2024-06-11 Q1(m=6)

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

⟨6th 5th 4th 3rd 2nd 1st 0th⟩

✅Match

4th → 4

⟦2,5⟧ ∋ 1,3,4,6

{p6, p5, p1} = ? + {0,1,2}

3rd → a, 0th → b, a+b=1+4n

⛔Avoid

⟨? ⋯ 3 ⋯ (?−1) ⋯⟩ (?≠3,4)

#125034_v2.5

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

Proof of 2024-06-11 Q1(m=6)
═══════════════════════════

Notation: if Nth -> a, then we write pN = a.

Plainly, our first step follows from ✅「4th → 4」.

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

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

To proceed, we consider ✅「{p6, p5, p1} = ? + {0,1,2}」. To match this pattern, we need p6, p5, p1 to be consecutive integers:

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

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

Observe that we cannot achieve this if 6 is one of them. Therefore,

(1) 6 = p3|p2|p0. We show that 6 = p2 actually.

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

For, ✅「⟦2,5⟧ ∋ 1,3,4,6」 implies that 1,3,4,6 are not in corners. A fortiori, 6 != p0. On the other hand, if 6 = p3, then ✅「3rd → a, 0th → b, a+b=1+4n」 implies that p0 = 3, but we have known that 3 is not in the right corner, so it is a contradiction.

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

We have verified our claim in (1). Accordingly, we get

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

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

Next, we consider where to place 5. Again, using ✅「{p6, p5, p1} = ? + {0,1,2}」, we see that 5 is not at 6th, 5th, or 1st. We have 5 != p3 as well, for otherwise we cannot match ✅「⟦2,5⟧ ∋ 1,3,4,6」. Therefore, 5 = p0:

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

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

Then, using ✅「3rd → a, 0th → b, a+b=1+4n」, we get p3 = 0:

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

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

Now, note that to match ✅「⟦2,5⟧ ∋ 1,3,4,6」, we need to put 2 at the left corner. So,

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

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

Finally, to avoid ⛔「⟨? ⋯ 3 ⋯ (?−1) ⋯⟩ (?≠3,4)」, we finish by

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

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

Q.E.D.

#125034_v2.5