Rearrange the digits in ⟨1263045⟩ to meet the rules below.
⟨6th 5th 4th 3rd 2nd 1st 0th⟩
✅Match
⟨⋯ 2 ⋯ 4 ⋯ 3 ⋯⟩
Jump(1,4) = 1
5th → a, 3rd → b, |a-b|=4
⛔Avoid
2nd → 2|3|4|5|6
5th|4th|3rd → 2
#125034_v2.5
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ 2 │ │ │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 2 │ 2 │ 5 │ │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 3 │ 2 │ 5 │ │ 1 │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 4 │ 2 │ 5 │ │ 1 │ 0 │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 5 │ 2 │ 5 │ │ 1 │ 0 │ 4 │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 6 │ 2 │ 5 │ │ 1 │ 0 │ 4 │ 3 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 7 │ 2 │ 5 │ 6 │ 1 │ 0 │ 4 │ 3 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
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Proof of 2024-05-28 Q1(m=6)
═══════════════════════════
Notation: if nth -> a, then we write [nth] = a.
Plainly, to avoid ⛔「5th|4th|3rd → 2」, we need
2 = [6th] | [2nd] | [1st] | [0th].
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ │ / │ / │ / │ │ │ │
└───┴───┴───┴───┴───┴───┴───┘
Actually, it is 2 = [6th] | [2nd], since ✅「⟨⋯ 2 ⋯ 4 ⋯ 3 ⋯⟩」 implies that there are at least two digits at the right of 2. Therefore, in view of ⛔「2nd → 2|3|4|5|6」, we get 2 = [6th].
┌───┬───┬───┬───┬───┬───┬───┐
│ 6■│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ 2 │ │ │ │ │ │ │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ │ 6 │ 3 │ 0 │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Next, we consider ✅「5th → a, 3rd → b, |a-b|=4」. There are exactly four ways to match it:
(1.1) ([5th], [3rd]) = (5,1);
(1.2) ([5th], [3rd]) = (1,5);
(1.3) ([5th], [3rd]) = (4,0);
(1.4) ([5th], [3rd]) = (0,4).
┌───┬───┬───┬───┬───┬───┬───┐
│6th│ 5▲│4th│ 3▲│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 2 │ ▬ │ │ ▬ │ │ │ │
└───┴───┴───┴───┴───┴───┴───┘
We show that (1.1) holds actually.
------------------------------
Observe that in order to match ✅「Jump(1,4) = 1」, neither (1.2) nor (1.3) holds. On the other hand, if (1.4) holds:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│ 5▲│4th│ 3▲│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 2 │ 0 │ │ 4 │ │ │ │
└───┴───┴───┴───┴───┴───┴───┘
then we cannot match ✅「Jump(1,4) = 1」 and avoid ⛔「2nd → 2|3|4|5|6」 at the same time. So, (1.4) does not hold as well.
------------------------------
We have verified that (1.1) holds. Accordingly, we get
┌───┬───┬───┬───┬───┬───┬───┐
│6th│ 5■│4th│ 3■│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 2 │ │ │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 2 │ 2 │ 5 │ │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 3 │ 2 │ 5 │ │ 1 │ │ │ │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ 6 │ 3 │ 0 │ 4 │ │
└───┴───┴───┴───┴───┴───┴───┘
Then, using ⛔「2nd → 2|3|4|5|6」 and ✅「Jump(1,4) = 1」, we get
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│ 2■│ 1■│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 2 │ 5 │ │ 1 │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 4 │ 2 │ 5 │ │ 1 │ 0 │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 5 │ 2 │ 5 │ │ 1 │ 0 │ 4 │ │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ 6 │ 3 │ │ │ │
└───┴───┴───┴───┴───┴───┴───┘
Finally, in view of ✅「⟨⋯ 2 ⋯ 4 ⋯ 3 ⋯⟩」, we finish by
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│ 4■│3rd│2nd│1st│ 0■│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 2 │ 5 │ │ 1 │ 0 │ 4 │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 6 │ 2 │ 5 │ │ 1 │ 0 │ 4 │ 3 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 7 │ 2 │ 5 │ 6 │ 1 │ 0 │ 4 │ 3 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.5