2024-04-02 Q1(m=6)

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

⟨6th 5th 4th 3rd 2nd 1st 0th⟩

✅Match

4th → a, 1st → b, ab=2+4n

Jump(2,3) = 1

⟨ ⁶ᵗʰc       ²ⁿᵈa ¹ˢᵗb   ⟩, a > b > c

⛔Avoid

⟨⋯ 3 ⋯ 0 ⋯⟩

⟨⋯ ? ⋯ 6 ⋯ (?+1)⟩ (?≠5)

min ⊢5⊣ ≥ 2

6th → a, 3rd → b, ab=0+6n

#125034_v2.4

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

Proof of 2024-04-02 Q1(m=6)
═══════════════════════════

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

We begin with the left corner, i.e. [6th]. Observe that

(i) By ✅「⟨ ⁶ᵗʰc       ²ⁿᵈa ¹ˢᵗb   ⟩, a > b > c」, it is not 6 or 5;

(ii) By ⛔「⟨⋯ 3 ⋯ 0 ⋯⟩」, it is not 3;

(iii) By ⛔「6th → a, 3rd → b, ab=0+6n」, it is not 0.

Therefore,

(1) [6th] = 4|2|1. We proceed to show that [6th] = 1 indeed.

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

(1.1) Assume [6th] = 4. Then, to match ✅「⟨ ⁶ᵗʰc       ²ⁿᵈa ¹ˢᵗb   ⟩, a > b > c」, we need "a"=6 and "b"=5:

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

Then, to avoid ⛔「min ⊢5⊣ ≥ 2」, we need [0th] = 1|0:

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

But it is a contradiction:

● If [0th] = 1, then we would match ⛔「⟨⋯ ? ⋯ 6 ⋯ (?+1)⟩ (?≠5)」;

● Else if [0th] = 0, then we would match ⛔「⟨⋯ 3 ⋯ 0 ⋯⟩」. 

Hence, our assumption "[6th] = 4" is not correct.

(1.2) Else, assume [6th] = 2. By ✅「Jump(2,3) = 1」, we get

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

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

Then, to match ✅「4th → a, 1st → b, ab=2+4n」, we need [1st] = 6:

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

But then ✅「⟨ ⁶ᵗʰc       ²ⁿᵈa ¹ˢᵗb   ⟩, a > b > c」 cannot be matched, which is a contradiction.

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

By (1.1) and (1.2), our claim in (1) is verified. Accordingly, we get

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

--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│   │ 2 │ 6 │ 3 │ 0 │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘

Next, we decide where to place 0. Note that:

(i) To avoid ⛔「⟨⋯ 3 ⋯ 0 ⋯⟩」, we need [0th] != 0;

(ii) By ✅「⟨ ⁶ᵗʰc       ²ⁿᵈa ¹ˢᵗb   ⟩, a > b > c」, 0 is not at 1st or 2nd;

(iii) To avoid ⛔「6th → a, 3rd → b, ab=0+6n」, we need [3rd] != 0;

(iv) To match ✅「4th → a, 1st → b, ab=2+4n」, we need [4th] != 0.

Consequently, 0 = [5th]:

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

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

Now, note that to avoid ⛔「min ⊢5⊣ ≥ 2」, we need 5 adjacent to 1 or 0. We have only one way to do so:

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

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

Then, to match ✅「4th → a, 1st → b, ab=2+4n」, we have [1st] = 6|2. Actually [1st] != 6, for we have to match ✅「⟨ ⁶ᵗʰc       ²ⁿᵈa ¹ˢᵗb   ⟩, a > b > c」 as well. Hence, we get

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

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

Then, by ✅「Jump(2,3) = 1」, plainly we have [3rd] = 3.

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

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

Finally, in view of ⛔「⟨⋯ ? ⋯ 6 ⋯ (?+1)⟩ (?≠5)」, we finish by

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

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

Q.E.D.

#125034_v2.4