Rearrange the digits in ⟨1263045⟩ to meet the rules below.
⟨6th 5th 4th 3rd 2nd 1st 0th⟩
✅Match
{p5, p3, p2, p1} = ? + {0,1,2,4}
⟨⋯ 2 ⋯ ? 6 ⋯ (?+4)⟩ (?≠2)
⛔Avoid
Jump(1,4) = 1
#125034_v2.8
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ │ │ │ │ │ 5 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ │ │ 6 │ │ │ 5 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 3 │ │ │ 1 │ 6 │ │ │ 5 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 4 │ │ 2 │ 1 │ 6 │ │ │ 5 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 5 │ 0 │ 2 │ 1 │ 6 │ │ │ 5 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 6 │ 0 │ 2 │ 1 │ 6 │ │ 4 │ 5 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 7 │ 0 │ 2 │ 1 │ 6 │ 3 │ 4 │ 5 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
Proof of 2024-12-17 Q1(m=6)
═══════════════════════════
Notation: if nth -> a, then we write [nth] = a.
By ✅「⟨⋯ 2 ⋯ ? 6 ⋯ (?+4)⟩ (?≠2)」, one of the following holds:
(A) ⟨⋯ 2 ⋯ 0 6 ⋯ 4⟩;
(B) ⟨⋯ 2 ⋯ 1 6 ⋯ 5⟩.
(1) We show that case (B) happens actually.
------------------------------
If on the contrary case (A) happens, then combining it with ✅「{p5, p3, p2, p1} = ? + {0,1,2,4}」, we have
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ │ ▬ │ │ ▬ │ ▬ │ ▬ │ 4 │
└───┴───┴───┴───┴───┴───┴───┘
where "▬" are occupied by 1,2,3,5. Note that to match (A), we need two consecutive boxes for placing 06. Now we are unable to do so, therefore it is a contradiction.
------------------------------
We have checked (1). Accordingly, we get
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│ 0■│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ │ │ │ │ │ 5 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ 6 │ 3 │ 0 │ 4 │ │
└───┴───┴───┴───┴───┴───┴───┘
Then, ✅「{p5, p3, p2, p1} = ? + {0,1,2,4}」 implies S := {p5,p3,p2,p1} = {0,1,2,4} or {2,3,4,6}.
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ │ ▬ │ │ ▬ │ ▬ │ ▬ │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Note that to match (B), we need to have the block "16" and place it to the right of 2. There is no way to do so if S = {0,1,2,4}. On the other hand, if S = {2,3,4,6}, there is exactly one way to meet the requirement:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│ 5■│ 4■│ 3■│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ │ │ │ │ │ 5 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ │ │ 6 │ │ │ 5 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 3 │ │ │ 1 │ 6 │ │ │ 5 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 4 │ │ 2 │ 1 │ 6 │ │ │ 5 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ │ 3 │ 0 │ 4 │ │
└───┴───┴───┴───┴───┴───┴───┘
We know that S = {2,3,4,6}, so {[2nd], [1st]} = {3,4}. It follows that [6th] = 0:
┌───┬───┬───┬───┬───┬───┬───┐
│ 6■│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ 2 │ 1 │ 6 │ │ │ 5 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 5 │ 0 │ 2 │ 1 │ 6 │ │ │ 5 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ │ 3 │ │ 4 │ │
└───┴───┴───┴───┴───┴───┴───┘
Finally, in view of ⛔「Jump(1,4) = 1」, we finish by
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│ 2■│ 1■│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 0 │ 2 │ 1 │ 6 │ │ │ 5 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 6 │ 0 │ 2 │ 1 │ 6 │ │ 4 │ 5 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 7 │ 0 │ 2 │ 1 │ 6 │ 3 │ 4 │ 5 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.8