2024-07-09 Q1(m=6)

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

⟨6th 5th 4th 3rd 2nd 1st 0th⟩

✅Match

Jump(3,5) = 1

⟨? ⋯ 0 (?+3) ⋯ 4 ⋯⟩ (?≠0)

Sim⟨ ⁶ᵗʰ0 ⁵ᵗʰ6 ⁴ᵗʰ5 ³ʳᵈ1 ²ⁿᵈ4 ¹ˢᵗ3 ⁰ᵗʰ2 ⟩ = 1

3rd → a, 0th → b, a+b=3+5n

#125034_v2.5

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

Proof of 2024-07-09 Q1(m=6)
═══════════════════════════

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

Note that by ✅「⟨? ⋯ 0 (?+3) ⋯ 4 ⋯⟩ (?≠0)」, we have

(1) [6th] = 1|2|3.

On the other hand, by ✅「3rd → a, 0th → b, a+b=3+5n」, we have

(2) a + b = 3|8, where a := [3rd] and b := [0th].

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

Note that ✅「Jump(3,5) = 1」 implies {a,b} != {3,5}. So, one of the following holds:

(i) {a,b} = {0,3};
(ii) {a,b} = {1,2};
(iii) {a,b} = {2,6}.

(3) We show that (i) holds actually.

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

(3.1) If (ii) holds, then by (1), we have [6th] = 3:

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

Then, using ✅「Jump(3,5) = 1」, we get [4th] = 5:

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

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

It is a contradiction, however, as we cannot match ✅「⟨? ⋯ 0 (?+3) ⋯ 4 ⋯⟩ (?≠0)」.

(3.2) Else, suppose (iii) holds. By (1), we have [6th] = 1|3. Actually [6th] != 3, for otherwise using ✅「Jump(3,5) = 1」, we get

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

and we cannot match ✅「⟨? ⋯ 0 (?+3) ⋯ 4 ⋯⟩ (?≠0)」 regardless of (a,b) = (2,6) or (6,2).

So, we have [6th] = 1 when {a,b} = {2,6}:

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

--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│   │   │   │ 3 │ 0 │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘

Then, to match ✅「⟨? ⋯ 0 (?+3) ⋯ 4 ⋯⟩ (?≠0)」, we need 0,4 to be adjacent. This cannot be satisfied, however, if ✅「Jump(3,5) = 1」 is matched as well. So, we again have a contradiction.

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

We have verified our claim in (3). Accordingly, we have {a,b} = {0,3}. As ✅「⟨? ⋯ 0 (?+3) ⋯ 4 ⋯⟩ (?≠0)」 implies that 0 is not at the right corner, we get

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

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

It then follows from ✅「Jump(3,5) = 1」 that [2nd] = 5:

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

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

Using ✅「⟨? ⋯ 0 (?+3) ⋯ 4 ⋯⟩ (?≠0)」, we get

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

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

Finally, to match ✅「Sim⟨ ⁶ᵗʰ0 ⁵ᵗʰ6 ⁴ᵗʰ5 ³ʳᵈ1 ²ⁿᵈ4 ¹ˢᵗ3 ⁰ᵗʰ2 ⟩ = 1」, we finish by

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

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

Q.E.D.

#125034_v2.5