2024-05-02 M3

P2 (老羅)

✅Match

max ⊢1⊣ ≤ 3

4th → a, 0th → b, a+b=5

⟨⋯ Perm(0,1,3) ⋯⟩

{p5, p2, p1} = ? + {0,1,2}

⛔Avoid

Sim⟨ ⁵ᵗʰ2 ⁴ᵗʰ4 ³ʳᵈ0 ²ⁿᵈ3 ¹ˢᵗ5 ⁰ᵗʰ1 ⟩ ≥ 1

Last HP: 0

Lowe the Lion defeated 老羅!

#125034_v2.5

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

Proof of 2024-05-02 M3
══════════════════════

Notation: if Nth -> a, then we write pN = a.

To match ✅「{p5, p2, p1} = ? + {0,1,2}」, we need p5,p2,p1 to be consecutive integers. 

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

On the other hand, by ✅「4th → a, 0th → b, a+b=5」, there are three possibilities of {p4,p0}:

(1) {p4,p0} = {2,3}; then p5,p2,p1 ∈ {0,1,4,5}.

(2) {p4,p0} = {1,4}; then p5,p2,p1 ∈ {0,2,3,5}.

(3) {p4,p0} = {0,5}; then p5,p2,p1 ∈ {1,2,3,4}.

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

Observe that only case (3) allows p5,p2,p1 to be consecutive integers. So, case (3) holds, and one of the following happens:

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

We show that (5) holds actually.

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

If (4) holds on the contrary, then by ✅「⟨⋯ Perm(0,1,3) ⋯⟩」, we have {p2,p1} = {1,3}:

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

Then, we need p5 = 2 in order that p5,p2,p1 are consecutive integers. This is a contradiction, however, in view of ⛔「Sim⟨ ⁵ᵗʰ2 ⁴ᵗʰ4 ³ʳᵈ0 ²ⁿᵈ3 ¹ˢᵗ5 ⁰ᵗʰ1 ⟩ ≥ 1」.

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

We have verified that (5) holds. Accordingly, we get

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

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

Next, we consider how to match ✅「⟨⋯ Perm(0,1,3) ⋯⟩」. We need three adjacent boxes with one of them at 4th. There are two ways to do so:

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

No matter which happens, we use the 3rd position. Therefore,

(8) p3 = 1|3.

We have p3 = 1 actually. For, if on the contraray p3 = 3:

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

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

then we do not have three consecutive integers left for matching ✅「{p5, p2, p1} = ? + {0,1,2}」. So, back to (8), we get:

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

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

We proceed to determine the value of p2. To avoid ⛔「Sim⟨ ⁵ᵗʰ2 ⁴ᵗʰ4 ³ʳᵈ0 ²ⁿᵈ3 ¹ˢᵗ5 ⁰ᵗʰ1 ⟩ ≥ 1」, we need p2 != 3; and to match ✅「max ⊢1⊣ ≤ 3」, we need p2 != 4. It follows that p2 = 2:

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

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

Finally, in view of ✅「⟨⋯ Perm(0,1,3) ⋯⟩」, we finish by

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

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

Q.E.D.

#125034_v2.5