Rearrange the digits in ⟨1263045⟩ to meet the rules below.
⟨6th 5th 4th 3rd 2nd 1st 0th⟩
✅Match
5th → a, 0th → b, |a-b|=1
⟨? 2 ⋯ (?+3) ⋯ 6 ⋯⟩ (?≠2)
Jump(2,3) ≥ 2
⛔Avoid
⟨? ⋯ 4 ⋯ (?+5) ⋯⟩
#125034_v2.7
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ 2 │ │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 2 │ 0 │ 2 │ │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 3 │ 0 │ 2 │ │ │ │ │ 1 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 4 │ 0 │ 2 │ │ │ 3 │ │ 1 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 5 │ 0 │ 2 │ │ │ 3 │ 6 │ 1 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 6 │ 0 │ 2 │ 5 │ │ 3 │ 6 │ 1 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 7 │ 0 │ 2 │ 5 │ 4 │ 3 │ 6 │ 1 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
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Proof of 2024-12-03 Q1(m=6)
═══════════════════════════
Notation: if nth -> a, then we write [nth] = a.
Our first step follows from ✅「⟨? 2 ⋯ (?+3) ⋯ 6 ⋯⟩ (?≠2)」:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ 2 │ │ │ │ │ │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ │ 6 │ 3 │ 0 │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
That pattern also implies
(1) [6th] = 0|1|3.
(2) We show that [6th] = 0 actually.
------------------------------
(2.1) For, if [6th] = 3:
┌───┬───┬───┬───┬───┬───┬───┐
│ 6▲│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 3 │ 2 │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┴───┘
then we cannot match ✅「Jump(2,3) ≥ 2」.
(2.2) Else if [6th] = 1:
┌───┬───┬───┬───┬───┬───┬───┐
│ 6▲│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 1 │ 2 │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┴───┘
then we cannot avoid ⛔「⟨? ⋯ 4 ⋯ (?+5) ⋯⟩」 and match ✅「⟨? 2 ⋯ (?+3) ⋯ 6 ⋯⟩ (?≠2)」 at the same time.
------------------------------
We have verified (2). Accordingly, we get
┌───┬───┬───┬───┬───┬───┬───┐
│ 6■│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ 2 │ │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 2 │ 0 │ 2 │ │ │ │ │ │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ │ 6 │ 3 │ │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Therefore, by ✅「⟨? 2 ⋯ (?+3) ⋯ 6 ⋯⟩ (?≠2)」, we have the following required pattern:
(3) ⟨0 2 ⋯ 3 ⋯ 6 ⋯⟩.
In particular, there is at least one digit at the right of 3. Hence, to match ✅「5th → a, 0th → b, |a-b|=1」, we need [0th] = 1:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│ 0■│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 0 │ 2 │ │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 3 │ 0 │ 2 │ │ │ │ │ 1 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ 6 │ 3 │ │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Then, combining ✅「Jump(2,3) ≥ 2」 with (3), we get
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│ 2■│ 1■│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 0 │ 2 │ │ │ │ │ 1 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 4 │ 0 │ 2 │ │ │ 3 │ │ 1 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 5 │ 0 │ 2 │ │ │ 3 │ 6 │ 1 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Finally, to avoid ⛔「⟨? ⋯ 4 ⋯ (?+5) ⋯⟩」, we finish by
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│ 4■│ 3■│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 0 │ 2 │ │ │ 3 │ 6 │ 1 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 6 │ 0 │ 2 │ 5 │ │ 3 │ 6 │ 1 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 7 │ 0 │ 2 │ 5 │ 4 │ 3 │ 6 │ 1 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.7