Rearrange the digits in ⟨1263045⟩ to meet the rules below.
⟨6th 5th 4th 3rd 2nd 1st 0th⟩
✅Match
4th → a, 2nd → b, ab=2+4n
Jump(0,2) ≤ 1
5th → a, 2nd → b, a+b=1+6n
⟨⋯ 6 ⋯ 4 ⋯⟩
⛔Avoid
6th|5th|4th|3rd|1st → 0
6th|5th|2nd|0th → 3
{p4, p3, p1} = ? + {0,2,3}
#125034_v2.6
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ │ │ │ │ │ 0 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ │ │ │ 2 │ │ 0 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 3 │ │ 5 │ │ │ 2 │ │ 0 │▒
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Step 4 │ 1 │ 5 │ │ │ 2 │ │ 0 │▒
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Step 5 │ 1 │ 5 │ 3 │ │ 2 │ │ 0 │▒
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Step 6 │ 1 │ 5 │ 3 │ 6 │ 2 │ │ 0 │▒
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Step 7 │ 1 │ 5 │ 3 │ 6 │ 2 │ 4 │ 0 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
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Proof of 2024-08-20 Q1(m=6)
═══════════════════════════
Notation: if nth -> a, then we write [nth] = a.
By ⛔「6th|5th|4th|3rd|1st → 0」, we have 0 = [2nd] | [0th]. We need [2nd] != 0 in order to match ✅「4th → a, 2nd → b, ab=2+4n」, so it is 0 = [0th].
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│ 0■│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ │ │ │ │ │ 0 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
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--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ 6 │ 3 │ │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Next, by ✅「Jump(0,2) ≤ 1」, we have 2 = [2nd] | [1st].
(1) We claim that 2 = [2nd].
------------------------------
For, suppose on the contrary 2 = [1st]:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│ 1▲│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ │ │ │ │ │ 2 │ 0 │
└───┴───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ │ 6 │ 3 │ │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Then, by ✅「5th → a, 2nd → b, a+b=1+6n」, one of the following holds:
(1.1) {[5th], [2nd]} = {3,4};
(1.2) {[5th], [2nd]} = {1,6}.
Observe that in order to avoid ⛔「6th|5th|2nd|0th → 3」, case (1.1) does not hold. Therefore, case (1.2) holds. Two possibilities follow:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│ 5▲│4th│3rd│ 2▲│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
(1.3) │ │ 1 │ │ │ 6 │ 2 │ 0 │
├───┼───┼───┼───┼───┼───┼───┤
(1.4) │ │ 6 │ │ │ 1 │ 2 │ 0 │
└───┴───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ │ 3 │ │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
However, this is a contradiction because if case (1.3) holds, then we cannot match ✅「⟨⋯ 6 ⋯ 4 ⋯⟩」, while if case (1.4) holds, then we cannot match ✅「4th → a, 2nd → b, ab=2+4n」.
------------------------------
We have verified our claim in (1). Accordingly, we get
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│ 2■│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ │ │ │ │ │ 0 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ │ │ │ 2 │ │ 0 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ │ 6 │ 3 │ │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Then, it follows from ✅「5th → a, 2nd → b, a+b=1+6n」 that [5th] = 5:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│ 5■│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ │ │ │ 2 │ │ 0 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 3 │ │ 5 │ │ │ 2 │ │ 0 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
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--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ │ 6 │ 3 │ │ 4 │ │
└───┴───┴───┴───┴───┴───┴───┘
Let S := {p4, p3, p1}. Observe that by ⛔「6th|5th|2nd|0th → 3」 and ✅「⟨⋯ 6 ⋯ 4 ⋯⟩」, we have 3 ∈ S and 4 ∈ S. So, to avoid ⛔「{p4, p3, p1} = ? + {0,2,3}」, we need 1 ∉ S. It follows that 1 = [6th]:
┌───┬───┬───┬───┬───┬───┬───┐
│ 6■│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ 5 │ │ │ 2 │ │ 0 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 4 │ 1 │ 5 │ │ │ 2 │ │ 0 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ 6 │ 3 │ │ 4 │ │
└───┴───┴───┴───┴───┴───┴───┘
Then, note that there is only one way to match ✅「4th → a, 2nd → b, ab=2+4n」:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│ 4■│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 1 │ 5 │ │ │ 2 │ │ 0 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 5 │ 1 │ 5 │ 3 │ │ 2 │ │ 0 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ 6 │ │ │ 4 │ │
└───┴───┴───┴───┴───┴───┴───┘
Finally, in view of ✅「⟨⋯ 6 ⋯ 4 ⋯⟩」, we finish by
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│ 3■│2nd│ 1■│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 1 │ 5 │ 3 │ │ 2 │ │ 0 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 6 │ 1 │ 5 │ 3 │ 6 │ 2 │ │ 0 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 7 │ 1 │ 5 │ 3 │ 6 │ 2 │ 4 │ 0 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.6