2025-01-28 WR

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

⟨5th 4th 3rd 2nd 1st 0th⟩

✅Match

5th → a, 2nd → b, a+b=2+5n

⟨     ³ʳᵈa ²ⁿᵈc   ⁰ᵗʰb ⟩, a > b > c

⟨? ⋯ 4 ⋯ (?+2) ⋯⟩ (?≠2)

⛔Avoid

⟨⋯ Perm(0,1) ⋯⟩

⟦3,4⟧ ∋ 1,2,5

#125034_v2.8

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

Proof of 2025-01-28 WR
══════════════════════

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

First, we consider which digit is placed to the left corner. By ✅「⟨? ⋯ 4 ⋯ (?+2) ⋯⟩ (?≠2)」, we have [5th] = 0|1. If it is 1, then we have no way to match ✅「5th → a, 2nd → b, a+b=2+5n」. Therefore, it is 0:

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

--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ 5 │   │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘

Then, ✅「5th → a, 2nd → b, a+b=2+5n」 gives [2nd] = 2:

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

--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │   │ 5 │   │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘

Next, we consider the value of [0th]. Given that [2nd] = 2, by ✅「⟨     ³ʳᵈa ²ⁿᵈc   ⁰ᵗʰb ⟩, a > b > c」 we have [0th] = 3|4. As 4 is not in the right corner by ✅「⟨? ⋯ 4 ⋯ (?+2) ⋯⟩ (?≠2)」, we get [0th] = 3:

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

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

We consider how to place 4. By ✅「⟨? ⋯ 4 ⋯ (?+2) ⋯⟩ (?≠2)」, we have 4 = [4th] | [3rd]. If it is at 4th, then we cannot avoid ⛔「⟦3,4⟧ ∋ 1,2,5」. Therefore, it is at 3rd:

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

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

Finally, to avoid ⛔「⟨⋯ Perm(0,1) ⋯⟩」, we finish by

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

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

Q.E.D.

#125034_v2.8