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2025-06-03 WR

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

⟨5th 4th 3rd 2nd 1st 0th⟩

✅Match
{p4, p2, p0} = ? + {0,1,2}
⟦1,4⟧ ∋ 2
⟨     ³ʳᵈb   ¹ˢᵗa   ⟩, a > b
{p5, p2, p1} = ? + {0,1,3}

#125034_v2.9


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

Proof of 2025-06-03 WR
══════════════════════

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

To match ✅「{p4, p2, p0} = ? + {0,1,2}」, we need

(1) {p4, p2, p0} consists of consecutive integers.

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

On the other hand, by ✅「{p5, p2, p1} = ? + {0,1,3}」, there are three possibilities for {p5, p2, p1}:

(A) {0,1,3};

(B) {1,2,4};

(C) {2,3,5}.

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

(2) We show by contradiction that (C) holds actually.

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

(2.1) Suppose (A) holds:

┌───┬───┬───┬───┬───┬───┐
│5th│ 4▲│3rd│ 2▲│1st│ 0▲│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│013│   │   │013│013│   │
└───┴───┴───┴───┴───┴───┘

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

We consider what [1st] is. To match ✅「⟨     ³ʳᵈb   ¹ˢᵗa   ⟩, a > b」, we need [1st] > [3rd]. Since [3rd] ∈ {2,4,5}, we have [3rd] >= 2. Therefore, [1st] = 3:

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

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

However, it follows that no matter whether [2nd] = 0 or 1, we cannot match (1). This shows a contradiction.

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

(2.2) On the other hand, suppose (B) holds:

┌───┬───┬───┬───┬───┬───┐
│5th│ 4▲│3rd│ 2▲│1st│ 0▲│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│124│   │   │124│124│   │
└───┴───┴───┴───┴───┴───┘

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

In view of ✅「⟦1,4⟧ ∋ 2」, we have [2nd] = 2:

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

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

However, observe that we again cannot match (1).

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

By (2.1) and (2.2), we have verified (2). Accordingly, we get

┌───┬───┬───┬───┬───┬───┐
│5th│ 4▲│3rd│ 2▲│1st│ 0▲│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│235│   │   │235│235│   │
└───┴───┴───┴───┴───┴───┘

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

We consider what [2nd] is. Observe that if it is 5 or 3, then we still cannot match (1). To match (1), there is only one possibility: [2nd] = 2, and {[4th], [0th]} = {0,1}:

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

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

Plainly, it follows that [3rd] = 4:

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

Then, to match ✅「⟨     ³ʳᵈb   ¹ˢᵗa   ⟩, a > b」, we need [1st] = 5, and [5th] = 3 follows:

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

Finally, to match ✅「⟦1,4⟧ ∋ 2」, we finish by

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

Q.E.D.

#125034_v2.9