Rearrange the digits in ⟨125034⟩ to meet the rules below.
⟨5th 4th 3rd 2nd 1st 0th⟩
✅Match
⟦3,5⟧ ∋ 0,2
4th → a, 1st → b, a+b=5
⟨⋯ Perm(2,5) ⋯⟩
⛔Avoid
⟨ ⁴ᵗʰa ²ⁿᵈb ¹ˢᵗc ⟩, (abc)₁₀ ≥ 152
⟦0,4⟧ ∋ 1,2
----- Information -----
🔲 「⟦3,5⟧ ∋ 0,2」
The closed interval given by 3 and 5 contains 0, 2.
120 permutations match this pattern.
Examples: ⟨340215⟩, ⟨302514⟩, ⟨132405⟩.
🔲 「4th → a, 1st → b, a+b=5」
144 permutations match this pattern.
Examples: ⟨420135⟩, ⟨251403⟩, ⟨134520⟩.
🔲 「⟨⋯ Perm(2,5) ⋯⟩」
2,5 are adjacent.
240 permutations match this pattern.
Examples: ⟨014325⟩, ⟨025143⟩, ⟨345210⟩.
🔳 「⟨ ⁴ᵗʰa ²ⁿᵈb ¹ˢᵗc ⟩, (abc)₁₀ ≥ 152」
(abc)₁₀ ∶= 100a + 10b + c.
498 permutations match this pattern.
Examples: ⟨345012⟩, ⟨531024⟩, ⟨520314⟩.
🔳 「⟦0,4⟧ ∋ 1,2」
The closed interval given by 0 and 4 contains 1, 2.
120 permutations match this pattern.
Examples: ⟨052134⟩, ⟨425103⟩, ⟨453120⟩.
#125034_v2.2
┌───┬───┬───┬───┬───┬───┐ │5th│4th│3rd│2nd│1st│0th│▒ ╞═══╪═══╪═══╪═══╪═══╪═══╡▒ Step 1 │ │ 0 │ │ │ │ │▒ ├───┼───┼───┼───┼───┼───┤▒ Step 2 │ │ 0 │ │ │ 5 │ │▒ ├───┼───┼───┼───┼───┼───┤▒ Step 3 │ 3 │ 0 │ │ │ 5 │ │▒ ├───┼───┼───┼───┼───┼───┤▒ Step 4 │ 3 │ 0 │ │ 2 │ 5 │ │▒ ├───┼───┼───┼───┼───┼───┤▒ Step 5 │ 3 │ 0 │ 4 │ 2 │ 5 │ │▒ ├───┼───┼───┼───┼───┼───┤▒ Step 6 │ 3 │ 0 │ 4 │ 2 │ 5 │ 1 │▒ └───┴───┴───┴───┴───┴───┘▒ ▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒ Proof of 2023-11-28 WR ══════════════════════ Notation: if nth -> a, then we write [nth] = a. By ⛔「⟨ ⁴ᵗʰa ²ⁿᵈb ¹ˢᵗc ⟩, (abc)₁₀ ≥ 152」, we have [4th] = 1 or 0. Together with ✅「4th → a, 1st → b, a+b=5」, there are only two possibilities: (i) [4th] = 1, [1st] = 4; (ii) [4th] = 0, [1st] = 5. (1) We show that (i) is not possible. Suppose on the contrary (i) holds: ┌───┬───┬───┬───┬───┬───┐ │5th│ 4▲│3rd│2nd│ 1▲│0th│ ╞═══╪═══╪═══╪═══╪═══╪═══╡ │ │ 1 │ │ │ 4 │ │ └───┴───┴───┴───┴───┴───┘ By ✅「⟨⋯ Perm(2,5) ⋯⟩」, we see that ┌───┬───┬───┬───┬───┬───┐ │5th│4th│*3 │*2 │1st│0th│ ╞═══╪═══╪═══╪═══╪═══╪═══╡ │ │ 1 │ x │ y │ 4 │ │ └───┴───┴───┴───┴───┴───┘ where {x,y} = {2,5}. To avoid ⛔「⟨ ⁴ᵗʰa ²ⁿᵈb ¹ˢᵗc ⟩, (abc)₁₀ ≥ 152」, we need (x,y) = (5,2): ┌───┬───┬───┬───┬───┬───┐ │5th│4th│ 3▲│ 2▲│1st│0th│ ╞═══╪═══╪═══╪═══╪═══╪═══╡ │ │ 1 │ 5 │ 2 │ 4 │ │ └───┴───┴───┴───┴───┴───┘ But then ✅「⟦3,5⟧ ∋ 0,2」 could not be matched, which is a contradiction. ------------------------------ We have verified our claim in (1). Accordingly, we obtain: ┌───┬───┬───┬───┬───┬───┐ │5th│ 4■│3rd│2nd│ 1■│0th│▒ ╞═══╪═══╪═══╪═══╪═══╪═══╡▒ Step 1 │ │ 0 │ │ │ │ │▒ ├───┼───┼───┼───┼───┼───┤▒ Step 2 │ │ 0 │ │ │ 5 │ │▒ └───┴───┴───┴───┴───┴───┘▒ ▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒ --- Idle --- ┌───┬───┬───┬───┬───┬───┐ │ 1 │ 2 │ │ │ 3 │ 4 │ └───┴───┴───┴───┴───┴───┘ Now, to match ✅「⟦3,5⟧ ∋ 0,2」, we need 0 to be between 5 and 3. Thus ┌───┬───┬───┬───┬───┬───┐ │ 5■│4th│3rd│2nd│1st│0th│▒ ╞═══╪═══╪═══╪═══╪═══╪═══╡▒ │ │ 0 │ │ │ 5 │ │▒ ├───┼───┼───┼───┼───┼───┤▒ Step 3 │ 3 │ 0 │ │ │ 5 │ │▒ └───┴───┴───┴───┴───┴───┘▒ ▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒ --- Idle --- ┌───┬───┬───┬───┬───┬───┐ │ 1 │ 2 │ │ │ │ 4 │ └───┴───┴───┴───┴───┴───┘ Next, we consider where to place 2. By ✅「⟨⋯ Perm(2,5) ⋯⟩」, 2,5 are adjacent, and by ✅「⟦3,5⟧ ∋ 0,2」, 2 is between 5 and 3. Therefore, we have ┌───┬───┬───┬───┬───┬───┐ │5th│4th│3rd│ 2■│1st│0th│▒ ╞═══╪═══╪═══╪═══╪═══╪═══╡▒ │ 3 │ 0 │ │ │ 5 │ │▒ ├───┼───┼───┼───┼───┼───┤▒ Step 4 │ 3 │ 0 │ │ 2 │ 5 │ │▒ └───┴───┴───┴───┴───┴───┘▒ ▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒ --- Idle --- ┌───┬───┬───┬───┬───┬───┐ │ 1 │ │ │ │ │ 4 │ └───┴───┴───┴───┴───┴───┘ Finally, note that there is only one way to avoid ⛔「⟦0,4⟧ ∋ 1,2」. We reach ┌───┬───┬───┬───┬───┬───┐ │5th│4th│ 3■│2nd│1st│ 0■│▒ ╞═══╪═══╪═══╪═══╪═══╪═══╡▒ │ 3 │ 0 │ │ 2 │ 5 │ │▒ ├───┼───┼───┼───┼───┼───┤▒ Step 5 │ 3 │ 0 │ 4 │ 2 │ 5 │ │▒ ├───┼───┼───┼───┼───┼───┤▒ Step 6 │ 3 │ 0 │ 4 │ 2 │ 5 │ 1 │▒ └───┴───┴───┴───┴───┴───┘▒ ▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒ --- Idle --- ┌───┬───┬───┬───┬───┬───┐ │ │ │ │ │ │ │ └───┴───┴───┴───┴───┴───┘ Q.E.D. #125034_v2.2
No comments:
Post a Comment