2025-01-07 WR

Rearrange the digits in ⟨125034⟩ to meet the rules below.

⟨5th 4th 3rd 2nd 1st 0th⟩

✅Match

⟨⋯ 2 ⋯ 3 ⋯ 5 ⋯⟩

4th|0th → 3

3rd → 0|1|2

⛔Avoid

⟨         ¹ˢᵗa ⁰ᵗʰb ⟩, (ab)₁₀ ≤ 30

1st → 1|4

⟨⋯ 3 ⋯ 0 ⋯ 4 ⋯⟩

#125034_v2.8

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

Proof of 2025-01-07 WR
══════════════════════

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

We consider where to place 3. Since ✅「⟨⋯ 2 ⋯ 3 ⋯ 5 ⋯⟩」 implies that there is at least one digit to the right of 3, we have 3 != [0th]. Therefore, it follows from ✅「4th|0th → 3」 that 3 = [4th]:

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

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

Then, by ✅「⟨⋯ 2 ⋯ 3 ⋯ 5 ⋯⟩」, we get [5th] = 2:

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

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

Then, we consider what [3rd] is. By ✅「3rd → 0|1|2」, it is 0 or 1. If it is 0, then we would match ⛔「⟨⋯ 3 ⋯ 0 ⋯ 4 ⋯⟩」. Therefore, it is not 0 but 1: 

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

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

Next, we consider what [1st] is. To avoid ⛔「⟨         ¹ˢᵗa ⁰ᵗʰb ⟩, (ab)₁₀ ≤ 30」, we need [1st] = 4|5. Combining this with ⛔「1st → 1|4」, we get [1st] = 5:

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

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

Finally, to avoid ⛔「⟨⋯ 3 ⋯ 0 ⋯ 4 ⋯⟩」, we finish by

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

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

Q.E.D.

#125034_v2.8