2023-10-25 Q1(m=6)

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

⟨6th 5th 4th 3rd 2nd 1st 0th⟩

✅Match

⟦1,4⟧ ∋ 0,3

0 ∾ 2 ∾ 4 ∾ 5 ∾ 6

⟨     ⁴ᵗʰc     ¹ˢᵗb ⁰ᵗʰa ⟩, a > b > c

⛔Avoid

⟨⋯ a ⋯ 4 ⋯⟩, a = 0|1|2|3

4th|3rd|2nd|1st|0th → 6

2nd → a, 1st → b, |a-b|=2

#125034_v2.1

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

Proof of 2023-10-25 Q1(m=6)
═══════════════════════════

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

We first note that ⛔「4th|3rd|2nd|1st|0th → 6」 is equivalent to 6 = [6th] or [5th]. If 6 = [6th], then 6 is itself a permutation cycle and ✅【0 ∾ 2 ∾ 4 ∾ 5 ∾ 6】 cannot be matched. Hence, we have 6 = [5th]:

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

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

Next, note that ⛔「⟨⋯ a ⋯ 4 ⋯⟩, a = 0|1|2|3」 is equivalent to the requirement that 4 is at the left of 0,1,2,3. Having at least four numbers at the right, 4 has to be placed at [6th] or [4th].

Same as the previous argument for 6, if 4 = [4th], then 4 is itself a permutation cycle and ✅【0 ∾ 2 ∾ 4 ∾ 5 ∾ 6】 cannot be matched. Hence, we have 4 = [6th]:

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

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

(1) Now, we consider where to place 1. By ✅【⟦1,4⟧ ∋ 0,3】, 0,3 are between 1 and 6. So 1 = [2nd] or [1st] or [0th].

(2) It is impossible that 1 = [0th], because by ✅【⟨     ⁴ᵗʰc     ¹ˢᵗb ⁰ᵗʰa ⟩, a > b > c】, [0th] is greater than two numbers, so it is at least 2.

(3) It is also impossible that 1 = [2nd]. For otherwise, by ✅【⟦1,4⟧ ∋ 0,3】, we have

┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│*4 │*3 │ 2▲│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 4 │ 6 │ X │ Y │ 1 │   │   │
└───┴───┴───┴───┴───┴───┴───┘

where {X,Y}={0,3}, so the idle are:

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

By ✅【⟨     ⁴ᵗʰc     ¹ˢᵗb ⁰ᵗʰa ⟩, a > b > c】, we need [1st] < [0th]. Accordingly, we have

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

But now 1,2 form a cycle, and ✅【0 ∾ 2 ∾ 4 ∾ 5 ∾ 6】 cannot be matched.

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

Therefore, by (1), (2), and (3), we have 1 = [1st]:

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

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

As ✅【⟨     ⁴ᵗʰc     ¹ˢᵗb ⁰ᵗʰa ⟩, a > b > c】 requires that [4th] < [1st] = 1, we see that [4th] = 0.

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

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

Using ✅【⟦1,4⟧ ∋ 0,3】 now, we have 3 = [3rd] or [2nd]. As ⛔「2nd → a, 1st → b, |a-b|=2」 rejects the latter case, we have 3 = [3rd].

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

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

Finally, to meet the cycle requirement ✅【0 ∾ 2 ∾ 4 ∾ 5 ∾ 6】, we have 2 ≠ [2nd]. As a result, we obtain

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

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

Q.E.D.

#125034_v2.1