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2025-03-18 WR

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

⟨5th 4th 3rd 2nd 1st 0th⟩

✅Match
⟨⋯ 3 ⋯ 4 ⋯ 5 ⋯⟩
max ⊢0⊣ ≤ 4
3rd → a, 0th → b, ab=0+4n

⛔Avoid
⟨⋯ ? ⋯ 1 ⋯ (?−1)⟩ (?≠1,2)
Sim⟨ ⁵ᵗʰ3 ⁴ᵗʰ5 ³ʳᵈ1 ²ⁿᵈ0 ¹ˢᵗ4 ⁰ᵗʰ2 ⟩ ≤ 1
5th → a, 1st → b, |a-b|=1

#125034_v2.8


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

Proof of 2025-03-18 WR
══════════════════════

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

Observe that by combining ✅「3rd → a, 0th → b, ab=0+4n」 with ✅「⟨⋯ 3 ⋯ 4 ⋯ 5 ⋯⟩」, we have

(1) one of the following holds:

(a) 0 = [3rd];
(b) 0 = [0th];
(c) 4 = [3rd].

(2) From this, we claim that 3 = [5th].

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

If on the contrary (2) does not hold, then by ✅「⟨⋯ 3 ⋯ 4 ⋯ 5 ⋯⟩」, there are three possible positions for 3:

      ┌───┬───┬───┬───┬───┬───┐
      │5th│ 4▲│ 3▲│ 2▲│1st│0th│
      ╞═══╪═══╪═══╪═══╪═══╪═══╡
(2.1) │   │   │   │ 3 │ 4 │ 5 │
      ├───┼───┼───┼───┼───┼───┤
(2.2) │   │   │ 3 │   │   │   │
      ├───┼───┼───┼───┼───┼───┤
(2.3) │   │ 3 │   │   │   │   │
      └───┴───┴───┴───┴───┴───┘

We show that all of them give contradictions. The following will be used frequently: To avoid ⛔「Sim⟨ ⁵ᵗʰ3 ⁴ᵗʰ5 ³ʳᵈ1 ²ⁿᵈ0 ¹ˢᵗ4 ⁰ᵗʰ2 ⟩ ≤ 1」, we need at least two agreed positional digits with ⟨351042⟩.

(3) If (2.1) holds, then by (1), we have

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

However, we will fail to have two agreed positional digits with ⟨351042⟩.

(4) Else if (2.2) holds, then by (1), we have

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

Using ✅「⟨⋯ 3 ⋯ 4 ⋯ 5 ⋯⟩」, we get

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

But again, we fail to have two agreed positional digits with ⟨351042⟩.

(5) Lastly, Suppose (2.3) holds:

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

By (1), it follows three cases:

      ┌───┬───┬───┬───┬───┬───┐
      │5th│4th│3rd│2nd│1st│0th│
      ╞═══╪═══╪═══╪═══╪═══╪═══╡
(5.1) │   │ 3 │ 0 │   │   │   │
      ├───┼───┼───┼───┼───┼───┤
(5.2) │   │ 3 │   │   │   │ 0 │
      ├───┼───┼───┼───┼───┼───┤
(5.3) │   │ 3 │ 4 │   │   │   │
      └───┴───┴───┴───┴───┴───┘

To have at least two agreed positional digits with ⟨351042⟩, (5.1) and (5.2) become

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

In both cases we cannot match ✅「⟨⋯ 3 ⋯ 4 ⋯ 5 ⋯⟩」. Therefore, (5.1) and (5.2) do not hold, and (5.3) holds instead:

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

Then, to have at least two agreed positional digits with ⟨351042⟩, we need

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

and to match ✅「⟨⋯ 3 ⋯ 4 ⋯ 5 ⋯⟩」, we reach

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

Now we fail to match ✅「max ⊢0⊣ ≤ 4」, which is a contradiction.

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

We have verified (2). Accordingly, we get

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

--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ 5 │ 0 │   │ 4 │
└───┴───┴───┴───┴───┴───┘

Next, we consider how to have one more agreed positional digit with ⟨351042⟩.

Observe that this digit cannot be 

# 5 (by ✅「⟨⋯ 3 ⋯ 4 ⋯ 5 ⋯⟩」)
# 4 (by ⛔「5th → a, 1st → b, |a-b|=1」)
# 2 (by ⛔「⟨⋯ ? ⋯ 1 ⋯ (?−1)⟩ (?≠1,2)」).

Therefore, one of the following holds:

    ┌───┬───┬───┬───┬───┬───┐
    │5th│4th│ 3▲│ 2▲│1st│0th│
    ╞═══╪═══╪═══╪═══╪═══╪═══╡
(6) │ 3 │   │ 1 │   │   │   │
    ├───┼───┼───┼───┼───┼───┤
(7) │ 3 │   │   │ 0 │   │   │
    └───┴───┴───┴───┴───┴───┘

(8) We show that (7) holds actually.

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

For, if (6) holds, then it follows from (1) that

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

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

And to avoid ⛔「5th → a, 1st → b, |a-b|=1」, we have

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

Note that it fails to match ✅「max ⊢0⊣ ≤ 4」, which is a contradiction.

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

We have verified (8). So we have

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

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

Plainly, (1) implies

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

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

We consider how to place 5. To match ✅「⟨⋯ 3 ⋯ 4 ⋯ 5 ⋯⟩」 and ✅「max ⊢0⊣ ≤ 4」 at the same time, we have

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

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

Finally, to avoid ⛔「5th → a, 1st → b, |a-b|=1」, we finish by

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

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

Q.E.D.

#125034_v2.8