2025-01-21 WR

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

⟨5th 4th 3rd 2nd 1st 0th⟩

✅Match

5th|3rd|1st → 3

⟨⋯ ? ⋯ 3 ⋯ (?+2)⟩ (?≠3,1)

5th → a, 4th → b, a+b=2

⛔Avoid

{p4, p2, p1} = ? + {0,1,2}

⟨       ²ⁿᵈa   ⁰ᵗʰb ⟩, max⟦a,b⟧ = 5

#125034_v2.8

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

Proof of 2025-01-21 WR
══════════════════════

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

Observe that ✅「5th → a, 4th → b, a+b=2」 implies

(1) {[5th], [4th]}  = {0,2}.

On the other hand, to avoid ⛔「⟨       ²ⁿᵈa   ⁰ᵗʰb ⟩, max⟦a,b⟧ = 5」, we need

(2) 5 = [5th] | [4th] | [3rd].

Combining this with (1), we get

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

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

Then, to match ✅「5th|3rd|1st → 3」, we have 3 = [5th] | [1st]. In view of (1), we have 3 = [1st] actually:

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

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

Now, by (1) we have

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

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

where "▬" are occupied by 0,2. Let us consider the value of [0th]. By ✅「⟨⋯ ? ⋯ 3 ⋯ (?+2)⟩ (?≠3,1)」, it is at least 2. Therefore, it is 4. Accordingly, we get [2nd] = 1 as well:

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

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

Finally, to avoid ⛔「{p4, p2, p1} = ? + {0,1,2}」, we finish by

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

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

Q.E.D.

#125034_v2.8