2023-11-14 Q1(m=6)

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

⟨6th 5th 4th 3rd 2nd 1st 0th⟩

✅Match

0th → 0|1|2|3

⟨? 2 ⋯ (?+3) ⋯ 6 ⋯⟩ (?≠2)

2nd → a, 0th → b, ab=12

⛔Avoid

⟨⋯ 2 ⋯ 5 ⋯ 0 ⋯ 3 ⋯⟩

#125034_v2.1

       ┌───┬───┬───┬───┬───┬───┬───┐
       │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 │   │   │ 4 │ 6 │ 3 │
       ├───┼───┼───┼───┼───┼───┼───┤
Step 6 │ 1 │ 2 │ 0 │   │ 4 │ 6 │ 3 │
       ├───┼───┼───┼───┼───┼───┼───┤
Step 7 │ 1 │ 2 │ 0 │ 5 │ 4 │ 6 │ 3 │
       └───┴───┴───┴───┴───┴───┴───┘

Proof of 2023-11-14 Q1(m=6)
═══════════════════════════

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

Plainly, by ✅【⟨? 2 ⋯ (?+3) ⋯ 6 ⋯⟩ (?≠2)】 and ✅【2nd → a, 0th → b, ab=12】, we have

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

--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │   │ 6 │ 3 │ 0 │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘

where ab = 12. It follows that {a,b} = {3,4}. In view of ✅【0th → 0|1|2|3】, we have b = 3 and thus a = 4.

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

--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │   │ 6 │   │ 0 │   │ 5 │
└───┴───┴───┴───┴───┴───┴───┘

Now we consider the value of ? := [6th] in ✅【⟨? 2 ⋯ (?+3) ⋯ 6 ⋯⟩ (?≠2)】. To have (?+3) <= 6, we need ? <= 3, so ? = 0 or 1. We cannot have ? = 0, for otherwise we need to match

⟨02 ⋯ 3 ⋯ 6 ⋯⟩,

which is not possible because we have shown that [0th] = 3.

Therefore, we have ? = 1 and ✅【⟨? 2 ⋯ (?+3) ⋯ 6 ⋯⟩ (?≠2)】 becomes

⟨12 ⋯ 4 ⋯ 6 ⋯⟩.

As a result, we get

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

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

Finally, to avoid ⛔「⟨⋯ 2 ⋯ 5 ⋯ 0 ⋯ 3 ⋯⟩」, we finish by

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

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

Q.E.D.

#125034_v2.1