Rearrange the digits in ⟨1263045⟩ to meet the rules below.
⟨6th 5th 4th 3rd 2nd 1st 0th⟩
✅Match
3rd → a, 0th → b, ab=4+6n
⟨⋯ Perm(0,3,5) ⋯⟩
⟨ ⁴ᵗʰc ³ʳᵈa ²ⁿᵈb ⟩, a > b > c
Jump(2,5) ≥ 3
#125034_v2.4
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ │ │ 4 │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ │ │ 4 │ │ │ 1 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 3 │ │ │ │ 4 │ │ 6 │ 1 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 4 │ │ │ │ 4 │ 2 │ 6 │ 1 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 5 │ 5 │ │ │ 4 │ 2 │ 6 │ 1 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 6 │ 5 │ │ 0 │ 4 │ 2 │ 6 │ 1 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 7 │ 5 │ 3 │ 0 │ 4 │ 2 │ 6 │ 1 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
Proof of 2024-04-16 Q1(m=6)
═══════════════════════════
Notation: if Nth -> a, then we write pN = a.
To begin with, observe that ✅「3rd → a, 0th → b, ab=4+6n」 implies (p3)(p0) = 4|10. Therefore, we have two cases:
i) {p3,p0} = {2,5};
ii) {p3,p0} = {1,4}.
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│ 3▲│2nd│1st│ 0▲│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ │ │ │ * │ │ │ * │
└───┴───┴───┴───┴───┴───┴───┘
Note that if the former case holds, then we cannot match ✅「Jump(2,5) ≥ 3」. Therefore, the latter case holds actually.
Also, in view of ✅「⟨ ⁴ᵗʰc ³ʳᵈa ²ⁿᵈb ⟩, a > b > c」, we cannot place 1 at 3rd. Consequently, we get (p3,p0) = (4,1):
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│ 3■│2nd│1st│ 0■│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ │ │ 4 │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ │ │ 4 │ │ │ 1 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ 2 │ 6 │ 3 │ 0 │ │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Next, note that we need three consecutive boxes to match ✅「⟨⋯ Perm(0,3,5) ⋯⟩」. Accordingly
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ ▬ │ ▬ │ ▬ │ 4 │ ● │ ● │ 1 │
└───┴───┴───┴───┴───┴───┴───┘
where "▬" ∈ {0,3,5} and "●" ∈ {2,6}. As 6 is the maximum digit, by ✅「⟨ ⁴ᵗʰc ³ʳᵈa ²ⁿᵈb ⟩, a > b > c」 we cannot place it at 2nd. Therefore, 6 = p1, and so 2 = p2:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│ 2■│ 1■│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ │ │ 4 │ │ │ 1 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 3 │ │ │ │ 4 │ │ 6 │ 1 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 4 │ │ │ │ 4 │ 2 │ 6 │ 1 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ │ 3 │ 0 │ │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Now there is only one way to match ✅「Jump(2,5) ≥ 3」:
┌───┬───┬───┬───┬───┬───┬───┐
│ 6■│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ │ │ 4 │ 2 │ 6 │ 1 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 5 │ 5 │ │ │ 4 │ 2 │ 6 │ 1 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ │ 3 │ 0 │ │ │
└───┴───┴───┴───┴───┴───┴───┘
Finally, to match ✅「⟨ ⁴ᵗʰc ³ʳᵈa ²ⁿᵈb ⟩, a > b > c」, we finish by
┌───┬───┬───┬───┬───┬───┬───┐
│6th│ 5■│ 4■│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 5 │ │ │ 4 │ 2 │ 6 │ 1 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 6 │ 5 │ │ 0 │ 4 │ 2 │ 6 │ 1 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 7 │ 5 │ 3 │ 0 │ 4 │ 2 │ 6 │ 1 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.4