WR 2023-10-17

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

⟨5th 4th 3rd 2nd 1st 0th⟩

✅Match

⟦4,5⟧ ∋ 0,3

⟨? ⋯ 3 ⋯ (?−2) ⋯⟩ (?≠3,5)

4th → a, 0th → b, |a-b|=2

3rd → a, 0th → b, |a-b|=1

⛔Avoid

⟨⋯ Perm(0,3,4) ⋯⟩

3rd → a, 2nd → b, a+b=2+5n

#125034_v2.0

WR 2023-10-17

✅⚪⚪⚪⚪⚪

✅⚪⚪⚪⚪✅

✅✅⚪⚪⚪✅

✅✅✅⚪⚪✅

✅✅✅✅⚪✅

✅✅✅✅✅✅

#125034_v2.0

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

Proof of 2023-10-17 WR
══════════════════════

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

We start with ✅【⟨? ⋯ 3 ⋯ (?−2) ⋯⟩ (?≠3,5)】. We need ?-2≥0, so ?=[5th] can only be 4 or 2.

We claim that [5th] = 4. Suppose on the contrary that [5th]=2. Then to match ✅【⟦4,5⟧ ∋ 0,3】, there are three possibilities:

    ┌───┬───┬───┬───┬───┬───┐
    │*5 │ 4 │ 3 │ 2 │ 1 │ 0 │
    ╞═══╪═══╪═══╪═══╪═══╪═══╡
(1) │ 2 │ [ │ ● │ ● │ ] │   │
    ├───┼───┼───┼───┼───┼───┤
(2) │ 2 │   │ [ │ ● │ ● │ ] │
    ├───┼───┼───┼───┼───┼───┤
(3) │ 2 │ [ │   │   │   │ ] │
    └───┴───┴───┴───┴───┴───┘

where the brackets indicate 4,5 and the dots are 0,3. But case (1) and (2) would match ⛔「⟨⋯ Perm(0,3,4) ⋯⟩」, while case (3) would not match ✅【4th → a, 0th → b, |a-b|=2】. This shows contradictions and verifies our claim. So, we have

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

--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ 5 │ 0 │ 3 │   │
└───┴───┴───┴───┴───┴───┘

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

Next, by ✅【⟦4,5⟧ ∋ 0,3】, we see that 0,3 cannot be at the boundary, so [0th]=1|2|5. Note that by ✅【3rd → a, 0th → b, |a-b|=1】, [0th] cannot be 5.

(4) We show by contradiction that [0th]=2 . If on the contrary [0th]=1, then ✅【4th → a, 0th → b, |a-b|=2】 gives

┌───┬───┬───┬───┬───┬───┐
│ 5 │ 4▲│*3 │ 2 │ 1 │ 0▲│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 4 │ 3 │ x │   │   │ 1 │
└───┴───┴───┴───┴───┴───┘

with x=0|2 by ✅【3rd → a, 0th → b, |a-b|=1】. As x≠0 by ⛔「⟨⋯ Perm(0,3,4) ⋯⟩」, we reach

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

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

It then unavoidably matches ⛔「3rd → a, 2nd → b, a+b=2+5n」, which is a contradiction. So, back to (4), we have

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

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

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

Now, using ✅【4th → a, 0th → b, |a-b|=2】, we get [4th]=0, and then using ✅【3rd → a, 0th → b, |a-b|=1】 with ⛔「⟨⋯ Perm(0,3,4) ⋯⟩」 in mind, we have [3rd]=1:

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

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

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

Finally, to match ✅【⟦4,5⟧ ∋ 0,3】, we finish by

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

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

Q.E.D.

#125034_v2.0