2024-09-10 Q1(m=6)

Rearrange the digits in ⟨1263045⟩ to meet the rules below.

⟨6th 5th 4th 3rd 2nd 1st 0th⟩

✅Match

⟨ ⁶ᵗʰa     ³ʳᵈb       ⟩, a > b

4th → a, 3rd → b, ab=2+4n

⟨         ²ⁿᵈa   ⁰ᵗʰb ⟩, min⟦a,b⟧ = 2

⟨         ²ⁿᵈa ¹ˢᵗb ⁰ᵗʰc ⟩, (abc)₁₀ ≤ 510

2nd → a, 1st → b, |a-b|=1

⛔Avoid

⟨ ⁶ᵗʰ↑       ²ⁿᵈ↑   ⁰ᵗʰ↓ ⟩ after 2×⟨←⟩

Sim⟨ ⁶ᵗʰ5 ⁵ᵗʰ0 ⁴ᵗʰ3 ³ʳᵈ4 ²ⁿᵈ1 ¹ˢᵗ2 ⁰ᵗʰ6 ⟩ ≥ 2

#125034_v2.6

       ┌───┬───┬───┬───┬───┬───┬───┐
       │6th│5th│4th│3rd│2nd│1st│0th│▒
       ╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │   │   │ 6 │   │   │   │   │▒
       ├───┼───┼───┼───┼───┼───┼───┤▒
Step 2 │   │ 0 │ 6 │   │   │   │   │▒
       ├───┼───┼───┼───┼───┼───┼───┤▒
Step 3 │   │ 0 │ 6 │ 1 │   │   │   │▒
       ├───┼───┼───┼───┼───┼───┼───┤▒
Step 4 │   │ 0 │ 6 │ 1 │   │   │ 2 │▒
       ├───┼───┼───┼───┼───┼───┼───┤▒
Step 5 │ 3 │ 0 │ 6 │ 1 │   │   │ 2 │▒
       ├───┼───┼───┼───┼───┼───┼───┤▒
Step 6 │ 3 │ 0 │ 6 │ 1 │ 4 │   │ 2 │▒
       ├───┼───┼───┼───┼───┼───┼───┤▒
Step 7 │ 3 │ 0 │ 6 │ 1 │ 4 │ 5 │ 2 │▒
       └───┴───┴───┴───┴───┴───┴───┘▒
        ▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒

Proof of 2024-09-10 Q1(m=6)
═══════════════════════════

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

Let S := {[2nd], [1st], [0th]}. By ✅「⟨         ²ⁿᵈa   ⁰ᵗʰb ⟩, min⟦a,b⟧ = 2」, we have

(1) 2 ∈ S; and

(2) 0,1 ∉ S.

It follows from (1) that 2 ∉ {[4th], [3rd]}. Therefore, ✅「4th → a, 3rd → b, ab=2+4n」 implies

(3) {[4th], [3rd]} = {1,6} | {3,6} | {5,6}.

In particular, 6 ∈ {[4th], [3rd]}. Since 6 is the maximum digit, it is not at 3rd in view of ✅「⟨ ⁶ᵗʰa     ³ʳᵈb       ⟩, a > b」. So, 6 = [4th]:

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

--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │   │ 3 │ 0 │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘

Next, we consider where to place 0. By (2) and (3), it is not at 3rd, 2nd, 1st, or 0th:

┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│   │   │ 6 │ / │ / │ / │ / │
└───┴───┴───┴───┴───┴───┴───┘

On the other hand, as 0 is the minimum digit, it is not at 6th in view of ✅「⟨ ⁶ᵗʰa     ³ʳᵈb       ⟩, a > b」. Therefore, 0 = [5th]:

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

--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │   │ 3 │   │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘

We can determine the position of 1 in a similar way. By (2), it is at 6th or 3rd:

┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│   │ 0 │ 6 │   │ / │ / │ / │
└───┴───┴───┴───┴───┴───┴───┘

--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │   │ 3 │   │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘

If 1 = [6th], then we cannot match ✅「⟨ ⁶ᵗʰa     ³ʳᵈb       ⟩, a > b」. Therefore, 1 = [3rd]:

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

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

Next, we consider ⛔「⟨ ⁶ᵗʰ↑       ²ⁿᵈ↑   ⁰ᵗʰ↓ ⟩ after 2×⟨←⟩」. Given that 0 = [5th] and 6 = [4th], we definitely have a "↓" at 0th and a "↑" at 6th after 2×⟨←⟩. So, to avoid the preceding pattern, we need [2nd] > [0th]. It follows that 2 != [2nd].

In view of ⛔「Sim⟨ ⁶ᵗʰ5 ⁵ᵗʰ0 ⁴ᵗʰ3 ³ʳᵈ4 ²ⁿᵈ1 ¹ˢᵗ2 ⁰ᵗʰ6 ⟩ ≥ 2」, we have 2 != [1st] as well. Therefore, by (1), we get 2 = [0th]:

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

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

Note that ⛔「Sim⟨ ⁶ᵗʰ5 ⁵ᵗʰ0 ⁴ᵗʰ3 ³ʳᵈ4 ²ⁿᵈ1 ¹ˢᵗ2 ⁰ᵗʰ6 ⟩ ≥ 2」 also implies 5 != [6th]. So, 5 ∈ {[2nd], [1st]}. Combining this with ✅「2nd → a, 1st → b, |a-b|=1」, we have

(4) {[2nd], [1st]} = {4,5}.

As a consequence, 3 = [6th]:

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

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

Finally, in view of ✅「⟨         ²ⁿᵈa ¹ˢᵗb ⁰ᵗʰc ⟩, (abc)₁₀ ≤ 510」, we finish by

       ┌───┬───┬───┬───┬───┬───┬───┐
       │6th│5th│4th│3rd│ 2■│ 1■│0th│▒
       ╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
       │ 3 │ 0 │ 6 │ 1 │   │   │ 2 │▒
       ├───┼───┼───┼───┼───┼───┼───┤▒
Step 6 │ 3 │ 0 │ 6 │ 1 │ 4 │   │ 2 │▒
       ├───┼───┼───┼───┼───┼───┼───┤▒
Step 7 │ 3 │ 0 │ 6 │ 1 │ 4 │ 5 │ 2 │▒
       └───┴───┴───┴───┴───┴───┴───┘▒
        ▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒

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

Q.E.D.

#125034_v2.6