2023-12-16 Q1(m=6)

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

⟨6th 5th 4th 3rd 2nd 1st 0th⟩

✅Match

⟨     ⁴ᵗʰa     ¹ˢᵗb   ⟩, (ab)₁₀ ≥ 52

⟨⋯ ? ⋯ 4 ⋯ (?+1)⟩ (?≠4,3)

⟨⋯ ? ⋯ 5 ⋯ (?−1)⟩ (?≠5,6)

6th → a, 2nd → b, a+b=6

⛔Avoid

6th|3rd|2nd|1st|0th → 1

#125034_v2.2

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

Proof of 2023-12-16 Q1(m=6)
═══════════════════════════

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

Note that by ✅「⟨     ⁴ᵗʰa     ¹ˢᵗb   ⟩, (ab)₁₀ ≥ 52」, we have

(1) [4th] = 5|6.

We consider where to place 1. By ⛔「6th|3rd|2nd|1st|0th → 1」, we have 1 = [5th] or [4th]. To match (1), we have 1 != [4th]. Therefore

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

--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│   │ 2 │ 6 │ 3 │ 0 │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘

Next, we determine the value of [0th]. As 1 has been used, it follows from ✅「⟨⋯ ? ⋯ 4 ⋯ (?+1)⟩ (?≠4,3)」 and ✅「⟨⋯ ? ⋯ 5 ⋯ (?−1)⟩ (?≠5,6)」 that [0th] = 2 or 3.

(2) We proceed to show that [0th] = 3.

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

If on the contrary [0th] = 2:

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

--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│   │   │ 6 │ 3 │ 0 │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘

then to match ✅「6th → a, 2nd → b, a+b=6」, we have

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

where {X,Y} = {0,6}.

It implies [4th] != 6, so to match (1), we have [4th] = 5:

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

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

But then we would never match ✅「⟨⋯ ? ⋯ 5 ⋯ (?−1)⟩ (?≠5,6)」, which is a contradiction.

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

We have verified (2). Accordingly, we have

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

--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│   │ 2 │ 6 │   │ 0 │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘

Now ✅「⟨⋯ ? ⋯ 4 ⋯ (?+1)⟩ (?≠4,3)」 and ✅「⟨⋯ ? ⋯ 5 ⋯ (?−1)⟩ (?≠5,6)」 becomes

⟨⋯ 2 ⋯ 4 ⋯ 3⟩, and
⟨⋯ 4 ⋯ 5 ⋯ 3⟩;

so we have to match the following pattern:

(3) ⟨⋯ 2 ⋯ 4 ⋯ 5 ⋯ 3⟩.

A fortiori, 5 != [4th]. Hence, by (1), we have [4th] = 6.

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

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

To match ✅「6th → a, 2nd → b, a+b=6」, we need {[6th], [2nd]} = {2,4}. Combining this with (3), we have

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

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

Finally, there is only one way to match (3). We finish by

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

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

Q.E.D.

#125034_v2.2