Rearrange the digits in ⟨1263045⟩ to meet the rules below.
⟨6th 5th 4th 3rd 2nd 1st 0th⟩
✅Match
⟨ ⁵ᵗʰd ⁴ᵗʰb ³ʳᵈa ²ⁿᵈc ⟩, a > b > c > d
⟨? ⋯ 1 ⋯ (?+2) ⋯⟩ (?≠1)
⛔Avoid
⟦0,2⟧ ∋ 5
⟨ ³ʳᵈ↓ ²ⁿᵈ↓ ⟩ after 3×⟨←⟩
⟨⋯ 6 ⋯ a ⋯⟩, a = 3|5
⟨⋯ 2 ⋯ a ⋯⟩, a = 1|3|6
⟨ ²ⁿᵈa ¹ˢᵗb ⟩, min⟦a,b⟧ = 0
#125034_v2.5
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ │ │ 6 │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ │ 5 │ 6 │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 3 │ 3 │ │ 5 │ 6 │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 4 │ 3 │ 1 │ 5 │ 6 │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 5 │ 3 │ 1 │ 5 │ 6 │ │ │ 0 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 6 │ 3 │ 1 │ 5 │ 6 │ 2 │ │ 0 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 7 │ 3 │ 1 │ 5 │ 6 │ 2 │ 4 │ 0 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
Proof of 2024-06-25 Q1(m=6)
═══════════════════════════
Notation: if nth -> a, then we write [nth] = a.
Firstly, we consider where to place 6. It is the maximum digit, so ✅「⟨ ⁵ᵗʰd ⁴ᵗʰb ³ʳᵈa ²ⁿᵈc ⟩, a > b > c > d」 implies that
(1) 6 = [6th] | [3rd] | [1st] | [0th].
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ ▬ │ │ │ ▬ │ │ ▬ │ ▬ │
└───┴───┴───┴───┴───┴───┴───┘
Note that by ✅「⟨? ⋯ 1 ⋯ (?+2) ⋯⟩ (?≠1)」, it is not at 6th. It is not at 0th as well, in view of ⛔「⟨⋯ 2 ⋯ a ⋯⟩, a = 1|3|6」. Therefore,
(2) 6 = [3rd] | [1st].
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ │ │ │ ▬ │ │ ▬ │ │
└───┴───┴───┴───┴───┴───┴───┘
(2.1) We show that 6 = [3rd] actually.
------------------------------
For, suppose on the contrary 6 = [1st]:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│ 1▲│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ │ │ │ │ │ 6 │ │
└───┴───┴───┴───┴───┴───┴───┘
Then, to avoid ⛔「⟨⋯ 2 ⋯ a ⋯⟩, a = 1|3|6」, we need
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│ 0▲│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ │ │ │ │ │ 6 │ 2 │
└───┴───┴───┴───┴───┴───┴───┘
Now we determine the position of 0. By ✅「⟨ ⁵ᵗʰd ⁴ᵗʰb ³ʳᵈa ²ⁿᵈc ⟩, a > b > c > d」, it is at 6th or 5th:
┌───┬───┬───┬───┬───┬───┬───┐
│*6 │*5 │4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ - │ - │ │ │ │ 6 │ 2 │
└───┴───┴───┴───┴───┴───┴───┘
As we need to avoid ⛔「⟦0,2⟧ ∋ 5」, we get 0 = [5th]. The preceding pattern then implies [6th] = 5:
┌───┬───┬───┬───┬───┬───┬───┐
│ 6▲│ 5▲│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ │ 0 │ │ │ │ 6 │ 2 │
├───┼───┼───┼───┼───┼───┼───┤
│ 5 │ 0 │ │ │ │ 6 │ 2 │
└───┴───┴───┴───┴───┴───┴───┘
This is a contradiction, however, as we cannot match ✅「⟨? ⋯ 1 ⋯ (?+2) ⋯⟩ (?≠1)」 now.
------------------------------
We have verified our claim in (2.1). Accordingly, we get our first step:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│ 3■│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ │ │ 6 │ │ │ │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ │ 3 │ 0 │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Next, note that by ⛔「⟨⋯ 6 ⋯ a ⋯⟩, a = 3|5」, 3 and 5 are at the left of 6. We proceed to consider their positions.
We start with 5. Using ✅「⟨ ⁵ᵗʰd ⁴ᵗʰb ³ʳᵈa ²ⁿᵈc ⟩, a > b > c > d」 and ✅「⟨? ⋯ 1 ⋯ (?+2) ⋯⟩ (?≠1)」, we get
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│ 4■│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ │ │ 6 │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ │ 5 │ 6 │ │ │ │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ │ 3 │ 0 │ 4 │ │
└───┴───┴───┴───┴───┴───┴───┘
For the position of 3, the two possibilities are:
(3) 3 = [6th] | [5th].
(3.1) We claim that indeed 3 = [6th].
For, if on the contrary 3 = [5th], then in view of ✅「⟨ ⁵ᵗʰd ⁴ᵗʰb ³ʳᵈa ²ⁿᵈc ⟩, a > b > c > d」, we get the position of 4 as well:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│ 5▲│4th│3rd│ 2▲│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ │ 3 │ 5 │ 6 │ │ │ │
├───┼───┼───┼───┼───┼───┼───┤
│ │ 3 │ 5 │ 6 │ 4 │ │ │
└───┴───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ │ │ 0 │ │ │
└───┴───┴───┴───┴───┴───┴───┘
This is a contradiction, however, as we cannot avoid ⛔「⟨ ³ʳᵈ↓ ²ⁿᵈ↓ ⟩ after 3×⟨←⟩」 now.
We have verified our claim in (3.1). Accordingly, we get
┌───┬───┬───┬───┬───┬───┬───┐
│ 6■│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ │ 5 │ 6 │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 3 │ 3 │ │ 5 │ 6 │ │ │ │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ │ │ 0 │ 4 │ │
└───┴───┴───┴───┴───┴───┴───┘
Then, by ✅「⟨? ⋯ 1 ⋯ (?+2) ⋯⟩ (?≠1)」, we get
┌───┬───┬───┬───┬───┬───┬───┐
│6th│ 5■│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 3 │ │ 5 │ 6 │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 4 │ 3 │ 1 │ 5 │ 6 │ │ │ │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ 2 │ │ │ 0 │ 4 │ │
└───┴───┴───┴───┴───┴───┴───┘
Note that to avoid ⛔「⟨ ²ⁿᵈa ¹ˢᵗb ⟩, min⟦a,b⟧ = 0」, we need
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│ 0■│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 3 │ 1 │ 5 │ 6 │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 5 │ 3 │ 1 │ 5 │ 6 │ │ │ 0 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ 2 │ │ │ │ 4 │ │
└───┴───┴───┴───┴───┴───┴───┘
Finally, to avoid ⛔「⟨ ³ʳᵈ↓ ²ⁿᵈ↓ ⟩ after 3×⟨←⟩」, we finish by
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│ 2■│ 1■│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 3 │ 1 │ 5 │ 6 │ │ │ 0 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 6 │ 3 │ 1 │ 5 │ 6 │ 2 │ │ 0 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 7 │ 3 │ 1 │ 5 │ 6 │ 2 │ 4 │ 0 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.5
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