2024-05-07 Q1(m=6)

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

⟨6th 5th 4th 3rd 2nd 1st 0th⟩

✅Match

⟦2,6⟧ ∋ 0,1,3

4th → a, 1st → b, a+b=1+6n

⟨? ⋯ 3 ⋯ (?+4) ⋯⟩

⟨     ⁴ᵗʰc   ²ⁿᵈa   ⁰ᵗʰb ⟩, a > b > c

3rd → 3|4|6

#125034_v2.5

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

Proof of 2024-05-07 Q1(m=6)
═══════════════════════════

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

By ✅「⟨? ⋯ 3 ⋯ (?+4) ⋯⟩」, we have [6th] = 0|1|2. Since ✅「⟦2,6⟧ ∋ 0,1,3」 implies 0 and 1 are not in corners, we have [6th] = 2.

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

--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │   │ 6 │ 3 │ 0 │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘

Next, we consider the value of [0th]. As ✅「⟦2,6⟧ ∋ 0,1,3」 implies that 0,1,3 are not in corners, we have

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

(2) In view of ✅「⟨     ⁴ᵗʰc   ²ⁿᵈa   ⁰ᵗʰb ⟩, a > b > c」, we have [0th] != 6. We claim that [0th] = 4 actually.

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

For, suppose on the contrary [0th] = 5. Then, using ✅「⟨     ⁴ᵗʰc   ²ⁿᵈa   ⁰ᵗʰb ⟩, a > b > c」, we get [2nd] = 6:

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

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

It then follows from ✅「⟦2,6⟧ ∋ 0,1,3」 that { [5th], [4th], [3rd] } = {0,1,3}, and that [1st] = 4:

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

Using ✅「3rd → 3|4|6」, we have [3rd] = 3:

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

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

But then we cannot not match ✅「4th → a, 1st → b, a+b=1+6n」, which is a contradiction.

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

We have verified our claim in (2). Accordingly, we get

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

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

Next, note that to match ✅「⟦2,6⟧ ∋ 0,1,3」, we cannot place 6 at 3rd. So, in view of ✅「3rd → 3|4|6」, we get [3rd] = 3:

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

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

We proceed to consider how to match ✅「4th → a, 1st → b, a+b=1+6n」. Observe that there are only two ways to do so:

(3) {[4th], [1st]} = {0,1} | {1,6}.

We need to use 1 in both cases, so 1 = [4th] | [1st]. To match ✅「⟦2,6⟧ ∋ 0,1,3」, we cannot have 1 = [1st]. It follows that 1 = [4th]:

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

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

Then, by (3), we have [1st] = 0|6. Again using ✅「⟦2,6⟧ ∋ 0,1,3」, we see that [1st] != 0. So, [1st] = 6:

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

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

Finally, to match ✅「⟨     ⁴ᵗʰc   ²ⁿᵈa   ⁰ᵗʰb ⟩, a > b > c」, we finish by

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

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

Q.E.D.

#125034_v2.5