Rearrange the digits in ⟨1263045⟩ to meet the rules below.
⟨6th 5th 4th 3rd 2nd 1st 0th⟩
✅Match
min ⊢3⊣ ≥ 2
⟦2,5⟧ ∋ 3,6
⟨⋯ 0 ⋯ 5 ⋯ 4 ⋯ 2 ⋯⟩
⟦0,3⟧ ∋ 1,6
{p5, p4, p1, p0} = ? + {0,1,2,3}
#125034_v2.7
┌───┬───┬───┬───┬───┬───┬───┐ │6th│5th│4th│3rd│2nd│1st│0th│▒ ╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒ Step 1 │ 0 │ │ │ │ │ │ │▒ ├───┼───┼───┼───┼───┼───┼───┤▒ Step 2 │ 0 │ │ │ │ │ │ 2 │▒ ├───┼───┼───┼───┼───┼───┼───┤▒ Step 3 │ 0 │ │ │ │ │ 3 │ 2 │▒ ├───┼───┼───┼───┼───┼───┼───┤▒ Step 4 │ 0 │ 5 │ │ │ │ 3 │ 2 │▒ ├───┼───┼───┼───┼───┼───┼───┤▒ Step 5 │ 0 │ 5 │ 4 │ │ │ 3 │ 2 │▒ ├───┼───┼───┼───┼───┼───┼───┤▒ Step 6 │ 0 │ 5 │ 4 │ │ 6 │ 3 │ 2 │▒ ├───┼───┼───┼───┼───┼───┼───┤▒ Step 7 │ 0 │ 5 │ 4 │ 1 │ 6 │ 3 │ 2 │▒ └───┴───┴───┴───┴───┴───┴───┘▒ ▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒ Proof of 2024-11-05 Q1(m=6) ═══════════════════════════ Notation: if nth -> a, then we write [nth] = a. Combining ✅「⟦2,5⟧ ∋ 3,6」 with ✅「⟨⋯ 0 ⋯ 5 ⋯ 4 ⋯ 2 ⋯⟩」, we see that one of the following holds: ⟨⋯ 0 ⋯ 5 ⋯ 3 ⋯ 6 ⋯ 2 ⋯⟩; or ⟨⋯ 0 ⋯ 5 ⋯ 6 ⋯ 3 ⋯ 2 ⋯⟩. The latter holds actually, for otherwise we cannot match ✅「⟦0,3⟧ ∋ 1,6」. This pattern then implies that one of the following holds: (i) ⟨⋯ 0 ⋯ 6 ⋯ 1 ⋯ 3 ⋯ 2 ⋯⟩; or (ii) ⟨⋯ 0 ⋯ 1 ⋯ 6 ⋯ 3 ⋯ 2 ⋯⟩. Together with ✅「⟨⋯ 0 ⋯ 5 ⋯ 4 ⋯ 2 ⋯⟩」, we see that no matter which case happens, we have (1) 0 is to the left of 1,2,3,4,5,6; and (2) 2 is to the right of 0,1,3,4,5,6. Therefore, we get ┌───┬───┬───┬───┬───┬───┬───┐ │ 6■│5th│4th│3rd│2nd│1st│ 0■│▒ ╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒ Step 1 │ 0 │ │ │ │ │ │ │▒ ├───┼───┼───┼───┼───┼───┼───┤▒ Step 2 │ 0 │ │ │ │ │ │ 2 │▒ └───┴───┴───┴───┴───┴───┴───┘▒ ▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒ --- Idle --- ┌───┬───┬───┬───┬───┬───┬───┐ │ 1 │ │ 6 │ 3 │ │ 4 │ 5 │ └───┴───┴───┴───┴───┴───┴───┘ Next, we consider ✅「{p5, p4, p1, p0} = ? + {0,1,2,3}」. ┌───┬───┬───┬───┬───┬───┬───┐ │6th│*5 │*4 │3rd│2nd│*1 │*0 │ ╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡ │ 0 │ │ │ │ │ │ 2 │ └───┴───┴───┴───┴───┴───┴───┘ We need p5,p4,p1,p0 to be consecutive integers. It follows that S := {p5, p4, p1, p0} is either {1,2,3,4} or {2,3,4,5}. As a consequence, (3) 3,4 ∈ S; and (4) 6 ∉ S; equivalently 6 = [3rd] | [2nd]. ┌───┬───┬───┬───┬───┬───┬───┐ │6th│*5 │*4 │3rd│2nd│*1 │*0 │ ╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡ │ 0 │ │ │ ▬ │ ▬ │ │ 2 │ └───┴───┴───┴───┴───┴───┴───┘ On the other hand, recall that case (i) or (ii) holds. A fortiori, we need to match (5) ⟨ ⋯ 6 ⋯ 3 ⋯ ⟩. Combining (3), (4), and (5), we get [1st] = 3: ┌───┬───┬───┬───┬───┬───┬───┐ │6th│5th│4th│3rd│2nd│ 1■│0th│▒ ╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒ │ 0 │ │ │ │ │ │ 2 │▒ ├───┼───┼───┼───┼───┼───┼───┤▒ Step 3 │ 0 │ │ │ │ │ 3 │ 2 │▒ └───┴───┴───┴───┴───┴───┴───┘▒ ▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒ --- Idle --- ┌───┬───┬───┬───┬───┬───┬───┐ │ 1 │ │ 6 │ │ │ 4 │ 5 │ └───┴───┴───┴───┴───┴───┴───┘ Next, we consider the position of 4. By (3), we have 4 ∈ S. Combining this with ✅「⟨⋯ 0 ⋯ 5 ⋯ 4 ⋯ 2 ⋯⟩」, we get ┌───┬───┬───┬───┬───┬───┬───┐ │6th│ 5■│ 4■│3rd│2nd│1st│0th│▒ ╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒ │ 0 │ │ │ │ │ 3 │ 2 │▒ ├───┼───┼───┼───┼───┼───┼───┤▒ Step 4 │ 0 │ 5 │ │ │ │ 3 │ 2 │▒ ├───┼───┼───┼───┼───┼───┼───┤▒ Step 5 │ 0 │ 5 │ 4 │ │ │ 3 │ 2 │▒ └───┴───┴───┴───┴───┴───┴───┘▒ ▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒ --- Idle --- ┌───┬───┬───┬───┬───┬───┬───┐ │ 1 │ │ 6 │ │ │ │ │ └───┴───┴───┴───┴───┴───┴───┘ Finally, in view of ✅「min ⊢3⊣ ≥ 2」, we finish by ┌───┬───┬───┬───┬───┬───┬───┐ │6th│5th│4th│ 3■│ 2■│1st│0th│▒ ╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒ │ 0 │ 5 │ 4 │ │ │ 3 │ 2 │▒ ├───┼───┼───┼───┼───┼───┼───┤▒ Step 6 │ 0 │ 5 │ 4 │ │ 6 │ 3 │ 2 │▒ ├───┼───┼───┼───┼───┼───┼───┤▒ Step 7 │ 0 │ 5 │ 4 │ 1 │ 6 │ 3 │ 2 │▒ └───┴───┴───┴───┴───┴───┴───┘▒ ▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒ --- Idle --- ┌───┬───┬───┬───┬───┬───┬───┐ │ │ │ │ │ │ │ │ └───┴───┴───┴───┴───┴───┴───┘ Q.E.D. #125034_v2.7
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