Perfect Dominating sets and Perfect Domination Polynomial of a Path

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ISSN: 2005-4238 IJAST Copyright ⓒ 2019 SERSC

Perfect Dominating sets and Perfect Domination Polynomial of a Path

A.M. Anto¹, P. Paul Hawkins² & T. Shyla Isac Mary³

1Assistant Professor, Department of Mathematics,Malankara Catholic College, Mariagiri, Tamil Nadu, India.

²Research Scholar,Reg.no 18223112091013, Research Department of Mathematics, Nesamony Memorial Christian College, Marthandam, Tamil Nadu, India.

3Assistant Professor, Research Department of Mathematics, Nesamony Memorial Christian College, Marthandam, Tamil Nadu, India

1,2,3Affiliated to Manonmaniam Sundaranar University, Abishekapatti, Tirunelveli – 627012,Tamilnadu,India.

Abstract

Let 𝐺 = (𝑉, 𝐸) be a simple graph. A set 𝑆 ⊆ 𝑉 is a perfect dominating set of G, if every vertex 𝑢 in 𝑉 − 𝑆 is adjacent to exactly one vertex in S. Let 𝛲_𝑝𝑓(𝑛, 𝑖) be the family of all perfect dominating sets of a path 𝑃_𝑛 with cardinality 𝑖, and let 𝑑_𝑝𝑓(𝑃_𝑛, 𝑖) =

|𝛲_𝑝𝑓(𝑛, 𝑖) |. In this paper, we construct 𝛲_𝑝𝑓(𝑛, 𝑖) for a path 𝑃_𝑛 with 𝑛 vertices and obtain a recursive formula of 𝑑_𝑝𝑓(𝑃_𝑛, 𝑖) . Using this recursive formula, we consider the polynomial 𝐷_𝑝𝑓(𝑃_𝑛, 𝑥) = ∑^𝑛_{𝑖=⌈𝑛/3⌉}𝑑_𝑝𝑓(𝑃_𝑛, 𝑖)𝑥^𝑖 , which we call perfect domination polynomial of paths and obtain some properties of this polynomial.

Keyword:. Dominating set, Minimal dominating set

1. Introduction

Let 𝐺 = (𝑉, 𝐸) be a simple graph of order |𝑉| = 𝑛. For any vertex 𝑢 ∈ 𝑉, the open neighborhood of 𝑢 is the set 𝑁(𝑢) = {𝑣 ∈ 𝑉|𝑢𝑣 ∈ 𝐸}. A set 𝑆 ⊆ 𝑉 is a dominating set of 𝐺, if every vertex 𝑢 ∈ 𝑉 is a element of 𝑆 or is adjacent to an element of 𝑆 [8].

The dominating set S is a perfect dominating set if |𝑁(𝑢) ∩ 𝑆| = 1 for each ∈ 𝑉 − 𝑆 [8], or equivalently, if every vertex 𝑢 in 𝑉 − 𝑆 is adjacent to exactly one vertex in S. The perfect domination number 𝛾_𝑝𝑓 is the minimum cardinality of a perfect dominating set in 𝐺. A path is a connected graph in which the end vertices have degree one and the remaining internal vertices have degree two. Let 𝑃_𝑛 be the path with 𝑛 vertices. Let Ρ_𝑝𝑓(𝑛, 𝑖) be the family of perfect dominating sets of a path 𝑃_𝑛 with cardinality 𝑖 and let 𝑑_𝑝𝑓(𝑃_𝑛, 𝑖) =

|Ρ_𝑝𝑓(𝑛, 𝑖) |. We call the polynomial 𝐷_𝑝𝑓(𝑃_𝑛, 𝑥) = ∑^𝑛_{𝑖=⌈𝑛/3⌉}𝑑_𝑝𝑓(𝑃_𝑛, 𝑖)𝑥^𝑖 , the perfect domination polynomial of the path 𝑃𝑛. We use ⌈𝑛⌉, for the smallest integer greater than or equal to 𝑛. In this article we denote the set {1,2, . . . , 𝑛} simply by [𝑛]

2. Perfect Dominating sets of a path

Let Ρ_𝑝𝑓(𝑛, 𝑖) be the family of perfect dominating sets of the path 𝑃_𝑛 with cardinality 𝑖. We are going to investigate perfect dominating sets of path 𝑃_𝑛.

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We required the following lemmas to prove our main results in this article.

Lemma 2.1 𝛾_𝑝𝑓(𝑃_𝑛) = ⌈^𝑛

3⌉.

By Lemma 2.1 and the definition of perfect domination number, we have the following lemma.

Lemma 2.2

Ρ_𝑝𝑓(𝑛, 𝑖) = ∅, if and only if 𝑖 > 𝑛 or 𝑖 < ⌈^𝑛

3⌉ . Proof

Let 𝑃_𝑛 be a path with 𝑛 vertices and any member of Ρ_𝑝𝑓(𝑛, 𝑖) contains atmost 𝑛 vertices.

Therefore Ρ_𝑝𝑓(𝑛, 𝑖) = ∅ for 𝑖 > 𝑛.

Since, ⌈^𝑛

3⌉ is a minimum cardinality of perfect dominating sets of path 𝑃_𝑛. Hence there is no perfect dominating set of cardinality less than ⌈^𝑛

3⌉.

Therefore Ρ_𝑝𝑓(𝑛, 𝑖) = ∅ for 𝑖 < ⌈^𝑛

3⌉.

Hence, Ρ_𝑝𝑓(𝑛, 𝑖) = ∅, if and only if 𝑖 > 𝑛 or 𝑖 < ⌈^𝑛

3⌉ . Conversely, if 𝑖 > 𝑛 or 𝑖 < ⌈^𝑛

3⌉ then by definition of perfect dominating set of path 𝑃_𝑛 we have Ρ_𝑝𝑓(𝑛, 𝑖) = ∅.

The following lemma follows from observation.

Lemma 2.3

If 𝐺 be a path of length 3𝑘 − 1, then every perfect dominating set of 𝐺 must contains at least 𝑘 vertices of 𝐺.

Lemma 2.4

𝑖) If Ρ_𝑝𝑓(𝑛 − 1, 𝑖 − 1) = Ρ_𝑝𝑓(𝑛 − 3, 𝑖 − 1) = ∅ then Ρ_𝑝𝑓(𝑛 − 2, 𝑖 − 1) = ∅.

𝑖𝑖) If Ρ_𝑝𝑓(𝑛 − 1, 𝑖 − 1) ≠ ∅ and Ρ_𝑝𝑓(𝑛 − 3, 𝑖 − 1) ≠ ∅ then Ρ_𝑝𝑓(𝑛 − 2, 𝑖 − 1) ≠ ∅.

𝑖𝑖𝑖) If Ρ_𝑝𝑓(𝑛 − 1, 𝑖 − 1) = Ρ_𝑝𝑓(𝑛 − 2, 𝑖 − 1) = Ρ_𝑝𝑓(𝑛 − 3, 𝑖 − 1) = ∅ then Ρ_𝑝𝑓(𝑛, 𝑖) = ∅.

Proof

𝑖)We have, Ρ_𝑝𝑓(𝑛 − 1, 𝑖 − 1) = Ρ_𝑝𝑓(𝑛 − 3, 𝑖 − 1) = ∅ then by lemma 2.2 𝑖 − 1 > 𝑛 − 1 or 𝑖 − 1 < ⌈^𝑛−1

3 ⌉ and 𝑖 − 1 > 𝑛 − 3 or 𝑖 − 1 < ⌈^𝑛−3

3 ⌉. Therefore, we get 𝑖 − 1 > 𝑛 − 1 or 𝑖 − 1 < ⌈^𝑛−3

3 ⌉ < ⌈^𝑛−2

3 ⌉. Hence, Ρ_𝑝𝑓(𝑛 − 2, 𝑖 − 1) = ∅.

𝑖𝑖)Suppose we assume that, Ρ_𝑝𝑓(𝑛 − 2, 𝑖 − 1) = ∅ then by lemma 2.2 we have 𝑖 − 1 > 𝑛 − 2 or 𝑖 − 1 < ⌈^𝑛−2

3 ⌉. If 𝑖 − 1 > 𝑛 − 2 then 𝑖 − 1 > 𝑛 − 2 > 𝑛 − 3 hence Ρ_𝑝𝑓(𝑛 − 3, 𝑖 − 1) = ∅ which is a contradiction. Therefore, we take 𝑖 − 1 <

⌈^𝑛−2

3 ⌉ ⟹ 𝑖 − 1 < ⌈^𝑛−2

3 ⌉ < ⌈^𝑛−1

3 ⌉. Which implies Ρ_𝑝𝑓(𝑛 − 1, 𝑖 − 1) = ∅, is also a contradiction. Hence Ρ_𝑝𝑓(𝑛 − 2, 𝑖 − 1) ≠ ∅.

𝑖𝑖𝑖) Suppose we assume that, Ρ_𝑝𝑓(𝑛, 𝑖) ≠ ∅. Let 𝑌 ∈ 𝑃_𝑝𝑓(𝑛, 𝑖) then at least one vertex labeled as 𝑛 or 𝑛 − 1 in 𝑌.

If 𝑛 ∈ 𝑌 then at least one vertex labeled as 𝑛 − 1 or 𝑛 − 3 in 𝑌. If 𝑛 − 1 ∈ 𝑌 then 𝑌 − {𝑛} ∈ Ρ_𝑝𝑓(𝑛 − 1, 𝑖 − 1) is a contradiction that Ρ_𝑝𝑓(𝑛 − 1, 𝑖 − 1) = ∅. If

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𝑛 − 3 ∈ 𝑌 then 𝑌 − {𝑛} ∈ Ρ_𝑝𝑓(𝑛 − 2, 𝑖 − 1) is also a contradiction.

Hence we take 𝑛 − 1 ∈ 𝑌 then 𝑛 − 2 ∈ 𝑌 or 𝑛 − 4 ∈ 𝑌. If 𝑛 − 2 ∈ 𝑌 then 𝑌 − {𝑛 − 1} ∈ Ρ_𝑝𝑓(𝑛 − 1, 𝑖 − 1) is a contradiction and if 𝑛 − 4 ∈ 𝑌 then 𝑌 − {𝑛 − 1} ∈ Ρ_𝑝𝑓(𝑛 − 3, 𝑖 − 1) is also leads to a contradiction. Hence Ρ_𝑝𝑓(𝑛, 𝑖) = ∅.

Lemma 2.5

If Ρ_𝑝𝑓(𝑛, 𝑖) ≠ ∅ then

𝑖)Ρ_𝑝𝑓(𝑛 − 1, 𝑖 − 1) = Ρ_𝑝𝑓(𝑛 − 2, 𝑖 − 1) = ∅ and Ρ_𝑝𝑓(𝑛 − 3, 𝑖 − 1) ≠ ∅ if and only if 𝑛 = 3𝑘 and 𝑖 = 𝑘 for some 𝑘 ∈ 𝑁;

𝑖𝑖)Ρ_𝑝𝑓(𝑛 − 2, 𝑖 − 1) = Ρ_𝑝𝑓(𝑛 − 3, 𝑖 − 1) = ∅ and Ρ_𝑝𝑓(𝑛 − 1, 𝑖 − 1) ≠ ∅ if and only if 𝑖 = 𝑛;

𝑖𝑖𝑖)Ρ_𝑝𝑓(𝑛 − 1, 𝑖 − 1) = ∅, Ρ_𝑝𝑓(𝑛 − 2, 𝑖 − 1) ≠ ∅ and Ρ_𝑝𝑓(𝑛 − 3, 𝑖 − 1) ≠ ∅ if and only if 𝑛 = 3𝑘 + 2 and 𝑖 = 𝑘 + 1 = ⌈^(3𝑘+2)

3 ⌉ for some 𝑘 ∈ 𝑁;

𝑖𝑣)Ρ_𝑝𝑓(𝑛 − 1, 𝑖 − 1) ≠ ∅, Ρ_𝑝𝑓(𝑛 − 2, 𝑖 − 1) ≠ ∅ and Ρ_𝑝𝑓(𝑛 − 3, 𝑖 − 1) = ∅ if and only if 𝑖 = 𝑛 − 1;

𝑣)Ρ_𝑝𝑓(𝑛 − 1, 𝑖 − 1) ≠ ∅, Ρ_𝑝𝑓(𝑛 − 2, 𝑖 − 1) ≠ ∅, Ρ_𝑝𝑓(𝑛 − 3, 𝑖 − 1) ≠ ∅ and Ρ_𝑝𝑓(𝑛 − 4, 𝑖 − 1) = ∅ if and only if 𝑖 = 𝑛 − 2.

𝑣𝑖) If Ρ_𝑝𝑓(𝑛 − 1, 𝑖 − 1) ≠ ∅, Ρ_𝑝𝑓(𝑛 − 2, 𝑖 − 1) ≠ ∅, Ρ_𝑝𝑓(𝑛 − 3, 𝑖 − 1) ≠ ∅ and Ρ_𝑝𝑓(𝑛 − 4, 𝑖 − 1) ≠ ∅ if and only if ⌈^𝑛−4

3 ⌉ + 1 ≤ 𝑖 ≤ 𝑛 − 3.

Proof

Since, Ρ_𝑝𝑓(𝑛 − 1, 𝑖 − 1) = Ρ_𝑝𝑓(𝑛 − 2, 𝑖 − 1) = ∅ by lemma 2.2 we have 𝑖 − 1 >

𝑛 − 1 or 𝑖 − 1 < ⌈^𝑛−2

3 ⌉. If 𝑖 − 1 > 𝑛 − 1 ⟹ 𝑖 > 𝑛 then again by lemma 2.2 Ρ_𝑝𝑓(𝑛, 𝑖) = ∅, is a contradiction. So, we take 𝑖 − 1 < ⌈^𝑛−2

3 ⌉ ⟹ 𝑖 < ⌈^𝑛−2

3 ⌉ + 1.

Also, since we have Ρ_𝑝𝑓(𝑛, 𝑖) ≠ ∅ implies 𝑖 ≥ ⌈^𝑛

3⌉. Hence we get ⌈^𝑛

3⌉ ≤ 𝑖 <

⌈^𝑛−2

3 ⌉ + 1.Suppose we take 𝑛 = 3𝑘 then ⌈^𝑛

3⌉ = ⌈^3𝑘

3⌉ = 𝑘 and ⌈^𝑛−2

3 ⌉ + 1 =

⌈^3𝑘−2

3 ⌉ + 1 = ⌈𝑘 −²

3⌉ + 1 = 𝑘 + 1 for some 𝑘 ∈ 𝑁. Hence we have 𝑘 ≤ 𝑖 < 𝑘 + 1. Since 𝑘 ∈ 𝑁 we concluded by 𝑖 = 𝑘. Suppose 𝑛 ≠ 3𝑘 then we have the inequality 𝑘 ≤ 𝑖 < 𝑘 is a contradiction. Therefore, 𝑛 = 3𝑘 and 𝑖 = 𝑘 for some 𝑘 ∈ 𝑁.

Conversely, if 𝑛 = 3𝑘 and 𝑖 = 𝑘 for some 𝑘 ∈ 𝑁.

If 𝑖 = 𝑘 =^𝑛

3 Then, 𝑖 − 1 =^𝑛

3− 1 =^𝑛−3

3

Hence, we get 𝑖 − 1 =^𝑛−3

3 < ^𝑛−2

3 < ⌈^𝑛−2

3 ⌉ and 𝑖 − 1 =^𝑛−3

3 <^𝑛−1

3 < ⌈^𝑛−1

3 ⌉ Therefore, Ρ_𝑝𝑓(𝑛 − 1, 𝑖 − 1) = Ρ_𝑝𝑓(𝑛 − 2, 𝑖 − 1) = ∅.

Also, 𝑖 − 1 =^𝑛

3− 1 =^𝑛−3

3 < 𝑛 − 3 ⟹ 𝑖 − 1 < 𝑛 − 3.

Therefore, Ρ_𝑝𝑓(𝑛 − 3, 𝑖 − 1) ≠ ∅.

𝑖𝑖) Since Ρ_𝑝𝑓(𝑛 − 2, 𝑖 − 1) = Ρ_𝑝𝑓(𝑛 − 3, 𝑖 − 1) = ∅ by lemma 2.2 we have 𝑖 − 1 > 𝑛 − 2 or 𝑖 − 1 < ⌈^𝑛−3

3 ⌉. If 𝑖 − 1 < ⌈^𝑛−3

3 ⌉ then 𝑖 − 1 < ⌈^𝑛−3

3 ⌉ < ⌈^𝑛−1

3 ⌉.

Therefore, Ρ_𝑝𝑓(𝑛 − 1, 𝑖 − 1) ≠ ∅ is a contradiction. So, 𝑖 − 1 > 𝑛 − 2 ⟹ 𝑖 >

𝑛 − 1. Also, Since Ρ_𝑝𝑓(𝑛 − 1, 𝑖 − 1) ≠ ∅ we have 𝑖 − 1 ≤ 𝑛 − 1 ⟹ 𝑖 ≤ 𝑛 together we get 𝑛 − 1 < 𝑖 ≤ 𝑛. Therefore 𝑖 = 𝑛.

Conversely, if 𝑖 = 𝑛 then by lemma 2.2 Ρ_𝑝𝑓(𝑛 − 2, 𝑖 − 1) = Ρ_𝑝𝑓(𝑛 − 2, 𝑛 − 1) =

∅, Ρ_𝑝𝑓(𝑛 − 3, 𝑖 − 1) = Ρ_𝑝𝑓(𝑛 − 3, 𝑛 − 1) = ∅ and Ρ_𝑝𝑓(𝑛 − 1, 𝑖 − 1) =

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Ρ_𝑝𝑓(𝑛 − 1, 𝑛 − 1) ≠ ∅.

𝑖𝑖𝑖)Since Ρ_𝑝𝑓(𝑛 − 1, 𝑖 − 1) = ∅ then by lemma 2.2 we have 𝑖 − 1 > 𝑛 − 1 or 𝑖 − 1 < ⌈^𝑛−1

3 ⌉. If 𝑖 − 1 > 𝑛 − 1 then 𝑖 − 1 > 𝑛 − 2 then by lemma 2.2 Ρ_𝑝𝑓(𝑛 − 2, 𝑖 − 1) = ∅ is a contradiction. So 𝑖 − 1 < ⌈^𝑛−1

3 ⌉ ⟹ 𝑖 < ⌈^𝑛−1

3 ⌉ + 1. Also, since Ρ_𝑝𝑓(𝑛 − 2, 𝑖 − 1) ≠ ∅ we have 𝑖 − 1 ≥ ⌈^𝑛−2

3 ⌉ ⟹ 𝑖 ≥ ⌈^𝑛−2

3 ⌉ + 1 together we have

⌈^𝑛−2

3 ⌉ + 1 ≤ 𝑖 < ⌈^𝑛−1

3 ⌉ + 1.

Suppose we take 𝑛 = 3𝑘 + 2 then ⌈^𝑛−2

3 ⌉ = ⌈^3𝑘

3⌉ = 𝑘 ⟹ ⌈^𝑛−2

3 ⌉ + 1 = 𝑘 + 1 and

⌈^𝑛−1

3 ⌉ + 1 = ⌈^3𝑘+1

3 ⌉ + 1 = ⌈𝑘 +¹

3⌉ + 1 = 𝑘 + 2 for some 𝑘 ∈ 𝑁. Hence we have 𝑘 + 1 ≤ 𝑖 < 𝑘 + 2. Since 𝑘 ∈ 𝑁 we concluded by 𝑖 = 𝑘 + 1. Suppose 𝑛 ≠ 3𝑘 + 2 then we have the inequality 𝑘 ≤ 𝑖 < 𝑘 is a contradiction. Therefore, 𝑛 = 3𝑘 + 2 and 𝑖 = 𝑘 + 1 = ⌈^(3𝑘+2)

3 ⌉ for some 𝑘 ∈ 𝑁.

Conversely, if 𝑛 = 3𝑘 + 2 and 𝑖 = 𝑘 + 1 = ⌈^(3𝑘+2)

3 ⌉ for some 𝑘 ∈ 𝑁. Now, 𝑛 = 3𝑘 + 2 ⟹ 𝑘 = ^𝑛−2

3 and 𝑖 = 𝑘 + 1 ⟹ 𝑖 − 1 = 𝑘. Therefore, 𝑖 − 1 = ^𝑛−2

3 <

𝑛−1

3 < ⌈^𝑛−1

3 ⌉. Hence, Ρ_𝑝𝑓(𝑛 − 1, 𝑖 − 1) = ∅. Also, 𝑖 − 1 =^𝑛−2

3 ≥

⌈^𝑛−2

3 ⌉ ⟹ Ρ_𝑝𝑓(𝑛 − 2, 𝑖 − 1) ≠ ∅ and 𝑖 − 1 =^𝑛−2

3 > ⌈^𝑛−3

3 ⌉ ⟹ Ρ_𝑝𝑓(𝑛 − 3, 𝑖 − 1) ≠ ∅.

𝑖𝑣)Since Ρ_𝑝𝑓(𝑛 − 3, 𝑖 − 1) = ∅ we have 𝑖 − 1 > 𝑛 − 3 or 𝑖 − 1 < ⌈^𝑛−3

3 ⌉ ⟹ 𝑖 <

⌈^𝑛−3

3 ⌉ + 1. Also, since Ρ_𝑝𝑓(𝑛 − 2, 𝑖 − 1) ≠ ∅ we have 𝑖 − 1 ≤ 𝑛 − 2 or 𝑖 − 1 ≥

⌈^𝑛−2

3 ⌉ hence we get ⌈^𝑛−2

3 ⌉ + 1 ≤ 𝑖 ≤ 𝑛 − 1. If 𝑖 < ⌈^𝑛−3

3 ⌉ + 1 then ⌈^𝑛−2

3 ⌉ + 1 ≤ 𝑖 ≤ 𝑛 − 1 < ⌈^𝑛−3

3 ⌉ + 1 is not possible. Therefore, 𝑖 − 1 > 𝑛 − 3 ⟹ 𝑖 > 𝑛 − 2.

Hence 𝑖 = 𝑛 or 𝑖 = 𝑛 − 1. If 𝑖 = 𝑛 then Ρ_𝑝𝑓(𝑛 − 2, 𝑖 − 1) = Ρ_𝑝𝑓(𝑛 − 2, 𝑛 − 1) = ∅ is a contradiction. Therefore 𝑖 = 𝑛 − 1.

Conversely, if 𝑖 = 𝑛 − 1 then by lemma 2.2 Ρ_𝑝𝑓(𝑛 − 1, 𝑖 − 1) = Ρ_𝑝𝑓(𝑛 − 1, 𝑛 − 2) ≠ ∅, Ρ_𝑝𝑓(𝑛 − 2, 𝑖 − 1) = Ρ_𝑝𝑓(𝑛 − 2, 𝑛 − 2) ≠ ∅ and Ρ_𝑝𝑓(𝑛 − 3, 𝑖 − 1) = Ρ_𝑝𝑓(𝑛 − 3, 𝑛 − 2) = ∅.

𝑣) Since, Ρ_𝑝𝑓(𝑛 − 4, 𝑖 − 1) = ∅ we have by lemma 2.2 𝑖 − 1 > 𝑛 − 4 or 𝑖 − 1 < ⌈^𝑛−4

3 ⌉ ⟹ 𝑖 < ⌈^𝑛−4

3 ⌉ + 1.Also since, Ρ_𝑝𝑓(𝑛 − 3, 𝑖 − 1) ≠ ∅ then we have 𝑖 − 1 ≥ ⌈^𝑛−3

3 ⌉ ⟹ 𝑖 − 1 > ⌈^𝑛−4

3 ⌉ ⟹ 𝑖 > ⌈^𝑛−4

3 ⌉ + 1. Hence, 𝑖 − 1 < ⌈^𝑛−4

3 ⌉ is not possible. Therefore, 𝑖 − 1 > 𝑛 − 4 ⟹ 𝑖 ≥ 𝑛 − 2. But, Ρ_𝑝𝑓(𝑛 − 2, 𝑖 − 1) ≠ ∅ implies 𝑖 ≤ 𝑛 − 2. Hence, 𝑖 = 𝑛 − 2.

Conversely, if 𝑖 = 𝑛 − 2 then, Ρ_𝑝𝑓(𝑛 − 1, 𝑖 − 1) = Ρ_𝑝𝑓(𝑛 − 1, 𝑛 − 3) ≠

∅, Ρ_𝑝𝑓(𝑛 − 2, 𝑖 − 1) = Ρ_𝑝𝑓(𝑛 − 2, 𝑛 − 3) ≠ ∅, Ρ_𝑝𝑓(𝑛 − 3, 𝑖 − 1) = Ρ_𝑝𝑓(𝑛 − 3, 𝑛 − 3) ≠ ∅ and Ρ_𝑝𝑓(𝑛 − 4, 𝑖 − 1) = Ρ_𝑝𝑓(𝑛 − 4, 𝑛 − 3) = ∅.

𝑣𝑖) Since, Ρ_𝑝𝑓(𝑛 − 1, 𝑖 − 1) ≠ ∅, Ρ_𝑝𝑓(𝑛 − 2, 𝑖 − 1) ≠ ∅, Ρ_𝑝𝑓(𝑛 − 3, 𝑖 − 1) ≠ ∅ and Ρ_𝑝𝑓(𝑛 − 4, 𝑖 − 1) ≠ ∅ by applying lemma 2.2 we have 𝑖 − 1 ≤ 𝑛 − 1 or 𝑖 − 1 ≥ ⌈^𝑛−1

3 ⌉ ⟹ ⌈^𝑛−1

3 ⌉ ≤ 𝑖 − 1 ≤ 𝑛 − 1, 𝑖 − 1 ≤ 𝑛 − 2 or 𝑖 − 1 ≥ ⌈^𝑛−2

3 ⌉ ⟹

⌈^𝑛−2

3 ⌉ ≤ 𝑖 − 1 ≤ 𝑛 − 2, 𝑖 − 1 ≤ 𝑛 − 3 or 𝑖 − 1 ≥ ⌈^𝑛−3

3 ⌉ ⟹ ⌈^𝑛−3

3 ⌉ ≤ 𝑖 − 1 ≤

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𝑛 − 3 and 𝑖 − 1 ≤ 𝑛 − 4 or 𝑖 − 1 ≥ ⌈^𝑛−4

3 ⌉ ⟹ ⌈^𝑛−4

3 ⌉ ≤ 𝑖 − 1 ≤ 𝑛 − 4. Hence, we have ⌈^𝑛−4

3 ⌉ ≤ 𝑖 − 1 ≤ 𝑛 − 4 which implies ⌈^𝑛−4

3 ⌉ + 1 ≤ 𝑖 ≤ 𝑛 − 3.

Conversely, if ⌈^𝑛−4

3 ⌉ + 1 ≤ 𝑖 ≤ 𝑛 − 3 then by lemma 2.2 Ρ_𝑝𝑓(𝑛 − 1, 𝑖 − 1) ≠

∅, Ρ_𝑝𝑓(𝑛 − 2, 𝑖 − 1) ≠ ∅,Ρ_𝑝𝑓(𝑛 − 3, 𝑖 − 1) ≠ ∅ and Ρ_𝑝𝑓(𝑛 − 4, 𝑖 − 1) ≠ ∅.

Theorem 2.6

For every 𝑛 ≥ 4 and 𝑖 ≥ ⌈^𝑛

3⌉

𝑖) If Ρ_𝑝𝑓(𝑛 − 1, 𝑖 − 1) = Ρ_𝑝𝑓(𝑛 − 2, 𝑖 − 1) = ∅ and Ρ_𝑝𝑓(𝑛 − 3, 𝑖 − 1) ≠ ∅, then Ρ_𝑝𝑓(𝑛, 𝑖) = {{2,5,8, … , 𝑛 − 4, 𝑛 − 1}};

𝑖𝑖) If Ρ_𝑝𝑓(𝑛 − 2, 𝑖 − 1) = Ρ_𝑝𝑓(𝑛 − 3, 𝑖 − 1) = ∅ and Ρ_𝑝𝑓(𝑛 − 1, 𝑖 − 1) ≠ ∅, then Ρ_𝑝𝑓(𝑛, 𝑖) = {[𝑛]};

𝑖𝑖𝑖) If Ρ_𝑝𝑓(𝑛 − 1, 𝑖 − 1) = ∅, Ρ_𝑝𝑓(𝑛 − 2, 𝑖 − 1) ≠ ∅ and Ρ_𝑝𝑓(𝑛 − 3, 𝑖 − 1) ≠ ∅, then Ρ_𝑝𝑓(𝑛, 𝑖) = {{1,4,7, … ,3𝑛 + 1}, {2,5,8, … ,3𝑛 + 2}};

𝑖𝑣) If Ρ_𝑝𝑓(𝑛 − 1, 𝑖 − 1) ≠ ∅, Ρ_𝑝𝑓(𝑛 − 2, 𝑖 − 1) ≠ ∅ and Ρ_𝑝𝑓(𝑛 − 3, 𝑖 − 1) = ∅, then Ρ_𝑝𝑓(𝑛, 𝑖) = {{1,2,3, … , 𝑛 − 1}, {2,3,4, … , 𝑛}};

𝑣) If Ρ_𝑝𝑓(𝑛 − 1, 𝑖 − 1) ≠ ∅, Ρ_𝑝𝑓(𝑛 − 2, 𝑖 − 1) ≠ ∅ and Ρ_𝑝𝑓(𝑛 − 3, 𝑖 − 1) ≠ ∅, Ρ_𝑝𝑓(𝑛 − 4, 𝑖 − 1) = ∅, then Ρ_𝑝𝑓(𝑛, 𝑖) = {{2,3,4, … , 𝑛 − 1}, {[𝑛] − {𝑥_𝑖, 𝑥_𝑖+1}|𝑖 = 2,3, … 𝑛 − 2}}

Proof

𝑖) If Ρ_𝑝𝑓(𝑛 − 1, 𝑖 − 1) = Ρ_𝑝𝑓(𝑛 − 2, 𝑖 − 1) = ∅ and Ρ_𝑝𝑓(𝑛 − 3, 𝑖 − 1) ≠ ∅, then by lemma 2.5 (𝑖), 𝑛 = 3𝑘 and 𝑖 = 𝑘 for some 𝑘 ∈ 𝑁. Therefore, 𝑖 = 𝑘 =^𝑛

3. Hence, Ρ_𝑝𝑓(𝑛, 𝑖) = Ρ_𝑝𝑓(𝑛, 𝑛/3). T. Hence, the only possible perfect dominating set of 𝑃_𝑛 with cardinality 𝑖 =^𝑛

3 is {2,5,8, … , 𝑛 − 4, 𝑛 − 1}.

𝑖𝑖) If Ρ_𝑝𝑓(𝑛 − 2, 𝑖 − 1) = Ρ_𝑝𝑓(𝑛 − 3, 𝑖 − 1) = ∅ and Ρ_𝑝𝑓(𝑛 − 1, 𝑖 − 1) ≠ ∅, then by lemma 2.5 (𝑖𝑖) 𝑖 = 𝑛. So, Ρ_𝑝𝑓(𝑛, 𝑖) = Ρ_𝑝𝑓(𝑛, 𝑛) = {{1,2,3, … , 𝑛}} = {[𝑛]}.

𝑖𝑖𝑖) If Ρ_𝑝𝑓(𝑛 − 1, 𝑖 − 1) = ∅, Ρ_𝑝𝑓(𝑛 − 2, 𝑖 − 1) ≠ ∅ and Ρ_𝑝𝑓(𝑛 − 3, 𝑖 − 1) ≠ ∅, then by lemma 2.5(𝑖𝑖𝑖) we have 𝑛 = 3𝑘 + 1 and 𝑖 = 𝑘 + 1. Therefore, 𝑖 =

𝑛+2

3 .Hence, Ρ_𝑝𝑓(𝑛, 𝑖) = Ρ_𝑝𝑓(𝑛,^𝑛+2

3 ). Clearly, {1,4,7, … ,3𝑛 + 1}, {2,5,8, … ,3𝑛 + 2} are the perfect dominating set with cardinality 𝑖 = ^𝑛+2

3 and no other sets with cardinality 𝑖 =^𝑛+2

3 is a perfect dominating set. Hence, Ρ_𝑝𝑓(𝑛, 𝑖) = {{1,4,7, … ,3𝑛 + 1}, {2,5,8, … ,3𝑛 + 2}}.

𝑖𝑣) If Ρ_𝑝𝑓(𝑛 − 1, 𝑖 − 1) ≠ ∅, Ρ_𝑝𝑓(𝑛 − 2, 𝑖 − 1) ≠ ∅ and Ρ_𝑝𝑓(𝑛 − 3, 𝑖 − 1) = ∅, then by lemma 2.5 (𝑖𝑣), 𝑖 = 𝑛 − 1. Therefore, Ρ_𝑝𝑓(𝑛, 𝑖) = Ρ_𝑝𝑓(𝑛, 𝑛 − 1). Clearly, in a path 𝑃_𝑛 with 𝑛 vertices the perfect dominating sets with cardinality 𝑛 − 1 is the removal of end vertices. Hence, Ρ_𝑝𝑓(𝑛, 𝑖) = {{1,2,3, … , 𝑛 − 1}, {2,3,4, … , 𝑛}}.

𝑣) If Ρ_𝑝𝑓(𝑛 − 1, 𝑖 − 1) ≠ ∅, Ρ_𝑝𝑓(𝑛 − 2, 𝑖 − 1) ≠ ∅ and Ρ_𝑝𝑓(𝑛 − 3, 𝑖 − 1) ≠ ∅, Ρ_𝑝𝑓(𝑛 − 4, 𝑖 − 1) = ∅, Then, by lemma 2.5 (𝑣) 𝑖 = 𝑛 − 2. Therefore,

Ρ_𝑝𝑓(𝑛, 𝑖) = Ρ_𝑝𝑓(𝑛, 𝑛 − 2). Hence the possible perfect dominating set is the removal of two vertices in 𝑛 vertices.Clearly, {2,3,4, … , 𝑛 − 1} and {[𝑛] − {𝑥_𝑖, 𝑥_𝑖+1}|𝑖 = 2,3, … 𝑛 − 2} are the possible perfect dominating sets with cardinality 𝑖 = 𝑛 − 2.

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3. Perfect Domination Polynomial of a Path

Definition 3.1

Let 𝑃_𝑛 be a path with 𝑛 vertices. Let Ρ_𝑝𝑓(𝑛, 𝑖) be the family of perfect dominating sets of a path 𝑃_𝑛 with cardinality 𝑖 and 𝑑_𝑝𝑓(𝑃_𝑛, 𝑖) = |Ρ_𝑝𝑓(𝑛, 𝑖) |. Then the Perfect dominating polynomial of a path 𝑃_𝑛 is given by 𝐷_𝑝𝑓(𝑃_𝑛, 𝑥) =

∑^𝑛_{𝑖=⌈𝑛/3⌉}𝑑_𝑝𝑓(𝑃_𝑛, 𝑖)𝑥^𝑖 . Theorem 3.2

If Ρ_𝑝𝑓(𝑛, 𝑖) be a family of perfect dominating sets with cardinality 𝑖 then, 𝑖) |Ρ_𝑝𝑓(𝑛, 𝑖)| = |Ρ_𝑝𝑓(𝑛 − 1, 𝑖 − 1)| + |Ρ_𝑝𝑓(𝑛 − 3, 𝑖 − 1)|

𝑖𝑖) For every 𝑛 ≥ 4, 𝐷_𝑝𝑓(𝑃_𝑛, 𝑥) = 𝑥[𝐷_𝑝𝑓(𝑃_𝑛−1, 𝑥) + 𝐷_𝑝𝑓(𝑃_𝑛−3, 𝑥)] with initial values 𝐷_𝑝𝑓(𝑃₁, 𝑥) = 𝑥, 𝐷_𝑝𝑓(𝑃₂, 𝑥) = 𝑥²+ 2𝑥, 𝐷_𝑝𝑓(𝑃₃, 𝑥) = 𝑥³+ 2𝑥²+ 𝑥.

Proof

𝑖) It follows from theorem 2.6

𝑖𝑖) It follows from the definition of perfect domination polynomial and part (𝑖) Using the above theorem we obtain 𝑑_𝑝𝑓(𝑃_𝑛, 𝑖) for 1 ≤ 𝑛 ≤ 15 as shown in the following Table 1

𝑖 𝑛

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

1 1 2 2 1

3 1 2 1

4 0 2 2 1

5 0 2 3 2 1

6 0 1 4 4 2 1

7 0 0 3 6 5 2 1

8 0 0 2 6 8 6 2 1

9 0 0 1 6 10 10 7 2 1

10 0 0 0 4 12 15 12 8 2 1

11 0 0 0 2 10 20 21 14 9 2 1

12 0 0 0 1 8 20 30 28 16 10 2 1 13 0 0 0 0 5 20 35 42 36 18 11 2 1 14 0 0 0 0 2 15 40 56 56 45 20 12 2 1 15 0 0 0 0 1 10 35 70 84 72 55 22 13 2 1

Table 1 Example 3.3

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Consider a path 𝑃₆ with 6 vertices. We construct a perfect dominating polynomial D_𝑝𝑓(𝑃₆, 𝑥) using Theorem 3.1 & Table 1

Then the Perfect dominating polynomial of a path 𝑃₆ is given by 𝐷_𝑝𝑓(𝑃₆, 𝑥) =

∑^𝑛_{𝑖=⌈𝑛/3⌉}𝑑_𝑝𝑓(𝑃₆, 𝑖)𝑥^𝑖 From the Table 1 we have 𝑑_𝑝𝑓(𝑃₆, 2) = 1, 𝑑_𝑝𝑓(𝑃₆, 3) = 4, 𝑑_𝑝𝑓(𝑃₆, 4) = 4, 𝑑_𝑝𝑓(𝑃₆, 5) = 2, 𝑑_𝑝𝑓(𝑃₆, 6) = 1. Hence, 𝐷_𝑝𝑓(𝑃₆, 𝑥) = 𝑥²+ 4𝑥³+ 4𝑥⁴+ 2𝑥⁵+ 𝑥⁶.

Theorem 3.4

The coefficients of 𝐷_𝑝𝑓(𝑃_𝑛, 𝑥) have the following properties 𝑖) 𝑑_𝑝𝑓(𝑃_3𝑛, 𝑛) = 1 for every 𝑛 ∈ 𝑁

𝑖𝑖) 𝑑_𝑝𝑓(𝑃_3𝑛+2, 𝑛 + 1) = 2 for every 𝑛 ∈ 𝑁 𝑖𝑖𝑖) 𝑑_𝑝𝑓(𝑃_3𝑛+1, 𝑛 + 1) = 𝑛 + 1 for every 𝑛 ∈ 𝑁 𝑖𝑣) 𝑑_𝑝𝑓(𝑃_3𝑛, 𝑛 + 1) = 2𝑛 for every 𝑛 ∈ 𝑁 𝑣) 𝑑_𝑝𝑓(𝑃_𝑛, 𝑛) = 1 for every 𝑛 ∈ 𝑁

𝑣𝑖) 𝑑_𝑝𝑓(𝑃_𝑛, 𝑛 − 1) = 2 for every 𝑛 ≥ 2 𝑣𝑖𝑖) 𝑑_𝑝𝑓(𝑃_𝑛, 𝑛 − 2) = 𝑛 − 2 for every 𝑛 ≥ 3 𝑣𝑖𝑖) 𝑑_𝑝𝑓(𝑃_𝑛, 𝑛 − 3) = 2(𝑛 − 4) for every 𝑛 ≥ 5 𝑖𝑥) 𝑑_𝑝𝑓(𝑃_𝑛, 𝑛 − 4) =^{𝑛2−9𝑛+20}

2 for every 𝑛 ≥ 6 Proof

𝑖) For every 𝑛 ∈ 𝑁, we have, Ρ_𝑝𝑓(3𝑛, 𝑛) = {{2,5,8, … , 𝑛 − 4, 𝑛 − 1}}. Hence 𝑑_𝑝𝑓(𝑃_3𝑛, 𝑛) = 1.

𝑖𝑖) By theorem 2.6 (𝑖𝑖𝑖) for every 𝑛 ∈ 𝑁, we have, Ρ_𝑝𝑓(3𝑛 + 2, 𝑛 + 1) = {{1,4,7, … ,3𝑛 + 1}, {2,5,8, … ,3𝑛 + 2}}. Hence, 𝑑_𝑝𝑓(𝑃_3𝑛+2, 𝑛 + 1) = 2.

𝑖𝑖𝑖) We prove this result by induction on 𝑛. When 𝑛 = 1, we have Ρ_𝑝𝑓(4,2) = {{1,4}, {2,3}}. So, 𝑑_𝑝𝑓(𝑃₄, 2) = 2. Therefore, the result is true for 𝑛 = 1.

Now, assume that the result is true for all natural numbers less than 𝑛. For 𝑛, we have by theorem 3.1 (𝑖) and by the induction hypothesis 𝑑_𝑝𝑓(𝑃_3𝑛+1, 𝑛 + 1) = 𝑑_𝑝𝑓(𝑃_3𝑛, 𝑛) + 𝑑_𝑝𝑓(𝑃_{3(𝑛−1)+1}, 𝑛) = 1 + 𝑛. Hence, by the principle of induction, the result is true for all 𝑛 ∈ 𝑁.

𝑖𝑣) We prove this result by induction on 𝑛. When 𝑛 = 1, we have Ρ_𝑝𝑓(3,2) = {{1,2}, {2,3}}. So, 𝑑_𝑝𝑓(𝑃₄, 2) = 2. Therefore, the result is true for 𝑛 = 1.

Now, assume that the result is true for all natural numbers less than 𝑛. For 𝑛, we have by theorem 3.1 (𝑖) 𝑑_𝑝𝑓(𝑃_3𝑛, 𝑛 + 1) = 𝑑_𝑝𝑓(𝑃_3𝑛−1, 𝑛) + 𝑑_𝑝𝑓(𝑃_3𝑛−3, 𝑛). Then by part (𝑖𝑖) and induction hypothesis 𝑑_𝑝𝑓(𝑃_3𝑛, 𝑛 + 1) = 𝑑_𝑝𝑓(𝑃_{3(𝑛−1)+2}, 𝑛) + 𝑑_𝑝𝑓(𝑃_3(𝑛−1), 𝑛) = 2 + 2(𝑛 − 1) = 2𝑛. Hence, by the principle of induction, the result is true for all 𝑛 ∈ 𝑁.

𝑣) We have, 𝑃_𝑝𝑓(𝑛, 𝑛) = {[𝑛]} therefore, 𝑑_𝑝𝑓(𝑃_𝑛, 𝑛) = |𝑃_𝑝𝑓(𝑛, 𝑛)| = 1.

𝑣𝑖) We prove this result by induction on 𝑛. The result is true for 𝑛 = 2, because, Ρ_𝑝𝑓(2,1) = {{1}, {2}}. So, 𝑑_𝑝𝑓(𝑃₂, 1) = 2.

Now, assume that the result is true for all natural numbers less than 𝑛. For 𝑛, we have by theorem 3.1 (𝑖) 𝑑_𝑝𝑓(𝑃_𝑛, 𝑛 − 1) = 𝑑_𝑝𝑓(𝑃_𝑛−1, 𝑛 − 2) + 𝑑_𝑝𝑓(𝑃_𝑛−3, 𝑛 − 2).

Then, by induction hypothesis and by lemma 1.2 𝑑_𝑝𝑓(𝑃_𝑛, 𝑛 − 1) = 2. Hence, by the principle of induction, the result is true for all 𝑛 ∈ 𝑁.

𝑣𝑖𝑖) We prove this result by induction on 𝑛. The result is true for 𝑛 = 3, because, Ρ_𝑝𝑓(3,1) = {{2}}. So, 𝑑_𝑝𝑓(𝑃₃, 1) = 1.

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Now, assume that the result is true for all natural numbers less than 𝑛. For 𝑛, we have by theorem 3.1 (𝑖) 𝑑_𝑝𝑓(𝑃_𝑛, 𝑛 − 2) = 𝑑_𝑝𝑓(𝑃_𝑛−1, 𝑛 − 3) + 𝑑_𝑝𝑓(𝑃_𝑛−3, 𝑛 − 3).

Then, by induction hypothesis and by part (𝑣), 𝑑_𝑝𝑓(𝑃_𝑛, 𝑛 − 2) = 𝑛 − 2. Hence, by the principle of induction, the result is true for all 𝑛 ∈ 𝑁.

𝑣𝑖𝑖𝑖) We prove this result by induction on 𝑛. The result is true for 𝑛 = 5, because, Ρ_𝑝𝑓(5,2) = {{1,4}, {2,5}}. So, 𝑑_𝑝𝑓(𝑃₅, 2) = 2.

Now, assume that the result is true for all natural numbers less than 𝑛. For 𝑛, we have by theorem 3.1 (𝑖) 𝑑_𝑝𝑓(𝑃_𝑛, 𝑛 − 3) = 𝑑_𝑝𝑓(𝑃_𝑛−1, 𝑛 − 4) + 𝑑_𝑝𝑓(𝑃_𝑛−3, 𝑛 − 4).

Then, by induction hypothesis and by part (𝑣𝑖) 𝑑_𝑝𝑓(𝑃_𝑛, 𝑛 − 3) = 2(𝑛 − 5) + 2 = 2(𝑛 − 4). Hence, by the principle of induction, the result is true for all 𝑛 ∈ 𝑁.

𝑖𝑥) We prove this result by induction on 𝑛. The result is true for 𝑛 = 6, because, Ρ_𝑝𝑓(6,2) = {{2,5}}. So, 𝑑_𝑝𝑓(𝑃₆, 2) = 1. Therefore, the result is true for 𝑛 = 6.

Now, assume that the result is true for all natural numbers less than 𝑛. For 𝑛, we have by theorem 3.1 (𝑖) 𝑑_𝑝𝑓(𝑃_𝑛, 𝑛 − 4) = 𝑑_𝑝𝑓(𝑃_𝑛−1, 𝑛 − 5) + 𝑑_𝑝𝑓(𝑃_𝑛−3, 𝑛 − 5).

Then, by induction hypothesis and by part (𝑣𝑖𝑖) 𝑑_𝑝𝑓(𝑃_𝑛, 𝑛 − 4) =

(𝑛−1)²−9(𝑛−1)+20

2 +^2(𝑛−5)

2 =^{𝑛2−9𝑛+20}

2 . Hence, by the principle of induction, the result is true for all 𝑛 ∈ 𝑁.

Theorem 3.5

For every 𝑛 ∈ 𝑁 and ⌈^𝑛

3⌉ ≤ 𝑖 ≤ 𝑛. |𝑃_𝑝𝑓(𝑛, 𝑖)| is the coefficient of 𝑢^𝑛𝑣^𝑖 in the expansion of the function 𝑓(𝑢, 𝑣) = ^{𝑢3𝑣(1+2𝑣+𝑣2+𝑢𝑣+2𝑢2𝑣+𝑢2𝑣2)}

(1−𝑢𝑣−𝑢³𝑣)

Proof

First we set 𝑓(𝑢, 𝑣) = ∑^∞_𝑛=3∑^∞_𝑖=1|𝑃_𝑝𝑓(𝑛, 𝑖)|𝑢^𝑛𝑣^𝑖. By using the recursive formula for |𝑃_𝑝𝑓(𝑛, 𝑖)| we can write 𝑓(𝑢, 𝑣) as follows:

𝑓(𝑢, 𝑣) = ∑ ∑(|𝑃_𝑝𝑓(𝑛 − 1, 𝑖 − 1)| + |𝑃_𝑝𝑓(𝑛 − 3, 𝑖 − 1)|)𝑢^𝑛𝑣^𝑖

∞

𝑖=1

∞

𝑛=3

= 𝑢𝑣 ∑^∞_𝑛=3∑^∞_𝑖=1|𝑃_𝑝𝑓(𝑛 − 1, 𝑖 − 1)|𝑢^𝑛−1𝑣^𝑖−1+ 𝑢³𝑣 ∑^∞_𝑛=3∑^∞_𝑖=1|𝑃_𝑝𝑓(𝑛 − 3, 𝑖 − 1)|𝑢^𝑛−3𝑣^𝑖−1

= 𝑢𝑣(|𝑃_𝑝𝑓(2,0)|𝑢²+ |𝑃_𝑝𝑓(2,1)|𝑢²𝑣 + |𝑃_𝑝𝑓(2,2)|𝑢²𝑣²) + 𝑢𝑣𝑓(𝑢, 𝑣) + 𝑢³𝑣(|𝑃_𝑝𝑓(0,0)| + |𝑃_𝑝𝑓(1,0)|𝑢 + |𝑃_𝑝𝑓(1,1)|𝑢𝑣 + |𝑃_𝑝𝑓(2,0)|𝑢²+

|𝑃_𝑝𝑓(2,1)|𝑢²𝑣 + |𝑃_𝑝𝑓(2,2)|𝑢²𝑣²) + 𝑢³𝑣𝑓(𝑢, 𝑣).

Substituting the values from Table 1 also for |𝑃_𝑝𝑓(𝑛, 0)| = 0 for all 𝑛 ∈ 𝑁 and

|𝑃_𝑝𝑓(0,0)| = 1 we have,

𝑓(𝑢, 𝑣) = 𝑢𝑣(2𝑢²𝑣 + 𝑢²𝑣²) + 𝑢𝑣𝑓(𝑢, 𝑣) + 𝑢³𝑣(1 + 𝑢𝑣 + 2𝑢²𝑣 + 𝑢²𝑣²) + 𝑢³𝑣𝑓(𝑢, 𝑣)

𝑓(𝑢, 𝑣) = 𝑢³𝑣(1 + 2𝑣 + 𝑣² + 𝑢𝑣 + 2𝑢²𝑣 + 𝑢²𝑣²) + 𝑢𝑣𝑓(𝑢, 𝑣) + 𝑢³𝑣𝑓(𝑢, 𝑣) 𝑓(𝑢, 𝑣)(1 − 𝑢𝑣 − 𝑢³𝑣) = 𝑢³𝑣(1 + 2𝑣 + 𝑣² + 𝑢𝑣 + 2𝑢²𝑣 + 𝑢²𝑣²)

Hence, 𝑓(𝑢, 𝑣) =^{𝑢3𝑣(1+2𝑣+𝑣2+𝑢𝑣+2𝑢2𝑣+𝑢2𝑣2)}

(1−𝑢𝑣−𝑢³𝑣)

References

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