2024-06-04 Q1(m=6)

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

⟨6th 5th 4th 3rd 2nd 1st 0th⟩

✅Match

⟨       ³ʳᵈa ²ⁿᵈb     ⟩, (ab)₁₀ ≤ 14

⟦1,6⟧ ∋ 0,3,4,5

⟨⋯ ? ⋯ 3 ⋯ (?+4)⟩

5th → a, 1st → b, ab=3+5n

⛔Avoid

⟨ ⁶ᵗʰa     ³ʳᵈb       ⟩, max⟦a,b⟧ = 5

{p6, p5, p4, p3} = ? + {0,1,2,4}

#125034_v2.5

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

Proof of 2024-06-04 Q1(m=6)
═══════════════════════════

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

To match ✅「⟨⋯ ? ⋯ 3 ⋯ (?+4)⟩」, we need

(1) [0th] = 4|5|6.

Since ✅「⟦1,6⟧ ∋ 0,3,4,5」 implies that 4 and 5 are not in the right corner, (1) gives [0th] = 6.

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

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

Now, 5 is the maximum idle digit. So, to avoid ⛔「⟨ ⁶ᵗʰa     ³ʳᵈb       ⟩, max⟦a,b⟧ = 5」, we need:

(2) 5 = [2nd] | [1st].

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

Observe that if 5 = [1st], then

(3) [5th] * [1st] will be divisible by 5.

This is a contradiction, however, if our answer matches ✅「5th → a, 1st → b, ab=3+5n」 as well. Therefore, 5 != [1st]. So, back to (2), we get 5 = [2nd]:

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

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

Next, note that ✅「⟨       ³ʳᵈa ²ⁿᵈb     ⟩, (ab)₁₀ ≤ 14」 determines the value of [3rd]:

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

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

We proceed to check the value of [1st]. There are four idle digits left:

(4) [1st] = 1|2|3|4.

We claim that [1st] = 4 actually.

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

In fact, note that:

(4.1) If [1st] = 1, then we cannot match ✅「⟦1,6⟧ ∋ 0,3,4,5」:

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

(4.2) Else if [1st] = 2, then we cannot match ✅「⟨⋯ ? ⋯ 3 ⋯ (?+4)⟩」:

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

(4.3) Else if [1st] = 3, then we would match ⛔「{p6, p5, p4, p3} = ? + {0,1,2,4}」:

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

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

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

Combining (4), (4.1), (4.2), and (4.3), we have verified that [1st] = 4. Accordingly, we get

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

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

Using ✅「5th → a, 1st → b, ab=3+5n」, we get [5th] = 2 as well:

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

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

Finally, in view of ✅「⟨⋯ ? ⋯ 3 ⋯ (?+4)⟩」, we finish by

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

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

Q.E.D.

#125034_v2.5