2026-03-03 WR

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

⟨ ⁵ᵗʰ▨ ⁴ᵗʰ▨ ³ʳᵈ▨ ²ⁿᵈ▨ ¹ˢᵗ▨ ⁰ᵗʰ▨ ⟩

✅Match

⟨ ⁵ᵗʰ↓ ⁴ᵗʰ↓ ³ʳᵈ↑ ²ⁿᵈ↓ ¹ˢᵗ↑ ⁰ᵗʰ↑ ⟩ after 5−⟨⋯⟩

{p5, p3, p1} = ? + {0,1,3}

⛔Avoid

min ⊢5⊣ ≥ 1

1st → a, 0th → b, a+b=3+4n

#125034_v2.12

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

Proof of 2026-03-03 WR
══════════════════════

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

By ✅「⟨ ⁵ᵗʰ↓ ⁴ᵗʰ↓ ³ʳᵈ↑ ²ⁿᵈ↓ ¹ˢᵗ↑ ⁰ᵗʰ↑ ⟩ after 5−⟨⋯⟩」, we have

(1.1) {[3rd], [1st], [0th]} = {0,1,2}; and

(1.2) {[5th], [4th], [2nd]} = {3,4,5}.

Note that (1.1) implies both [1st] and [0th] are in {0,1,2}:

┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│   │   │   │   │012│012│
└───┴───┴───┴───┴───┴───┘

If {[1st], [0th]} = {1,2}, then we shall fail to avoid ⛔「1st → a, 0th → b, a+b=3+4n」. Therefore, we need

(2) 0 = [1st] or [0th].

On the other hand, observe that ⛔「min ⊢5⊣ ≥ 1」 implies 5 and 0 are adjacent. Combining this with (2) and (1.2), we get

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

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

It then follows from ✅「{p5, p3, p1} = ? + {0,1,3}」 that {[5th], [3rd]} = {1,3}. Using (1.1) and (1.2), we get

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

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

Finally, using (1.1) or (1.2) once more, we finish by

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

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

Q.E.D.

#125034_v2.12