2024-10-29 Q1(m=6)

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

⟨6th 5th 4th 3rd 2nd 1st 0th⟩

✅Match

{p6, p4, p3, p2} = ? + {0,1,3,4}

2nd → 1|2|6

⟨? ⋯ 5 (?+4) ⋯ 3 ⋯⟩ (?≠1)

⛔Avoid

⟨     ⁴ᵗʰa   ²ⁿᵈc   ⁰ᵗʰb ⟩, a > b > c

#125034_v2.7

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

Proof of 2024-10-29 Q1(m=6)
═══════════════════════════

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

By ✅「⟨? ⋯ 5 (?+4) ⋯ 3 ⋯⟩ (?≠1)」, we have

(1) [6th] = 0|2.

(1.1) We show that [6th] = 0 actually.

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

For, suppose on the contrary [6th] = 2:

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

Then, in view of ✅「{p6, p4, p3, p2} = ? + {0,1,3,4}」, one of the following holds:

(2.1) {[4th], [3rd], [2nd]} = {1,4,5};

(2.2) {[4th], [3rd], [2nd]} = {3,5,6}.

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

If (2.1) holds, then [2nd] = 1|4|5. Using ✅「2nd → 1|2|6」, we get [2nd] = 1:

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

But then we cannot match ✅「⟨? ⋯ 5 (?+4) ⋯ 3 ⋯⟩ (?≠1)」, which is a contradiction.

Therefore, (2.1) does not hold and (2.2) holds instead. A fortiori [2nd] = 3|5|6. Using ✅「2nd → 1|2|6」, we get [2nd] = 6:

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

Nevertheless, this again contradicts ✅「⟨? ⋯ 5 (?+4) ⋯ 3 ⋯⟩ (?≠1)」.

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

We have verified (1.1). Accordingly, we get

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

--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ 6 │ 3 │   │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘

So, in view of ✅「{p6, p4, p3, p2} = ? + {0,1,3,4}」, we have

(3) {[4th], [3rd], [2nd]} = {1,3,4}.

┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 0 │   │134│134│134│   │   │
└───┴───┴───┴───┴───┴───┴───┘

On the other hand, by ✅「⟨? ⋯ 5 (?+4) ⋯ 3 ⋯⟩ (?≠1)」 with ?=0, we need to have the block "54". Therefore, we obtain:

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

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

Combining ✅「2nd → 1|2|6」 with (3), we find [2nd] and [3rd] as well:

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

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

Finally, to avoid ⛔「⟨     ⁴ᵗʰa   ²ⁿᵈc   ⁰ᵗʰb ⟩, a > b > c」, we finish by

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

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

Q.E.D.

#125034_v2.7