Properties of Partial Dominating Sets of Graphs

TL;DR

This paper explores properties of partial dominating sets in graphs, focusing on graph products, locating sets, and introducing p-influencing sets, advancing understanding of partial domination concepts.

Contribution

It introduces the concept of p-influencing sets and investigates their properties, extending the theory of partial dominating sets in graph products.

Findings

Properties of p-influencing sets analyzed

Relations between partial domination and graph products established

New bounds and characterizations provided for p-domination

Abstract

A set $S\subseteq V$ is a dominating set of $G$ if every vertex in $V - S$ is adjacent to at least one vertex in $S$. The domination number $\gamma(G)$ of $G$ equals the minimum cardinality of a dominating set $S$ in $G$; we say that such a set $S$ is a $\gamma$-set. A generalization of this is partial domination which was introduced in 2017 by Case, Hedetniemi, Laskar, and Lipman [3,2] . In partial domination a set $S$ is a $p$-dominating set if it dominates a proportion $p$ of the vertices in $V$. The p-domination number $\gamma_{p}(G)$ is the minimum cardinality of a $p$-dominating set in $G$. In this paper, we investigate further properties of partial dominating sets, particularly ones related to graph products and locating partial dominating sets. We also introduce the concept of a $p$-influencing set as the union of all $p$-dominating sets for a fixed $p$ and investigate some of its properties.

Equations

$$\frac{|N[S]|}{|V|}\ge p.$$

$$\gamma (G\square H)\ge \gamma (G)\gamma (H).$$

$${\gamma }_p(G\square H)\ge {\gamma }_p(G){\gamma }_p(H).$$

$${\gamma }_{1/2}(P_m\square P_n)\ge {\gamma }_{1/2}(P_m){\gamma }_{1/2}(P_n).$$

$${\gamma }_{1/2}(K_m\square K_n)=\lceil \frac{m+n-\sqrt{m^2+n^2}}{2}\rceil .$$

$$(1)v_2=(v,u^{\prime })(2)v_2=(u,v^{\prime })(3)v_2=(u,u^{\prime })$$

$$\min \{k|\sum _{i=1}^k\left(m+n-(2i-1)\right)\ge \frac{mn}{2}\}.$$

$$k(m+n)-k^2\ge \frac{mn}{2}$$

$$k=\lceil \frac{m+n-\sqrt{m^2+n^2}}{2}\rceil .$$

$${\gamma }_{1/2}(P_n\square K_m)=\lceil \frac{mn}{2(m+2)}\rceil .$$

$$\lceil \frac{n-2}{2}\rceil =k-1.$$

$${\gamma }_{1/2}(P_n\square K_m)=\lceil \frac{mn}{2(m+2)}\rceil .$$

$$k(m+2)>\frac{mn}{2}>(k-1)(m+2)⟹k=\lceil \frac{mn}{2(m+2)}\rceil .$$

$${\gamma }_{1/2}(G\square P_2)\ge {\gamma }_{1/2}(G).$$

$$S^{\prime }=S_1\cup \{v|v\in N[S_2]\cap G_1\}.$$

$${\gamma }_{1/2}(G\square P_m)\ge {\gamma }_{1/2}(P_m).$$

$${\gamma }_{1/2}(G\square P_m)\ge \lceil \frac{mn}{n+2}\rceil \ge \lceil \frac{m}{3}\rceil \ge \lceil \frac{m}{6}\rceil ={\gamma }_{1/2}(P_m).$$

$${\gamma }_{1/2}(G\square P_m)\ge 2{\gamma }_{1/2}(P_m).$$

$${\gamma }_{1/2}(G\square P_m)\ge \lceil \frac{mn}{2\left(\lceil \frac{n}{2}\rceil +1\right)}\rceil \ge \lceil \frac{mn}{n+3}\rceil .$$

$${\gamma }_{1/2}(P_m)=\lceil \frac{m}{6}\rceil =\{\begin{array}{ll}a& b=0\\ a+1& b\ne 0\end{array}.$$

$${\gamma }_{1/2}(G\square P_m)\ge \lceil 0.7m\rceil =\lceil 4.2a+0.7b\rceil \ge 2(a+1)\ge 2{\gamma }_{1/2}(P_m).$$

$${\gamma }_{1/2}(G\square P_m)\ge 3{\gamma }_{1/2}(P_m).$$

$${\gamma }_{1/2}(G\square P_m)\ge \lceil \frac{mn}{2\left(\lceil \frac{n}{2}\rceil -1\right)}\rceil \ge \lceil \frac{mn}{n-1}\rceil \ge m.$$

$${\gamma }_{1/2}(P_m)=\lceil \frac{m}{6}\rceil =\{\begin{array}{ll}a& b=0\\ a+1& b\ne 0\end{array}.$$

$${\gamma }_{1/2}(G\square P_m)\ge m=6a+b\ge 3(a+1)\ge 3{\gamma }_{1/2}(P_m).$$

$${\gamma }_{1/2}(G\square P_2)\ge \lceil \frac{n}{\lceil \frac{n}{2}\rceil -1}\rceil =3=3{\gamma }_{1/2}(P_2).$$

$$S^{\prime }=\{v\}\cup (S\setminus \{s\})$$

$$1/n,2/n,3/n,...,1.$$

$$V_1=\{v_1,v_4,\dots ,v_{n-2}\},V_2=\{v_2,v_5,\dots ,v_{n-1}\},V_3=\{v_3,v_6,\dots ,v_n\}.$$

$$V_1=\{v_1,v_4,\dots ,v_n\},V_2=\{v_2,v_5,\dots ,v_{n-2}\},V_3=\{v_3,v_6,\dots ,v_{n-1}\}.$$

Full text

Keywords: partial domination, dominating set, partial domination number, domination number, influencing set, graph parameters, Vizing’s conjecture

Benjamin M. Case was partially supported by the National Science Foundation under grants DMS-1403062 and DMS-1547399.

1. Introduction

Let $G=(V,E)$ be a graph with vertex set $V=\{v_{1},v_{2},...,v_{n}\}$ and order $n=|V|$. The open neighborhood of a vertex $v$ is the set $N(v):=\{u\>|\>uv\in E\}$ of vertices $u$ that are adjacent to $v$; the closed neighborhood of $v$ is $N[v]:=N(v)\cup\{v\}.$ A set $S\subseteq V$ is a dominating set of $G$ if every vertex in $V-S$ is adjacent to at least one vertex in $S$, or equivalently, if $N[S]:=\bigcup_{u\in S}N[u]=V$. The domination number $\gamma(G)$ of $G$ equals the minimum cardinality of a dominating set $S$ in $G$; we say that such a set $S$ is a $\gamma$-set. Domination has been a well studied area for many years [1, 4, 7, 8, 9].

For any graph $G=(V,E)$ and proportion $p\in[0,1]$, a set $S\subseteq V$ is a p-dominating set if

[TABLE]

The p-domination number $\gamma_{p}(G)$ equals the minimum cardinality of a $p$-dominating set in $G$. Partial domination was first introduced by Case, Hedetniemi, Laskar, and Lipman [3, 2] in 2017. Around the same time, in an independent work of Das the same concept was introduced [6, 5].

As noted in [3], a $\gamma_{p}$-set is not in general related to a $\gamma$-set. In particular, a $\gamma$-set does not necessarily contain a $\gamma_{p}$-set. Equivalently, a $\gamma_{p}$-set cannot necessarily be extended to a $\gamma$-set. To see this, it is helpful to revisit the subdivided star graph in Figure 1, where the $\gamma$-set denoted by triangles is disjoint from $\gamma_{1/2}$-set consisting of just the square vertex.

Organization of the paper. In section 2 we generalize Vizing’s conjecture about domination in graph products to the setting of partial domination and prove some special cases. In Section 3 we investigate some results related to finding a partial dominating set in a graph. In Section 4 we introduce $p$-influencing sets and consider some properties and examples.

2. Graph Products

We investigate properties of partial dominating sets related to graph products. In 1986, Vadim G. Vizing conjectured that for domination

[TABLE]

Here we conjecture that

[TABLE]

We will primarily be interested in the case when $p=1/2.$

Proposition 2.1**.**

For paths $P_{m}$ and $P_{n},$ with $n\geq m\geq 2$

[TABLE]

Proof.

The claim follows directly from Theorem 2.8 in [2], which says

1. (1)

$\gamma_{1/2}(P_{n})=\lceil n/6\rceil,$ 2. (2)

for $m=2,$ $\gamma_{1/2}(P_{2}\square P_{n})=\lceil n/4\rceil,$ 3. (3)

for $m\geq 3,$ $\gamma_{1/2}(P_{m}\square P_{n})=\lceil mn/10\rceil.$

∎

Note that if $\gamma_{p}(G)=\gamma_{p}(H)=1,$ then the conjecture holds trivially (e.g., if $G$ and $H$ are both complete graphs).

Proposition 2.2**.**

For complete graphs, $K_{m}$ and $K_{n},$

[TABLE]

Proof.

We consider choosing vertices for a $\gamma_{1/2}$-set $S.$ Say the first vertex added to $S$ is $v_{1}=(v,v^{\prime}).$ Note that $v$ dominates exactly $m+n-1$ vertices. If $m+n-1\geq mn/2,$ we are done. If not, we must add another vertex $u$ to $S.$ Note we essentially have three options here:

[TABLE]

where $u\neq v$ and $u^{\prime}\neq v^{\prime}.$ In the first case, $v_{2}$ dominates exactly $m-1$ vertices not already dominated by $S.$ In the second case $v_{2}$ dominates $n-1$ such vertices, and in the last case $v_{2}$ dominates $m+n-3$ such vertices.

Now in general, when adding the next vertex to $S,$ it is helpful to consider which copies of $K_{m}$ and $K_{n}$ the vertex is coming from relative to the last vertex chosen. We have these three options: (1) choose the same $K_{m}$ but a different $K_{n},$ (2) choose a different $K_{m}$ but the same $K_{n},$ (3) choose a different $K_{m}$ and $K_{n}.$ In general option (3) is optimal, followed by option (1), then option (2). When option (3) is used every time in such a way that each vertex added to $S$ comes from a copy of $K_{m}$ and $K_{n}$ not previously used, then the $k$th vertex dominates $m+n-2k+1$ vertices not already dominated by $S.$ We claim this is the optimal strategy in choosing vertices for a $\gamma_{1/2}$-set. We refer to it as $*$

To see this, consider alternative choice of vertices, where we execute option (2) optimally $\ell$ times, and then execute option (3) optimally until we have a $1/2$-dominating set. We refer to this as $**.$ Now compare the number of vertices dominated by $*$ and $**$ at each step. There is no difference for the first vertex. For the second, $**$ is dominates $n-2$ fewer than $*.$ In general, for the $k$th vertex added in $**$ with $2\leq k\leq\ell+1,$ $**$ dominates $(k-1)(n-k)$ fewer vertices than $*$. So at step $\ell+1,$ $**$ dominates $\ell(n-\ell-1)$ fewer vertices than $*.$ Now for each vertex added after step $\ell+1,$ $**$ dominates $\ell$ more vertices than $*.$ Thus, it will take $**$ $(n-\ell-1)+1+\ell=n$ steps to dominate the same number of vertices as $*.$ However, $*$ will have produced a $1/2$-dominating set well before the $n$th step. Using a similar argument, we can show that any choice of vertices different than $*$ will be suboptimal.

Now the cardinality of the $\gamma_{1/2}$-set found by $*$ is

[TABLE]

But note that $\sum_{i=1}^{k}\left(m+n-(2i-1)\right)=k(m+n)-k^{2}.$ So the inequality to solve is

[TABLE]

which has minimum positive integer solution

[TABLE]

∎

Proposition 2.3**.**

For a path $P_{n}$ and complete graph $K_{m},$

[TABLE]

Proof.

The maximum degree of a vertex in $P_{n}\square K_{m}$ is $m+2.$ Consider the set of vertices $S$ of maximum cardinality such that each vertex in $S$ dominates exactly $m+2$ vertices, and no two vertices in $S$ dominate the same vertex. The cardinality of $S$ is $\left\lceil\frac{n-2}{2}\right\rceil.$ Now write $n=2k$ $(n=2k-1)$ for even (odd) $n$ and $k\in\mathbb{N}.$ Note that in either case

[TABLE]

That is $S$ has cardinality $k-1.$

Now suppose $(m+2)(k-1)\geq mn/2.$ Then either $S$ or a subset of $S$ is a $\gamma_{1/2}$-set, and

[TABLE]

On the other hand, suppose $(m+2)(k-1)<mn/2.$ Then neither $S$ nor any subset of $S$ is a $\gamma_{1/2}$-set. However, we may add a vertex $v$ to $S$ so $S\cup\{v\}$ dominates at least $k$ copies of $K_{m}.$ That is, $S\cup\{v\}$ is a $\gamma_{1/2}$-set. Moreover, we have

[TABLE]

∎

Proposition 2.4**.**

Let $G$ be a graph of order $n,$

[TABLE]

Proof.

Let $S$ be a $\gamma_{1/2}(G\square P_{2})$-set. Now think of $G\square P_{2}$ as two copies of $G,$ say $G_{1}$ and $G_{2}.$ Then since $|N[S]|\geq n,$ we have WLOG that $|N[S]\cap G_{1}|\geq n/2.$ Denote $S\cap G_{i}$ as $S_{i}.$ Now consider the vertex set

[TABLE]

Then $S^{\prime}$ dominates at least half of $G,$ and $|S^{\prime}|\leq|S|.$

Thus, given a $\gamma_{1/2}(G\square P_{2})$-set, we can find a vertex set of $G$ of size at most $\gamma_{1/2}(G\square P_{2})$ which dominates $1/2$ of $G.$ So $\gamma_{1/2}(G\square P_{2})\geq\gamma_{1/2}(G).$ ∎

Proposition 2.5**.**

If $\gamma_{1/2}(G)=1,$ then

[TABLE]

Proof.

Let $|V(G)|=n.$ Since $\gamma_{1/2}(G)=1,$ there exists some $v\in V(G)$ such that $|N[v]|\geq\left\lceil\frac{n}{2}\right\rceil.$ Now at best, $|N[v]|=n,$ in which case, $\max\{|N[v]|:v\in V(G\square P_{m})\}=n+2.$ Thus we have

[TABLE]

∎

Proposition 2.6**.**

If $\gamma_{1/2}(G)=2,$ then

[TABLE]

Proof.

Let $|V(G)|=n.$ Since $\gamma_{1/2}(G)=2,$ then $\max\{|N[v]|:v\in V(G)\}\leq\lceil\frac{n}{2}\rceil-1.$ Therefore, $\max\{|N[v]|:v\in V(G\square P_{m})\}\leq\lceil\frac{n}{2}\rceil+1.$

Thus we have

[TABLE]

Now, since $\gamma_{1/2}(G)=2,$ we must have $n\geq 7.$ Thus $\gamma_{1/2}(G\square P_{m})\geq\lceil 0.7m\rceil$.

Now say $m=6a+b.$ Then

[TABLE]

Thus for all $m\geq 2$,

[TABLE]

∎

Proposition 2.7**.**

If $\gamma_{1/2}(G)=3,$ then

[TABLE]

Proof.

Let $|V(G)|=n.$ Since $\gamma_{1/2}(G)=3,$ we must have $n\geq 13.$ Moreover, we have $\max\{|N[v]|:v\in V(G)\}\leq\left\lceil\frac{n}{2}\right\rceil-3.$ Thus, $\max\{|N[v]|:v\in V(G\square P_{m})\}\leq\left\lceil\frac{n}{2}\right\rceil-1.$

Therefore we have

[TABLE]

Now say $m=6a+b.$ Then

[TABLE]

Thus for $m\geq 3,$

[TABLE]

Lastly, when $m=2,$ we have

[TABLE]

∎

3. Locating Partial Dominating Sets in Graphs

When we look a graph we want some tools (theorems and/or algorithms) that will help us locate a partial dominating set or elements of a partial dominating set. A $\gamma_{p}$ set does not have to be unique; and for many applications having anyone of them would work. The first intuitive idea in looking for a partial dominating set is to consider vertices with high degrees. The following several results will explore what can and cannot be achieved following this idea.

Lemma 3.1**.**

For any $p\in[0,1]$, if a vertex $v\in G$ has the highest degree, then there is a $\gamma_{p}$-set that contains at least one of the following:

$v$,

• 2)

a neighbor of $v$, i.e. an element of $N(v)$,

• 3)

a distance two neighbor of $v$, i.e. an element of $N(N[v])$.

Proof.

Suppose none of these vertices is in any $\gamma_{p}$ set. Let $v\in G$ be a vertex of highest degree. If $S\subseteq G$ is a $p$-dominating set of $G$, consider any $s\in S$. The degree of $s$ is less than or equal to the degree of $v$. So the set

[TABLE]

dominates at least as many vertices as $S$. Thus, $S^{\prime}$ is a $p$-dominating set that contains $v$. ∎

Now we illustrate with some examples that in the preceding theorem, it may be the case that 2) or 3) hold and not 1).

Example 3.2**.**

Let $p=8/9$ consider the graph in Figure 2. The set of boxed vertices is the only $p-$dominating set. This shows that the highest degree vertex may not be in any of the $p$-dominating sets of the graph but that instead some of its distance 2 neighbors are.

Example 3.3**.**

Let $p=7/9$ and consider the graph in Figure 3. The triangle vertex, $v$, is a maximum degree vertex in the graph. Any two vertices chosen from $N(v)$ form a $\gamma_{7/9}$-set. Thus it may be that a highest degree vertex is not in any $p$-dominating set, but that instead some its neighbors are.

Lemma 3.4**.**

Let $p\in[0,1]$ be fixed. If $v\in V$ has highest degree and $v$ is not in any $p$-dominating set, then $|S\cap N[N[v]]|\geq 2$ for every $\gamma_{p}$-set $S.$

Proof.

Prove by contradiction; suppose $|S\cap N[N[v]]|<2$ for some $\gamma_{p}$-set $S.$ In the case $|S\cap N[N[v]]|=0$, we can swap any $s\in S$ with $v$ to make a $p$-dominating set since $v$ has highest degree and none of its neighbors would be dominated. This contradicts $v$ not being in any $p$-dominating set.

In the case $|S\cap N[N[v]]|=1$, suppose $\{u\}=S\cap N[N[v]]$. In the case $u\in N(v)$, we can swap $u$ and $v$ and still have a $p$-dominating set. In the case $u\in N(N(v))$, we can again swap $u$ and $v$ and still have a $p$-dominating set. In either case this again contradicts $v$ not being in any $p$-dominating set. ∎

These results and examples together show that being greedy for the highest degree vertices does not work by itself in finding you elements from a $p$-dominating set, but that this greedy mindset can get you looking in the right area of the graph. Observe that neither of graphs in Figures 2 or 3 were trees. The graph if Figure 4 also shows that in a tree, a highest degree vertex need not be in a $p$-dominating set.

4. $p$-Influencing Set

Now we introduce a related definition by considering the union of all $p$-dominating sets of a graph. We will call this union the influencing set.

Definition 4.1**.**

Let $p\in[0,1]$ be fixed. The union of all $\gamma_{p}$-sets of $G$ is called the $p-$influencing set of $G$.

Note that there are $n=|V|$ interesting proportions $p$ that can be considered for a graph

[TABLE]

Also we’ve allowed $p=0$, but here the $p$-dominating set is just $\emptyset$. For the smallest of these above $p=1/n$ the $p$-influencing set is all of $V$.

Lemma 4.2**.**

Let $p=1/n$, then the $p$-influencing set of $G$ is all of $V$. If $G$ is a connected graph and $p=2/n$, then the $p$-influencing set of $G$ is all of $V$.

Proof.

For the first, every individual vertex in $V$ is a $1/n$-dominating set. Thus the union of all $1/n$-dominating sets contains every vertex in $V$.

For the second, every individual vertex in $V$ is a $2/n$-dominating set, since it dominates itself and at least one neighbor. Thus the union of all $2/n$-dominating sets is all of $V$. ∎

This lemma can be generalized further as follows.

Lemma 4.3**.**

Let $\delta(G)$ be the smallest degree of any vertex in $G$. If $p\leq\frac{\delta(G)+1}{n}$, then the $p$-influencing set of $G$ is all of $V$.

Proof.

Every vertex is a $\frac{\delta(G)+1}{n}$ dominating set, thus the $\frac{\delta(G)+1}{n}$-influencing set is all of $G$. ∎

As we see above, for the smaller interesting proportions the $p$-influencing set is as large as possible. One might ask if the other $p$-influencing sets have any containment properties as the proportion increases or decreases, or if the size of $p$-influencing set only decreases as $p$ increases. In general, this does not happen. Considering the graph in Figure 2, one can see that as $p$ runs from $1/9$ to 1 the $p-$influencing sets change from all the vertices down to one vertex and then back to all the vertices. Also in this example the intersection of all the $p$-influencing sets with $p>0$ is $\emptyset$.

Lemma 4.4**.**

Let $\Delta(G)$ be the highest degree of any vertex in $G$. If $p=\frac{\Delta(G)+1}{n}$, then the $p$-influencing set is made up of exactly those vertices in $G$ with degree equal to $\Delta(G)$.

Proof.

When $p=\frac{\Delta(G)+1}{n}$, each vertex of degree $\Delta(G)$ is a $p$-dominating set. Furthermore, there are no $p$-dominating sets with more than one vertex, since they would not be of minimum size. Thus, the $p$-influencing set consists exactly of the degree $\Delta(G)$ vertices. ∎

The previous results gave the $p$-influencing sets of any graph for a fixed $p.$ We now find all $p$-influencing sets for two common graphs, namely complete bipartite graphs and paths.

Proposition 4.5**.**

Consider the complete bipartite graph $K_{m,n}=(V_{1},V_{2},E)$ with $|V_{1}|=m,|V_{2}|=n,$ and $m>n.$ The $p$-influencing sets of $K_{m,n}$ are

• •

$V_{1}\cup V_{2}$* for $0<p\leq\frac{n+1}{m+n}$ and $p\geq\frac{m+2}{m+n}$*

• •

$V_{2}$* for $\frac{n+2}{m+n}\leq p\leq\frac{m+1}{m+n}$*

Moreover, if $m=n,$ then the $p-$influencing set of $K_{m,n}$ is $V_{1}\cup V_{2}$ for all $p>0.$

Proof.

Suppose $m>n.$ If $0<p\leq\frac{n+1}{m+n},$ then any vertex of $K_{m,n}$ is a $\gamma_{p}$-set. Thus the $p$-influencing set is $V_{1}\cup V_{2}.$ If $p\geq\frac{m+2}{m+n},$ then $\{v_{1},v_{2}\}$ where $v_{1}\in V_{1}$ and $v_{2}\in V_{2},$ is a $\gamma_{p}$-set. Thus the $p$-influencing set is $V_{1}\cup V_{2}.$ Lastly, if $\frac{n+2}{m+n}\leq p\leq\frac{m+1}{m+n},$ then any single vertex from $V_{2}$ is a $\gamma_{p}$-set, and no vertex from $V_{1}$ can be in a $\gamma_{p}$-set.

The argument is similar when $m=n.$ ∎

Corollary 4.6**.**

Consider the complete bipartite graph $K_{m,n}=(V_{1},V_{2},E)$ with $|V_{1}|=m,|V_{2}|=n,$ and $m>n.$ The intersection of all the $p$-influencing sets ($p>0$) of $K_{m,n}$ is $V_{2}.$ Moreover, if $m=n,$ the intersection is $V_{1}\cup V_{2}.$

For the following proposition and corollary, we consider a path $P_{n}$ where the vertices are labeled $v_{i}$, $i\in\{1,2,\dots,n\}$ with $v_{1},v_{n}$ as leaves and $N(v_{i})=\{v_{i-1},v_{i+1}\}$ for $i\in\{2,3,\dots,n-1\}.$

Proposition 4.7**.**

Consider a path $P_{n}$ with the vertices labeled as described above. The $p$-influencing sets of $P_{n}$ are given below. Note $k$ is a nonnegative integer chosen so that $0<p\leq 1$ unless otherwise stated.

1. (1)

For $n\equiv 0\mod{3}$

• •

$V(P_{n})$* if $p=\frac{3k+1}{n}$ or $p=\frac{3k+2}{n}$*

• •

$\{v_{2},v_{3},\dots,v_{n-1}\}$* if $p=\frac{3k}{n}<1$*

• •

$\{v_{2},v_{5},v_{8},\dots,v_{n-1}\}$* if $p=1$* 2. (2)

For $n\equiv 1\mod{3}$

• •

$V(P_{n})$* if $p=\frac{3k+1}{n}$ or $p=\frac{3k+2}{n}$*

• •

$\{v_{2},v_{3},\dots,v_{n-1}\}$* if $p=\frac{3k}{n}$ and $3k\neq n-1$*

• •

$V(P_{n})\setminus\{v_{1},v_{4},v_{7},\dots,v_{n}\}$* if $p=\frac{n-1}{n}$* 3. (3)

For $n\equiv 2\mod{3}$

• •

$V(P_{n})$* if $p=\frac{3k+1}{n}$ or $p=\frac{3k+2}{n}<1$*

• •

$\{v_{2},v_{3},\dots,v_{n-1}\}$* if $p=\frac{3k}{n}$*

• •

$V(P_{n})\setminus\{v_{3},v_{6},v_{9},\dots,v_{n-2}\}$* if $p=1$*

Proof.

Suppose $n\equiv 0\mod{3}.$ Partition $V(P_{n})$ into three sets

[TABLE]

Now consider $p=\frac{3k+1}{n}$ or $p=\frac{3k+2}{n}.$ Then $\gamma_{p}(P_{n})=k+1,$ and any $k+1$ vertices of $V_{1},V_{2},$ or $V_{3}$ comprise a $\gamma_{p}$-set.

Now consider $p=\frac{3k}{n}<1.$ Then $\gamma_{p}(P_{n})=k,$ and any $k$ vertices of $V_{1}\setminus\{v_{1}\},V_{2}$ or $V_{3}\setminus\{v_{n}\}$ comprises a $\gamma_{p}$-set.

Lastly, if $p=1,$ then $\gamma_{1}(P_{n})=\frac{n}{3},$ and $V_{2}$ is the only $\gamma_{1}$-st.

Suppose $n\equiv 1\mod{3}.$ Partition $V(P_{n})$ into three sets

[TABLE]

Note that $|V_{1}|=\frac{n-1}{3}+1,$ $|V_{2}|=|V_{3}|=\frac{n-1}{3}.$

Now consider $p=\frac{3k+1}{n}$ or $p=\frac{3k+2}{n},$ then $\gamma_{p}(P_{n})=k+1.$ Now if $p<1,$ so $k<\frac{n-1}{3},$ then any $k+1$ vertices of $V_{1},V_{2},$ or $V_{3}$ comprise a $\gamma_{p}$-set. If $p=1,$ so $k=\frac{n-1}{3},$ then $V_{1},$ $V_{2}\cup\{v_{n}\},$ and $V_{3}\cup\{v_{1}\}$ are $\gamma_{1}(P_{n})$-sets.

Now if $p=\frac{3k}{n}$ and $3k\neq n-1,$ so $k<\frac{n-1}{3}.$ Then $\gamma_{p}(P_{n})=k,$ and any $k$ vertices from $V_{1}\setminus\{v_{1},v_{n}\},V_{2},$ or $V_{3}$ comprise a $\gamma_{p}$-set.

Lastly, if $p=\frac{n-1}{n},$ then $\gamma_{p}(P_{n})=\frac{n-1}{3}.$ Then note that $V_{2}$ and $V_{3}$ are the only $\gamma_{p}(P_{n})$-sets.

Suppose $n\equiv 2\mod{3}.$ Partition $V(P_{n})$ into three sets

[TABLE]

Note that $|V_{1}|=|V_{2}|=\frac{n+1}{3},$ and $|V_{3}|=\frac{n-2}{3}.$

Now consider $p=\frac{3k+1}{n}$ or $p=\frac{3k+2}{n}<1,$ so $k<\frac{n-2}{3}.$ Then $\gamma_{p}(P_{n})=k+1$ and any $k+1$ vertices from $V_{1},V_{2},$ or $V_{3}$ comprise a $\gamma_{p}$-set.

Now if $p=\frac{3k}{n},$ so $k\leq\frac{n-2}{3},$ then $\gamma_{p}(P_{n})=k.$ Any $k$ vertices from $V_{1}\setminus{v_{1}},V_{2}\setminus{v_{n}},$ or $V_{3}$ will be a $\gamma_{p}$-set.

Lastly, if $p=1,$ then $\gamma_{1}(P_{n})=\frac{n+1}{3}.$ Note that $V_{1}$ and $V_{2}$ are the only $\gamma_{1}(P_{n})$-sets. ∎

Corollary 4.8**.**

The intersection of all $p$-influencing sets ($p>0$) for a path $P_{n}$ is

1. (1)

$\{v_{2+3k}|0\leq k\leq\frac{n-3}{3}\}$* if $n\equiv 0\mod{3}$* 2. (2)

$\{v_{2+3k},v_{3+3k}|0\leq k\leq\frac{n-4}{3}\}$* if $n\equiv 1\mod{3}$* 3. (3)

$\{v_{1+3k},v_{2+3j}|0<k\leq\frac{n-2}{3},0\leq j<\frac{n-2}{3}\}$* if $n\equiv 2\mod{3}$.*