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│▒
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Step 1 │ │ │ │ │ │ 5 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ 0 │ │ │ │ 5 │▒
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Step 3 │ │ 0 │ 1 │ │ │ 5 │▒
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Step 4 │ │ 0 │ 1 │ 2 │ │ 5 │▒
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Step 5 │ 3 │ 0 │ 1 │ 2 │ │ 5 │▒
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Step 6 │ 3 │ 0 │ 1 │ 2 │ 4 │ 5 │▒
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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│
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(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