2024-03-19 Q1(m=6)

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

⟨6th 5th 4th 3rd 2nd 1st 0th⟩

✅Match

⟨     ⁴ᵗʰa     ¹ˢᵗb   ⟩, max⟦a,b⟧ = 6

⟨   ⁵ᵗʰc   ³ʳᵈb     ⁰ᵗʰa ⟩, a > b > c

⟨⋯ ? ⋯ 5 ⋯ (?+2)⟩ (?≠3)

⛔Avoid

6th|5th|4th|2nd|1st → 4

⟨ ⁶ᵗʰa   ⁴ᵗʰb   ²ⁿᵈc     ⟩, (abc)₁₀ ≤ 362

5th → a, 0th → b, |a-b|=3

#125034_v2.4

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

Proof of 2024-03-19 Q1(m=6)
═══════════════════════════

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

To begin with, we consider where to place 4. On the one hand, to avoid ⛔「6th|5th|4th|2nd|1st → 4」, we need

(1) 4 = [3rd] | [0th].

On the other hand, by ✅「⟨⋯ ? ⋯ 5 ⋯ (?+2)⟩ (?≠3)」, we have [0th] = 2|3|4|6. Actually [0th] != 6 by ✅「⟨     ⁴ᵗʰa     ¹ˢᵗb   ⟩, max⟦a,b⟧ = 6」. Therefore,

(2) [0th] = 2|3|4.

Combining (2) with ✅「⟨   ⁵ᵗʰc   ³ʳᵈb     ⁰ᵗʰa ⟩, a > b > c」, we have

(3) [3rd] < 4.

So, it follows from (1) and (3) that 4 = [0th]:

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

--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ 6 │ 3 │ 0 │   │ 5 │
└───┴───┴───┴───┴───┴───┴───┘

Next, we determine the value of [6th]. By ⛔「⟨ ⁶ᵗʰa   ⁴ᵗʰb   ²ⁿᵈc     ⟩, (abc)₁₀ ≤ 362」, we need

[6th] = 3|5|6.

Note that ✅「⟨     ⁴ᵗʰa     ¹ˢᵗb   ⟩, max⟦a,b⟧ = 6」 implies [6th] != 6, and ✅「⟨⋯ ? ⋯ 5 ⋯ (?+2)⟩ (?≠3)」 implies [6th] != 5. It follows that [6th] = 3:

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

--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ 6 │   │ 0 │   │ 5 │
└───┴───┴───┴───┴───┴───┴───┘

Then, to avoid ⛔「⟨ ⁶ᵗʰa   ⁴ᵗʰb   ²ⁿᵈc     ⟩, (abc)₁₀ ≤ 362」, we need [4th] = 6:

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

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

We proceed to determine the value of [5th]. It is the "c" in ✅「⟨   ⁵ᵗʰc   ³ʳᵈb     ⁰ᵗʰa ⟩, a > b > c」. Hence, it cannot be 5 or 2. It is not 1 as well, by ⛔「5th → a, 0th → b, |a-b|=3」. Consequently, [5th] = 0:

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

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

Then, note that there is only one way to avoid ⛔「⟨ ⁶ᵗʰa   ⁴ᵗʰb   ²ⁿᵈc     ⟩, (abc)₁₀ ≤ 362」:

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

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

Finally, to match ✅「⟨⋯ ? ⋯ 5 ⋯ (?+2)⟩ (?≠3)」, we finish by

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

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

Q.E.D.

#125034_v2.4