Effects on fractional domination in graphs

Abstract

A graph G is an undirected finite connected graph. A function f : V (G) → [0, 1] is called a fractional dominating function if, ∑_u∈N[v]f (u) ≥1, for all v ∈ V, where N [v] is the closed neighborhood of v. The weight of a fractional dominating function is w (f) = ∑_v∈V(G)f (v). The fractional domination number γ_f (G) has the least weight of all the fractional dominating functions of G. In this paper, we analyze the effects on γ_f (G) of deleting a vertex from G. Additionally, some bounds on γ_f (G) are discussed, and provide the exactness of some bounds. If we remove any leaves from any tree T, then the resulting graphs are , where |l| is the number of leaves. Some of the results are proved by the eccentricity value of a vertex e (v).

Keywords

Domination number edge domination number fractional domination number

1 Introduction

Allow G = (V, E) to be a simple connected graph with a vertex set V (G) and an edge set E (G). The maximum degree of a vertex in a graph G is denoted by Δ (G). Similarly, the minimum degree of a vertex in a graph G is denoted by δ (G). The distance to the vertex farthest from a vertex v is its eccentricity, denoted by the symbol e (v). As a result, e (v) = max {d (u, v) : ∀ u ∈ V}. The length of the shortest cycle equals the girth of G. A graph is considered self-centered if the eccentricity of all of its vertices is the same. A wheel graph is a graph built by connecting all the vertices of a cycle to a single universal vertex. The crown graph is a complete bipartite graph with the edges of a perfect matching omitted. A tree is a connected graph that has no cycles. A vertex having a degree of one is known as a pendant vertex. Support is a vertex that is neighbour of pendant vertices. Weak support is one that is adjacent to precisely one pendant vertex, whereas strong support is one that is adjacent to at least two pendant vertices. The open neighbourhood of v is N (v) = {u ∈ V/u is adjacent to v} for each v ∈ V (G), while the closed neighbourhood of v is N [v] = N (v) ∪ {v}, see [12]. If every vertex u ∈ V - D is adjacent to at least one vertex in D, then the vertex set D is called a dominating set. The minimum cardinality among all dominating sets in G is the domination number and it is denoted by γ (G). If every edge not in X ⊆ E (G) is adjacent to some edge in X, then the set X is termed as an edge dominating set of G. The smallest cardinality considered overall edge dominating sets of G is the edge domination number . The concept of fractional domination are discussed in [5 , 11]. Define a dominating function f to be any f : V (G) → [0, 1] with f (N [v]) = ∑_u∈N[v]f (u) ≥1 for every v ∈ V (G). The fractional domination number γ_f (G) is the minimum value of ∑_v∈V(G)f (v), see [10]. Similarly, we define a function f : E (G) → [0, 1] with f (N [x]) = ∑_e∈N[x]f (e) ≥1 for every x ∈ E (G). The fractional edge domination number $γ_{f}^{'} (G)$ is the minimum weight of ∑_e∈E(G)f (e), see [2]. Harary [6] was the one who proposed the changing and unchanging domination terminalogy. The findings of a study that attempted to categorize graph G into six types are surveyed in [5]. [1 , 7] provides some information on changing and unchanging domination. Many authors have addressed the impacts of changing and not changing of deleting a vertex from G on various other known dominating parameters. There is no one who has explored the effects on the fractional domination number in graphs when removing a vertex or removing or adding an edge from G. In this study, we predict the effects on fractional dominating parameters and provide some bounds on the fractional domination number γ_f (G). Let G - v (or G - e) be the graph created when vertex v (or an edge e) is removed from G. Here, C stands for Changing; U stands for Unchanging; V stands for Vertex; E stands for Edge; R stands for Removal; and A stands for Addition.(CVR) γ_f (G - v) ≠ γ_f (G) for all v ∈ V;

(CER) γ_f (G - e) ≠ γ_f (G) for all e ∈ E;(CEA) γ_f (G + e) ≠ γ_f (G) for all $e \in E (\bar{G})$ ; (UVR) γ_f (G - v) = γ_f (G) for all v ∈ V; (UER) γ_f (G - e) = γ_f (G) for all e ∈ E; (UEA) γ_f (G + e) = γ_f (G) for all $e \in E (\bar{G})$ . In this paper, the effects of the removal of a vertex on the fractional domination number of graphs and trees are discussed when using the following sets.V_fD⁰ = {v ∈ V (G)/γ_f (G - v) = γ_f (G)} V_fD⁺ = {v ∈ V (G)/γ_f (G - v) > γ_f (G)} V_fD^- = {v ∈ V (G)/γ_f (G - v) < γ_f (G)}. Also, we covered certain bounds on the fractional domination number in graphs that yield the exactness. For any tree T, if we remove any leaves from T, then the resultant graphs are , where |l| is the number of leaves.

2 CVR and UVR on fractional domination numbers

In this section, we discussed the effects on fractional domination numbers due to removing a single vertex from some standard graphs, like cycles and paths.

Observation 2.1. [2] For any graph G, we have $γ_{f}^{'} (G) = γ (L (G))$ . Hence it follows that $γ_{f}^{'} (C_{n}) = γ_{f} (C_{n}) = \frac{n}{3}$ and $γ_{f}^{'} (P_{n}) = γ_{f} (P_{n - 1}) = ⌈ \frac{n - 1}{3} ⌉$ .

Theorem 2.2. If G is a path on at least three vertices, removing any pendant vertex v from G then

(i) v ∈ V_fD^- for n ≡ 1 (mod 3)

(ii) v ∈ V_fD⁰ for n ≡ 0, 2 (mod 3).

Proof. Let G be a path on at least two vertices and |V (G) | = n and |E (G) | = n - 1 and v be any pendant vertex of G, and by observation 2.1, $γ_{f} (G) = ⌈ \frac{n}{3} ⌉$ . If n = 2 and we remove any v, then the graph becomes isolated. Here, we have three cases.

Case(i) When n ≡ 1 (mod 3). We define a function f : (V (G) - v) → [0, 1] by $f (v_{i}) (1 \leq i \leq n - 1) = {\begin{matrix} 1 & if v_{3 k + 2} \\ 0 & otherwise, \end{matrix}$ where $(0 \leq k \leq ⌊ \frac{n - 1}{2} ⌋)$ .

Here, f (N [v_i]) =1 for all i, then the weight of f is $γ_{f} (G - v) = \sum_{v_{3 k + 2}} f (v_{i}) + 0 = ⌈ \frac{n - 2}{3} ⌉ < ⌈ \frac{n}{3} ⌉ = γ_{f} (G)$ .

Thus γ_f (G - v) < γ_f (G) ⇒ v ∈ V_fD^-.

Case(ii)Whenn ≡ 0 (mod 3).

Subcase(i)We define a function f : (V (G) - v) → [0, 1] by $f (v_{i}) (1 \leq i \leq n - 1) = {\begin{matrix} 1 & if v_{3 k + 2} or v_{3 k + 1} \\ 0 & otherwise . \end{matrix}$ Here, f (N [v_i]) =1 for all i, then the weight of f is either $γ_{f} (G - v) = \sum_{v_{3 k + 2}} f (v_{i}) + 0 = ⌈ \frac{n}{3} ⌉ = γ_{f} (G)$ (or) $γ_{f} (G - v) = \sum_{v_{3 k + 1}} f (v_{i}) + 0 = ⌈ \frac{n}{3} ⌉ = γ_{f} (G)$ . Thus γ_f (G - v) = γ_f (G) ⇒ v ∈ V_fD⁰.

Case(iii)When n ≡ 2 (mod 3).

Subcase(i)We define a function f : (V (G) - v) → [0, 1] by $f (v_{i}) (1 \leq i \leq n - 1) = {\begin{matrix} 1 & if v_{3 k + 1} or v_{3 k + 2} and v_{n} \\ 0 & otherwise . \end{matrix}$ Here, f (N [v_i]) =1 for all i, then the weight of f is either $γ_{f} (G - v) = \sum_{v_{3 k + 1}} f (v_{i}) + 0 = ⌈ \frac{n}{3} ⌉ = γ_{f} (G)$ (or) $γ_{f} (G - v) = \sum_{v_{3 k + 2}} f (v_{i}) + f (v_{n}) + 0 = ⌈ \frac{n}{3} ⌉ = γ_{f} (G)$ . Thus γ_f (G - v) = γ_f (G) ⇒ v ∈ V_fD⁰.

Subcase(ii) We define a function to give another way of giving values to vertices, f : (V (G) - v) → [0, 1] by $f (v_{i}) (1 \leq i \leq n - 1) = {\begin{matrix} \frac{1}{t} & if v_{3 k + 2} or v_{3 k + 1} \\ 1 - \frac{1}{t} & if v_{3 k + 1} or v_{3 k + 2} \\ 0 & otherwise . \end{matrix}$ or $f (v_{i}) (1 \leq i \leq n - 1) = {\begin{matrix} 1 & if v_{3 k + 4} \\ \frac{1}{t} & if v_{3 k + 2} or v_{3 k + 1} \\ 1 - \frac{1}{t} & if v_{3 k + 1} or v_{3 k + 2} \\ 0 & otherwise . \end{matrix}$ or $f (v_{i}) (1 \leq i \leq n - 1) = {\begin{matrix} 1 & if v_{3 k + 2} or v_{3 k + 1} \\ \frac{1}{t} & if v_{3 k + 4} \\ 1 - \frac{1}{t} & if v_{3 k + 5} \\ 0 & otherwise . \end{matrix}$ or $f (v_{i}) (1 \leq i \leq n - 1) = {\begin{matrix} 1 & if v_{n - 1} or v_{n} or v_{2} or v_{1} \\ \frac{1}{t} & if v_{3 k + 1} or v_{3 k + 4} or v_{3 k + 2} \\ 1 - \frac{1}{t} & if v_{3 k + 1} or v_{3 k + 2} or v_{3 k + 4} \\ 0 & otherwise . \end{matrix}$

In all cases, f (N [v_i]) =1 for all i, when we put the values in all possible combinations, we have the weight of f as the minimum among all fractional dominating functions of G. Thus $γ_{f} (G - v) = ⌈ \frac{n}{3} ⌉ = γ_{f} (G) \Rightarrow v \in V_{f} D^{0}$ .

Theorem 2.3. If G is a cycle on at least 3 vertices, removing any vertex v from G, then the following holds:

(i) v ∈ V_fD⁰ for n ≡ 0 (mod 3)

(ii)v ∈ V_fD^- for n ≡ 1 (mod 3)

(iii) v ∈ V_fD⁺ for n ≡ 2 (mod 3).

Proof.Let G be cycle graph with n ≥ 3 and v be any vertex of G. By observation 2.1, $γ_{f} (C_{n}) = \frac{n}{3}$ .

Case(i)When n ≡ 0 (mod 3). For any integer $k, (0 \leq k \leq ⌊ \frac{n - 1}{3} ⌋)$ .

Subcase(i)We define a function f : (V (G) - v) → [0, 1] by $f (v_{i}) (1 \leq i \leq n - 1) = {\begin{matrix} 1 & if v_{3 k + 2} or v_{3 k + 1} \\ 0 & otherwise . \end{matrix}$

Here, f (N [v_i]) =1 for all i (1 ≤ i ≤ n - 1). This value gives the minimum weight for all fractional dominating functions of G. Then γ_f (G - v) is either $\sum_{v_{i} \in v_{3 k + 2}} f (v_{i}) + 0 = \frac{n}{3} = γ_{f} (G)$ or $\sum_{v_{i} \in v_{3 k + 1}} f (v_{i}) + 0 = \frac{n}{3} = γ_{f} (G)$ . Thus v ∈ V_fD⁰.

Subcase(ii)We define a function f : (V (G) - v) → [0, 1] by $f (v_{i}) (1 \leq i \leq n - 1) = {\begin{matrix} \frac{1}{t} & if v_{3 k + 1} or v_{3 k + 2} \\ 1 - \frac{1}{t} & if v_{3 k + 2} or v_{3 k + 1} \\ 0 & otherwise, \end{matrix}$ where t is any positive integer. Here, f (N [v_i]) =1 for all i (1 ≤ i ≤ n - 1), then the weight of f is either γ_f (G - v)= $\sum_{v_{i} \in v_{3 k + 2}} f (v_{i}) + \sum_{v_{i} \in v_{3 k + 1}} f (v_{i}) + 0 = \frac{⌈ \frac{2 n - 3}{5} ⌉}{t} + ⌈ \frac{2 n - 3}{5} ⌉ - \frac{⌈ \frac{2 n - 3}{5} ⌉}{t} = ⌈ \frac{2 n - 3}{5} ⌉ = \frac{n}{3} = γ_{f} (G)$ (or) γ_f (G - v)= $\sum_{v_{i} \in v_{3 k + 1}} f (v_{i}) + \sum_{v_{i} \in v_{3 k + 2}} f (v_{i}) + 0 = \frac{⌈ \frac{2 n - 3}{5} ⌉}{t} + ⌈ \frac{2 n - 3}{5} ⌉ - \frac{⌈ \frac{2 n - 3}{5} ⌉}{t} = ⌈ \frac{2 n - 3}{5} ⌉ = \frac{n}{3} = γ_{f} (G)$ . Thus v ∈ V_fD⁰.

Case(ii)When n ≡ 1 (mod 3). We define a function f : (V (G) - v) → [0, 1] by $f (v_{i}) (1 \leq i \leq n - 1) = {\begin{matrix} 1 & if v_{3 k + 2} \\ 0 & otherwise . \end{matrix}$ Here, f (N [v_i]) =1 for all i, then the weight of f is γ_f (G - v)= $\sum_{v_{i} \in v_{3 k + 2}} f (v_{i}) + 0 = ⌊ \frac{n}{3} ⌋ < \frac{n}{3} = γ_{f} (G)$ . Thus v ∈ V_fD^-.

Case(iii)When n ≡ 2 (mod 3).

Subcase(i)We define a function f : (V (G) - v) → [0, 1] by $f (v_{i}) (1 \leq i \leq n - 1) = {\begin{matrix} 1 & if v_{3 k + 1} \\ 0 & otherwise . \end{matrix}$ Here, f (N [v_i]) =1 for all i (1 ≤ i ≤ n - 1), then the weight is γ_f (G - v)= $\sum_{v_{i} \in v_{3 k + 1}} f (v_{i}) + 0 = ⌈ \frac{n}{3} ⌉ > \frac{n}{3} = γ_{f} (G)$ . Thus v ∈ V_fD⁺.

Here, f (N [v_i]) =1 for all i (1 ≤ i ≤ n - 1), then the weight of f is either γ_f (G - v)= $\sum_{v_{i} \in v_{3 k + 2}} f (v_{i}) + \sum_{v_{i} \in v_{3 k + 1}} f (v_{i}) + 0 = ⌈ \frac{n}{3} ⌉ > \frac{n}{3} = γ_{f} (G)$ (or) γ_f (G - v)= $\sum_{v_{i} \in v_{3 k + 1}} f (v_{i}) + \sum_{v_{i} \in v_{3 k + 2}} f (v_{i}) + 0 = ⌈ \frac{n}{3} ⌉ > \frac{n}{3} = γ_{f} (G)$ . This is the minimum weight for all fractional dominating functions of G. Thus v ∈ V_fD⁺.

Theorem 2.4. If G is n - 2 regular graph with n ≥ 4 and n is even, and if G contains only one vertex with e (v) =1 where e (v) is eccentricity value of a vertex, and remove any vertex v from G then v ∈ V_fD^-.

Proof. Let G be an n - 2 regular graph with n ≥ 4 and n even, and G contains e (v_i) =1 for exactly one i, (1 ≤ i ≤ n), where e (v) is eccentricity value of a vertex. Here, first we compute the value of γ_f (G). We define a function f : V (G) → [0, 1] by $f (v_{i}) = \frac{1}{n - 1}$ for all i (1 ≤ i ≤ n). In this case, f (N [v_i]) =1 for all i, then the weight is $γ_{f} (G) = \sum_{i = 1}^{n} f (v_{i}) = \frac{n}{n - 1}$ . This is the minimum weight among all fractional dominating functions of G. Next, we take v to be any vertex of G. Suppose we remove any v, then we define a function f : (V (G) - v) → [0, 1] by $f (v_{i}) (1 \leq i \leq n - 1) = {\begin{matrix} 1 & if e (v_{i}) = 1 \\ 0 & otherwise . \end{matrix}$ Then the weight is $γ_{f} (G - v) = v \in e (v_{i}) + \sum_{i = 1}^{n - 2} f (v_{i}) = 1 + 0 = 1 < \frac{n}{n - 1} = γ_{f} (G)$ . Thus γ_f (G - v) < γ_f (G) ⇒ v ∈ V_fD^-.

Theorem 2.5. If G is a self-centered graph with n ≥ 3 and $γ_{f} (G) = \frac{n}{3}$ and remove any vertex y from G, then the followings are true:

(i) y ∈ V_fD⁰ if n ≡ 0 (mod 3)

(ii) y ∈ V_fD^- if n ≡ 1 (mod 3)

(iii) y ∈ V_fD⁺ if n ≡ 2 (mod 3).

Proof.Let G be a self centered graph with n ≥ 3 and $γ_{f} (G) = \frac{n}{3}$ and y be any vertex of G.

Case(i) When n ≡ 0 (mod 3). For any integer $k, (0 \leq k \leq ⌊ \frac{n - 1}{3} ⌋)$ .

Subcase(i)We define a function f : (V (G) - y) → [0, 1] by $f (v_{i}) (1 \leq i \leq n - 1) = {\begin{matrix} 1 & if v_{3 k + 2} or v_{3 k + 1} \\ 0 & otherwise . \end{matrix}$ Here, f (N [v_i]) =1 for all i (1 ≤ i ≤ n - 1), then the weight γ_f (G - y) is either $\sum_{v_{i} \in v_{3 k + 2}} f (v_{i}) + 0 = \frac{n}{3} = γ_{f} (G)$ or $\sum_{v_{i} \in v_{3 k + 1}} f (v_{i}) + 0 = \frac{n}{3} = γ_{f} (G)$ . This is the minimum weight for all fractional dominating functions of G. Thus y ∈ V_fD⁰.

Subcase(ii)We define a function to give another way of giving values to vertices, f : (V (G) - v) → [0, 1] by $f (v_{i}) (1 \leq i \leq n - 1) = {\begin{matrix} \frac{1}{t} & if v_{3 k + 1} or v_{3 k + 2} \\ 1 - \frac{1}{t} & if v_{3 k + 2} or v_{3 k + 1} \\ 0 & otherwise, \end{matrix}$ where t is any positive integer. Here, f (N [v_i]) =1 for all i (1 ≤ i ≤ n - 1), then the weight of f is either γ_f (G - y)= $\sum_{v_{i} \in v_{3 k + 2}} f (v_{i}) + \sum_{v_{i} \in v_{3 k + 1}} f (v_{i}) + 0 = \frac{⌈ \frac{2 n - 3}{5} ⌉}{t} + ⌈ \frac{2 n - 3}{5} ⌉ - \frac{⌈ \frac{2 n - 3}{5} ⌉}{t} = ⌈ \frac{2 n - 3}{5} ⌉ = \frac{n}{3} = γ_{f} (G)$ (or) γ_f (G - y)= $\sum_{v_{i} \in v_{3 k + 1}} f (v_{i}) + \sum_{v_{i} \in v_{3 k + 2}} f (v_{i}) + 0 = \frac{⌈ \frac{2 n - 3}{5} ⌉}{t} + ⌈ \frac{2 n - 3}{5} ⌉ - \frac{⌈ \frac{2 n - 3}{5} ⌉}{t} = ⌈ \frac{2 n - 3}{5} ⌉ = \frac{n}{3} = γ_{f} (G)$ . This is the minimum weight for all fractional dominating functions of G. Thus y ∈ V_fD⁰.

Case(ii)When n ≡ 1 (mod 3). We define a function f : (V (G) - y) → [0, 1] by $f (v_{i}) (1 \leq i \leq n - 1) = {\begin{matrix} 1 & if v_{3 k + 2} \\ 0 & otherwise . \end{matrix}$ Here, f (N [v_i]) =1 for all i, then the weight γ_f (G - y)= $\sum_{v_{i} \in v_{3 k + 2}} f (v_{i}) + 0 = ⌊ \frac{n}{3} ⌋ < \frac{n}{3} = γ_{f} (G)$ . This is the minimum weight for all fractional dominating functions of G. Thus y ∈ V_fD^-.

Case(iii)When n ≡ 2 (mod 3).

Subcase(i)We define a function f : (V (G) - y) → [0, 1] by $f (v_{i}) (1 \leq i \leq n - 1) = {\begin{matrix} 1 & if v_{3 k + 1} \\ 0 & otherwise . \end{matrix}$ Here, f (N [v_i]) =1 for all i (1 ≤ i ≤ n - 1), then the weight γ_f (G - y)= $\sum_{v_{i} \in v_{3 k + 1}} f (v_{i}) + 0 = ⌈ \frac{n}{3} ⌉ > \frac{n}{3} = γ_{f} (G)$ . This is the minimum weight for all fractional dominating functions of G. Thus y ∈ V_fD⁺.

Subcase(ii)We define a function to give another way of giving values to vertices, f : (V (G) - y) → [0, 1] by $f (v_{i}) (1 \leq i \leq n - 1) = {\begin{matrix} \frac{1}{t} & if v_{3 k + 1} or v_{3 k + 2} \\ 1 - \frac{1}{t} & if v_{3 k + 2} or v_{3 k + 1} \\ 0 & otherwise, \end{matrix}$ where t is any positive integer. Here, f (N [v_i]) =1 for all i (1 ≤ i ≤ n - 1), then the weight of f is either γ_f (G - y)= $\sum_{v_{i} \in v_{3 k + 2}} f (v_{i}) + \sum_{v_{i} \in v_{3 k + 1}} f (v_{i}) + 0 = ⌈ \frac{n}{3} ⌉ > \frac{n}{3} = γ_{f} (G)$ (or) γ_f (G - y)= $\sum_{v_{i} \in v_{3 k + 1}} f (v_{i}) + \sum_{v_{i} \in v_{3 k + 2}} f (v_{i}) + 0 = ⌈ \frac{n}{3} ⌉ > \frac{n}{3} = γ_{f} (G)$ . This is the minimum weight for all fractional dominating functions of G. Thus y ∈ V_fD⁺.

Theorem 2.6. If G is a graph with n ≥ 3 and exactly two vertices with degree n - 1, and if γ_f (G) =1 and remove any vertex v from G, then γ_f (G - v) = γ_f (G).

Proof. Let G be a graph with n ≥ 3 and exactly two vertices having degree n - 1. We take v to be any vertex of G and γ_f (G) =1.

Case(i)Suppose we remove any vertex having an eccentricity value is one, then the function f : (V (G) - v) → [0, 1] by $f (v_{i}) (1 \leq i \leq n - 1) = {\begin{matrix} 1 & if e (v_{i}) = 1 \\ 0 & otherwise . \end{matrix}$ Here, f (N [v_i]) =1 for all i, then $γ_{f} (G - v) = e (v_{i}) + \sum_{i = 1}^{n - 2} f (v_{i}) = 1 = γ_{f} (G)$ .

Case(ii) Suppose we remove any vertex having an eccentricity value is two, then the function f : (V (G) - v) → [0, 1] by $f (v_{i}) (1 \leq i \leq n - 1) = {\begin{matrix} 1 & if e (v_{i}) = 1 \\ 0 & otherwise . \end{matrix}$ Here, f (N [v_i]) =1 for all i, then the weight $γ_{f} (G - v) = e (v_{i}) + \sum_{i = 1}^{n - 2} f (v_{i}) = 1 = γ_{f} (G)$ .

Theorem 2.7. If G is a graph with n ≥ 4 and there is exactly one vertex of degree n - 1, also if γ_f (G) =1 and remove any vertex v, then the following holds:

(i) γ_f (G - v) > γ_f (G) if {v ∈ V (G)/e (v) =1}

(ii) γ_f (G - v) = γ_f (G) if {v ∈ V (G)/e (v) =2}.

Proof. Let G be a graph with n ≥ 4 and exactly one vertex with the degree n - 1 and we know that γ_f (G) =1. Let v be any vertex of G.

Case(i) Let v be any vertex having e (v) =1.

Subcase(i) We define a function f : (V (G) - v) → [0, 1] by $f (v_{i}) (1 \leq i \leq n - 1) = {\begin{matrix} 1 & if \deg (v_{i}) = 0 and \deg (v_{i}) = 1 \\ 0 & otherwise . \end{matrix}$ (or) $f (v_{i}) (1 \leq i \leq n - 1) = {\begin{matrix} 1 & if \deg (v_{i}) = 0 and v_{3 k + 2} \\ 0 & otherwise . \end{matrix}$ where $(0 \leq k \leq ⌊ \frac{n - 1}{3} ⌋)$ . In this case, f (N [v_i]) =1 for all i, then the weight of f is either $γ_{f} (G - v) = (v \in \deg (v) = 0) + (v \in \deg (v) = 1) + \sum_{i = 1}^{n - 3} f (v_{i}) = 2 > 1 = γ_{f} (G)$ (or) $γ_{f} (G - v) = (v \in \deg (v) = 0) + (v \in v_{3 k + 2}) + \sum_{i = 1}^{n - 3} f (v_{i}) = 2 > 1 = γ_{f} (G)$ .

Subcase(ii)For t is any positive integer. Here, we define a function f : (V (G) - v) → [0, 1] by $f (v_{i}) (1 \leq i \leq n - 1) = {\begin{matrix} 1 & if \deg (v_{i}) = 0 \\ \frac{1}{t} & if \deg (v_{i}) = 1 \\ 1 - \frac{1}{t} & if {u \in V (G) / u is adjacent to \deg (v) = 1} \\ 0 & otherwise . \end{matrix}$ (or) $f (v_{i}) (1 \leq i \leq n - 1) = {\begin{matrix} \frac{1}{t} & if \deg (v_{i}) = 1 \\ 1 - \frac{1}{t} & if {u \in V (G) / u is adjacent to \deg (v) = 1} \\ 0 & otherwise . \end{matrix}$ (or) $f (v_{i}) (1 \leq i \leq n - 1) = {\begin{matrix} \frac{1}{t} & if \deg (v_{i}) = 1 \\ 1 - \frac{1}{t} & if {u \in V (G) / u is adjacent to \deg (v) = 1} \end{matrix}$ In above all cases, f (N [v_i]) =1 for some i, then the weight of f is γ_f (G - v) = (v ∈ deg (v) =0) + ∑_deg(v)=1f (v_i) + ∑_u∈deg(v)=1f (v_i) +0 = 2 >1 = γ_f (G) or γ_f (G - v) = ∑_deg(v)=1f (v_i) + ∑_u∈deg(v)=1f (v_i) +0 = 2 >1 = γ_f (G) or γ_f (G - v) = ∑_deg(v)=1f (v_i) + ∑_u∈deg(v)=1f (v_i) =2 > 1 = γ_f (G). Thus v ∈ V_fD⁺.

Case(ii) Let v be any vertex of {v ∈ V (G)/e (v) =2}. In this case we define a function f : (V (G) - v) → [0, 1] by $f (v_{i}) (1 \leq i \leq n - 1) = {\begin{matrix} 1 & if e (v_{i}) = 1 \\ 0 & otherwise . \end{matrix}$

Here, f (N [v_i]) =1 for all i, then the weight of f is $γ_{f} (G - v) = (v \in e (v) = 1) + \sum_{i = 1}^{n - 2} f (v_{i}) = 1 = γ_{f} (G)$

Thus v ∈ V_fD⁰.

Theorem 2.8. If G is a graph with n ≥ 3 and at least three vertices of degree n - 1, and γ_f (G) =1, and remove any vertex v from G, then γ_f (G - v) = γ_f (G).

Proof. Let G be a graph with n ≥ 3 and having at least three vertices with degree n - 1, and we take v to be any vertex of G. We know that γ_f (G) =1. Here, graph G is neither a complete graph with a radius of 1 nor a complete graph with a radius of 1. For n = 3, 4, 5 we have G as a complete graph with a radius of one, and n ≥ 6 then the graph may or may not be a complete graph with its radius of one. Suppose we remove any vertex from G, then we define a function f : (V (G) - v) → [0, 1] by $f (v_{i}) (1 \leq i \leq n - 1) = {\begin{matrix} 1 & if e (v_{i}) = 1 \\ 0 & otherwise . \end{matrix}$

Therefore f (N [v_i]) =1 for all i, then the weight $γ_{f} (G - v) = (v \in e (v_{i}) = 1) + \sum_{i = 1}^{n - 2} f (v_{i}) = 1 + 0 = 1 = γ_{f} (G)$ . Thus v ∈ V_fD⁰.

Theorem 2.9. If H_n,n is a crown graph with n ≥ 2 and remove any vertex v, then the following holds: (i) v ∈ V_fD⁰ for n = 2. (ii) v ∈ V_fD⁺ for n ≥ 3.

Proof. Let x₁, x₂, . . . , x_n be vertices of the first partition and y₁, y₂, . . . , y_n be vertices of the second partition. Let G = H_n,n be a crown graph obtained from the complete bipartite graph K_n,n by removing a perfect matching. Let us consider the vertex set to be V (H_n,n) = {x₁, x₂, . . . , x_n, y₁, y₂, . . . , y_n} and an edge set E (H_n,n) = {(x_iy_j)/0 ≤ i, j ≤ n, i ≠ j}. In this graph, we put the value $\frac{1}{n}$ to all vertices of G then f (N [x_i]) =1 and f (N [y_j]) =1 for all i, j. This is the minimum weight among all fractional dominating functions of G. Thus γ_f (G) =2. Take v to be any vertex of the first partition or the second partition. We remove any vertex from G.

Case(i) If n = 2, then γ_f (G - v) =2 = γ_f (G). Thus v ∈ V_fD⁰.

Here, Fig. 1 gives the unchanged vertex removal of the fractional domination number of the crown graph with two vertices. Thus, γ_f (G) =2 and γ_f (G - y₂) =2.

Fig. 1

UVR of fractional domination number with H_2,2.

Case(ii) If n = 3 then $γ_{f} (G - v) = \frac{7}{3} > 2 = γ_{f} (G)$ .

Thus v ∈ V_fD⁺. For n > 3, if we assign the value $\frac{1}{n - 1}$ to all vertices of G - v, then f (N [x_i]) =1 and f (N [y_j]) =1 for some i, j. This is the minimum weight among all fractional dominating functions of G - v. Thus $γ_{f} (G - v) = \frac{2 n - 1}{n - 1} > 2 = γ_{f} (G) \Rightarrow v \in V_{f} D^{+}$ .

Theorem 2.10. If G is a self-centered graph of girth 4 with 2n vertices where n ≥ 3 and all vertices having degree 3, and if remove any vertex from G then the followings are held:

(i) V_fD⁺ for n = 3

(ii) V_fD⁰ otherwise.

Proof. Consider G as a self-centered graph with a girth of 4 and a vertex set V (G) = {x_i/1 ≤ i ≤ 2n} and |E (G) |=3n. This graph is nothing but a prism graph, which is an example of a generalized Peterson graph. We define a function f : V (G) → [0, 1] by $f (x_{i}) (1 \leq i \leq 2 n) = \frac{1}{4}$ for all i. Here, f (N [x_i]) =1 for all i, then the weight of f is minimum among all fractional dominating functions of G. Thus $γ_{f} (G) = \sum_{i = 1}^{2 n} f (x_{i}) = \frac{2 n}{4}$ . Let x be any vertex of G, we have two cases.

Case(i) For n = 3, then the graph becomes $γ_{f} (G - x) = \frac{7}{4} > \frac{6}{4} = γ_{f} (G)$ . Thus x ∈ V_fD⁺.

Case(ii) For n > 3, we define a function f : (V (G) - x) → [0, 1] by $f (x_{i}) (1 \leq i \leq 2 n - 1) = {\begin{matrix} \frac{2}{4} & if x_{i} is adjacent to δ (G) \\ 0 & if minimum degree vertex \\ \frac{1}{4} & otherwise . \end{matrix}$ Here, f (N [x_i]) =1 for some i, then the weight of f is minimum among all fractional dominating functions of G. Thus $γ_{f} (G - x) = \frac{2 n}{4} = γ_{f} (G) \Rightarrow x \in V_{f} D^{0}$ .

Theorem 2.11. If G is a wheel graph with more than three vertices and any vertex with e (v) =1 is removed, then (i) V_fD⁰ for n = 4, (ii) V_fD⁺ for n > 4. Also, for n > 4, remove any peripheral vertex and then V_fD⁰ for n > 4.

Proof. Let G be a wheel graph with n > 3. Now, we define a function f : V (G) → [0, 1] by $f (v_{i}) (1 \leq i \leq n) = {\begin{matrix} 1 & if e (v_{i}) = 1 \\ 0 & otherwise . \end{matrix}$ Here, f (N [v_i]) =1 for all i, then the weight of f is minimum among all fractional dominating functions of G. Thus $γ_{f} (G) = (v \in e (v) = 1) + \sum_{i = 1}^{n - 1} f (v_{i}) = 1 + 0 = 1$ . We have two cases

Case(i) We take a vertex v with an eccentric value is one.

Subcase(i) For n = 4. If we remove a vertex then γ_f (G - v) =1 = γ_f (G) ⇒ v ∈ V_fD⁰.

Subcase(ii) For n > 4, we define a function f : (V (G) - v) → [0, 1] by $f (v_{i}) (1 \leq i \leq n - 1) = \frac{1}{3}$ for all i. Here, f (N [v_i]) =1 for all i, then the weight of f is minimum among all fractional dominating functions of G - v. Thus $γ_{f} (G - v) = \sum_{i = 1}^{n - 1} f (v_{i}) = \frac{n - 1}{3} > 1 = γ_{f} (G)$ . Thus v ∈ V_fD⁺.

Case(ii) Let v be a peripheral vertex of G. That is e (v) = diam (G). For n ≥ 4, we put the value 1 to the central vertex of G. Then $γ_{f} (G - v) = e (v) + \sum_{i = 1}^{n - 1} f (v_{i}) = 1 = γ_{f} (G)$ . Thus v ∈ V_fD⁰.

Theorem 2.12. If G is a complete bipartite graph with m, n ≥ 3 and remove any vertex v with a maximum degree, then V_fD^- also remove v with a minimum degree, then V_fD⁰.

Proof. Let G be a complete bipartite graph with m, n ≥ 3, and V(G) to be two set partitions. Let x_i, (1 ≤ i ≤ m) be a vertices of the first partition and y_j, (1 ≤ j ≤ n) be a vertices of the second partition. We denote x_r = {(x₁, x₂, . . . , x_i, y₁, y₂, . . . , y_j)/1 ≤ r ≤ m + n}.

Case(i) If m > n, we must first we find γ_f (G). We define a function f : V (G) → [0, 1] by $f (x_{r}) (1 \leq r \leq m + n) = {\begin{matrix} \frac{1}{n} & if for all y_{j} \in Δ (G) \\ 1 - \frac{1}{n} & if for any one x_{i} \in δ (G) \\ 0 & otherwise . \end{matrix}$ Here, f (N [x_r]) =1 for some r (1 ≤ r ≤ m + n). This is the minimum weight for all fractional dominating functions of K_m,n. Then the weight of f is γ_f (G)= $\sum_{j = 1}^{n} \frac{1}{n} + \sum_{i = 1}^{m - 1} f (x_{i}) + f (x) = \frac{n}{n} + 0 + (1 - \frac{1}{n}) = \frac{2 n - 1}{n}$ . Thus $γ_{f} (G) = \frac{2 n - 1}{n}$ .

Subcase(i) If m is even and n is even. Let v be any vertex of first partition and also second partition. Suppose we remove any v ∈ Δ (G). Here, we put the value $\frac{1}{n - 1}$ to all y_j-1 and $1 - \frac{1}{n - 1}$ for any one x_i and 0 to all other vertices. Then γ_f (G - v)= $\sum_{j = 1}^{n - 1} \frac{1}{n - 1} + \sum_{i = 1}^{m - 1} f (x_{i}) + f (x) = \frac{n - 1}{n - 1} + 0 + (1 - \frac{1}{n - 1}) = \frac{2 n - 3}{n - 1} < \frac{2 n - 1}{n} = γ_{f} (G)$ . Thus v ∈ V_fD^-. Suppose we remove any v ∈ δ (G). Here, we put the value $\frac{1}{n}$ to all y_j and $1 - \frac{1}{n}$ for any one x_i and 0 to all other vertices. Then γ_f (G - v)= $\sum_{j = 1}^{n} \frac{1}{n} + \sum_{i = 1}^{m - 2} f (x_{i}) + f (x) = \frac{n}{n} + 0 + (1 - \frac{1}{n}) = \frac{2 n - 1}{n} = γ_{f} (G)$ . Thus v ∈ V_fD⁰.

Subcase(ii) If m is odd and n is odd. Suppose we remove any v ∈ Δ (G). Here, we put the value $\frac{1}{n - 1}$ to all y_j-1 and $1 - \frac{1}{n - 1}$ for any one x_i and 0 to all other vertices. Then γ_f (G - v)= $\sum_{j = 1}^{n - 1} \frac{1}{n - 1} + \sum_{i = 1}^{m - 1} f (x_{i}) + f (x) = \frac{n - 1}{n - 1} + 0 + (1 - \frac{1}{n - 1}) = \frac{2 n - 3}{n - 1} < \frac{2 n - 1}{n} = γ_{f} (G)$ . Thus v ∈ V_fD^-. Suppose we remove any v ∈ δ (G). Here, we put the value $\frac{1}{n}$ to all y_j and $1 - \frac{1}{n}$ for any one x_i and 0 to all other vertices. Then γ_f (G - v)= $\sum_{j = 1}^{n} \frac{1}{n} + \sum_{i = 1}^{m - 2} f (x_{i}) + f (x) = \frac{n}{n} + 0 + (1 - \frac{1}{n}) = \frac{2 n - 1}{n} = γ_{f} (G)$ . Thus v ∈ V_fD⁰.

Subcase(iii) If m is odd and n is even. Suppose we remove any v ∈ Δ (G). Here, we put the value $\frac{1}{n - 1}$ to all y_j-1 and $1 - \frac{1}{n - 1}$ for any one x_i and 0 to all other vertices. Then γ_f (G - v)= $\sum_{j = 1}^{n - 1} \frac{1}{n - 1} + \sum_{i = 1}^{m - 1} f (x_{i}) + f (x) = \frac{n - 1}{n - 1} + 0 + (1 - \frac{1}{n - 1}) = \frac{2 n - 3}{n - 1} < \frac{2 n - 1}{n} = γ_{f} (G)$ . Thus v ∈ V_fD^-. Suppose we remove any v ∈ δ (G). Here, we put the value $\frac{1}{n}$ to all y_j and $1 - \frac{1}{n}$ for any one x_i and 0 to all other vertices. Then γ_f (G - v)= $\sum_{j = 1}^{n} \frac{1}{n} + \sum_{i = 1}^{m - 2} f (x_{i}) + f (x) = \frac{n}{n} + 0 + (1 - \frac{1}{n}) = \frac{2 n - 1}{n} = γ_{f} (G)$ . Thus v ∈ V_fD⁰.

Subcase(iv) If m is even and n is odd. Suppose we remove any v ∈ Δ (G). Here, we put the value $\frac{1}{n - 1}$ to all y_j-1 and $1 - \frac{1}{n - 1}$ for any one x_i and 0 to all other vertices. Then γ_f (G - v)= $\sum_{j = 1}^{n - 1} \frac{1}{n - 1} + \sum_{i = 1}^{m - 1} f (x_{i}) + f (x) = \frac{n - 1}{n - 1} + 0 + (1 - \frac{1}{n - 1}) = \frac{2 n - 3}{n - 1} < \frac{2 n - 1}{n} = γ_{f} (G)$ . Thus v ∈ V_fD^-. Suppose we remove any v ∈ δ (G). Here, we put the value $\frac{1}{n}$ to all y_j and $1 - \frac{1}{n}$ for any one x_i and 0 to all other vertices. Then γ_f (G - v)= $\sum_{j = 1}^{n} \frac{1}{n} + \sum_{i = 1}^{m - 2} f (x_{i}) + f (x) = \frac{n}{n} + 0 + (1 - \frac{1}{n}) = \frac{2 n - 1}{n} = γ_{f} (G)$ . Thus v ∈ V_fD⁰.

Case(ii) If m < n, we must first we find γ_f (G). We define a function f : V (G) → [0, 1] by $f (x_{r}) (1 \leq r \leq m + n) = {\begin{matrix} \frac{1}{m} & if for all x_{i} \in Δ (G) \\ 1 - \frac{1}{m} & if for any one y_{j} \in δ (G) \\ 0 & otherwise . \end{matrix}$ Here, f (N [x_r]) =1 for some r (1 ≤ r ≤ m + n). This is the minimum weight for all fractional dominating functions of K_m,n. Then weight of f is γ_f (G)= $\sum_{i = 1}^{m} \frac{1}{m} + \sum_{j = 1}^{n - 1} f (y_{j}) + f (y) = \frac{m}{m} + 0 + (1 - \frac{1}{m}) = \frac{2 m - 1}{m}$ . Thus $γ_{f} (G) = \frac{2 m - 1}{m}$ .

Subcase(i) If m is even and n is even. Let v be any vertex of first partition and also second partition. Suppose we remove any v ∈ Δ (G). Here, we put the value $\frac{1}{m - 1}$ to all x_i-1 and $1 - \frac{1}{n - 1}$ for any one y_j and 0 to all other vertices. Then γ_f (G - v)= $\sum_{i = 1}^{m - 1} \frac{1}{m - 1} + \sum_{j = 1}^{n - 1} f (y_{j}) + f (y) = \frac{m - 1}{m - 1} + 0 + (1 - \frac{1}{m - 1}) = \frac{2 m - 3}{m - 1} < \frac{2 m - 1}{m} = γ_{f} (G)$ . Thus v ∈ V_fD^-. Suppose we remove any v ∈ δ (G). Here, we put the value $\frac{1}{m}$ to all x_i and $1 - \frac{1}{m}$ for any one y_j and 0 to all other vertices. Then γ_f (G - v)= $\sum_{i = 1}^{m} \frac{1}{m} + \sum_{j = 1}^{n - 2} f (y_{j}) + f (y) = \frac{m}{m} + 0 + (1 - \frac{1}{m}) = \frac{2 m - 1}{m} = γ_{f} (G)$ . Thus v ∈ V_fD⁰.

Subcase(ii) If m is odd and n is odd. Suppose we remove any v ∈ Δ (G). Here, we put the value $\frac{1}{m - 1}$ to all x_i-1 and $1 - \frac{1}{m - 1}$ for any one y_j and 0 to all other vertices. Then γ_f (G - v)= $\sum_{i = 1}^{m - 1} \frac{1}{m - 1} + \sum_{j = 1}^{n - 1} f (y_{j}) + f (y) = \frac{m - 1}{m - 1} + 0 + (1 - \frac{1}{m - 1}) = \frac{2 m - 3}{m - 1} < \frac{2 m - 1}{m} = γ_{f} (G)$ . Thus v ∈ V_fD^-. Suppose we remove any v ∈ δ (G). Here, we put the value $\frac{1}{m}$ to all x_i and $1 - \frac{1}{m}$ for any one y_j and 0 to all other vertices. Then γ_f (G - v)= $\sum_{i = 1}^{m} \frac{1}{m} + \sum_{j = 1}^{n - 2} f (y_{j}) + f (y) = \frac{m}{m} + 0 + (1 - \frac{1}{m}) = \frac{2 m - 1}{m} = γ_{f} (G)$ . Thus v ∈ V_fD⁰.

Subcase(iii) If m is odd and n is even. Suppose we remove any v ∈ Δ (G). Here, we put the value $\frac{1}{m - 1}$ to all x_i-1 and $1 - \frac{1}{m - 1}$ for any one y_j and 0 to all other vertices. Then γ_f (G - v)= $\sum_{i = 1}^{m - 1} \frac{1}{m - 1} + \sum_{j = 1}^{n - 1} f (y_{j}) + f (y) = \frac{m - 1}{m - 1} + 0 + (1 - \frac{1}{m - 1}) = \frac{2 m - 3}{m - 1} < \frac{2 m - 1}{m} = γ_{f} (G)$ .

Thus v ∈ V_fD^-. Suppose we remove any v ∈ δ (G). Here, we put the value $\frac{1}{m}$ to all x_i and $1 - \frac{1}{m}$ for any one y_j and 0 to all other vertices. Then γ_f (G - v)= $\sum_{i = 1}^{m} \frac{1}{m} + \sum_{j = 1}^{n - 2} f (y_{j}) + f (y) = \frac{m}{m} + 0 + (1 - \frac{1}{m}) = \frac{2 m - 1}{m} = γ_{f} (G)$ .

Thus v ∈ V_fD⁰.

Subcase(iv) If m is even and n is odd. Suppose we remove any v ∈ Δ (G). Here, we put the value $\frac{1}{m - 1}$ to all x_i-1 and $1 - \frac{1}{m - 1}$ for any one y_j and 0 to all other vertices. Then γ_f (G - v)= $\sum_{i = 1}^{m - 1} \frac{1}{m - 1} + \sum_{j = 1}^{n - 1} f (y_{j}) + f (y) = \frac{m - 1}{m - 1} + 0 + (1 - \frac{1}{m - 1}) = \frac{2 m - 3}{m - 1} < \frac{2 m - 1}{m} = γ_{f} (G)$ . Thus v ∈ V_fD^-. Suppose we remove any v ∈ δ (G). Here, we put the value $\frac{1}{m}$ to all x_i and $1 - \frac{1}{m}$ for any one y_j and 0 to all other vertices. Then γ_f (G - v)= $\sum_{i = 1}^{m} \frac{1}{m} + \sum_{j = 1}^{n - 2} f (y_{j}) + f (y) = \frac{m}{m} + 0 + (1 - \frac{1}{m}) = \frac{2 m - 1}{m} = γ_{f} (G)$ . Thus v ∈ V_fD⁰. This completes the proof of this theorem.

Corollary 2.13. If G is a complete bipartite graph with m = n ≥ 3 and remove any vertex v from G, then v ∈ V_fD^-.

3 Bounds

In this section, we obtain some bounds that relate to fractional domination numbers.

Definition 3.1. [9] Define a family of trees ⊤ as follows: A tree T∈ ⊤ if the following conditions are satisfied. (i) T does not have a strong support. (ii) Each weak support is of degree two. (iii) Every non leaf vertex is either a support or adjacent to a support. (iv) Every non leaf vertex which is not a support is of degree at least three.

Theorem 3.2. For any tree T, remove any pendant vertex which is adjacent to weak support, then γ_f (T - v) < γ_f (T). Also, remove any pendant vertex that is adjacent to strong support, then γ_f (T - v) = γ_f (T).

Proof. Take T to be a tree. We give the value of 1 to the strong support vertex or weak support vertex or a leaf that is adjacent to a weak support of degree two and 0 to the remaining vertices. Then γ_f (T) is how many strong and weak supports occur in that tree.

Case(i) Let v be any pendant vertex which is adjacent to weak support, then the fractional domination number of (T - v) must be decrease. Thus v ∈ V_fD^-.

Case(ii) Let v be any pendant vertex which is adjacent to strong support, then the fractional domination number of (T - v) must increase its value. Thus v ∈ V_fD⁰.

Corollary 3.3. For any tree T and remove any strong support vertex of T, then γ_f (T - v) > γ_f (T).

Theorem 3.4. For any tree T, and remove any pendant vertex which is adjacent to strong support, then , where |l| is the number of leaves.

Proof. Take T to be a tree, and v to be any pendant vertex that is adjacent to strong support. Suppose we remove v from T, then to prove: . Suppose . Let be an edge dominating set of T and remove v and so also be a dominating set of T - v, then . We have enough to prove that |l| < Δ. Since, by definition of any tree, the number of leaves is always greater than or equal to the maximum degree of T. Which is contradiction to . Hence .

Theorem 3.5. For a graph with Δ (G) =2 and if removing any pendant vertex, then γ_f (G - v) + δ ≥ γ (G) ≥ γ (G - v) + Δ - 2δ.

Proof. Consider G as a graph with Δ (G) =2. Let v be any pendant vertex of G. Claim 1: γ_f (G - v) + δ ≥ γ (G). Let D be a γ-set of G and we remove any v, we define a function f : V (G) → [0, 1] by either v = 1 or v = 0 or $v = \frac{1}{t}$ or $t = 1 - \frac{1}{t}$ , where t is any integer. If v = 0 then γ_f (G - v) = |D| = γ (G) ⇒ γ_f (G - v) + δ > γ (G). If v = 1 then γ_f (G - v) = |D|-1 = γ (G) - δ ⇒ γ_f (G - v) + δ = γ (G). If $v = \frac{1}{t}$ then exactly f (N [v_i]) <1 for any one i. So the adjacent vertex of the pendant vertex value is one. Thus γ_f (G - v) = |D| = γ (G) ⇒ γ_f (G - v) + δ > γ (G). If $v = 1 - \frac{1}{t}$ then exactly f (N [v_i]) <1 for any one i. So the adjacent vertex of the pendant vertex value is one. Thus γ_f (G - v) = |D| = γ (G) ⇒ γ_f (G - v) + δ > γ (G). Therefore γ_f (G - v) + δ ≥ γ (G). Claim 2: γ (G) ≥ γ (G - v) + Δ - 2δ. Let D be a dominating set of G and we remove any v that is either v ∈ D or v does not belong to D. Suppose v ∈ D then γ_f (G - v) ≤ |D| and Δ (G) =2. This implies that γ (G - v) ≤ γ (G) ⇒ γ (G - v) + Δ - 2δ ≤ γ (G). Suppose v does not belong to D then γ (G - v) = |D| = γ (G) ⇒ γ (G - v) + Δ - 2δ = γ (G). Hence γ_f (G - v) + δ ≥ γ (G) ≥ γ (G - v) + Δ - 2δ.

Theorem 3.6. If any graph G contains no cycle with Δ (G) ≥2 and any pendant vertex v is removed from G, then $γ_{f} (G) + Δ (G) \geq γ_{f} (G - v) + 2 \geq γ_{f}^{'} (G - v) + 2 δ$ .

Proof. A graph G in which there is no cycle with a maximum degree of at least two. Assume that v is any pendant vertex of G. Claim 1: γ_f (G) + Δ (G) ≥ γ_f (G - v) +2. Case(i): When Δ (G) =2. Let D be a γ-set of G and we remove any pendant vertex v. Here, γ_f (G) = γ (G). In this case, we define a function f : V (G) → [0, 1] by either f (v) =1 (or) f (v) =0 (or) $f (v) = \frac{1}{t}$ (or) $f (v) = 1 - \frac{1}{t}$ , where $t \in ℕ$ . If f (v) =0 then γ_f (G - v) = |D| = γ_f (G) ⇒ γ_f (G) + Δ (G) = γ_f (G - v) +2. If f (v) =1 then γ_f (G - v) = |D|-1 = γ_f (G) -1 ⇒ γ_f (G) +2 > γ_f (G) -1 + 2 ⇒ γ_f (G) + Δ (G) > γ_f (G - v) +2. If $f (v) = \frac{1}{t}$ then exactly f (N [u]) <1 for any u which is adjacent to the pendant vertex and its value is one. Thus γ_f (G - v) is either γ_f (G - v) = γ_f (G) ⇒ γ_f (G) + Δ (G) = γ_f (G - v) +2 (or) γ_f (G - v) < γ_f (G) ⇒ γ_f (G) + Δ (G) > γ_f (G - v) +2. If $f (v) = 1 - \frac{1}{t}$ then exactly f (N [u]) <1 for any u which is adjacent to the pendant vertex and its value is one. Thus γ_f (G - v) is either γ_f (G - v) = γ_f (G) ⇒ γ_f (G) + Δ (G) = γ_f (G - v) +2 (or) γ_f (G - v) < γ_f (G) ⇒ γ_f (G) + Δ (G) > γ_f (G - v) +2. In path graph, we have $n = 3 k + 1, k \in ℕ$ then we get γ_f (G) + Δ (G) > γ_f (G - v) +2 otherwise γ_f (G) + Δ (G) = γ_f (G - v) +2. In star graph, γ_f (G) + Δ (G) = γ_f (G - v) +2. Thus γ_f (G) + Δ (G) ≥ γ_f (G - v) +2. Case(ii): If Δ (G) >2, then clearly we get γ_f (G) + Δ (G) ≥ γ_f (G - v) +2. Claim 2: $γ_{f} (G - v) + 2 \geq γ_{f}^{'} (G - v) + 2 δ$ . By our assumption, the minimum degree must be one that is, δ (G) =1. Here, γ (G) = γ_f (G). In path graph, if we have $n = 3 k + 2, k \in ℕ$ then we get $γ_{f} (G - v) + 2 > γ_{f}^{'} (G - v) + 2 δ$ otherwise $γ_{f} (G - v) + 2 = γ_{f}^{'} (G - v) + 2 δ$ . Suppose if we take the star graph, then clearly we have $γ_{f} (G - v) + 2 = γ_{f}^{'} (G - v) + 2 δ$ . For any tree graph, if we remove any pendant vertex that is adjacent to the strong support of degree two, then $γ_{f} (G - v) + 2 \geq γ_{f}^{'} (G - v) + 2 δ$ and if we remove any pendant vertex which is adjacent to the strong support of degree at least two, then $γ_{f} (G - v) + 2 > γ_{f}^{'} (G - v) + 2 δ$ . Thus $γ_{f} (G - v) + 2 \geq γ_{f}^{'} (G - v) + 2 δ$ . Hence we have $γ_{f} (G) + Δ (G) \geq γ_{f} (G - v) + 2 \geq γ_{f}^{'} (G - v) + 2 δ$ . For example, the above bound is sharp when the path graph n = 3k, $k \in ℕ$ , and the star graph k_1,2 give the exactness of this bound.

In Fig. 2, if we remove any pendant vertex v, then it will show the sharpness of this bound. Hence, Fig. 2 gives the exact bound for theorem 3.6.

Fig. 2

Sharpness with Path on 3k vertices.

4 Conclusion

In this paper, the concept of changing and unchanging domination on a fractional domination number of some standard graphs like paths, cycles, trees are derived by eliminating a vertex from graph G. We also derived the changing and unchanging of fractional domination numbers on some specific graphs, such as the wheel graph, the crown graph, and the self-centered graph. In addition, we obtained the bounds for fractional domination numbers on trees. For any graph G that contains no cycle with a maximum degree of at least two, if we remove any pendant vertex v from G, then we get the inequality $γ_{f} (G) + Δ (G) \geq γ_{f} (G - v) + 2 \geq γ_{f}^{'} (G - v) + 2 δ$ and also discussed some examples in terms of the bound which yield the exactness. Future considerations could include removing or adding an edge on graphs on some other fractional dominated related parameters. Furthermore, we examine this concept in linear programming problems (Lpp) also.

Footnotes

Acknowledgments

This article has been written with the joint financial support of the RUSA Phase 2.0 (F 24-51/2014-U), DST-FIST (SR/FIST/MS-I/2018/17), DST-PURSE 2nd Phase programme (SR/PURSE Phase 2/38), Govt. of India.

References

Amutha

and Sridharan

, A note on sets

V_{t}^{-}, V_{t}^{0}, V_{t}^{+}

of a simple graph G with δ (G) ≥2, Journal of Pure and Applied Mathematics: Advances and Applications 9 (2013), 69–79.

Arumugam

Sithara Jerry

Fractional Edge Domination in Graphs, Applicable Analysis and Discrete Mathematics 3 (2009), 359–370.

Balamurugan

, Changing and unchanging isolate domination: Edge removal, Discrete Mathematics Algorithms and Applications 9(1) (2017).

Bhanumathi

Sudhasenthil , Changing and unchanging of eccentric domination number in graphs, International Journal of Pure and Applied Mathematics 118(23) (2018), 373–381.

Haynes

T.W.

, Hedetniemi

S.T.

, Slater

P.J.

Fundamentals of domination in Graphs, Marcel Dekker, Inc., (1998).

Harary

Graph Theory, Addison wesley Publishing Company Reading, Massachuretts, (1972).

Janakiraman

T.N.

, Alphonse

P.J.A.

and Sangeetha

, Changing and Unchanging of distance closed domination number in graphs, International Journal of Engineering Science, Advanced Computing and Bio-Technology 3(2) (2012), 67–84.

Muhiuddin

, Sridharan

, Al-kadi

, Amutha

and Elnair

M.E.

, Reinforcement Number of a Graph with respect to Half-Domination, Hindawi Journal of Mathematics 2021 (2021), 7 Article ID 6689816.

Roushini Leely Pushpam

and Shanthi

P.A.

, Changing and unchanging in m-eternal total domination in graphs, Electronic Notes in Discrete Mathematics 53 (2016), 181–198.

10.

Scheinerman

E.R.

, Ullman

D.H.

Fractional Graph Theory: A rational approach to the theory of graphs, JohnWiley and Sons Inc., (1997).

11.

Sridharan

, Amutha

and Ramesh

, Babu, Bounds for λ- domination number γ _ λ (G) of a graph, Journal of Computational Mathematica 1(2) (2017), 79–90.

12.

Suriya Prabha

Amutha

Anbazhagan

Sankara Gomathi

, Neighborhood outer split domination in graphs, Journal of Discrete Mathematical Sciences and Cryptography 22(5) (2019), 787–799.