Rearrange the digits in ⟨1263045⟩ to meet the rules below.
⟨6th 5th 4th 3rd 2nd 1st 0th⟩
✅Match
⟨⋯ 2 ⋯ 4 ⋯ 3 ⋯⟩
Jump(1,4) = 1
5th → a, 3rd → b, |a-b|=4
⛔Avoid
2nd → 2|3|4|5|6
5th|4th|3rd → 2
#125034_v2.5
┌───┬───┬───┬───┬───┬───┬───┐ │6th│5th│4th│3rd│2nd│1st│0th│▒ ╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒ Step 1 │ 2 │ │ │ │ │ │ │▒ ├───┼───┼───┼───┼───┼───┼───┤▒ Step 2 │ 2 │ 5 │ │ │ │ │ │▒ ├───┼───┼───┼───┼───┼───┼───┤▒ Step 3 │ 2 │ 5 │ │ 1 │ │ │ │▒ ├───┼───┼───┼───┼───┼───┼───┤▒ Step 4 │ 2 │ 5 │ │ 1 │ 0 │ │ │▒ ├───┼───┼───┼───┼───┼───┼───┤▒ Step 5 │ 2 │ 5 │ │ 1 │ 0 │ 4 │ │▒ ├───┼───┼───┼───┼───┼───┼───┤▒ Step 6 │ 2 │ 5 │ │ 1 │ 0 │ 4 │ 3 │▒ ├───┼───┼───┼───┼───┼───┼───┤▒ Step 7 │ 2 │ 5 │ 6 │ 1 │ 0 │ 4 │ 3 │▒ └───┴───┴───┴───┴───┴───┴───┘▒ ▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒ Proof of 2024-05-28 Q1(m=6) ═══════════════════════════ Notation: if nth -> a, then we write [nth] = a. Plainly, to avoid ⛔「5th|4th|3rd → 2」, we need 2 = [6th] | [2nd] | [1st] | [0th]. ┌───┬───┬───┬───┬───┬───┬───┐ │6th│5th│4th│3rd│2nd│1st│0th│ ╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡ │ │ / │ / │ / │ │ │ │ └───┴───┴───┴───┴───┴───┴───┘ Actually, it is 2 = [6th] | [2nd], since ✅「⟨⋯ 2 ⋯ 4 ⋯ 3 ⋯⟩」 implies that there are at least two digits at the right of 2. Therefore, in view of ⛔「2nd → 2|3|4|5|6」, we get 2 = [6th]. ┌───┬───┬───┬───┬───┬───┬───┐ │ 6■│5th│4th│3rd│2nd│1st│0th│▒ ╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒ Step 1 │ 2 │ │ │ │ │ │ │▒ └───┴───┴───┴───┴───┴───┴───┘▒ ▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒ --- Idle --- ┌───┬───┬───┬───┬───┬───┬───┐ │ 1 │ │ 6 │ 3 │ 0 │ 4 │ 5 │ └───┴───┴───┴───┴───┴───┴───┘ Next, we consider ✅「5th → a, 3rd → b, |a-b|=4」. There are exactly four ways to match it: (1.1) ([5th], [3rd]) = (5,1); (1.2) ([5th], [3rd]) = (1,5); (1.3) ([5th], [3rd]) = (4,0); (1.4) ([5th], [3rd]) = (0,4). ┌───┬───┬───┬───┬───┬───┬───┐ │6th│ 5▲│4th│ 3▲│2nd│1st│0th│ ╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡ │ 2 │ ▬ │ │ ▬ │ │ │ │ └───┴───┴───┴───┴───┴───┴───┘ We show that (1.1) holds actually. ------------------------------ Observe that in order to match ✅「Jump(1,4) = 1」, neither (1.2) nor (1.3) holds. On the other hand, if (1.4) holds: ┌───┬───┬───┬───┬───┬───┬───┐ │6th│ 5▲│4th│ 3▲│2nd│1st│0th│ ╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡ │ 2 │ 0 │ │ 4 │ │ │ │ └───┴───┴───┴───┴───┴───┴───┘ then we cannot match ✅「Jump(1,4) = 1」 and avoid ⛔「2nd → 2|3|4|5|6」 at the same time. So, (1.4) does not hold as well. ------------------------------ We have verified that (1.1) holds. Accordingly, we get ┌───┬───┬───┬───┬───┬───┬───┐ │6th│ 5■│4th│ 3■│2nd│1st│0th│▒ ╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒ │ 2 │ │ │ │ │ │ │▒ ├───┼───┼───┼───┼───┼───┼───┤▒ Step 2 │ 2 │ 5 │ │ │ │ │ │▒ ├───┼───┼───┼───┼───┼───┼───┤▒ Step 3 │ 2 │ 5 │ │ 1 │ │ │ │▒ └───┴───┴───┴───┴───┴───┴───┘▒ ▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒ --- Idle --- ┌───┬───┬───┬───┬───┬───┬───┐ │ │ │ 6 │ 3 │ 0 │ 4 │ │ └───┴───┴───┴───┴───┴───┴───┘ Then, using ⛔「2nd → 2|3|4|5|6」 and ✅「Jump(1,4) = 1」, we get ┌───┬───┬───┬───┬───┬───┬───┐ │6th│5th│4th│3rd│ 2■│ 1■│0th│▒ ╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒ │ 2 │ 5 │ │ 1 │ │ │ │▒ ├───┼───┼───┼───┼───┼───┼───┤▒ Step 4 │ 2 │ 5 │ │ 1 │ 0 │ │ │▒ ├───┼───┼───┼───┼───┼───┼───┤▒ Step 5 │ 2 │ 5 │ │ 1 │ 0 │ 4 │ │▒ └───┴───┴───┴───┴───┴───┴───┘▒ ▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒ --- Idle --- ┌───┬───┬───┬───┬───┬───┬───┐ │ │ │ 6 │ 3 │ │ │ │ └───┴───┴───┴───┴───┴───┴───┘ Finally, in view of ✅「⟨⋯ 2 ⋯ 4 ⋯ 3 ⋯⟩」, we finish by ┌───┬───┬───┬───┬───┬───┬───┐ │6th│5th│ 4■│3rd│2nd│1st│ 0■│▒ ╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒ │ 2 │ 5 │ │ 1 │ 0 │ 4 │ │▒ ├───┼───┼───┼───┼───┼───┼───┤▒ Step 6 │ 2 │ 5 │ │ 1 │ 0 │ 4 │ 3 │▒ ├───┼───┼───┼───┼───┼───┼───┤▒ Step 7 │ 2 │ 5 │ 6 │ 1 │ 0 │ 4 │ 3 │▒ └───┴───┴───┴───┴───┴───┴───┘▒ ▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒ --- Idle --- ┌───┬───┬───┬───┬───┬───┬───┐ │ │ │ │ │ │ │ │ └───┴───┴───┴───┴───┴───┴───┘ Q.E.D. #125034_v2.5
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