Rearrange the digits in ⟨1263045⟩ to meet the rules below.
⟨6th 5th 4th 3rd 2nd 1st 0th⟩
✅Match
⟨? ⋯ 3 (?−1) ⋯ 6 ⋯⟩ (?≠3,4)
⟨⋯ 0 ⋯ ? ⋯ 4 (?−1)⟩ (?≠4,5)
⛔Avoid
⟨⋯ a ⋯ 0 ⋯⟩, a = 1|3|4|5
#125034_v2.2
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ 2 │ │ │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 2 │ 2 │ 0 │ │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 3 │ 2 │ 0 │ │ │ │ 4 │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 4 │ 2 │ 0 │ │ │ │ 4 │ 5 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 5 │ 2 │ 0 │ 3 │ │ │ 4 │ 5 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 6 │ 2 │ 0 │ 3 │ 1 │ │ 4 │ 5 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 7 │ 2 │ 0 │ 3 │ 1 │ 6 │ 4 │ 5 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
Proof of 2024-01-02 Q1(m=6)
═══════════════════════════
Notation: if nth -> a, then we write [nth] = a.
We begin with determining the value of [6th]. By ⛔「⟨⋯ a ⋯ 0 ⋯⟩, a = 1|3|4|5」, it is not 1|3|4|5.
Therefore, [6th] = 0|2|6. It is not 0 or 6 by ✅「⟨? ⋯ 3 (?−1) ⋯ 6 ⋯⟩ (?≠3,4)」. Accordingly, we have
┌───┬───┬───┬───┬───┬───┬───┐
│ 6■│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ 2 │ │ │ │ │ │ │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ │ 6 │ 3 │ 0 │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Next, we consider where to place 0. To avoid ⛔「⟨⋯ a ⋯ 0 ⋯⟩, a = 1|3|4|5」, 0 is at the left of 1,3,4,5, so it is at 5th or 4th.
If 0 = [4th], then the preceding pattern implies that [5th] = 6:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│ 5▲│ 4▲│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 2 │ 6 │ 0 │ - │ - │ - │ - │
└───┴───┴───┴───┴───┴───┴───┘
("-" are occupied by 1|3|4|5)
However, this contradicts ✅「⟨? ⋯ 3 (?−1) ⋯ 6 ⋯⟩ (?≠3,4)」. Therefore, we have 0 = [5th]:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│ 5■│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 2 │ │ │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 2 │ 2 │ 0 │ │ │ │ │ │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ │ 6 │ 3 │ │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Next, to match ✅「⟨⋯ 0 ⋯ ? ⋯ 4 (?−1)⟩ (?≠4,5)」, plainly we need [1st] = 4.
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│ 1■│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 2 │ 0 │ │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 3 │ 2 │ 0 │ │ │ │ 4 │ │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ │ 6 │ 3 │ │ │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
We proceed to determine the value of [0th]. By ✅「⟨? ⋯ 3 (?−1) ⋯ 6 ⋯⟩ (?≠3,4)」, we need to match
(1) ⟨2 ⋯ 3 1 ⋯ 6 ⋯⟩
A fortiori, 3,1 are not in the right corner. We have [0th] != 6 as well, because by ✅「⟨⋯ 0 ⋯ ? ⋯ 4 (?−1)⟩ (?≠4,5)」 we have [0th] <= 5.
Consequently, we have [0th] = 5.
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│ 0■│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 2 │ 0 │ │ │ │ 4 │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 4 │ 2 │ 0 │ │ │ │ 4 │ 5 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ │ 6 │ 3 │ │ │ │
└───┴───┴───┴───┴───┴───┴───┘
Finally, there is only one way to match (1). We finish by
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│ 4■│ 3■│ 2■│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 2 │ 0 │ │ │ │ 4 │ 5 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 5 │ 2 │ 0 │ 3 │ │ │ 4 │ 5 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 6 │ 2 │ 0 │ 3 │ 1 │ │ 4 │ 5 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 7 │ 2 │ 0 │ 3 │ 1 │ 6 │ 4 │ 5 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.2