2024-11-19 Q1(m=6)

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

⟨6th 5th 4th 3rd 2nd 1st 0th⟩

✅Match

6th|5th|2nd|0th → 4

⟨ ⁶ᵗʰ↓ ⁵ᵗʰ= ⁴ᵗʰ↑ ³ʳᵈ= ²ⁿᵈ↑ ¹ˢᵗ= ⁰ᵗʰ↓ ⟩ after 6−⟨⇌⟩

⟨⋯ 0 ⋯ ? 5 ⋯ (?+2)⟩ (?≠3)

⛔Avoid

⟨ ⁶ᵗʰa     ³ʳᵈb   ¹ˢᵗc   ⟩, (abc)₁₀ ≤ 346

#125034_v2.7

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

Proof of 2024-11-19 Q1(m=6)
═══════════════════════════

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

Consider the "=" positions of ✅「⟨ ⁶ᵗʰ↓ ⁵ᵗʰ= ⁴ᵗʰ↑ ³ʳᵈ= ²ⁿᵈ↑ ¹ˢᵗ= ⁰ᵗʰ↓ ⟩ after 6−⟨⇌⟩」. Plainly, we need [3rd] = 3.

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

--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ 6 │   │ 0 │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘

The preceding pattern also implies

(1) {[5th], [1st]} = {0,6} | {1,5} | {2,4}.

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

On the other hand, by ✅「⟨⋯ 0 ⋯ ? 5 ⋯ (?+2)⟩ (?≠3)」, one of the following holds:

(2) ⟨⋯ 05 ⋯ 2⟩;

(3) ⟨⋯ 25 ⋯ 4⟩;

(4) ⟨⋯ 45 ⋯ 6⟩.

(5) Based on (1), we proceed to show that {[5th], [1st]} = {1,5}.

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

(5.1) If {[5th], [1st]} = {0,6}, then to match ✅「⟨⋯ 0 ⋯ ? 5 ⋯ (?+2)⟩ (?≠3)」, we need

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

Then, noting that cases (3) and (4) are not possible, case (2) holds and we get

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

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

But then it cannot match the 4th position of ✅「⟨ ⁶ᵗʰ↓ ⁵ᵗʰ= ⁴ᵗʰ↑ ³ʳᵈ= ²ⁿᵈ↑ ¹ˢᵗ= ⁰ᵗʰ↓ ⟩ after 6−⟨⇌⟩」, which is a contradiction.

(5.2) Else, suppose {[5th], [1st]} = {2,4}:

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

then (2) and (3) are not possible, so (4) holds and we get

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

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

In view of ✅「⟨⋯ 0 ⋯ ? 5 ⋯ (?+2)⟩ (?≠3)」, 0 is to the left of 5, so we reach

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

It does not match the 6th position of ✅「⟨ ⁶ᵗʰ↓ ⁵ᵗʰ= ⁴ᵗʰ↑ ³ʳᵈ= ²ⁿᵈ↑ ¹ˢᵗ= ⁰ᵗʰ↓ ⟩ after 6−⟨⇌⟩」, however.

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

We have verified (5). Consequently, one of the following holds:

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

Case (6) holds actually, for otherwise by ✅「⟨⋯ 0 ⋯ ? 5 ⋯ (?+2)⟩ (?≠3)」, we have

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

and it does not match the 6th position of ✅「⟨ ⁶ᵗʰ↓ ⁵ᵗʰ= ⁴ᵗʰ↑ ³ʳᵈ= ²ⁿᵈ↑ ¹ˢᵗ= ⁰ᵗʰ↓ ⟩ after 6−⟨⇌⟩」.

Therefore, we get

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

--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│   │ 2 │ 6 │   │ 0 │ 4 │   │
└───┴───┴───┴───┴───┴───┴───┘

Next, we consider where to place 0. It cannot be at a "↓" position of ✅「⟨ ⁶ᵗʰ↓ ⁵ᵗʰ= ⁴ᵗʰ↑ ³ʳᵈ= ²ⁿᵈ↑ ¹ˢᵗ= ⁰ᵗʰ↓ ⟩ after 6−⟨⇌⟩」, so

(8) 0 = [4th] | [2nd].

(9) We show that 0 = [4th] indeed.

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

For, suppose on the contrary 0 = [2nd]:

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

Then ✅「⟨⋯ 0 ⋯ ? 5 ⋯ (?+2)⟩ (?≠3)」 gives [0th] = 2:

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

and ✅「6th|5th|2nd|0th → 4」 gives [6th] = 4:

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

This is a contradiction because it does not match the 6th position of ✅「⟨ ⁶ᵗʰ↓ ⁵ᵗʰ= ⁴ᵗʰ↑ ³ʳᵈ= ²ⁿᵈ↑ ¹ˢᵗ= ⁰ᵗʰ↓ ⟩ after 6−⟨⇌⟩」.

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

We have verified (9) and get

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

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

Now, recall to one of (2), (3), (4) holds. Plainly, case (2) does not hold. If case (4) holds, then we cannot avoid ⛔「⟨ ⁶ᵗʰa     ³ʳᵈb   ¹ˢᵗc   ⟩, (abc)₁₀ ≤ 346」. Therefore, case (3) holds. We finish by

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

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

Q.E.D.

#125034_v2.7