2024-10-22 Q1(m=6)

Rearrange the digits in ⟨1263045⟩ to meet the rules below.

⟨6th 5th 4th 3rd 2nd 1st 0th⟩

✅Match

⟨⋯ 1 ⋯ ? 0 ⋯ (?+3)⟩ (?≠0)

6th → a, 3rd → b, ab=0+4n

median {p5, p1, p0} = 1

⟨ ⁶ᵗʰb ⁵ᵗʰc ⁴ᵗʰa         ⟩, a > b > c

⛔Avoid

⟨⋯ a ⋯ 1 ⋯⟩, a = 3|4|6

#125034_v2.7

       ┌───┬───┬───┬───┬───┬───┬───┐
       │6th│5th│4th│3rd│2nd│1st│0th│▒
       ╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │   │   │   │   │   │ 0 │   │▒
       ├───┼───┼───┼───┼───┼───┼───┤▒
Step 2 │   │ 1 │   │   │   │ 0 │   │▒
       ├───┼───┼───┼───┼───┼───┼───┤▒
Step 3 │ 2 │ 1 │   │   │   │ 0 │   │▒
       ├───┼───┼───┼───┼───┼───┼───┤▒
Step 4 │ 2 │ 1 │   │   │ 3 │ 0 │   │▒
       ├───┼───┼───┼───┼───┼───┼───┤▒
Step 5 │ 2 │ 1 │   │   │ 3 │ 0 │ 6 │▒
       ├───┼───┼───┼───┼───┼───┼───┤▒
Step 6 │ 2 │ 1 │   │ 4 │ 3 │ 0 │ 6 │▒
       ├───┼───┼───┼───┼───┼───┼───┤▒
Step 7 │ 2 │ 1 │ 5 │ 4 │ 3 │ 0 │ 6 │▒
       └───┴───┴───┴───┴───┴───┴───┘▒
        ▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒

Proof of 2024-10-22 Q1(m=6)
═══════════════════════════

Notation: if Nth -> a, then we write pN = a.

Note that ✅「median {p5, p1, p0} = 1」 implies

(1) 0,1 ∈ {p5, p1, p0}.

A fortiori, 0 ∈ {p5, p1, p0}. Let us consider where to place it.

┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│   │ ▬ │   │   │   │ ▬ │ ▬ │
└───┴───┴───┴───┴───┴───┴───┘

By ✅「⟨⋯ 1 ⋯ ? 0 ⋯ (?+3)⟩ (?≠0)」, 0 is not at the right corner; and if 0 = p5, then 1 = p6:

┌───┬───┬───┬───┬───┬───┬───┐
│ 6▲│ 5▲│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 1 │ 0 │   │   │   │   │   │
└───┴───┴───┴───┴───┴───┴───┘

But this contradicts (1). Therefore, 0 = p1:

       ┌───┬───┬───┬───┬───┬───┬───┐
       │6th│5th│4th│3rd│2nd│ 1■│0th│▒
       ╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │   │   │   │   │   │ 0 │   │▒
       └───┴───┴───┴───┴───┴───┴───┘▒
        ▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒

--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ 6 │ 3 │   │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘

Combining (1) and ✅「⟨⋯ 1 ⋯ ? 0 ⋯ (?+3)⟩ (?≠0)」, we get 1 = p5 as well:

       ┌───┬───┬───┬───┬───┬───┬───┐
       │6th│ 5■│4th│3rd│2nd│1st│0th│▒
       ╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
       │   │   │   │   │   │ 0 │   │▒
       ├───┼───┼───┼───┼───┼───┼───┤▒
Step 2 │   │ 1 │   │   │   │ 0 │   │▒
       └───┴───┴───┴───┴───┴───┴───┘▒
        ▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒

--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│   │ 2 │ 6 │ 3 │   │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘

Then, by ⛔「⟨⋯ a ⋯ 1 ⋯⟩, a = 3|4|6」, we have

(2) p6 ∈ {2,5}.

┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ ▬ │ 1 │   │   │   │ 0 │   │
└───┴───┴───┴───┴───┴───┴───┘

(2.1) We claim that p6 = 2 actually.

------------------------------

For, if on the contrary p6 = 5:

┌───┬───┬───┬───┬───┬───┬───┐
│ 6▲│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 5 │ 1 │   │   │   │ 0 │   │
└───┴───┴───┴───┴───┴───┴───┘

--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│   │ 2 │ 6 │ 3 │   │ 4 │   │
└───┴───┴───┴───┴───┴───┴───┘

then to match ✅「⟨ ⁶ᵗʰb ⁵ᵗʰc ⁴ᵗʰa         ⟩, a > b > c」, we need p4 = 6:

┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│ 4▲│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 5 │ 1 │ 6 │   │   │ 0 │   │
└───┴───┴───┴───┴───┴───┴───┘

--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│   │ 2 │   │ 3 │   │ 4 │   │
└───┴───┴───┴───┴───┴───┴───┘

But then we cannot match ✅「⟨⋯ 1 ⋯ ? 0 ⋯ (?+3)⟩ (?≠0)」, which is a contradiction.

------------------------------

We have verified our claim in (2.1). Accordingly, we get

       ┌───┬───┬───┬───┬───┬───┬───┐
       │ 6■│5th│4th│3rd│2nd│1st│0th│▒
       ╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
       │   │ 1 │   │   │   │ 0 │   │▒
       ├───┼───┼───┼───┼───┼───┼───┤▒
Step 3 │ 2 │ 1 │   │   │   │ 0 │   │▒
       └───┴───┴───┴───┴───┴───┴───┘▒
        ▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒

--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│   │   │ 6 │ 3 │   │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘

Next, we consider the value of p0. In view of ✅「⟨⋯ 1 ⋯ ? 0 ⋯ (?+3)⟩ (?≠0)」, we get p0 = 6 and the "?" in that pattern is 3. It follows

       ┌───┬───┬───┬───┬───┬───┬───┐
       │6th│5th│4th│3rd│ 2■│1st│ 0■│▒
       ╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
       │ 2 │ 1 │   │   │   │ 0 │   │▒
       ├───┼───┼───┼───┼───┼───┼───┤▒
Step 4 │ 2 │ 1 │   │   │ 3 │ 0 │   │▒
       ├───┼───┼───┼───┼───┼───┼───┤▒
Step 5 │ 2 │ 1 │   │   │ 3 │ 0 │ 6 │▒
       └───┴───┴───┴───┴───┴───┴───┘▒
        ▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒

--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│   │   │   │   │   │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘

Finally, to match ✅「6th → a, 3rd → b, ab=0+4n」, we finish by

       ┌───┬───┬───┬───┬───┬───┬───┐
       │6th│5th│ 4■│ 3■│2nd│1st│0th│▒
       ╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
       │ 2 │ 1 │   │   │ 3 │ 0 │ 6 │▒
       ├───┼───┼───┼───┼───┼───┼───┤▒
Step 6 │ 2 │ 1 │   │ 4 │ 3 │ 0 │ 6 │▒
       ├───┼───┼───┼───┼───┼───┼───┤▒
Step 7 │ 2 │ 1 │ 5 │ 4 │ 3 │ 0 │ 6 │▒
       └───┴───┴───┴───┴───┴───┴───┘▒
        ▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒

--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│   │   │   │   │   │   │   │
└───┴───┴───┴───┴───┴───┴───┘

Q.E.D.

#125034_v2.7