Rearrange the digits in ⟨1263045⟩ to meet the rules below.
⟨6th 5th 4th 3rd 2nd 1st 0th⟩
✅Match
⟨ ⁶ᵗʰd ⁵ᵗʰa ⁴ᵗʰc ²ⁿᵈb ⟩, a > b > c > d
⟦1,4⟧ ∋ 2,5
4th → 2|6
3rd → a, 1st → b, ab=3+4n
#125034_v2.7
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ │ 2 │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ │ 2 │ │ 4 │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 3 │ │ │ 2 │ │ 4 │ 3 │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 4 │ 1 │ │ 2 │ │ 4 │ 3 │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 5 │ 1 │ │ 2 │ 5 │ 4 │ 3 │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 6 │ 1 │ 6 │ 2 │ 5 │ 4 │ 3 │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 7 │ 1 │ 6 │ 2 │ 5 │ 4 │ 3 │ 0 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
Proof of 2024-11-12 Q1(m=6)
═══════════════════════════
Notation: if nth -> a, then we write [nth] = a.
Firstly, by combining ✅「4th → 2|6」 with ✅「⟨ ⁶ᵗʰd ⁵ᵗʰa ⁴ᵗʰc ²ⁿᵈb ⟩, a > b > c > d」, we get
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│ 4■│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ │ 2 │ │ │ │ │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ │ 6 │ 3 │ 0 │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Then, observe that ✅「3rd → a, 1st → b, ab=3+4n」 implies
(1) {[3rd], [1st]} = {1,3} | {3,5}.
A fortiori
(2) 3 ∈ {[3rd], [1st]}.
On the other hand, to match ✅「⟨ ⁶ᵗʰd ⁵ᵗʰa ⁴ᵗʰc ²ⁿᵈb ⟩, a > b > c > d」, we need
(3) [2nd] = 3|4|5.
In view of (2), it is [2nd] = 4|5 actually.
(4) Below we show that [2nd] = 4.
------------------------------
If on the contrary [2nd] = 5:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│ 2▲│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ │ │ 2 │ │ 5 │ │ │
└───┴───┴───┴───┴───┴───┴───┘
then ✅「⟨ ⁶ᵗʰd ⁵ᵗʰa ⁴ᵗʰc ²ⁿᵈb ⟩, a > b > c > d」 implies [5th] = 6 and [6th] = 0|1:
┌───┬───┬───┬───┬───┬───┬───┐
│ 6▲│ 5▲│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ │ 6 │ 2 │ │ 5 │ │ │
└───┴───┴───┴───┴───┴───┴───┘
However, this is a contradiction. For, if [6th] = 0, then we cannot match ✅「⟦1,4⟧ ∋ 2,5」, else if [6th] = 1, then we cannot match (1).
------------------------------
We have verified in (4). Accordingly, we get
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│ 2■│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ │ 2 │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ │ 2 │ │ 4 │ │ │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ │ 6 │ 3 │ 0 │ │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Then, to match ✅「⟦1,4⟧ ∋ 2,5」, we need
(5) 1 and 5 are to the left of 4.
Combining (5) with (1), we get [1st] = 3:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│ 1■│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ │ 2 │ │ 4 │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 3 │ │ │ 2 │ │ 4 │ 3 │ │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ │ 6 │ │ 0 │ │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Next, by ✅「⟨ ⁶ᵗʰd ⁵ᵗʰa ⁴ᵗʰc ²ⁿᵈb ⟩, a > b > c > d」, we have [6th] = 0|1.
(6) We show that [6th] = 1 actually.
------------------------------
If instead [6th] = 0, then to match ✅「⟦1,4⟧ ∋ 2,5」 we need [5th] = 1:
┌───┬───┬───┬───┬───┬───┬───┐
│ 6▲│ 5▲│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 0 │ │ 2 │ │ 4 │ 3 │ │
├───┼───┼───┼───┼───┼───┼───┤
│ 0 │ 1 │ 2 │ │ 4 │ 3 │ │
└───┴───┴───┴───┴───┴───┴───┘
But then we cannot match ✅「⟨ ⁶ᵗʰd ⁵ᵗʰa ⁴ᵗʰc ²ⁿᵈb ⟩, a > b > c > d」, which is a contradiction.
------------------------------
It follows from (6) that
┌───┬───┬───┬───┬───┬───┬───┐
│ 6■│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ │ 2 │ │ 4 │ 3 │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 4 │ 1 │ │ 2 │ │ 4 │ 3 │ │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ 6 │ │ 0 │ │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Then, (1) gives
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│ 3■│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 1 │ │ 2 │ │ 4 │ 3 │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 5 │ 1 │ │ 2 │ 5 │ 4 │ 3 │ │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ 6 │ │ 0 │ │ │
└───┴───┴───┴───┴───┴───┴───┘
Finally, using ✅「⟨ ⁶ᵗʰd ⁵ᵗʰa ⁴ᵗʰc ²ⁿᵈb ⟩, a > b > c > d」, we finish by
┌───┬───┬───┬───┬───┬───┬───┐
│6th│ 5■│4th│3rd│2nd│1st│ 0■│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 1 │ │ 2 │ 5 │ 4 │ 3 │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
│ 1 │ 6 │ 2 │ 5 │ 4 │ 3 │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 7 │ 1 │ 6 │ 2 │ 5 │ 4 │ 3 │ 0 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.7