Rearrange the digits in ⟨1263045⟩ to meet the rules below.
⟨6th 5th 4th 3rd 2nd 1st 0th⟩
✅Match
⟨? 6 ⋯ (?−2) ⋯ 1 ⋯⟩ (?≠6)
⟨ ⁶ᵗʰb ⁵ᵗʰa ¹ˢᵗc ⟩, a > b > c
⟨Perm(1,3,5,6) Perm(0,2,4)⟩
2nd → 1|2|4
⛔Avoid
max ⊢1⊣ ≤ 4
3rd → 4|5|6
3rd → a, 0th → b, ab=2+4n
⟨⋯ 3 ⋯ ? ⋯ 2 (?−6)⟩
#125034_v2.4
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ 6 │ │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ 6 │ │ 1 │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 3 │ │ 6 │ 5 │ 1 │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 4 │ 3 │ 6 │ 5 │ 1 │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 5 │ 3 │ 6 │ 5 │ 1 │ │ 0 │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 6 │ 3 │ 6 │ 5 │ 1 │ 2 │ 0 │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 7 │ 3 │ 6 │ 5 │ 1 │ 2 │ 0 │ 4 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
Proof of 2024-04-09 Q1(m=6)
═══════════════════════════
Notation: if nth -> a, then we write [nth] = a.
Plainly, our first step follows from ✅「⟨? 6 ⋯ (?−2) ⋯ 1 ⋯⟩ (?≠6)」:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│ 5■│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ 6 │ │ │ │ │ │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ │ 3 │ 0 │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Next, we determine the value of [3rd]. By ✅「⟨Perm(1,3,5,6) Perm(0,2,4)⟩」 and ⛔「3rd → 4|5|6」, we see that
(1) [3rd] = 1|3. We claim that [3rd] = 1.
------------------------------
For, suppose on the contrary [3rd] = 3:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│ 3▲│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ │ 6 │ │ 3 │ │ │ │
└───┴───┴───┴───┴───┴───┴───┘
Then, by ✅「⟨Perm(1,3,5,6) Perm(0,2,4)⟩」, we have [6th] = 1|5. As ✅「⟨? 6 ⋯ (?−2) ⋯ 1 ⋯⟩ (?≠6)」 implies [6th] != 1, we get [6th] = 5:
┌───┬───┬───┬───┬───┬───┬───┐
│ 6▲│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 5 │ 6 │ │ 3 │ │ │ │
└───┴───┴───┴───┴───┴───┴───┘
But then we cannot match ✅「⟨Perm(1,3,5,6) Perm(0,2,4)⟩」 and ✅「⟨? 6 ⋯ (?−2) ⋯ 1 ⋯⟩ (?≠6)」 at the same time, which is a contradiction.
------------------------------
Our claim in (1) is thus verified. Accordingly, we get
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│ 3■│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ 6 │ │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ 6 │ │ 1 │ │ │ │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ 2 │ │ 3 │ 0 │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Next, observe that there is only one way to match ✅「⟨Perm(1,3,5,6) Perm(0,2,4)⟩」 and avoid ⛔「max ⊢1⊣ ≤ 4」 at the same time. That is
┌───┬───┬───┬───┬───┬───┬───┐
│ 6■│5th│ 4■│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ 6 │ │ 1 │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 3 │ │ 6 │ 5 │ 1 │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 4 │ 3 │ 6 │ 5 │ 1 │ │ │ │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ 2 │ │ │ 0 │ 4 │ │
└───┴───┴───┴───┴───┴───┴───┘
We proceed to determine the value of [1st]. By ✅「⟨ ⁶ᵗʰb ⁵ᵗʰa ¹ˢᵗc ⟩, a > b > c」, we have
(2) [1st] = 2|0. We claim that [1st] = 0 actually.
------------------------------
For, if on the contrary [1st] = 2:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│ 1▲│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 3 │ 6 │ 5 │ 1 │ │ 2 │ │
└───┴───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ │ │ 0 │ 4 │ │
└───┴───┴───┴───┴───┴───┴───┘
Then, to match ✅「2nd → 1|2|4」, we reach
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│ 2▲│1st│ 0▲│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 3 │ 6 │ 5 │ 1 │ 4 │ 2 │ 0 │
└───┴───┴───┴───┴───┴───┴───┘
But then we match ⛔「⟨⋯ 3 ⋯ ? ⋯ 2 (?−6)⟩」, which is a contradiction.
------------------------------
We have verified our claim in (2). As a result, we get
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│ 1■│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 3 │ 6 │ 5 │ 1 │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 5 │ 3 │ 6 │ 5 │ 1 │ │ 0 │ │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ 2 │ │ │ │ 4 │ │
└───┴───┴───┴───┴───┴───┴───┘
Finally, noting that to avoid ⛔「3rd → a, 0th → b, ab=2+4n」 we cannot have [0th] = 2, we finish by
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│ 2■│1st│ 0■│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 3 │ 6 │ 5 │ 1 │ │ 0 │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 6 │ 3 │ 6 │ 5 │ 1 │ 2 │ 0 │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 7 │ 3 │ 6 │ 5 │ 1 │ 2 │ 0 │ 4 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.4