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2025-08-26 WR

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

⟨5th 4th 3rd 2nd 1st 0th⟩

✅Match
⟨⋯ a ⋯ 1 ⋯⟩, a = 2|5
⟨   ⁴ᵗʰa ³ʳᵈb     ⁰ᵗʰc ⟩, (abc)₁₀ ≥ 123
⟨ − ▧ ▧ ▢ ▢ − ⟩, Σ▧ ≤ Σ▢

⛔Avoid
⟨   ⁴ᵗʰa     ¹ˢᵗb   ⟩, (ab)₁₀ ≥ 14
⟨ − ▧ ▧ ▢ − − ⟩, Σ▧ ≤ ▢
⟨⋯ 2 ⋯ 3 ⋯⟩

#125034_v2.10


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

Proof of 2025-08-26 WR
══════════════════════

To avoid ⛔「⟨   ⁴ᵗʰa     ¹ˢᵗb   ⟩, (ab)₁₀ ≥ 14」, we need [4th] = 1 or 0. If it is 0, then we cannot match ✅「⟨   ⁴ᵗʰa ³ʳᵈb     ⁰ᵗʰc ⟩, (abc)₁₀ ≥ 123」. Therefore, [4th] = 1.

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

--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│   │ 2 │ 5 │ 0 │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘

Then, to avoid ⛔「⟨   ⁴ᵗʰa     ¹ˢᵗb   ⟩, (ab)₁₀ ≥ 14」, we need

(1) [1st] = 0,2,3.

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

(1.1) We show that [1st] = 2 actually.

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

(2.1) Note that if [1st] = 0:

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

then we cannot match ✅「⟨ − ▧ ▧ ▢ ▢ − ⟩, Σ▧ ≤ Σ▢」 and avoid ⛔「⟨ − ▧ ▧ ▢ − − ⟩, Σ▧ ≤ ▢」 at the same time. This shows a contradiction.

(2.2) Else if [1st] = 3:

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

then to avoid ⛔「⟨⋯ 2 ⋯ 3 ⋯⟩」, we need 2 = [0th]:

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

It follows that to match ✅「⟨⋯ a ⋯ 1 ⋯⟩, a = 2|5」, we need [5th] = 5:

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

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

In view of ✅「⟨   ⁴ᵗʰa ³ʳᵈb     ⁰ᵗʰc ⟩, (abc)₁₀ ≥ 123」, we reach

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

However, it fails to match ✅「⟨ − ▧ ▧ ▢ ▢ − ⟩, Σ▧ ≤ Σ▢」, which shows a contradiction.

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

We have verified (1.1). Accordingly, we get

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

--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│   │   │ 5 │ 0 │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘

Then, there is only one way to match ✅「⟨⋯ a ⋯ 1 ⋯⟩, a = 2|5」:

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

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

Next, we consider where to place 0. Note that if it is at 2nd:

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

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

then we would fail to match ✅「⟨ − ▧ ▧ ▢ ▢ − ⟩, Σ▧ ≤ Σ▢」. On the other hand, by ✅「⟨   ⁴ᵗʰa ³ʳᵈb     ⁰ᵗʰc ⟩, (abc)₁₀ ≥ 123」, 0 is not at 3rd. Therefore, 0 = [0th]:

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

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

Finally, to avoid ⛔「⟨ − ▧ ▧ ▢ − − ⟩, Σ▧ ≤ ▢」, we finish by

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

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

Q.E.D.

#125034_v2.10