Bound for the k-Fault-Tolerant Power-Domination Number

Abstract

A set $S\subseteq V$ is referred to as a k-fault-tolerant power-dominating set of a given graph $G=(V,E)$ if the difference $S\setminus F$ remains a power-dominating set of G for any $F\subseteq S$ with $\left|F\right|\le k$, where k is an integer with $0\le k<\left|V\right|$. The lowest cardinality of a k-fault-tolerant power-dominating set is the k-fault-tolerant power-domination number of G, denoted by ${\gamma }_P^k\left(G\right)$. Generalized Petersen graphs $GP(m,k)$ and generalized cylinders $SG$ are two well-known graph classes. In this paper, we calculate the k-fault-tolerant power-domination number of the generalized Petersen graphs $GP(m,1)$ and $GP(m,2)$. Also, we obtain ${\gamma }_P^k\left(G\right)$ for the subclasses of cylinders $S{C}_m$ and $S{B}_m$.

1. Introduction

The concept of power-domination in graphs emerged from a monitoring problem in the electric power system industry [1]. Electric power companies must continuously monitor their systems, understanding their state through variables like voltage magnitude and machine phase angle. Phasor Measurement Units (PMUs) placed at strategic points measure these variables, which are then used to calculate variables in other parts of the network using Ohm’s and Kirchoff’s laws. To minimize costs, it is desirable to monitor the entire system with the fewest number of PMUs. This monitoring problem in electric power systems was transposed to a graph-domination problem by Haynes et al. [2].

Definition 1. 

A vertex or edge of a graph is said to be monitored or observed using the following rules: Initially, each vertex with a PMU and the edges incident to it are said to be observed or monitored. Then, the set of monitored edges and vertices is expanded in an iterative manner using the following rules:

1. 

A vertex which is incident to a monitored edge whose other end is monitored is said to be monitored.

2. 

An edge connecting two monitored vertices is monitored.

3. 

If all but one incident edge of a vertex are monitored, this one edge is also monitored.

Definition 2. 

In a finite and simple undirected graph, $G=(V,E)$, with vertex set V and edge set E, a subset $S\subseteq V$ is defined as a power-dominating set of a graph if each vertex and edge in the graph is monitored or observed by S.

The minimum cardinality of a power-dominating set of G is called the power-domination number, denoted by ${\gamma }_P\left(G\right)$. Different concepts of domination and its variations have been studied in graph theory [3,4,5,6,7]. Various algorithms and their time complexity are shown in [8].

Once a minimal number of PMUs are placed in an electric network, the next challenge is to detect, locate, and replace faulty PMUs. However, identifying, finding, and replacing faulty PMUs in a timely manner may be challenging, time-demanding, and financially burdensome. As a result, Pai et al. [9] proposed the concept of k-fault-tolerant power-domination. The k-fault-tolerant power-domination problem is a generalization of the power-domination problem, and it remains NP-complete for general graphs.

Definition 3. 

A set $S\subseteq V$ is a k-fault-tolerant power-dominating set if $S\setminus F$ is still a power-dominating set of G for any subset $F\subset S$ with $\left|F\right|\le k$. A k-fault-tolerant power-dominating set is abbreviated as k-FPDS.

The minimum cardinality of a k-fault-tolerant power-dominating set is the k-fault-tolerant power-domination number of G, denoted by ${\gamma }_P^k\left(G\right)$.

A k-fault-tolerant power-dominating set S whose cardinality is ${\gamma }_P^k\left(G\right)$ is also called a ${\gamma }_P^k\left(G\right)$-set. The k-fault-tolerant power-domination problem is to find the value of ${\gamma }_P^k\left(G\right)$ and find a ${\gamma }_P^k\left(G\right)$-set. This set actually gives the bare minimum number of PMUs needed to monitor the system, taking into account faults in k PMUs. This is a boon for monitoring electrical power systems since it saves the time required to locate and replace PMUs when they are found to be faulty. It is easy to see the following:

• ${\gamma }_P^0\left(G\right)={\gamma }_P\left(G\right)$
• ${\gamma }_P\left(G\right)+k\le {\gamma }_P^k\left(G\right)$
• ${\gamma }_P^k\left(G\right)<{\gamma }_P^{k+1}\left(G\right).$

It is difficult to establish the equality in (3). In this paper, we have modeled electric power networks using the graphs $GP(m,1)$, $GP(m,2)$, $S{C}_n$, and $S{B}_m$, where the first two are the subclasses of the generalized Petersen graphs and the last two are the subclasses of cylinders, and the k-fault-tolerant power-domination number for these graphs is established.

The following assumptions and definitions are considered to be implicit and are consistently applied throughout the text:

• A graph G is said to be ‘power-dominated by a set S’ or ‘observed by a set S’ if all the vertices and edges of the graph are monitored by the set S.
• When there is no ambiguity, the notations $\gamma \left(G\right)$, ${\gamma }_p\left(G\right)$, ${\gamma }_p^k\left(G\right)$, $\overline{{\gamma }_P}\left(G\right)$, etc., will be denoted as $\gamma$, ${\gamma }_p$, ${\gamma }_p^k$, $\overline{{\gamma }_P}$, etc., for brevity.
• The operation denoted by the addition symbol “+” refers to the operation of addition modulo m.
• When we refer to a subset A as a candidate for a k-FPDS, it implies that PMUs are to be positioned at the vertices of set A to monitor the mentioned graph.
• When discussing the deletion of vertices from a candidate k-FPDS, we are referring to the PMUs at those vertices being identified as faulty.

Following are a few additional definitions used in this paper for various purposes.

The Cartesian product of two graphs G and H, denoted as $G\square H$, has the vertex set $V\left(G\right)\times V\left(H\right)$, and the two vertices $(u,v)$ and $(x,y)$ in $G\square H$ are adjacent if, and only if, either $u=x$ and $vy$ is an edge in H, or $v=y$ and $ux$ is an edge in G.

The corona product of two graphs G and H, denoted as $G\circ H$, is a graph obtained by taking n copies of graph H (where n is the number of vertices in G and each vertex of the $i^{th}$ copy of H is adjacent to the $i^{th}$ vertex of G.

2. $\mathit{k}$-Fault-Tolerant Power-Domination Number for Generalized Petersen Graph

The computational power of a multicomputer system crucially relies on the functionality of all nodes. The failure of a node not only harms the computational power but changes the network topology. Network fault tolerance quantifies the robustness of the system and measures the number of failures it can tolerate before a catastrophic breakdown happens [10]. In this context, the k-fault-tolerant power-domination number (${\gamma }_p^k\left(G\right)$) is an important parameter.

The generalized Petersen graphs are cubic graphs that are formed by joining the vertices of a regular polygon to the corresponding vertices of a star polygon. The generalized Petersen graphs show some symmetry in their structure. Various graph parameters, such as chromatic and domination numbers, are obtained for this type of graph. They have lots of applications in network science and engineering. For example, the Petersen graph pattern has been applied in phonocardiogram (PCG) signals in the automated classification of normal, aortic stenosis, mitral valve prolapse, regurgitation, and stenosis [11,12].

Let m and n be positive integers. For $m\ge 3$ and $m>n\ge 1$, the generalized Petersen graph $GP(m,n)$ is a 3-regular graph with $V\left(GP(m,n)\right)=\{v_0,v_1,\cdots ,v_{m-1}\}\cup \{w_0,w_1,\cdots ,w_{m-1}\}$ and $E\left(GP(m,n)\right)=\left\{v_i{w}_i\right\}\cup \left\{v_i{v}_{i+1}\right\}\cup \left\{w_i{w}_{i+n}\right\}$, where $i=0,1,\cdots ,$$m-1$, and the subscripts are taken as modulo m. Here, $v_0,v_1,\cdots ,v_{m-1}$ are called the outer vertices and $w_0,w_1,\cdots ,w_{m-1}$ are called the inner vertices. The edges $v_i{w}_i$ are called spokes. For example, the Durer graph $GP(6,2)$ is shown in Figure 1. The generalized Petersen graph with $n=1$ is called a prism.

Lemma 1. 

For $m\ge 4$, $GP(m,1)$ is observed if, and only if, every vertex of an induced subgraph $P_2\square P_2$ is observed.

Proof. 

It is easy to see that the necessity is true. To prove the sufficiency, we suppose that every vertex of an induced subgraph $P_2\square P_2$, namely, $v_i,v_{i+1},w_i$, and $w_{i+1}$, are observed. Each of these observed vertices will now have a unique unobserved neighbour, namely, $v_{i-1},v_{i+2},w_{i-1}$, and $w_{i+2}$, respectively. Hence, these are observed by the propagation rule. Now the subgraphs induced by the two different vertex sets $\{v_{i+1},v_{i+2},w_{i+2},w_{i+1}\}$ and $\{v_i,v_{i-1},w_{i-1},w_i\}$ form two distinct $P_2\square P_2$, whose vertices are all observed. A reasoning similar to the above can show that $v_{i-2},w_{i-2},v_{i+2}$ and $w_{i+2}$ can be observed by the propagation rule. Consequently, all vertices in the entire prism $GP(m,1)$ can be observed. An example is shown in Figure 2; the darkened vertices induce a $P_2\square P_2$ and the arrows show the propagation steps.  □

Lemma 2. 

A prism $GP(m,1)$, for $m\ge 5$, is power-dominated by two vertices if, and only if, the geodesic distance between them is less than or equal to three.

Proof. 

Suppose $GP(m,1)$ is power-dominated by two vertices. When the two PMUs are placed so that one is on an inner vertex $w_i$ and the other is on an outer vertex $v_j$, where the distance between them is one or two (i.e., $d(v_i,w_j)=1$ or 2), or when both PMUs are placed on the outer vertices or both on the inner vertices, where the distance between them is one or two or three (i.e., $d(v_i,u_j)=1$ or 2 or 3), the vertices $v_i,v_{i+1},w_i,w_{i+1}$ form a $P_2\square P_2$. From Lemma 1, the graph $GP(m,1)$ becomes fully observed. When the two PMUs are placed so that one is on an inner vertex $w_i$ and the other is on an outer vertex $v_j$, where the distance between them is three, then the subgraph induced by the vertices $v_{i+2},v_{i+3},w_{i+2},w_{i+3}$ forms a $P_2\square P_2$, and, therefore, the result follows by Lemma 1. Various cases arising are illustrated in $2\left(a\right)$, $2\left(b\right)$, $2\left(c\right)$, $2\left(e\right)$, $2\left(f\right)$, and $2\left(g\right)$ of Figure 3.

Also, whenever the distance between the two PMUs placed is greater than or equal to four, the neighbours of the nodes with PMU are all left with two unobserved neighbours, meaning propagation cannot occur. Hence, in these cases, the graph is not observed. This is illustrated in $2\left(d\right)$ and $2\left(h\right)$ of Figure 3.  □

The following corollary is an easy consequence of Lemma 2:

Corollary 1. 

${\gamma }_P\left(GP(m,1)\right)=2$, for $m\ge 5$.

Theorem 1. 

The k-fault-tolerant power-domination number of the generalized Petersen graph $GP(m,1)$ is given by

1. 

${\gamma }_P^k\left(GP(3,1)\right)={\gamma }_P\left(GP(3,1)\right)+k, \mathit{when} 0\le k\le 5$

2. 

${\gamma }_P^k\left(GP(4,1)\right)=\left\{\begin{array}{cc}{\gamma }_P\left(GP(4,1)\right)+k,\hfill & \mathit{when} 0\le k\le 3\hfill \\ {\gamma }_P\left(GP(4,1)\right)+k+1,\hfill & \mathit{when} 4\le k\le 6\hfill \end{array}\right.$

3. 

${\gamma }_P^k\left(GP(5,1)\right)={\gamma }_P\left(GP(5,1)\right)+k, \mathit{when} 0\le k\le 8.$

Proof. 

The sets with cardinalities equal to the lower bound of ${\gamma }_P^k$ corresponding to the graphs $GP(m,1)$ for $m=3,4$ are presented below. Consequently, this implies that ${\gamma }_P^k\left(GP(m,1)\right)$ assumes its lowest bound in each of these cases, resulting in ${\gamma }_P^k\left(GP(m,1)\right)={\gamma }_P\left(GP(m,1)\right)+k$.

Table 1. ${\gamma }_P^k\left(GP(m,1)\right)$ for different k and m.

k	m	${\mathit{\gamma }}_{\mathit{P}}^{\mathit{k}}$-Set
1	3, 4	$\{v_0,w_0\}$
2	3, 4	$\{v_0,w_0,v_{m-1}\}$
3	3, 4	$\{v_0,w_0,v_{m-1},w_{m-1}\}$
4	3	$\{v_0,w_0,v_2,w_2,v_1\}$
5	3	$\{v_0,w_0,v_2,w_2,v_1,w_1\}$

shows the values of ${\gamma }_P^k\left(GP(m,1)\right)$ for different k and m.

One can further notice that in $GP(4,1)$, a PMU placed solely at $\{v_1,v_2,w_1,w_2\}$ cannot observe the entire graph. However, if PMUs are placed at any two of these vertices, they will collectively power-dominate the graph. Consequently, the ${\gamma }_P^4$-set of $GP(4,1)$ can only be achieved by adding two of $\{v_1,v_2,w_1,w_2\}$ to the ${\gamma }_P^3$-set of $GP(4,1)$. It is then straightforward to determine the ${\gamma }_P^5$-set and ${\gamma }_P^6$-set of $GP(4,1)$ as follows:

$$\begin{array}{c}{\gamma }_P^5\text{-}\mathrm{set}:\{v_0,w_0,v_3,w_3,v_1,w_1,v_2\}\\ {\gamma }_P^6\text{-}\mathrm{set}:\{v_0,w_0,v_3,w_3,v_1,w_1,v_2,w_2\}.\end{array}$$

In the graph $GP(5,1)$, observe that the distance between any two vertices is always less than 4. Consequently, it directly follows from Lemma 2 that any set with $2+k$ vertices in $GP(5,1)$ is a k-FPDS of $GP(5,1)$.  □

It is easy to see that the value of ${\gamma }_P^k\left(G(m,1)\right)={\gamma }_P\left(G(m,1)\right)+k$ for $m,k\le 4$. Before obtaining the value of ${\gamma }_P^k\left(G(m,1)\right)$ for $m,k\ge 6$, we look into the definition of failed power-domination and a related result, as this will be useful in the proof that follows.

Definition 4. 

Let $G=(V,E)$ be a graph with vertex set V and edge set E. A subset $S\subseteq V$ is said to be a Failed Power-Dominating Set (FPDS) of the graph if S is not a power-dominating set of G.

The failed power-domination number of a graph G, denoted by $\overline{{\gamma }_P}\left(G\right)$, is defined as the maximum cardinality of any FPDS of G. This parameter signifies the maximum number of PMUs that can be placed on a power network represented by G, while still failing to fully observe the network/graph. The parameter $\overline{{\gamma }_P}\left(G\right)+1$ represents the minimum number of PMUs required to ensure complete observation of G, regardless of their placement.

Theorem 2. 

For $m\ge 6$,

$$\overline{{\gamma }_P}\left(GP(m,1)\right)=\left\{\begin{array}{cc}2\lfloor \frac{m}{6}\rfloor ,\hfill & m\equiv 0,1,2,3\left(mod6\right)\hfill \\ 2\lfloor \frac{m}{6}\rfloor +1,\hfill & m\equiv 4,5\left(mod6\right).\hfill \end{array}\right.$$

Proof. 

According to Lemma 2, two PMUs within a distance of less than four can always monitor $GP(m,1)$. Therefore, our objective is to construct the largest subset of vertices of $GP(m,1)$ such that the pairwise distance between any two vertices in the set is greater than or equal to four. To achieve this, we begin by selecting an arbitrary vertex of the graph (without loss of generality, we start from an outer vertex) and then proceed to select vertices that are exactly distance four apart from each other. The distance is taken as the minimum, specifically four, to maximize the number of vertices, thus obtaining the largest required set. This process results in the following sets: For any $s\in \mathbb{Z}$ $A=\{v_{s+0},w_{s+3},v_{s+6},w_{s+9},\cdots ,w_{s+6\lfloor \frac{m}{6}\rfloor +3}\},\mathrm{while}m\equiv 0,1,2,3\left(\mathrm{mod}6\right)$ and

$B=\{v_{s+0},w_{s+3},v_{s+6},w_{s+9},\cdots v_{s+6\lfloor \frac{m}{6}\rfloor }\},\mathrm{while}m\equiv 4,5\left(\mathrm{mod}6\right).$

Consequently, no subsets of these sets can power-dominate $GP(m,1)$. Additionally, adding any other vertex from $GP(m,1)$ to these sets will inevitably create a pair of vertices that are less than distance four apart, resulting in the power-domination of the entire graph. This demonstrates that sets A and B constitute the largest failed power-domination set for the graph $GP(m,1)$. As the cardinality of these sets is $\left|A\right|=2\lfloor \frac{m}{6}\rfloor ,\&m\equiv 0,1,2,3\left(mod6\right)$, and $\left|B\right|=2\lfloor \frac{m}{6}\rfloor +1,\&m\equiv 4,5\left(mod6\right)$, the theorem is established.  □

Without loss of generality and for simplicity, throughout this paper, we regard the following sets as the largest failed power-domination sets of $GP(m,1)$, wherever necessary, $A=\{v_0,w_3,v_6,w_9,\cdots ,w_{6\lfloor \frac{m}{6}\rfloor +3}\},\mathrm{while}m\equiv 0,1,2,3\left(\mathrm{mod}6\right)$ and $B=\{v_0,w_3,v_6,w_9,\cdots v_{6\lfloor \frac{m}{6}\rfloor }\},\mathrm{while}m\equiv 4,5\left(\mathrm{mod}6\right).$

Theorem 3. 

For $m\ge 6$,

$${\gamma }_P^k\left(GP(m,1)\right)=\left\{\begin{array}{cc}k+2+\lfloor \frac{k}{5}\rfloor ,\hfill & k\le 5\overline{{\gamma }_p}\left(GP(m,1)\right)-1\hfill \\ k+1+\lfloor \frac{k}{5}\rfloor ,\hfill & 5\overline{{\gamma }_p}\left(GP(m,1)\right)\le k\le 2m-1-\overline{{\gamma }_p}\left(GP(m,1)\right).\hfill \end{array}\right.$$

Proof. 

We begin the proof by establishing an upper bound for k. The maximum value of k will be attained when $2m$ vertices generate a k-FPDS for $GP(m,1)$. Thus, we seek the maximum number of vertices that can be removed from the $2m$ vertices in any order, while ensuring the remaining vertices still power-dominate $GP(m,1)$. As per the previous Theorem 2, any subset with $\overline{{\gamma }_P}\left(GP(m,1)\right)+1$ or more vertices of $GP(m,1)$ will power-dominate $GP(m,1)$. Hence, removing any collection of $2m-\overline{{\gamma }_P}-1$ vertices from $2m$ vertices will still result in a power-dominating set of $GP(m,1)$. This means that the vertex set V of $GP(m,1)$ is a $2m-\overline{{\gamma }_P}-1$-FPDS of $GP(m,1)$. Hence, $k\le 2m-\overline{{\gamma }_P}-1$.

Lemma 2 establishes that two PMUs within a distance of less than four units can always monitor $GP(m,1)$. Consequently, to create a k-FPDS with the smallest possible cardinality, we need to construct a subset of vertices where the pairwise distance between at least one pair of vertices is less than four, even after the removal of k vertices. This implies that in order to achieve a k-FPDS with minimum cardinality, the placement of PMUs should be such that the pairwise distance between any two PMUs is minimized. Therefore, without loss of generality, for the rest of the proof, in each attempt to construct a k-FPDS with t vertices, the PMUs are placed at the vertices $v_0,w_0,v_1,w_1,\cdots v_{\frac{t-1}{2}}$ when t is odd, and when t is even, we place the PMUs in the vertices of $v_0,w_0,v_1,w_1,\cdots v_{\frac{t}{2}-1},w_{\frac{t}{2}-1}.$ This will ensure that the pairwise distance between any two PMUs is minimal. We call this the optimized placement technique.

Because the pairwise distance between any two vertices in set A is less than four, any pair of vertices from set $\{v_0,w_0,v_1,w_1,v_2,w_2\}$ can power-dominate $GP(m,1)$. This guarantees that any subset of $\{v_0,w_0,v_1,w_1,v_2,w_2\}$ with cardinality 3, 4, 5, and 6 will form the ${{\gamma }_p}^1$, ${{\gamma }_p}^2$, ${{\gamma }_p}^3$, and ${{\gamma }_p}^4$ sets, respectively. Therefore, ${{\gamma }_p}^k={\gamma }_p+k$, for $m\ge 6$ and $0\le k\le 4$. In other words, we have

$${{\gamma }_p}^k={\gamma }_p+k;5\times 0\le k\le 5\times 0+4.$$

(1)

Consider the case when $k=5$. Since we know that the lower bound is ${\gamma }_p+k=2+5=7$, we place seven PMUs as mentioned above (i.e., $\{v_0,w_0,v_1,w_1,v_2,w_2,v_3\}$ ). Out of the ${}^{7}C_5=21$ different ways in which five PMUs can fail and leave behind a pair of working PMUs, there is the possibility that working PMUs are on $w_0$ and $v_3$, which are distance four apart. When this occurs, $GP(m,1)$ cannot be completely monitored. However, placing just one additional vertex (on $w_3$) ensures that the removal of five vertices will leave behind three vertices, of which at least a pair of them is less than distance four apart. Therefore, ${\gamma }_P^5\ge 8$. Further, we have indeed generated the set $\{v_0,w_0,v_1,w_1,v_2,w_2,v_3,w_3\}$ with exactly eight vertices with the required property, implying ${\gamma }_P^5=8$.

Similarly, if we observe, as k runs from 5 to 9, the lower bound of ${\gamma }_p^k$ will vary between 7 to 11, respectively. But in each of these cases, out of the ${}^{{\gamma }_p+k}C_k$ different ways in which k PMUs can fail and leave behind ${\gamma }_p=2$ working PMUs, there are possibilities that working PMUs are distance four or more apart (e.g., on $w_0$ and $v_3$). When this occurs, $GP(m,1)$ cannot be completely monitored. However, placing just one additional vertex ensures that the removal of k vertices will leave behind three vertices, of which at least a pair of them is less than the distance four apart. Therefore, ${\gamma }_P^k\ge {\gamma }_p+k+1=2+k+1$. Further, we can indeed generate sets with exactly $2+k+1$ vertices with the required property, implying ${\gamma }_P^k=2+k+1$, while $5\le k\le 9$. In other words, we have

$${{\gamma }_p}^k={\gamma }_p+k+1;5\times 1\le k\le 5\times 1+4.$$

(2)

Likewise, for k ranging from 10 to 14, the lower bound of ${\gamma }_p^k$ will vary from 13 to 17, respectively. However, in each of these cases, among the $\left(\genfrac{}{}{0pt}{}{{\gamma }_p+k+1}{k}\right)$ different ways in which k PMUs can fail and leave behind ${\gamma }_p+1=3$ working PMUs, there are instances where the working PMUs are distance four or more apart (e.g., on $w_0$, $v_3$, $w_6$). In such cases, $GP(m,1)$ cannot be entirely monitored. Yet, adding just one additional vertex ensures that the removal of k vertices will still leave behind four vertices, with at least one pair less than distance four apart. Thus, ${\gamma }_P^k\ge {\gamma }_p+k+2=2+k+2$. Moreover, sets with exactly $2+k+1$ vertices can indeed be constructed as described above.

An identical pattern emerges for all values of k from 0 to $5\times (\overline{{\gamma }_P}-1)$. Hence, through iterative repetition of the same reasoning, we have

$${{\gamma }_p}^k={\gamma }_p+k+i;5\times i\le k\le 5\times i+4;where0\le i\le \overline{{\gamma }_p}-1.$$

(3)

Alternatively, we can write,

$${{\gamma }_p}^k=2+k+\lfloor \frac{k}{5}\rfloor ;5\times i\le k\le 5\times i+4;where0\le i\le \overline{{\gamma }_p}-1.$$

(4)

This is the same as

$${{\gamma }_p}^k=2+k+\lfloor \frac{k}{5}\rfloor ;where0\le k\le 5\times \overline{{\gamma }_p}-1.$$

(5)

Since ${\gamma }_P^k<{\gamma }_P^{k+1}$, we have the lower bound for ${\gamma }_p^k$ is ${\gamma }_p+k+\overline{{\gamma }_p}-1$, for $5\times \overline{{\gamma }_p}\le k\le 2m-\overline{{\gamma }_P}-1$. Removing k vertices from this set of ${\gamma }_p+k+\overline{{\gamma }_p}-1$ vertices will leave behind ${\gamma }_p+\overline{{\gamma }_p}-1=2+\overline{{\gamma }_p}-1=\overline{{\gamma }_p}+1$. By Theorem 2, any subset with $\overline{{\gamma }_p}+1$ vertices is a power-dominating set of $GP(m,1)$. Hence, this lower bound is the required value. So, we have

$${{\gamma }_p}^k=1+k+\lfloor \frac{k}{5}\rfloor ;where0\le k\le 5\times \overline{{\gamma }_p}-1.$$

(6)

By combining Equations (5) and (6), we obtain the result stated in the theorem.  □

Lemma 3. 

$GP(m,2)$ is observed whenever the graph induced by the vertices of the set $\{w_i,w_{i+1},w_{i+2},w_{i+3},u_i,u_{i+1},u_{i+2},u_{i+3}\}$ is observed.

Proof. 

Observe that once the vertices $\{w_i,w_{i+1},w_{i+2},w_{i+3},v_i,v_{i+1},v_{i+2},v_{i+3}\}$ are observed, then each of $w_i,w_{i+1},w_{i+2}$, $w_{i+3},v_i$ and $v_{i+3}$ have a unique unobserved neighbour, namely, $w_{i-2},w_{i-1},w_{i+4},w_{i+5},v_{i-1}$ and $v_{i+4}$, respectively. These can be observed by propagation step 1. Further, by similar reasoning, propagation steps can be followed to observe the entire graph. An illustration is shown in Figure 4. The darkened vertices are $\{w_i,w_{i+1},w_{i+2},w_{i+3},v_i,v_{i+1},v_{i+2},v_{i+3}\}$ and the arrows show the propagation steps.  □

Lemma 4. 

$GP(m,2)$ is power-dominated by two vertices whenever the geodesic distance between them is exactly three.

Proof. 

There are three possibilities as follows: First case, both PMUs are at the outer vertices. Secondly, one PMU is on the outer vertex and the other on the inner vertex. Third, both PMUs are at the inner vertices. When the distance is three, using Lemma 3, we can guarantee that the whole graph is observed. An illustration is shown in Figure 5.  □

Corollary 2. 

${\gamma }_p\left(GP(m,2)\right)=2.$

Lemma 5. 

If $S_0=\{w_3,w_4\}$ and $S_k=\left\{\begin{array}{cc}{S}_{k-1}\cup \left\{v_{k-1}\right\},\hfill & kisodd\hfill \\ S_{k-1}\cup \{w_{k+3},w_{k+4}\},\hfill & kiseven\hfill \end{array}\right.$, for all $k\ge 1$, then $|S_k|=2+k+\lfloor \frac{k}{2}\rfloor$.

Proof. 

Claim: $|S_k|=2+k+\lfloor \frac{k}{2}\rfloor$. To prove our claim, we use mathematical induction.

Clearly, $|S_1|=3=2+k+\lfloor \frac{k}{2}\rfloor$. Hence, the claim is obviously true for $k=1$.

Let us assume that, whenever $t<k$,

$$|S_t|=2+t+\lfloor \frac{t}{2}\rfloor .$$

(7)

Using the above induction hypothesis and definition of $S_k$, we have, $|S_k|=\left\{\begin{array}{cc}|S_{k-1}|+1,\hfill & kisodd\hfill \\ |S_{k-1}|+2,\hfill & kiseven.\hfill \end{array}\right.$

Substituting (7) in the above equation, we have, $|S_k|=\left\{\begin{array}{cc}2+(k-1)+\lfloor \frac{k-1}{2}\rfloor +1,\hfill & kisodd\hfill \\ 2+(k-1)+\lfloor \frac{k-1}{2}\rfloor +2,\hfill & kiseven\hfill \end{array}\right.$

$=\left\{\begin{array}{cc}2+k+\lfloor \frac{k-1}{2}\rfloor ,\hfill & kisodd\hfill \\ 3+k+\lfloor \frac{k-1}{2}\rfloor ,\hfill & kiseven\hfill \end{array}\right.$

$=\left\{\begin{array}{cc}2+k+\frac{k-1}{2},\hfill & kisodd\hfill \\ 3+k+\frac{k-2}{2},\hfill & kiseven\hfill \end{array}\right.$

$=\left\{\begin{array}{cc}2+k+\frac{k-1}{2},\hfill & kisodd\hfill \\ 2+k+(1+\frac{k-2}{2}),\hfill & kiseven\hfill \end{array}\right.$

$=2+k+\lfloor \frac{k}{2}\rfloor .$ □

Theorem 4. 

For $m\ge 3$, ${\gamma }_P\left(GP(m,2)\right)+k\le {\gamma }_P^k\left(GP(m,2)\right)\le {\gamma }_P\left(GP(m,2)\right)+k+\lfloor \frac{k}{2}\rfloor$.

Proof. 

Let $S_0=\{w_3,w_4\}$ and for all $k\ge 1$, $S_k=\left\{\begin{array}{cc}{S}_{k-1}\cup \left\{v_{k-1}\right\},\hfill & kisodd\hfill \\ S_{k-1}\cup \{w_{k+3},w_{k+4}\},\hfill & kiseven.\hfill \end{array}\right.$

From Lemma 5, $|S_k|=2+k+\lfloor \frac{k}{2}\rfloor$.

Claim: $S_k$ is a k-fault-tolerant power-dominating set of $GP(m,2)$.

To prove our claim, we use mathematical induction to show that any combination of $|S_k|-k=\lfloor \frac{k}{2}\rfloor +2$ vertices from $S_k$ will always power-dominate $GP(m,2)$, whenever $m\ge 7$.

Let $C_k$ be the collection of all possible subsets of $S_k$ whose cardinality is $|S_k|-k=\lfloor \frac{k}{2}\rfloor +2$. Now, it is enough to show that every element of $C_k$ power-dominates $GP(m,2)$.

For $k=1$, we have $S_1=\{w_3,w_4,v_0\}$. Because these vertices are, pairwise, distance three from each other, removing any one among them still results in two vertices that will power-dominate $GP(m,2)$ by Lemma 4.

Induction Hypothesis: Assume that $S_t$ is a t-fault-tolerant power-dominating set of $GP(m,2)$, whenever $t<k$. In other words, assume that any combination of $\lfloor \frac{t}{2}\rfloor +2$ vertices from $t+\lfloor \frac{t}{2}\rfloor +2$ vertices of $S_t$ will power-dominate $GP(m,2)$, whenever $t<k$.

Case 1: k is even, i.e., $k=2k^{\prime }$, for some positive integer $k^{\prime }$.

Here, $S_k=S_{2k^{\prime }}=S_{2k^{\prime }-1}\cup \{w_{2k^{\prime }+3},w_{2k^{\prime }+4}\}$. We observe that $C_k$ can be completely partitioned into 4 mutually disjoint subsets $C_1^{\prime },C_2^{\prime }$, $C_3^{\prime }$, and $C_4^{\prime }$, where,

• $C_1^{\prime }$ contains subsets of $S_k$ with $(k^{\prime }+2)$ elements from $S_{2k^{\prime }-1}$.
• $C_2^{\prime }$ contains subsets of $S_k$ with the element $w_{2k^{\prime }+3}$, along with $(k^{\prime }+1)$ elements from $S_{2k^{\prime }-1}$.
• $C_3^{\prime }$ contains subsets of $S_k$ with the element $w_{2k^{\prime }+4}$, along with $(k^{\prime }+1)$ elements from $S_{2k^{\prime }-1}$.
• $C_4^{\prime }$ contains subsets of $S_k$ with the elements $w_{2k^{\prime }+3}$ and $w_{2k^{\prime }+4}$, along with $\left(k^{\prime }\right)$ elements from $S_{2k^{\prime }-1}$.

Observe that every element of $C_k^{\prime }$, $k=1,2,3$ contains $k^{\prime }+1$ elements from $S_{2k^{\prime }-1}$. Also, from the induction hypothesis, any combination of $\lfloor \frac{2k^{\prime }-1}{2}\rfloor +2=k^{\prime }+1$ vertices from $2k^{\prime }-1+\lfloor \frac{2k^{\prime }-1}{2}\rfloor +2=3k^{\prime }$ vertices of $S_{2k^{\prime }-1}$ will power-dominate $GP(m,2)$.

Every element of $C_4^{\prime }$ has $w_{2k^{\prime }+3}$ and $w_{2k^{\prime }+4}$ in it and $d(w_{2k^{\prime }+3},w_{2k^{\prime }+4})=3$. Hence, every element of $C_4^{\prime }$ power-dominates $GP(m,2)$.

And since $C_k=C_1^{\prime }\cup C_2^{\prime }\cup C_3^{\prime }\cup C_4^{\prime }$, we conclude that every element of $C_k$ power-dominates $GP(m,2)$ here.

Case 2: k is odd, i.e., $k=2k^{\prime }+1$, for some positive integer $k^{\prime }$.

Here, $S_k=S_{2k^{\prime }+1}=S_{2k^{\prime }}\cup \left\{v_{2k^{\prime }}\right\}=S_{2k^{\prime }-1}\cup \{w_{2k^{\prime }+3},w_{2k^{\prime }+4}\cup \left\{v_{2k^{\prime }}\right\}$. We observe that

• $C_k$ can be completely partitioned into five mutually disjoint subsets $C_1^{\prime },C_2^{\prime }$, $C_3^{\prime }$, $C_4^{\prime }$, and $C_5^{\prime }$ where,
• $C_1^{\prime }$ contains subsets of $S_k$ with the element $v_{2k^{\prime }}$, along with $(k^{\prime }+1)$ elements from $S_{2k^{\prime }-1}$.
  $C_2^{\prime }$ contains subsets of $S_k$ with the element $w_{2k^{\prime }+3}$ and $v_{2k^{\prime }}$, along with $\left(k^{\prime }\right)$ elements from $S_{2k^{\prime }-1}$.
• $C_3^{\prime }$ contains subsets of $S_k$ with the element $w_{2k^{\prime }+4}$ and $v_{2k^{\prime }}$, along with $\left(k^{\prime }\right)$ elements from $S_{2k^{\prime }-1}$.
• $C_4^{\prime }$ contains subsets of $S_k$ with the elements $v_{2k^{\prime }}$, $w_{2k^{\prime }+3}$ and $w_{2k^{\prime }+4}$, along with $(k^{\prime }-1)$ elements from $S_{2k^{\prime }-1}$.
• $C_5^{\prime }$ contains subsets of $S_k$ with $(k^{\prime }+2)$ number of elements from $S_{2k^{\prime }}$.

From the induction hypothesis, any combination of $\lfloor \frac{2k^{\prime }-1}{2}\rfloor +2=k^{\prime }+1$ vertices from $2k^{\prime }-1+\lfloor \frac{2k^{\prime }-1}{2}\rfloor +2=3k^{\prime }$ vertices of $S_{2k^{\prime }-1}$ will power-dominate $GP(m,2)$. And every element of $C_1^{\prime }$ has $(k^{\prime }+1)$ number of elements from $S_{2k^{\prime }-1}$. Therefore, every element of $C_1^{\prime }$ power-dominates $GP(m,2)$.

Every element of $C_2^{\prime }$ has $w_{2k^{\prime }+3}$ and $u_{2k^{\prime }}$ in it and $d(w_{2k^{\prime }+3},u_{2k^{\prime }})=3$. Hence, every element of $C_2^{\prime }.$ power-dominates $GP(m,2)$. Every element of $C_3^{\prime }$ has $w_{2k^{\prime }+4}$ and $u_{2k^{\prime }}$ in it and $d(w_{2k^{\prime }+4},u_{2k^{\prime }})=3$. Hence, every element of $C_3^{\prime }.$ power-dominates $GP(m,2)$. Every element of $C_4^{\prime }$ has $w_{2k^{\prime }+3}$ and $w_{2k^{\prime }+4}$ in it and $d(w_{2k^{\prime }+3},w_{2k^{\prime }+4})=3$. Hence, every element of $C_4^{\prime }.$ power-dominates $GP(m,2)$.

From the induction hypothesis, any combination of $k^{\prime }+2$ vertices from $3k^{\prime }+2$ vertices of $S_{2k^{\prime }}$ will power-dominate $GP(m,2)$. And every element of $C_5^{\prime }$ has $(k^{\prime }+2)$ number of elements from $S_{2k^{\prime }}$. Therefore, every element of $C_5^{\prime }$ power-dominates $GP(m,2)$.

And since $C_k=C_1^{\prime }\cup C_2^{\prime }\cup C_3^{\prime }\cup C_4^{\prime }\cup C_5^{\prime }$, we conclude that every element of $C_k$ power-dominates $GP(m,2)$ here.

From cases $\left(1\right)$ and $\left(2\right)$, $S_k$ is a k-fault-tolerant power-dominating set of $GP(m,2)$. Therefore, ${\gamma }_P^k\left(GP(m,2)\right)\le \left|S_k\right|=2+k+\lfloor \frac{k}{2}\rfloor ={\gamma }_P+k+\lfloor \frac{k}{2}\rfloor$.  □

3. $\mathit{k}$-Fault-Tolerant Power-Domination Number for Cylinders

In this section, we obtain the k-fault-tolerant power-domination number for some subclasses of cylinders, namely, spiked cycle and spiked band [13,14]. The spiked cycle is a graph G with $V\left(G\right)=\{v_0,v_1,\cdots ,v_{m-1}\}\cup \{u_0,u_1,\cdots ,u_{m-1}\}$ and $E\left(G\right)=\left\{v_i{v}_{i+1}\right\}\cup \left\{v_i{u}_i\right\}$, where $i=0,1,\cdots m-1$, and the subscript are addition modulo m. It is denoted as $S{C}_m$. In other words, the spiked cycle is isomorphic to the corona product of a cycle $C_m$ and a vertex. The spiked band is a graph, denoted by $S{B}_m$, with $V\left(S{B}_m\right)=\{a_0,a_1,\cdots ,a_{m-1}\}\cup \{b_0,b_1,\cdots ,b_{m-1}\}\cup \{c_0,c_1,\cdots ,c_{m-1}\}$ $\cup \{d_0,d_1,\cdots ,d_{m-1}\}$ and $E\left(S{B}_m\right)=\left\{a_i{b}_i\right\}\cup \left\{b_i{c}_i\right\}\cup \left\{c_i{d}_i\right\}\cup \left\{b_i{b}_{i+1}\right\}\cup \left\{c_i{c}_{i+1}\right\},$ where $i=0,1,2,\cdots ,m-1$. In other words, the spiked band is isomorphic to the corona product of a ladder and a vertex.

Theorem 5. 

${\gamma }_P^1\left(S{C}_m\right)=m-1$ and ${\gamma }_P^2\left(S{C}_m\right)=m$.

Proof. 

It is easy to see that any subset, S, of $\{v_0,v_1,\cdots v_{m-1}\}$ with cardinality $m-1$ is a 1-fault-tolerant power-dominating set of $S{C}_m$. We now show that any subset of vertices of $S{C}_m$ with cardinality less than or equal to $m-2$ cannot be a 1-fault-tolerant power-dominating set of $S{C}_m$. Any arrangement, denoted as $S^{\prime }$, of $m-2$ PMUs on the vertices of $S{C}_m$ will necessarily contain at least one vertex whose removal leads to three consecutive vertices within the set $\{v_0,v_1,\dots ,v_{m-1}\}$ without any operational PMUs among them. Let these three vertices be denoted as $\{v_{i-1},v_i,v_{i+1}\}$. It is easy to see that if at least one edge incident to $v_i$ is observed, then $v_i$ is observed. Also, if the edges $v_i{v}_{i+1}$ and $v_i{v}_{i-1}$ are observed, then the edge $v_i{u}_i$ can be observed. But the edges $v_i{v}_{i+1}$ and $v_i{v}_{i-1}$ are observed if, and only if, $v_{i+1}{u}_{i+1}$ and $v_{i-1}{u}_{i-1}$ are observed. This cannot happen without working PMUs on $v_{i-1}$ and/or $v_{i+1}$. Hence, $S^{\prime }$ fails to power-dominate $S{C}_m$.

Further, placing a PMU on each vertex of $\{v_0,v_1,\dots ,v_{m-1}\}$ forms a 2-fault-tolerant power-dominating set of $S{C}_m$ because this set already consists of m vertices. Even if two PMUs fail, a subset of $m-2$ vertices with operational PMUs remains, constituting a PDS. Also, as previously discussed, in any arrangement of $m-1$ PMUs on the vertices of $\{v_0,v_1,\dots ,v_{m-1}\}$, there exists at least one pair of PMUs such that their removal or failure will lead to three consecutive vertices within the set $\{v_0,v_1,\dots ,v_{m-1}\}$ without any operational PMUs among them. As previously discussed, this set is, hence, unable to fully power-dominate $S{C}_m$. Therefore, any subset of vertices of $S{C}_m$ with a cardinality less than or equal to $m-1$ cannot serve as a 2-fault-tolerant power-dominating set of $S{C}_m$, implying ${\gamma }_P^2\left(S{C}_m\right)=m$.  □

Theorem 6. 

${\gamma }_P^1\left(S{B}_m\right)=\left\{\begin{array}{cc}m+1,\hfill & m\le 3\hfill \\ m+2,\hfill & m\ge 4.\hfill \end{array}\right.$

Proof. 

It is easy to see that the sets $\{b_0,c_0\}$, $\{b_0,c_0,b_1\}$, and $\{b_0,c_0,b_1,c_1\}$ are the 1-fault-tolerant power-dominating sets of $S{B}_1$, $S{B}_2$, and $S{B}_3$, respectively. Furthermore, as the cardinalities of these sets match the lower bound of the ${\gamma }_P^1$ values for $S{B}_m$, for $m\le 3$, we can conclude that $S{B}_m=m+1$, for $m\le 3$.

For $m\ge 4$, consider a set S with a cardinality of $m+2$ such that $b_0,c_0,b_{m-1},c_{m-1}\in S$, and for each $i=1,2,\dots ,m-2$, either $b_i$ or $c_i$ (or both) is an element of S. One can apply the rules of domination and propagation to see that this indeed is a 1-fault-tolerant power-dominating set of $S{B}_m$, for $m\ge 4$.

We now claim that S is the smallest subset that will act as a 1-fault-tolerant power-dominating set for $S{B}_m$.

Let $S^{\prime }$ be any power-dominating set of $S{B}_m$ with cardinality $m+1$. There exists at least one vertex $u\in S^{\prime }$ whose deletion from $S^{\prime }$ will result in one of the following conditions:

• There is no PMU on the 0^th row; however, there is one in either $b_1$ or $c_1$. But this will leave either $a_0$ or $d_0$ unobserved.
• There is no PMU on the ${(m-1)}^{th}$ row, but there is one in either $b_{m-2}$ or $c_{m-2}$. But this will leave either $a_{m-2}$ or $d_{m-2}$ unobserved.
• There are two successive rows i and $i+1$ without any PMUs, and two rows, ${(i-1)}^{th}$ and ${(i+2)}^{th}$, containing a single PMU each. Here, $a_i,a_{i+1},d_i,d_{i+1}$ will be left unobserved.

Hence, any set smaller than S cannot be a 1-FPDS of $S{B}_m$, for $m\ge 4$.  □