Rearrange the digits in ⟨1263045⟩ to meet the rules below.
⟨6th 5th 4th 3rd 2nd 1st 0th⟩
✅Match
⟨ ⁶ᵗʰa ²ⁿᵈb ⟩, min⟦a,b⟧ = 1
⟨? 6 ⋯ (?+4) ⋯ 5 ⋯⟩ (?≠2)
Jump(2,4) = 0
⟦1,2⟧ ∋ 3,4,6
⛔Avoid
3rd → 3|4
#125034_v2.3
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ 6 │ │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 2 │ 1 │ 6 │ │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 3 │ 1 │ 6 │ │ │ 4 │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 4 │ 1 │ 6 │ │ │ 4 │ 2 │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 5 │ 1 │ 6 │ │ │ 4 │ 2 │ 0 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 6 │ 1 │ 6 │ 3 │ │ 4 │ 2 │ 0 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 7 │ 1 │ 6 │ 3 │ 5 │ 4 │ 2 │ 0 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
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Proof of 2024-02-06 Q1(m=6)
═══════════════════════════
Notation: if nth -> a, then we write [nth] = a.
Plainly, our first step follows from ✅「⟨? 6 ⋯ (?+4) ⋯ 5 ⋯⟩ (?≠2)」:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│ 5■│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ 6 │ │ │ │ │ │▒
└───┴───┴───┴───┴───┴───┴───┘▒
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--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ │ 3 │ 0 │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Next, in view of ✅「⟦1,2⟧ ∋ 3,4,6」, we need to place 1 or 2 at 6th, otherwise 6 would not be contained in ⟦1,2⟧. Since [6th] != 2 by ✅「⟨? 6 ⋯ (?+4) ⋯ 5 ⋯⟩ (?≠2)」, we have [6th] = 1.
┌───┬───┬───┬───┬───┬───┬───┐
│ 6■│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ 6 │ │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 2 │ 1 │ 6 │ │ │ │ │ │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ 2 │ │ 3 │ 0 │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Then, it follows from ✅「⟦1,2⟧ ∋ 3,4,6」 and ✅「Jump(2,4) = 0」 that we have to match the following pattern:
(1) ⟨16 ⋯ 3 ⋯ 42 ⋯⟩.
Accordingly, there are only four ways to place 2,4:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
(2) │ 1 │ 6 │ 4 │ 2 │ │ │ │
├───┼───┼───┼───┼───┼───┼───┤
(3) │ 1 │ 6 │ │ 4 │ 2 │ │ │
├───┼───┼───┼───┼───┼───┼───┤
(4) │ 1 │ 6 │ │ │ 4 │ 2 │ │
├───┼───┼───┼───┼───┼───┼───┤
(5) │ 1 │ 6 │ │ │ │ 4 │ 2 │
└───┴───┴───┴───┴───┴───┴───┘
Case (4) holds actually, for:
● If (2) holds, then we could not match ✅「⟦1,2⟧ ∋ 3,4,6」 wherever 3 is placed;
● If (3) holds, then we match ⛔「3rd → 3|4」;
● If (5) holds, then we could not match ✅「⟨ ⁶ᵗʰa ²ⁿᵈb ⟩, min⟦a,b⟧ = 1」 wherever 0 is placed.
Hence, we get
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│ 2■│ 1■│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 1 │ 6 │ │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 3 │ 1 │ 6 │ │ │ 4 │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 4 │ 1 │ 6 │ │ │ 4 │ 2 │ │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ │ 3 │ 0 │ │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Next, to match ✅「⟨ ⁶ᵗʰa ²ⁿᵈb ⟩, min⟦a,b⟧ = 1」, we have 0 != [4th] and 0 != [3rd]. Therefore, 0 = [0th]:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│ 0■│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 1 │ 6 │ │ │ 4 │ 2 │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 5 │ 1 │ 6 │ │ │ 4 │ 2 │ 0 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ │ 3 │ │ │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Finally, to avoid ⛔「3rd → 3|4」, we finish by
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│ 4■│ 3■│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 1 │ 6 │ │ │ 4 │ 2 │ 0 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 6 │ 1 │ 6 │ 3 │ │ 4 │ 2 │ 0 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 7 │ 1 │ 6 │ 3 │ 5 │ 4 │ 2 │ 0 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.3