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2026-09-15 WR

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

⟨ ⁵ᵗʰ▨ ⁴ᵗʰ▨ ³ʳᵈ▨ ²ⁿᵈ▨ ¹ˢᵗ▨ ⁰ᵗʰ▨ ⟩

✅Match
⟨ ⁵ᵗʰ↓ ⁴ᵗʰ↑ ³ʳᵈ↓ ²ⁿᵈ↑ ¹ˢᵗ↓ ⁰ᵗʰ↑ ⟩ after ⟨⇌⟩
⟨⋯ Perm(0,1) ⋯⟩

⛔Avoid
⟨⋯ ᵃb ⋯⟩, |a-b|=0
⟨⋯ 3 ⋯ a ⋯⟩, a = 0|1|2|5

#125034_v2.14



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

Proof of 2026-09-15 WR
══════════════════════

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

To avoid ⛔「⟨⋯ 3 ⋯ a ⋯⟩, a = 0|1|2|5」, we need 

(1) 3 = [1st] or [0th].

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

(1.1) We show that actually 3=[0th].

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

For, if 3=[1st], then ⛔「⟨⋯ 3 ⋯ a ⋯⟩, a = 0|1|2|5」 implies that [0th]=4:

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

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

To match the 0th position of ✅「⟨ ⁵ᵗʰ↓ ⁴ᵗʰ↑ ³ʳᵈ↓ ²ⁿᵈ↑ ¹ˢᵗ↓ ⁰ᵗʰ↑ ⟩ after ⟨⇌⟩」, we then need [5th]=5:

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

We have matched ⛔「⟨⋯ ᵃb ⋯⟩, |a-b|=0」, which is a contradiction.

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

We have verified (1.1). Accordingly, we get

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

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

Then, to match the 0th position of ✅「⟨ ⁵ᵗʰ↓ ⁴ᵗʰ↑ ³ʳᵈ↓ ²ⁿᵈ↑ ¹ˢᵗ↓ ⁰ᵗʰ↑ ⟩ after ⟨⇌⟩」, we need [5th] = 4 or 5. To avoid ⛔「⟨⋯ ᵃb ⋯⟩, |a-b|=0」, it has to be 4.

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

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

Next, we consider where to place 0. It can only be at the "↑" positions of ✅「⟨ ⁵ᵗʰ↓ ⁴ᵗʰ↑ ³ʳᵈ↓ ²ⁿᵈ↑ ¹ˢᵗ↓ ⁰ᵗʰ↑ ⟩ after ⟨⇌⟩」, so 

(2) 0 = [4th] or [2nd].

(2.1) We show that 0 = [2nd].

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

If on the contrary 0=[4th], then using ✅「⟨⋯ Perm(0,1) ⋯⟩」, we have

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

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

Consider [2nd], however. If [2nd] = 2, then we match ⛔「⟨⋯ ᵃb ⋯⟩, |a-b|=0」, else if [2nd]=5, then we cannot match ✅「⟨ ⁵ᵗʰ↓ ⁴ᵗʰ↑ ³ʳᵈ↓ ²ⁿᵈ↑ ¹ˢᵗ↓ ⁰ᵗʰ↑ ⟩ after ⟨⇌⟩」. Therefore, we have reached a contradiction.

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

Now (2.1) has been verified and we get

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

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

Then, to match ✅「⟨⋯ Perm(0,1) ⋯⟩」 and avoid ⛔「⟨⋯ ᵃb ⋯⟩, |a-b|=0」 at the same time, we need 1=[3rd]:

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

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

Finally, in view of ✅「⟨ ⁵ᵗʰ↓ ⁴ᵗʰ↑ ³ʳᵈ↓ ²ⁿᵈ↑ ¹ˢᵗ↓ ⁰ᵗʰ↑ ⟩ after ⟨⇌⟩」, we finish by

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

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

Q.E.D.

#125034_v2.14