │ 6 │ │ │ 4 │ 2 │ 0 │▒ ├───┼───┼───┼───┼───┼───┼───┤▒ Step 6 │ 1 │ 6 │ 3 │ │ 4 │ 2 │ 0 │▒ ├───┼───┼───┼───┼───┼───┼───┤▒ Step 7 │ 1 │ 6 │ 3 │ 5 │ 4 │ 2 │ 0 │▒ └───┴───┴───┴───┴───┴───┴───┘▒ ▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒ Proof of 2024-02-06 Q1(m=6) ═══════════════════════════ Notation: if nth -> a, then we write [nth] = a. Plainly, our first step follows from ✅「⟨? 6 ⋯ (?+4) ⋯ 5 ⋯⟩ (?≠2)」: ┌───┬───┬───┬───┬───┬───┬───┐ │6th│ 5■│4th│3rd│2nd│1st│0th│▒ ╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒ Step 1 │ │ 6 │ │ │ │ │ │▒ └───┴───┴───┴───┴───┴───┴───┘▒ ▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒ --- Idle --- ┌───┬───┬───┬───┬───┬───┬───┐ │ 1 │ 2 │ │ 3 │ 0 │ 4 │ 5 │ └───┴───┴───┴───┴───┴───┴───┘ Next, in view of ✅「⟦1,2⟧ ∋ 3,4,6」, we need to place 1 or 2 at 6th, otherwise 6 would not be contained in ⟦1,2⟧. Since [6th] != 2 by ✅「⟨? 6 ⋯ (?+4) ⋯ 5 ⋯⟩ (?≠2)」, we have [6th] = 1. ┌───┬───┬───┬───┬───┬───┬───┐ │ 6■│5th│4th│3rd│2nd│1st│0th│▒ ╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒ │ │ 6 │ │ │ │ │ │▒ ├───┼───┼───┼───┼───┼───┼───┤▒ Step 2 │ 1 │ 6 │ │ │ │ │ │▒ └───┴───┴───┴───┴───┴───┴───┘▒ ▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒ --- Idle --- ┌───┬───┬───┬───┬───┬───┬───┐ │ │ 2 │ │ 3 │ 0 │ 4 │ 5 │ └───┴───┴───┴───┴───┴───┴───┘ Then, it follows from ✅「⟦1,2⟧ ∋ 3,4,6」 and ✅「Jump(2,4) = 0」 that we have to match the following pattern: (1) ⟨16 ⋯ 3 ⋯ 42 ⋯⟩. Accordingly, there are only four ways to place 2,4: ┌───┬───┬───┬───┬───┬───┬───┐ │6th│5th│4th│3rd│2nd│1st│0th│ ╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡ (2) │ 1 │ 6 │ 4 │ 2 │ │ │ │ ├───┼───┼───┼───┼───┼───┼───┤ (3) │ 1 │ 6 │ │ 4 │ 2 │ │ │ ├───┼───┼───┼───┼───┼───┼───┤ (4) │ 1 │ 6 │ │ │ 4 │ 2 │ │ ├───┼───┼───┼───┼───┼───┼───┤ (5) │ 1 │ 6 │ │ │ │ 4 │ 2 │ └───┴───┴───┴───┴───┴───┴───┘ Case (4) holds actually, for: ● If (2) holds, then we could not match ✅「⟦1,2⟧ ∋ 3,4,6」 wherever 3 is placed; ● If (3) holds, then we match ⛔「3rd → 3|4」; ● If (5) holds, then we could not match ✅「⟨ ⁶ᵗʰa ²ⁿᵈb ⟩, min⟦a,b⟧ = 1」 wherever 0 is placed. Hence, we get ┌───┬───┬───┬───┬───┬───┬───┐ │6th│5th│4th│3rd│ 2■│ 1■│0th│▒ ╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒ │ 1 │ 6 │ │ │ │ │ │▒ ├───┼───┼───┼───┼───┼───┼───┤▒ Step 3 │ 1 │ 6 │ │ │ 4 │ │ │▒ ├───┼───┼───┼───┼───┼───┼───┤▒ Step 4 │ 1 │ 6 │ │ │ 4 │ 2 │ │▒ └───┴───┴───┴───┴───┴───┴───┘▒ ▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒ --- Idle --- ┌───┬───┬───┬───┬───┬───┬───┐ │ │ │ │ 3 │ 0 │ │ 5 │ └───┴───┴───┴───┴───┴───┴───┘ Next, to match ✅「⟨ ⁶ᵗʰa ²ⁿᵈb ⟩, min⟦a,b⟧ = 1」, we have 0 != [4th] and 0 != [3rd]. Therefore, 0 = [0th]: ┌───┬───┬───┬───┬───┬───┬───┐ │6th│5th│4th│3rd│2nd│1st│ 0■│▒ ╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒ │ 1 │ 6 │ │ │ 4 │ 2 │ │▒ ├───┼───┼───┼───┼───┼───┼───┤▒ Step 5 │ 1 │ 6 │ │ │ 4 │ 2 │ 0 │▒ └───┴───┴───┴───┴───┴───┴───┘▒ ▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒ --- Idle --- ┌───┬───┬───┬───┬───┬───┬───┐ │ │ │ │ 3 │ │ │ 5 │ └───┴───┴───┴───┴───┴───┴───┘ Finally, to avoid ⛔「3rd → 3|4」, we finish by ┌───┬───┬───┬───┬───┬───┬───┐ │6th│5th│ 4■│ 3■│2nd│1st│0th│▒ ╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒ │ 1 │ 6 │ │ │ 4 │ 2 │ 0 │▒ ├───┼───┼───┼───┼───┼───┼───┤▒ Step 6 │ 1 │ 6 │ 3 │ │ 4 │ 2 │ 0 │▒ ├───┼───┼───┼───┼───┼───┼───┤▒ Step 7 │ 1 │ 6 │ 3 │ 5 │ 4 │ 2 │ 0 │▒ └───┴───┴───┴───┴───┴───┴───┘▒ ▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒ --- Idle --- ┌───┬───┬───┬───┬───┬───┬───┐ │ │ │ │ │ │ │ │ └───┴───┴───┴───┴───┴───┴───┘ Q.E.D. #125034_v2.3
No comments:
Post a Comment