Rearrange the digits in ⟨1263045⟩ to meet the rules below.
⟨6th 5th 4th 3rd 2nd 1st 0th⟩
✅Match
⟨⋯ 2 ⋯ ? 3 ⋯ (?−1)⟩ (?≠3,4)
⟦0,2⟧ ∋ 1,3,4,5
⛔Avoid
⟨⋯ Perm(4,5) ⋯⟩
⟦2,6⟧ ∋ 3,4
⟨⋯ ? ⋯ 1 ⋯ (?−6)⟩
⟨ ⁶ᵗʰc ⁴ᵗʰd ³ʳᵈb ²ⁿᵈa ⟩, a > b > c > d
#125034_v2.5
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ 2 │ │ │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 2 │ 2 │ │ │ │ │ │ 0 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 3 │ 2 │ │ │ │ 6 │ │ 0 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 4 │ 2 │ │ │ │ 6 │ 4 │ 0 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 5 │ 2 │ 1 │ │ │ 6 │ 4 │ 0 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 6 │ 2 │ 1 │ 3 │ │ 6 │ 4 │ 0 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 7 │ 2 │ 1 │ 3 │ 5 │ 6 │ 4 │ 0 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
Proof of 2024-07-16 Q1(m=6)
═══════════════════════════
Notation: if nth -> a, then we write [nth] = a.
By ✅「⟨⋯ 2 ⋯ ? 3 ⋯ (?−1)⟩ (?≠3,4)」, there are at least two digits at the right of 2. Combining this with ✅「⟦0,2⟧ ∋ 1,3,4,5」, we have 2 = [6th] | [5th]:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ ▬ │ ▬ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┴───┘
(1) We claim that 2 = [6th].
------------------------------
For, if on the contrary 2 = [5th], then to match ✅「⟦0,2⟧ ∋ 1,3,4,5」, we need
┌───┬───┬───┬───┬───┬───┬───┐
│6th│ 5▲│4th│3rd│2nd│1st│ 0▲│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ │ 2 │ / │ / │ / │ / │ 0 │
└───┴───┴───┴───┴───┴───┴───┘
where "/" are occupied by 1,3,4,5. It follows that [6th] = 6:
┌───┬───┬───┬───┬───┬───┬───┐
│ 6▲│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 6 │ 2 │ / │ / │ / │ / │ 0 │
└───┴───┴───┴───┴───┴───┴───┘
So we would match ⛔「⟨⋯ ? ⋯ 1 ⋯ (?−6)⟩」, which is a contradiction.
------------------------------
We have verified our claim in (1). Accordingly, we get
┌───┬───┬───┬───┬───┬───┬───┐
│ 6■│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ 2 │ │ │ │ │ │ │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ │ 6 │ 3 │ 0 │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Then, there are two ways to match ✅「⟦0,2⟧ ∋ 1,3,4,5」:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
(2) │ 2 │ │ │ │ │ 0 │ 6 │
├───┼───┼───┼───┼───┼───┼───┤
(3) │ 2 │ │ │ │ │ │ 0 │
└───┴───┴───┴───┴───┴───┴───┘
Since ✅「⟨⋯ 2 ⋯ ? 3 ⋯ (?−1)⟩ (?≠3,4)」 requires that [0th] < 6, case (2) does not hold. Therefore, case (3) holds, and we get
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│ 0■│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 2 │ │ │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 2 │ 2 │ │ │ │ │ │ 0 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ │ 6 │ 3 │ │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Now, to match ✅「⟨⋯ 2 ⋯ ? 3 ⋯ (?−1)⟩ (?≠3,4)」 and avoid ⛔「⟨⋯ ? ⋯ 1 ⋯ (?−6)⟩」 at the same time, we need to match the following pattern:
(4) ⟨2 ⋯ 1 3 ⋯ 6 ⋯ 0⟩.
We consider how to place 6. By (4), the digits 1,3 are at the left of 6, so 6 = [3rd] | [2nd] | [1st]:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 2 │ │ │ ▬ │ ▬ │ ▬ │ 0 │
└───┴───┴───┴───┴───┴───┴───┘
(5) We proceed to show that 6 = [2nd].
------------------------------
If 6 = [3rd], then (4) gives
┌───┬───┬───┬───┬───┬───┬───┐
│6th│ 5▲│ 4▲│ 3▲│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 2 │ 1 │ 3 │ 6 │ │ │ 0 │
└───┴───┴───┴───┴───┴───┴───┘
and we would match ⛔「⟨⋯ Perm(4,5) ⋯⟩」. Else if 6 = [1st]:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│ 1▲│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 2 │ │ │ │ │ 6 │ 0 │
└───┴───┴───┴───┴───┴───┴───┘
then we would match ⛔「⟦2,6⟧ ∋ 3,4」.
------------------------------
We have verified (5). As a result, we get
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│ 2■│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 2 │ │ │ │ │ │ 0 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 3 │ 2 │ │ │ │ 6 │ │ 0 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ │ │ 3 │ │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
In view of (4), 3 is at the left of 6. So, in order to avoid ⛔「⟦2,6⟧ ∋ 3,4」, we need 4 to be at the right of 6:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│ 1■│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 2 │ │ │ │ 6 │ │ 0 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 4 │ 2 │ │ │ │ 6 │ 4 │ 0 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ │ │ 3 │ │ │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Now, there is only one way to match (4) and avoid ⛔「⟨ ⁶ᵗʰc ⁴ᵗʰd ³ʳᵈb ²ⁿᵈa ⟩, a > b > c > d」 at the same time. We finish by
┌───┬───┬───┬───┬───┬───┬───┐
│6th│ 5■│ 4■│ 3■│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 2 │ │ │ │ 6 │ 4 │ 0 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 5 │ 2 │ 1 │ │ │ 6 │ 4 │ 0 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 6 │ 2 │ 1 │ 3 │ │ 6 │ 4 │ 0 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 7 │ 2 │ 1 │ 3 │ 5 │ 6 │ 4 │ 0 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.5