Rearrange the digits in ⟨1263045⟩ to meet the rules below.
⟨6th 5th 4th 3rd 2nd 1st 0th⟩
✅Match
⟨ ⁶ᵗʰ↑ ²ⁿᵈ↓ ¹ˢᵗ↓ ⟩ after 2×⟨←⟩
3rd → a, 0th → b, ab=3+4n
⟦3,4⟧ ∋ 0,5
#125034_v2.6
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ 0 │ │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 2 │ 4 │ 0 │ │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 3 │ 4 │ 0 │ │ │ 2 │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 4 │ 4 │ 0 │ │ 3 │ 2 │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 5 │ 4 │ 0 │ │ 3 │ 2 │ │ 1 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 6 │ 4 │ 0 │ 5 │ 3 │ 2 │ │ 1 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 7 │ 4 │ 0 │ 5 │ 3 │ 2 │ 6 │ 1 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
Proof of 2024-08-06 Q1(m=6)
═══════════════════════════
Notation: if nth -> a, then we write [nth] = a.
We first consider where to place 0. We cannot place 0 at a "↓" position of ✅「⟨ ⁶ᵗʰ↑ ²ⁿᵈ↓ ¹ˢᵗ↓ ⟩ after 2×⟨←⟩」, so 0 != [2nd] and [1st]. To have a "↑" at 6th, we need 0 != [4th] as well.
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│ 4▲│3rd│ 2▲│ 1▲│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ │ │ / │ │ / │ / │ │
└───┴───┴───┴───┴───┴───┴───┘
Moreover, by ✅「3rd → a, 0th → b, ab=3+4n」, we have 0 != [3rd] and [0th]:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│ 3▲│2nd│1st│ 0▲│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ │ │ / │ / │ / │ / │ / │
└───┴───┴───┴───┴───┴───┴───┘
and by ✅「⟦3,4⟧ ∋ 0,5」, 0 is not at the left corner:
┌───┬───┬───┬───┬───┬───┬───┐
│ 6▲│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ / │ │ / │ / │ / │ / │ / │
└───┴───┴───┴───┴───┴───┴───┘
It follows that 0 = [5th]:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│ 5■│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ 0 │ │ │ │ │ │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ 6 │ 3 │ │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Then, to match ✅「⟦3,4⟧ ∋ 0,5」, we need
(1) [6th] = 3|4.
Observe that ✅「3rd → a, 0th → b, ab=3+4n」 implies
(2) {[3rd], [0th]} = {1,3} | {3,5}.
A fortiori, 3 ∈ {[3rd], [0th]}. Combining this with (1), we get
┌───┬───┬───┬───┬───┬───┬───┐
│ 6■│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ 0 │ │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 2 │ 4 │ 0 │ │ │ │ │ │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ 6 │ 3 │ │ │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Next, we use ✅「⟨ ⁶ᵗʰ↑ ²ⁿᵈ↓ ¹ˢᵗ↓ ⟩ after 2×⟨←⟩」. Given that [6th] = 4, to have a "↑" at 6th and a "↓" at 1st, we need [4th] > 4 and [1st] > 4. This means
(3) {[4th], [1st]} = {5,6}.
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│ 4▲│3rd│2nd│ 1▲│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 4 │ 0 │ ▬ │ │ │ ▬ │ │
└───┴───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ │ 3 │ │ │ │
└───┴───┴───┴───┴───┴───┴───┘
(where "▬" are occupied by 5,6)
And combining (3) with (2), we get
(4) {[3rd], [0th]} = {1,3}.
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 4 │ 0 │ ▬ │ # │ │ ▬ │ # │
└───┴───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ 2 │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┴───┘
(where "#" are occupied by 1,3)
Plainly, it follows that [2nd] = 2:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│ 2■│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 4 │ 0 │ │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 3 │ 4 │ 0 │ │ │ 2 │ │ │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ │ 6 │ 3 │ │ │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Now, in view of (4), there are two possibilities:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│ 3▲│2nd│1st│ 0▲│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
(5) │ 4 │ 0 │ │ 1 │ 2 │ │ 3 │
├───┼───┼───┼───┼───┼───┼───┤
(6) │ 4 │ 0 │ │ 3 │ 2 │ │ 1 │
└───┴───┴───┴───┴───┴───┴───┘
If case (5) holds, then we cannot have a "↓" at 2nd, contradicting ✅「⟨ ⁶ᵗʰ↑ ²ⁿᵈ↓ ¹ˢᵗ↓ ⟩ after 2×⟨←⟩」. Therefore, case (6) holds:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│ 3■│2nd│1st│ 0■│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 4 │ 0 │ │ │ 2 │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 4 │ 4 │ 0 │ │ 3 │ 2 │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 5 │ 4 │ 0 │ │ 3 │ 2 │ │ 1 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ 6 │ │ │ │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Finally, in view of ✅「⟦3,4⟧ ∋ 0,5」, we finish by
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│ 4■│3rd│2nd│ 1■│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 4 │ 0 │ │ 3 │ 2 │ │ 1 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 6 │ 4 │ 0 │ 5 │ 3 │ 2 │ │ 1 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 7 │ 4 │ 0 │ 5 │ 3 │ 2 │ 6 │ 1 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.6