2023-12-19 Q1(m=6)

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

⟨6th 5th 4th 3rd 2nd 1st 0th⟩

✅Match

6th → a, 0th → b, ab=0

3rd → a, 2nd → b, ab=0+6n

5th → a, 2nd → b, a+b=0+5n

⛔Avoid

⟨   ⁵ᵗʰb ⁴ᵗʰa         ⟩, a > b

Sim⟨ ⁶ᵗʰ3 ⁵ᵗʰ5 ⁴ᵗʰ2 ³ʳᵈ6 ²ⁿᵈ0 ¹ˢᵗ1 ⁰ᵗʰ4 ⟩ ≥ 1

⟨⋯ 4 ⋯ 1 ⋯ 3 ⋯ 2 ⋯⟩

⟨⋯ 1 ⋯ ? ⋯ 5 (?−1)⟩ (?≠5,6)

⟨⋯ 2 ⋯ a ⋯⟩, a = 0|1|3|4|5

#125034_v2.2

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

Proof of 2023-12-19 Q1(m=6)
═══════════════════════════

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

To match ✅「6th → a, 0th → b, ab=0」, we need 0 = [6th] or [0th]. We need to avoid ⛔「⟨⋯ 2 ⋯ a ⋯⟩, a = 0|1|3|4|5」 too, so 0 = [6th].

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

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

(1) Note that ⛔「⟨⋯ 2 ⋯ a ⋯⟩, a = 0|1|3|4|5」 implies 2 is at the right of 0,1,3,4,5. Therefore, 2 = [1st] or [0th].

Also, note that ✅「5th → a, 2nd → b, a+b=0+5n」 implies

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

(2) where A + B = 5 or 10.

In view of (1), there are two possibilities:

(2.1) {A,B} = {1,4};

(2.2) {A,B} = {4,6}.

We show that case (2.2) holds actually.

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

If on the contrary case (2.1) holds, then to avoid ⛔「⟨   ⁵ᵗʰb ⁴ᵗʰa         ⟩, a > b」, we need (A,B) = (4,1):

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

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

Then, to match ✅「3rd → a, 2nd → b, ab=0+6n」, we need [3rd] = 6:

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

However, it follows that we would match ⛔「Sim⟨ ⁶ᵗʰ3 ⁵ᵗʰ5 ⁴ᵗʰ2 ³ʳᵈ6 ²ⁿᵈ0 ¹ˢᵗ1 ⁰ᵗʰ4 ⟩ ≥ 1」, which is a contradiction.

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

We have verified that case (2.2) holds. In particular, 6 = [5th] or [2nd]. Combining this with (1), we see that 2 is at the right of all other numbers. Consequently, 2 = [0th].

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

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

Furthermore, as (2.2) holds, we have

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

where (A,B) = (4,6) or (6,4).

(3) We claim that (A,B) = (4,6) actually.

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

For, if on the contrary (A,B) = (6,4):

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

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

then there is only one way to match ✅「3rd → a, 2nd → b, ab=0+6n」:

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

No matter how we place 1,5, we would match ⛔「Sim⟨ ⁶ᵗʰ3 ⁵ᵗʰ5 ⁴ᵗʰ2 ³ʳᵈ6 ²ⁿᵈ0 ¹ˢᵗ1 ⁰ᵗʰ4 ⟩ ≥ 1」 or ⛔「⟨⋯ 1 ⋯ ? ⋯ 5 (?−1)⟩ (?≠5,6)」. This shows a contradiction.

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

Our claim in (3) is thus verified. Accordingly

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

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

We consider where to place 1. To avoid ⛔「Sim⟨ ⁶ᵗʰ3 ⁵ᵗʰ5 ⁴ᵗʰ2 ³ʳᵈ6 ²ⁿᵈ0 ¹ˢᵗ1 ⁰ᵗʰ4 ⟩ ≥ 1」, it is not at 1st, and to avoid ⛔「⟨⋯ 4 ⋯ 1 ⋯ 3 ⋯ 2 ⋯⟩」, it is not at 4th. Therefore, 1 = [3rd].

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

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

Finally, to avoid ⛔「⟨⋯ 4 ⋯ 1 ⋯ 3 ⋯ 2 ⋯⟩」, we have 3 = [4th]. We finish by

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

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

Q.E.D.

#125034_v2.2