Rearrange the digits in ⟨125034⟩ to meet the rules below.
⟨5th 4th 3rd 2nd 1st 0th⟩
✅Match
1st → 1|5
⟨ ⁴ᵗʰ↑ ³ʳᵈ↑ ⟩ after 2×⟨←⟩
5th → a, 2nd → b, a+b=0+4n
⟨⋯ 4 ⋯ a ⋯⟩, a = 1|2|5
{p5, p3, p0} = ? + {0,1,2}
⛔Avoid
⟨ ⁵ᵗʰa ²ⁿᵈb ¹ˢᵗc ⟩, (abc)₁₀ ≤ 231
#125034_v2.6
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ │ │ │ 5 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ 3 │ │ │ │ 5 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ 3 │ │ │ 1 │ 5 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ 3 │ 0 │ │ 1 │ 5 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ 3 │ 0 │ 4 │ 1 │ 5 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 3 │ 0 │ 4 │ 1 │ 5 │ 2 │▒
└───┴───┴───┴───┴───┴───┘▒
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Proof of 2024-07-26 WR
══════════════════════
Notation: if nth -> a, then we write [nth] = a.
(1) We first show that [1st] = 5.
------------------------------
If this is not the case, then by ✅「1st → 1|5」, we have [1st] = 1. It then follows from ✅「⟨ ⁴ᵗʰ↑ ³ʳᵈ↑ ⟩ after 2×⟨←⟩」 that [3rd] = 0:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│ 3▲│2nd│ 1▲│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ │ │ │ 1 │ │
├───┼───┼───┼───┼───┼───┤
│ │ │ 0 │ │ 1 │ │
└───┴───┴───┴───┴───┴───┘
Combining this with ✅「{p5, p3, p0} = ? + {0,1,2}」, we have { [5th], [3rd], [0th] } = {0,1,2}:
┌───┬───┬───┬───┬───┬───┐
│*5 │4th│*3 │2nd│1st│*0 │
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ ▬ │ │ 0 │ │ 1 │ ▬ │
└───┴───┴───┴───┴───┴───┘
A fortiori, 1 ∈ { [5th], [3rd], [0th] }. This is a contradiction, however, as we have already placed 1 at 1st.
------------------------------
We have verified (1). Accordingly, we get
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│ 1■│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ │ │ │ 5 │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ │ 0 │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘
To proceed, observe that ✅「5th → a, 2nd → b, a+b=0+4n」 implies [5th] + [2nd] = 4. We also need [5th] >= 2, in view of ⛔「⟨ ⁵ᵗʰa ²ⁿᵈb ¹ˢᵗc ⟩, (abc)₁₀ ≤ 231」. So, there are two possibilities:
(2) ([5th], [2nd]) = (4,0);
(3) ([5th], [2nd]) = (3,1).
Note that if case (2) holds:
┌───┬───┬───┬───┬───┬───┐
│ 5▲│4th│3rd│ 2▲│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 4 │ │ │ 0 │ 5 │ │
└───┴───┴───┴───┴───┴───┘
then we cannot match ✅「⟨ ⁴ᵗʰ↑ ³ʳᵈ↑ ⟩ after 2×⟨←⟩」. Therefore, case (3) holds, and we get
┌───┬───┬───┬───┬───┬───┐
│ 5■│4th│3rd│ 2■│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ │ │ │ 5 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ 3 │ │ │ │ 5 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ 3 │ │ │ 1 │ 5 │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ 2 │ │ 0 │ │ 4 │
└───┴───┴───┴───┴───┴───┘
Then, to match ✅「⟨ ⁴ᵗʰ↑ ³ʳᵈ↑ ⟩ after 2×⟨←⟩」, we need [4th] = 0:
┌───┬───┬───┬───┬───┬───┐
│5th│ 4■│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 3 │ │ │ 1 │ 5 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ 3 │ 0 │ │ 1 │ 5 │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ 2 │ │ │ │ 4 │
└───┴───┴───┴───┴───┴───┘
Finally, as ✅「⟨⋯ 4 ⋯ a ⋯⟩, a = 1|2|5」 implies 4 is not at the right corner, we finish by
┌───┬───┬───┬───┬───┬───┐
│5th│4th│ 3■│2nd│1st│ 0■│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 3 │ 0 │ │ 1 │ 5 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ 3 │ 0 │ 4 │ 1 │ 5 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 3 │ 0 │ 4 │ 1 │ 5 │ 2 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.6