2025-12-23 WR

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

⟨ ⁵ᵗʰ▨ ⁴ᵗʰ▨ ³ʳᵈ▨ ²ⁿᵈ▨ ¹ˢᵗ▨ ⁰ᵗʰ▨ ⟩

✅Match

min {p5, p1, p0} = 1

{p5, p4, p1} = ? + {0,1,2}

{p5, p4, p2} = ? + {0,1,2}

⛔Avoid

{p4, p2, p1} = ? + {0,2,3}

⟨⋯ 2 ⋯ 5 ⋯⟩

#125034_v2.11

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

Proof of 2025-12-23 WR
══════════════════════

Notation: if Nth -> a, then we write pN = a.

By ✅「{p5, p4, p1} = ? + {0,1,2}」, the following are consecutive integers:

• p5, p4, p1.

Similarly, by ✅「{p5, p4, p2} = ? + {0,1,2}」, the following are also consecutive integers:

• p5, p4, p2.

Combining them, we see that

(1) p5, p4, p2, p1 are consecutive integers. Their minimum and maximum values are in {p2, p1}.

Then, there are two cases: whether p5>p4 or p4>p5. If it is p4>p5, then in view of (1), we will match ⛔「{p4, p2, p1} = ? + {0,2,3}」, which is a contradiction. Therefore, we have

(2) p5 > p4.

Now we consider what consecutive integers p5, p4, p2, p1 are. There are three cases:

• {0,1,2,3};
• {1,2,3,4};
• {2,3,4,5}.

Combining (1) with (2), they become

(3.1) {p1,p2} = {0,3} and p4 = 1 and p5 = 2;

(3.2) {p1,p2} = {1,4} and p4 = 2 and p5 = 3;

(3.3) {p1,p2} = {2,5} and p4 = 3 and p5 = 4.

If case (3.1) holds, then to match ✅「min {p5, p1, p0} = 1」 we cannot have p1=0, so we have

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

Since 1∉{p5,p1,p0}, we fail to match ✅「min {p5, p1, p0} = 1」.

Else if case (3.2) holds, then we have

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

and we fail to avoid ⛔「⟨⋯ 2 ⋯ 5 ⋯⟩」.

Consequently, case (3.3) holds. We get

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

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

and {p2,p1} = {2,5}. A fortiori, {p3,p0} = {0,1}. To match ✅「min {p5, p1, p0} = 1」, 0 cannot be p0. Therefore, we have

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

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

Finally, to avoid ⛔「⟨⋯ 2 ⋯ 5 ⋯⟩」, we finish by

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

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

Q.E.D.

#125034_v2.11