2024-04-23 Q1(m=6)

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

⟨6th 5th 4th 3rd 2nd 1st 0th⟩

✅Match

⟨⋯ Perm(0,1,4) ⋯⟩

⟨⋯ 2 ⋯ 5 ⋯ 1 ⋯⟩

6th → a, 5th → b, a+b=5

⟨⋯ Perm(1,6) ⋯⟩

⛔Avoid

⟨⋯ 4 ⋯ a ⋯⟩, a = 0|5

6th → a, 0th → b, a+b=8

2nd → 1

#125034_v2.4

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

Proof of 2024-04-23 Q1(m=6)
═══════════════════════════

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

To begin with, note that by ✅「⟨⋯ 2 ⋯ 5 ⋯ 1 ⋯⟩」, ✅「⟨⋯ Perm(0,1,4) ⋯⟩」, and ✅「⟨⋯ Perm(1,6) ⋯⟩」, we need to match

(1) ⟨⋯ 2 ⋯ 5 ⋯ Perm(0,1,4,6) ⋯⟩.

Four consecutive boxes are needed for Perm(0,1,4,6). In view of (1), there are two possibilities:

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

Actually (2) is not possible, for otherwise it follows from (1) that

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

and we cannot match ✅「6th → a, 5th → b, a+b=5」, which is a contradiction.

Therefore, (3) holds. A fortiori,

(4) 1 = [3rd] | [2nd] | [1st] | [0th].

By ⛔「2nd → 1」, plainly 1 != [2nd]. On the other hand, if 1 is at the boundary of the four consecutive boxes (i.e. 3rd or 0th), then by ✅「⟨⋯ Perm(1,6) ⋯⟩」 and (3), we get

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

It follows that we cannot match ✅「⟨⋯ Perm(0,1,4) ⋯⟩」, which is a contradiction. Therefore, 1 = [1st]:

       ┌───┬───┬───┬───┬───┬───┬───┐
       │6th│5th│4th│3rd│2nd│ 1■│0th│▒
       ╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │   │   │   │ ▬ │ ▬ │ 1 │ ▬ │▒
       └───┴───┴───┴───┴───┴───┴───┘▒
        ▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
        
Then, to match both ✅「⟨⋯ Perm(1,6) ⋯⟩」 and ✅「⟨⋯ Perm(0,1,4) ⋯⟩」, we have to place 6 at 0th.

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

Next, we determine where to place 2. By (1), we have

(5) 2 = [6th] | [5th].

Since we have to avoid ⛔「6th → a, 0th → b, a+b=8」, so 2 = [5th]:

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

Then, by (1), we get

       ┌───┬───┬───┬───┬───┬───┬───┐
       │ 6■│5th│ 4■│3rd│2nd│1st│0th│▒
       ╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 4 │   │ 2 │ 5 │ ▬ │ ▬ │ 1 │ 6 │▒
       ├───┼───┼───┼───┼───┼───┼───┤▒
Step 5 │ 3 │ 2 │ 5 │ ▬ │ ▬ │ 1 │ 6 │▒
       └───┴───┴───┴───┴───┴───┴───┘▒
        ▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
        
Finally, in view of ⛔「⟨⋯ 4 ⋯ a ⋯⟩, a = 0|5」, we finish by

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

#125034_v2.4