Rearrange the digits in ⟨1263045⟩ to meet the rules below.
⟨6th 5th 4th 3rd 2nd 1st 0th⟩
✅Match
min ⊢3⊣ ≥ 2
⟦2,5⟧ ∋ 3,6
⟨⋯ 0 ⋯ 5 ⋯ 4 ⋯ 2 ⋯⟩
⟦0,3⟧ ∋ 1,6
{p5, p4, p1, p0} = ? + {0,1,2,3}
#125034_v2.7
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ 0 │ │ │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 2 │ 0 │ │ │ │ │ │ 2 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 3 │ 0 │ │ │ │ │ 3 │ 2 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 4 │ 0 │ 5 │ │ │ │ 3 │ 2 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 5 │ 0 │ 5 │ 4 │ │ │ 3 │ 2 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 6 │ 0 │ 5 │ 4 │ │ 6 │ 3 │ 2 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 7 │ 0 │ 5 │ 4 │ 1 │ 6 │ 3 │ 2 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
Proof of 2024-11-05 Q1(m=6)
═══════════════════════════
Notation: if nth -> a, then we write [nth] = a.
Combining ✅「⟦2,5⟧ ∋ 3,6」 with ✅「⟨⋯ 0 ⋯ 5 ⋯ 4 ⋯ 2 ⋯⟩」, we see that one of the following holds:
⟨⋯ 0 ⋯ 5 ⋯ 3 ⋯ 6 ⋯ 2 ⋯⟩; or
⟨⋯ 0 ⋯ 5 ⋯ 6 ⋯ 3 ⋯ 2 ⋯⟩.
The latter holds actually, for otherwise we cannot match ✅「⟦0,3⟧ ∋ 1,6」. This pattern then implies that one of the following holds:
(i) ⟨⋯ 0 ⋯ 6 ⋯ 1 ⋯ 3 ⋯ 2 ⋯⟩; or
(ii) ⟨⋯ 0 ⋯ 1 ⋯ 6 ⋯ 3 ⋯ 2 ⋯⟩.
Together with ✅「⟨⋯ 0 ⋯ 5 ⋯ 4 ⋯ 2 ⋯⟩」, we see that no matter which case happens, we have
(1) 0 is to the left of 1,2,3,4,5,6; and
(2) 2 is to the right of 0,1,3,4,5,6.
Therefore, we get
┌───┬───┬───┬───┬───┬───┬───┐
│ 6■│5th│4th│3rd│2nd│1st│ 0■│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ 0 │ │ │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 2 │ 0 │ │ │ │ │ │ 2 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ │ 6 │ 3 │ │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Next, we consider ✅「{p5, p4, p1, p0} = ? + {0,1,2,3}」.
┌───┬───┬───┬───┬───┬───┬───┐
│6th│*5 │*4 │3rd│2nd│*1 │*0 │
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 0 │ │ │ │ │ │ 2 │
└───┴───┴───┴───┴───┴───┴───┘
We need p5,p4,p1,p0 to be consecutive integers. It follows that S := {p5, p4, p1, p0} is either {1,2,3,4} or {2,3,4,5}. As a consequence,
(3) 3,4 ∈ S; and
(4) 6 ∉ S; equivalently 6 = [3rd] | [2nd].
┌───┬───┬───┬───┬───┬───┬───┐
│6th│*5 │*4 │3rd│2nd│*1 │*0 │
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 0 │ │ │ ▬ │ ▬ │ │ 2 │
└───┴───┴───┴───┴───┴───┴───┘
On the other hand, recall that case (i) or (ii) holds. A fortiori, we need to match
(5) ⟨ ⋯ 6 ⋯ 3 ⋯ ⟩.
Combining (3), (4), and (5), we get [1st] = 3:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│ 1■│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 0 │ │ │ │ │ │ 2 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 3 │ 0 │ │ │ │ │ 3 │ 2 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ │ 6 │ │ │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Next, we consider the position of 4. By (3), we have 4 ∈ S. Combining this with ✅「⟨⋯ 0 ⋯ 5 ⋯ 4 ⋯ 2 ⋯⟩」, we get
┌───┬───┬───┬───┬───┬───┬───┐
│6th│ 5■│ 4■│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 0 │ │ │ │ │ 3 │ 2 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 4 │ 0 │ 5 │ │ │ │ 3 │ 2 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 5 │ 0 │ 5 │ 4 │ │ │ 3 │ 2 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ │ 6 │ │ │ │ │
└───┴───┴───┴───┴───┴───┴───┘
Finally, in view of ✅「min ⊢3⊣ ≥ 2」, we finish by
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│ 3■│ 2■│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 0 │ 5 │ 4 │ │ │ 3 │ 2 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 6 │ 0 │ 5 │ 4 │ │ 6 │ 3 │ 2 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 7 │ 0 │ 5 │ 4 │ 1 │ 6 │ 3 │ 2 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.7