Rearrange the digits in ⟨1263045⟩ to meet the rules below.
⟨6th 5th 4th 3rd 2nd 1st 0th⟩
✅Match
Jump(2,3) ≥ 3
3rd|0th → 4
⟨? ⋯ 6 (?−2) ⋯ 4 ⋯⟩ (?≠6)
⛔Avoid
4th → a, 1st → b, |a-b|=1
⟨ ⁶ᵗʰa ⁵ᵗʰb ²ⁿᵈc ⟩, (abc)₁₀ ≥ 435
⟨⋯ a ⋯ 1 ⋯⟩, a = 0|4|6
min ⊢5⊣ = 1
#125034_v2.8
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ 2 │ │ │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 2 │ 2 │ │ │ │ │ │ 4 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 3 │ 2 │ 1 │ │ │ │ │ 4 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 4 │ 2 │ 1 │ 6 │ │ │ │ 4 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 5 │ 2 │ 1 │ 6 │ 0 │ │ │ 4 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 6 │ 2 │ 1 │ 6 │ 0 │ 5 │ │ 4 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 7 │ 2 │ 1 │ 6 │ 0 │ 5 │ 3 │ 4 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
Proof of 2024-12-24 Q1(m=6)
═══════════════════════════
Notation: if nth -> a, then we write [nth] = a.
We first consider what [6th] is. By ⛔「⟨ ⁶ᵗʰa ⁵ᵗʰb ²ⁿᵈc ⟩, (abc)₁₀ ≥ 435」 and ✅「⟨? ⋯ 6 (?−2) ⋯ 4 ⋯⟩ (?≠6)」, we have
(1) [6th] = 3|2.
If [6th] = 3, then ✅「⟨? ⋯ 6 (?−2) ⋯ 4 ⋯⟩ (?≠6)」 implies that we have the block "61". But then we cannot avoid ⛔「⟨⋯ a ⋯ 1 ⋯⟩, a = 0|4|6」, which is a contradiction. Therefore, [6th] = 2:
┌───┬───┬───┬───┬───┬───┬───┐
│ 6■│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ 2 │ │ │ │ │ │ │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ │ 6 │ 3 │ 0 │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Then, by combining ✅「⟨? ⋯ 6 (?−2) ⋯ 4 ⋯⟩ (?≠6)」 and ⛔「⟨⋯ a ⋯ 1 ⋯⟩, a = 0|4|6」, we have the following required pattern:
(2) ⟨2 ⋯ 1 ⋯ 60 ⋯ 4 ⋯⟩.
To match (2), we need to reserve boxes to the right of 4 for placing 1,6,0. Therefore, ✅「3rd|0th → 4」 implies
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│ 0■│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 2 │ │ │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 2 │ 2 │ │ │ │ │ │ 4 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ │ 6 │ 3 │ 0 │ │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Then, we consider where to place 1. By (2), there are three possibilities:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│ 5▲│ 4▲│ 3▲│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
(3) │ 2 │ │ │ 1 │ │ │ 4 │
├───┼───┼───┼───┼───┼───┼───┤
(4) │ 2 │ │ 1 │ │ │ │ 4 │
├───┼───┼───┼───┼───┼───┼───┤
(5) │ 2 │ 1 │ │ │ │ │ 4 │
└───┴───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ 6 │ 3 │ 0 │ │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
(6) We show that case (5) holds actually.
------------------------------
Note that by ✅「Jump(2,3) ≥ 3」, we need
(7) 3 = [2nd] | [1st].
Therefore, case (3) cannot happen, for otherwise we cannot match (7) and (2) at the same time. On the other hand, if case (4) happens, then (7) and (2) gives
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 2 │ 5 │ 1 │ ▬ │ ▬ │ ▬ │ 4 │
└───┴───┴───┴───┴───┴───┴───┘
where "▬" are occupied by 3,6,0. It has matched ⛔「min ⊢5⊣ = 1」, which is a contradiction.
------------------------------
We have verified (6). Accordingly, we get
┌───┬───┬───┬───┬───┬───┬───┐
│6th│ 5■│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 2 │ │ │ │ │ │ 4 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 3 │ 2 │ 1 │ │ │ │ │ 4 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ 6 │ 3 │ 0 │ │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Then, we consider how to place 6. Combining (2) and (7), we see that there are two possibilities:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│ 4▲│ 3▲│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
(8) │ 2 │ 1 │ │ 6 │ 0 │ 3 │ 4 │
├───┼───┼───┼───┼───┼───┼───┤
(9) │ 2 │ 1 │ 6 │ 0 │ │ │ 4 │
└───┴───┴───┴───┴───┴───┴───┘
Actually case (8) does not hold, for otherwise it would match ⛔「min ⊢5⊣ = 1」. Therefore, case (9) holds and we get
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│ 4■│ 3■│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 2 │ 1 │ │ │ │ │ 4 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 4 │ 2 │ 1 │ 6 │ │ │ │ 4 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 5 │ 2 │ 1 │ 6 │ 0 │ │ │ 4 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ │ 3 │ │ │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Finally, in view of ⛔「4th → a, 1st → b, |a-b|=1」, we finish by
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│ 2■│ 1■│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 2 │ 1 │ 6 │ 0 │ │ │ 4 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 6 │ 2 │ 1 │ 6 │ 0 │ 5 │ │ 4 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 7 │ 2 │ 1 │ 6 │ 0 │ 5 │ 3 │ 4 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.8