2024-04-09 Q1(m=6)

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

⟨6th 5th 4th 3rd 2nd 1st 0th⟩

✅Match

⟨? 6 ⋯ (?−2) ⋯ 1 ⋯⟩ (?≠6)

⟨ ⁶ᵗʰb ⁵ᵗʰa       ¹ˢᵗc   ⟩, a > b > c

⟨Perm(1,3,5,6) Perm(0,2,4)⟩

2nd → 1|2|4

⛔Avoid

max ⊢1⊣ ≤ 4

3rd → 4|5|6

3rd → a, 0th → b, ab=2+4n

⟨⋯ 3 ⋯ ? ⋯ 2 (?−6)⟩

#125034_v2.4

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

Proof of 2024-04-09 Q1(m=6)
═══════════════════════════

Notation: if nth -> a, then we write [nth] = a.

Plainly, our first step follows from ✅「⟨? 6 ⋯ (?−2) ⋯ 1 ⋯⟩ (?≠6)」:

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

--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │   │ 3 │ 0 │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘

Next, we determine the value of [3rd]. By ✅「⟨Perm(1,3,5,6) Perm(0,2,4)⟩」 and ⛔「3rd → 4|5|6」, we see that

(1) [3rd] = 1|3. We claim that [3rd] = 1.

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

For, suppose on the contrary [3rd] = 3:

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

Then, by ✅「⟨Perm(1,3,5,6) Perm(0,2,4)⟩」, we have [6th] = 1|5. As ✅「⟨? 6 ⋯ (?−2) ⋯ 1 ⋯⟩ (?≠6)」 implies [6th] != 1, we get [6th] = 5:

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

But then we cannot match ✅「⟨Perm(1,3,5,6) Perm(0,2,4)⟩」 and ✅「⟨? 6 ⋯ (?−2) ⋯ 1 ⋯⟩ (?≠6)」 at the same time, which is a contradiction.

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

Our claim in (1) is thus verified. Accordingly, we get

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

--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│   │ 2 │   │ 3 │ 0 │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘

Next, observe that there is only one way to match ✅「⟨Perm(1,3,5,6) Perm(0,2,4)⟩」 and avoid ⛔「max ⊢1⊣ ≤ 4」 at the same time. That is

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

--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│   │ 2 │   │   │ 0 │ 4 │   │
└───┴───┴───┴───┴───┴───┴───┘

We proceed to determine the value of [1st]. By ✅「⟨ ⁶ᵗʰb ⁵ᵗʰa       ¹ˢᵗc   ⟩, a > b > c」, we have

(2)  [1st] = 2|0. We claim that [1st] = 0 actually.

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

For, if on the contrary [1st] = 2:

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

--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│   │   │   │   │ 0 │ 4 │   │
└───┴───┴───┴───┴───┴───┴───┘

Then, to match ✅「2nd → 1|2|4」, we reach

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

But then we match ⛔「⟨⋯ 3 ⋯ ? ⋯ 2 (?−6)⟩」, which is a contradiction.

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

We have verified our claim in (2). As a result, we get

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

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

Finally, noting that to avoid ⛔「3rd → a, 0th → b, ab=2+4n」 we cannot have [0th] = 2, we finish by

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

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

Q.E.D.

#125034_v2.4