2024-07-26 WR

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

⟨5th 4th 3rd 2nd 1st 0th⟩

✅Match

1st → 1|5

⟨   ⁴ᵗʰ↑ ³ʳᵈ↑       ⟩ after 2×⟨←⟩

5th → a, 2nd → b, a+b=0+4n

⟨⋯ 4 ⋯ a ⋯⟩, a = 1|2|5

{p5, p3, p0} = ? + {0,1,2}

⛔Avoid

⟨ ⁵ᵗʰa     ²ⁿᵈb ¹ˢᵗc   ⟩, (abc)₁₀ ≤ 231

#125034_v2.6

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

Proof of 2024-07-26 WR
══════════════════════

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

(1) We first show that [1st] = 5.

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

If this is not the case, then by ✅「1st → 1|5」, we have [1st] = 1. It then follows from ✅「⟨   ⁴ᵗʰ↑ ³ʳᵈ↑       ⟩ after 2×⟨←⟩」 that [3rd] = 0:

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

Combining this with ✅「{p5, p3, p0} = ? + {0,1,2}」, we have { [5th], [3rd], [0th] } = {0,1,2}:

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

A fortiori, 1 ∈ { [5th], [3rd], [0th] }. This is a contradiction, however, as we have already placed 1 at 1st.

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

We have verified (1). Accordingly, we get

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

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

To proceed, observe that ✅「5th → a, 2nd → b, a+b=0+4n」 implies [5th] + [2nd] = 4. We also need [5th] >= 2, in view of ⛔「⟨ ⁵ᵗʰa     ²ⁿᵈb ¹ˢᵗc   ⟩, (abc)₁₀ ≤ 231」. So, there are two possibilities:

(2) ([5th], [2nd]) = (4,0);

(3) ([5th], [2nd]) = (3,1).

Note that if case (2) holds:

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

then we cannot match ✅「⟨   ⁴ᵗʰ↑ ³ʳᵈ↑       ⟩ after 2×⟨←⟩」. Therefore, case (3) holds, and we get

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

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

Then, to match ✅「⟨   ⁴ᵗʰ↑ ³ʳᵈ↑       ⟩ after 2×⟨←⟩」, we need [4th] = 0:

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

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

Finally, as ✅「⟨⋯ 4 ⋯ a ⋯⟩, a = 1|2|5」 implies 4 is not at the right corner, we finish by

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

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

Q.E.D.

#125034_v2.6