2024-11-12 Q1(m=6)

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