Rearrange the digits in ⟨1263045⟩ to meet the rules below.
⟨6th 5th 4th 3rd 2nd 1st 0th⟩
✅Match
⟦4,5⟧ ∋ 2,6
⟨⋯ Perm(0,3,4) ⋯⟩
2nd → a, 1st → b, a+b=7
4th → 3|4
⛔Avoid
⟨ ²ⁿᵈa ⁰ᵗʰb ⟩, (ab)₁₀ ≤ 60
5th → 0|1|3|5
#125034_v2.2
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ │ │ │ 6 │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ │ │ │ 6 │ 1 │ │▒
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Step 3 │ │ │ │ │ 6 │ 1 │ 5 │▒
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Step 4 │ │ │ │ 2 │ 6 │ 1 │ 5 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 5 │ │ 4 │ │ 2 │ 6 │ 1 │ 5 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 6 │ │ 4 │ 3 │ 2 │ 6 │ 1 │ 5 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 7 │ 0 │ 4 │ 3 │ 2 │ 6 │ 1 │ 5 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
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Proof of 2023-12-20 Q1(m=6)
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Notation: if nth -> a, then we write [nth] = a.
Plainly, to avoid ⛔「⟨ ²ⁿᵈa ⁰ᵗʰb ⟩, (ab)₁₀ ≤ 60」, we need 6 = [2nd]. Then, to match ✅「2nd → a, 1st → b, a+b=7」, we need 1 = [1st].
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│ 2■│ 1■│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ │ │ │ 6 │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ │ │ │ 6 │ 1 │ │▒
└───┴───┴───┴───┴───┴───┴───┘▒
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--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ 2 │ │ 3 │ 0 │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Next, we determine the value of [0th]. By ✅「⟨⋯ Perm(0,3,4) ⋯⟩」, 0,3,4 are adjacent, so they cannot be placed at 0th. On the other hand, by ✅「⟦4,5⟧ ∋ 2,6」, 2 is between 4,5. A fortiori 2 is not in corners. Therefore, [0th] = 5.
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│ 0■│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ │ │ │ 6 │ 1 │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 3 │ │ │ │ │ 6 │ 1 │ 5 │▒
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--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ 2 │ │ 3 │ 0 │ 4 │ │
└───┴───┴───┴───┴───┴───┴───┘
Since 0,3,4 are adjacent by ✅「⟨⋯ Perm(0,3,4) ⋯⟩」, we have 2 = [6th] or [3rd]:
┌───┬───┬───┬───┬───┬───┬───┐
│*6 │5th│4th│*3 │2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 2 │ ▬ │ ▬ │ ▬ │ 6 │ 1 │ 5 │
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│ ▬ │ ▬ │ ▬ │ 2 │ 6 │ 1 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
As 2 is not in corners by ✅「⟦4,5⟧ ∋ 2,6」, the latter case holds. It then follows from ⛔「5th → 0|1|3|5」 that [5th] = 4:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│ 5■│4th│ 3■│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ │ │ │ 6 │ 1 │ 5 │▒
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Step 4 │ │ │ │ 2 │ 6 │ 1 │ 5 │▒
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Step 5 │ │ 4 │ │ 2 │ 6 │ 1 │ 5 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
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--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ │ 3 │ 0 │ │ │
└───┴───┴───┴───┴───┴───┴───┘
Finally, to match ✅「4th → 3|4」, we finish by
┌───┬───┬───┬───┬───┬───┬───┐
│ 6■│5th│ 4■│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ 4 │ │ 2 │ 6 │ 1 │ 5 │▒
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Step 6 │ │ 4 │ 3 │ 2 │ 6 │ 1 │ 5 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 7 │ 0 │ 4 │ 3 │ 2 │ 6 │ 1 │ 5 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.2