Rearrange the digits in ⟨1263045⟩ to meet the rules below.
⟨6th 5th 4th 3rd 2nd 1st 0th⟩
✅Match
2nd → a, 0th → b, a+b=3
⟨ ⁶ᵗʰa ⁵ᵗʰd ⁴ᵗʰb ³ʳᵈc ⟩, a > b > c > d
⟨ ⁴ᵗʰa ²ⁿᵈb ¹ˢᵗc ⟩, (abc)₁₀ ≥ 531
⟨⋯ Perm(4,5) ⋯⟩
⛔Avoid
1 ∾ 2 ∾ 5
#125034_v2.2
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ │ 5 │ │ │ │ │▒
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Step 2 │ │ │ 5 │ 4 │ │ │ │▒
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Step 3 │ 6 │ │ 5 │ 4 │ │ │ │▒
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Step 4 │ 6 │ │ 5 │ 4 │ 3 │ │ │▒
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Step 5 │ 6 │ │ 5 │ 4 │ 3 │ │ 0 │▒
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Step 6 │ 6 │ │ 5 │ 4 │ 3 │ 1 │ 0 │▒
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Step 7 │ 6 │ 2 │ 5 │ 4 │ 3 │ 1 │ 0 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
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Proof of 2023-12-08 Q1(m=6)
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Notation: if nth -> a, then we write [nth] = a.
To begin with, note that to match ✅「⟨ ⁴ᵗʰa ²ⁿᵈb ¹ˢᵗc ⟩, (abc)₁₀ ≥ 531」, we need [4th] = 5 or 6. It cannot be 6, because by ✅「⟨ ⁶ᵗʰa ⁵ᵗʰd ⁴ᵗʰb ³ʳᵈc ⟩, a > b > c > d」, [4th] cannot be the maximum number. Therefore, [4th] = 5.
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│ 4■│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ │ 5 │ │ │ │ │▒
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--- Idle ---
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│ 1 │ 2 │ 6 │ 3 │ 0 │ 4 │ │
└───┴───┴───┴───┴───┴───┴───┘
By ✅「⟨⋯ Perm(4,5) ⋯⟩」, 4 is adjacent to 5, so 4 = [5th] or [3rd]. The former is not possible, for otherwise by ✅「⟨ ⁶ᵗʰa ⁵ᵗʰd ⁴ᵗʰb ³ʳᵈc ⟩, a > b > c > d」, we have 5 > [3rd] > 4, which is a contradiction. It follows that 4 = [3rd]. The preceding pattern also implies [6th] > 5, so [6th] = 6.
┌───┬───┬───┬───┬───┬───┬───┐
│ 6■│5th│4th│ 3■│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ │ 5 │ │ │ │ │▒
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Step 2 │ │ │ 5 │ 4 │ │ │ │▒
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Step 3 │ 6 │ │ 5 │ 4 │ │ │ │▒
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--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ │ 3 │ 0 │ │ │
└───┴───┴───┴───┴───┴───┴───┘
Next, we determine the value of [2nd]. To match ✅「⟨ ⁴ᵗʰa ²ⁿᵈb ¹ˢᵗc ⟩, (abc)₁₀ ≥ 531」, we can only take [2nd] = 3. Then, to match ✅「2nd → a, 0th → b, a+b=3」, we need [0th] = 0.
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│ 2■│1st│ 0■│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 6 │ │ 5 │ 4 │ │ │ │▒
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Step 4 │ 6 │ │ 5 │ 4 │ 3 │ │ │▒
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Step 5 │ 6 │ │ 5 │ 4 │ 3 │ │ 0 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
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--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┴───┘
Finally, to avoid ⛔「1 ∾ 2 ∾ 5」, we simply place 1 at the 1st position so that 1 is itself a permutation circle. We finish by
┌───┬───┬───┬───┬───┬───┬───┐
│6th│ 5■│4th│3rd│2nd│ 1■│0th│▒
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│ 6 │ │ 5 │ 4 │ 3 │ │ 0 │▒
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Step 6 │ 6 │ │ 5 │ 4 │ 3 │ 1 │ 0 │▒
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Step 7 │ 6 │ 2 │ 5 │ 4 │ 3 │ 1 │ 0 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.2