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Generalized power domination of graphs
Gerard Jennhwa Chang, Paul Dorbec, Mickael Montassier, and Andr´ e Raspaud
July 10, 2011
Generalized power domination of graphs
Gerard Jennhwa Changa,b,1, Paul Dorbecc,2, Mickael Montassierc,2, Andr´e Raspaudc,2
aDepartment of Mathematics and Taida Institute for Mathematical Sciences, National Taiwan University, Taipei 10617, Taiwan
bNational Center for Theoretical Sciences, Taipei Office, Taiwan
cUniversit´e de Bordeaux – CNRS, LaBRI, 351 cours de la Lib´eration, 33405 TALENCE CEDEX, FRANCE
Abstract
In this paper, we introduce the concept of k-power domination which is a common generalization of domination and power domination. We extend several known results for power domination to k-power domination. Con- cerning the complexity of the k-power domination problem, we first show that deciding whether a graph admits a k-power dominating set of size at most t is NP-complete for chordal graphs and for bipartite graphs. We then give a linear algorithm for the problem on trees. Finally, we propose sharp upper bounds for the power domination number of connected graphs and of connected claw-free (k + 2)-regular graphs.
Keywords: Power domination, electrical network monitoring, domination
1. Introduction
In this paper we only consider simple graphs, that are graphs without multiple edges or loops. Let G be a graph with vertex set V (G) and edge set E(G). We use the following definitions.
Email addresses: gjchang@math.ntu.edu.tw (Gerard Jennhwa Chang), paul.dorbec@u-bordeaux1.fr (Paul Dorbec), montassi@labri.fr (Mickael Montassier), andre.raspaud@labri.fr (Andr´e Raspaud)
1Supported in part by the National Science Council under grants NSC98-2115-M-002- 013-MY3 and NSC99-2923-M-110-001-MY3.
2Supported in part by Agence Nationale de la Recherche under grant ANR-09-blan- 0373-01.
The open neighbourhood of a vertex v, denoted NG(v), is the set of vertices adjacent to v, namely NG(v) ={u ∈ V (G) : uv ∈ E(G)}. The closed neigh- bourhood of v is NG[v] = NG(v)∪ {v}. The open (resp. closed) neighbourhood of a set S ⊆ V is the union of the open (resp. closed) neighbourhoods of its elements: NG(S) = ∪
v∈SNG(v) and NG[S] = ∪
v∈SNG[v] = NG(S)∪ S.
When G is clear from context, we use N instead of NG. The degree of a vertex v, denoted d(v), is the size of its open neighbourhood |N(v)|. The maximum degree of the graph G is denoted by ∆(G).
A dominating set of a graph G is a set of vertices S such that N [S] = V (G). The domination number of a graph G, denoted γ(G), is the minimum cardinality of a dominating set of G. The dominating set problem, that is deciding whether a graph admits a dominating set of size at most t, is NP- complete [4, 7] (even restricted to bipartite or chordal graphs).
Power domination was introduced in [3, 13] to model a problem of moni- toring electrical networks. It was then described as a graph theoretical prob- lem in [10]. The problem is similar to a problem of domination, in which, additionally, the possibility of some propagation according to Kirschoff laws is considered. The definition of power domination, originally asking to monitor both edges and vertices, was simplified to the following, introduced indepen- dently in [5, 6].
Let G be a graph and S a subset of its vertices. The set monitored by S, denoted by M (S), is defined algorithmically as follows:
• (domination) M(S) ← S ∪ N(S),
• (propagation) as long as there exists v ∈ M(S) such that N(v)∩(V (G)−
M (S)) ={w}, set M(S) ← M(S) ∪ {w}.
In other words, the set M (S) is obtained from S as follows. First put into M (S) the vertices from the closed neighbourhood of S. Then repeatedly add to M (S) vertices w that have a neighbour v in M (S) such that all the other neighbours of v are already in M (S). After no such vertex w exists, the set monitored by S has been constructed. The set S is called a power dominating set of G if M (S) = V (G) and the power domination number γP(G) is the minimum cardinality of a power dominating set.
This definition implies some propagating behaviour of the set of moni- tored vertices, a phenomenon very different from the standard domination parameter. Different works were started on the topic. From the algorithmic point of view, the power domination problem was known to be NP-complete
[1, 2, 8, 9, 10] and approximation algorithms were given in [2]. On the other hand, linear-time algorithms for the power domination problem were given for trees [10], for interval graphs [12] and for block graphs [15]. Parameter- ized results were given in [11]. The exact values for the power domination numbers are determined for various products of graphs [5, 6]. Bounds for the power domination numbers of connected graphs and of claw-free cubic graphs are given in [17] and for planar or outerplanar graphs with bounded diameter in [16].
Let k be a nonnegative integer. Let us introduce k-power domination, using a definition of monitored set close to what Aazami proposed in [1].
Definition 1 (Monitored set). Let G be a graph and S ⊆ V (G). We define the sets(
PGi(S))
i≥0 of vertices monitored by S at step i by the following rules.
• PG0(S) = N [S].
• PGi+1(S) = ∪{N[v]: v ∈ PGi(S) such that N [v]\ PGi(S) ≤k}.
Remark that PGi (S)⊆ PGi+1(S) ⊆ V (G) for any i. Moreover, every time a vertex of the set PGi (S) has at most k neighbours outside the set, we add its neighbours to the next generation PGi+1(S).
If PGi0(S) = PGi0+1(S) for some i0, then PGj(S) = PGi0(S) for any j ≥ i0. We thus define PG∞(S) =PGi0(S).
Definition 2 (k-power dominating set). A set S such that PG∞(S) = V (G) is a k-power dominating set of G. The least cardinality of such a set is called the k-power domination number of G, written γP,k(G). A γP,k(G)-set is a minimum k-power dominating set of G.
When G is clear from context, we will use Pi(S) to denote PGi(S). From this definition, the following observation clearly holds.
Observation 3. Let G be a graph and S, S′ ⊆ V (G). If Pi(S)⊆ Pj(S′) for some integers i and j, then Pi+1(S)⊆ Pj+1(S′) and so P∞(S)⊆ P∞(S′) by extension. In particular, if S is a k-power dominating set of G, then so is S′.
We introduce k-power domination as a natural generalization of power domination, with correspondence when k = 1. It also generalizes usual dom- ination, for k = 0. Whereas power domination seemed a problem of very different type than domination, it is remarkable how this generalization uni- fies the two problems. First, the algorithm we propose here for computing a k-power dominating set of a tree is closely related to the algorithm proposed for domination in [14], whereas the algorithm proposed for power domina- tion in [10] was identifying spiders in the tree, a very different technique.
Moreover, the bound in Theorem 11 is a very natural generalization of the original classical bound for domination in graphs with no isolated vertices, and the examples of graphs reaching the bound are similar to the coronas, which are graph reaching the bound for domination. Similarly, Lemma 10 is a non trivial generalization of the fact that a dominating set in a graph is minimal only if each of its vertices has a private neighour.
These similarities make generalized power domination shed a new light on power domination. It seems that one might use this new approach to extend some of the numerous results known for domination to the applied problem of power domination.
The paper is organized as follows. Section 2 is dedicated to complexity results; we show that the k-power domination number is an NP-complete problem even in the case of chordals graphs or bipartite graphs and we give a linear time algorithm computing γP,k(G) when G is a tree.
In Section 3, we prove that, for any connected graph G of order n, γP,k(G) ≤ k+2n , and we show the sharpness of the bound. Also we show that, for any claw-free (k + 2)-regular graph of order n, γP,k(G)≤ k+3n , and we characterize the graphs achieving this last bound.
2. Complexity results
For any graph G and any nonnegative integer k, let Gk be the graph obtained from G by adding k new neighbours to each vertex of G. By the construction of Gk, we get γP,k(Gk) = γ(G). Notice that if G is chordal (or bipartite), then so is Gk. The NP-completeness of the domination problem [7] gives the following result.
Theorem 4. The k-power domination problem, that is deciding whether a graph admits a k-power dominating set of size at most t, is NP-complete for chordal graphs and for bipartite graphs.
A linear algorithm for trees
We now give a linear-time algorithm for the k-power domination problem on trees. We in fact solve a slightly more general problem.
The algorithm we propose is largely inspired by the algorithm proposed in [14] for domination. The main idea is to put labels on the vertices, then it- eratively deleting the leafs, recording all useful information about the deleted subtree within the labels on the parent vertex.
We use two labels for each vertex v, denoted av and bv. The first label av conveys the same partition idea as in [14]. The vertex set of the graph is partitioned into three sets R, B and F corresponding to labels R, B and F.
We assign a vertex the label R (for required) when the optimal choice is to put the corresponding vertex in the k-power dominating set, B (for bound) when the vertex is not yet monitored in any way, and F (for free) when the vertex is already monitored.
The second label bv is an integer, useful only for states F and B. When a vertex gets monitored, it may be used for propagation, but to at most k neighbours. This second label bv denotes how many neighbours the corre- sponding vertex may still propagate to. This label may only decrease one by one when we require a propagation from the corresponding vertex.
Formally, to each vertex v of G we associate a label L(v) = (av, bv) where av ∈ {B, F, R} and bv ∈ {0, 1, . . . , k}. An L-power dominating set of G = (V, E) is a subset S of V such that v ∈ S for all vertices v with av = R, where the sets (
PG,Li (S))
i≥0 of vertices monitored by S at step i are defined by the following rules.
• PG,L0 (S) = N [S]∪ {v : av = F}.
• PG,Li+1(S) = PG,Li (S)∪ {N[v]: v ∈ PG,Li (S) such that |N[v] \ PG,Li (S)| ≤ bv}.
When v ∈ PG,Li (S) satisfies |N[v] \ PG,Li (S)| ≤ bv, we say that v satis- fies the L-propagating condition (for short L-PC). The L-power domination number of G, denoted γP,L(G), is defined as the least cardinality of a L-power dominating set of G. A γP,L(G)-set denotes a minimum L-power dominating set of G.
Remark that when all L(v) = (B, k), L-power domination is the same as the k-power domination. Having this general setting in mind, we now have the following theorem from which one can infer a linear-time algorithm for computing a minimum L-power dominating set of a tree.
Theorem 5. Consider a graph G = (V, E) and a vertex labeling L. Suppose x is a vertex adjacent to only one other vertex y. If G′ = G− x and L′ is the restriction of L on V′ = V \ {x} with modification on L′(y) as indicated below, then the following statements hold.
(1) If ax = R, then γP,L(G) = γP,L′(G′) + 1, where a′y is redefined to be F when ay = B.
Otherwise, we assume ax̸= R for (2) to (5).
(2) If (ay = R) or (ax = F and bx= 0), then γP,L(G) = γP,L′(G′).
(3) If ax = B and by > 0, then γP,L(G) = γP,L′(G′), where b′y is redefined to be by − 1.
(4) If ax = B and by = 0, then γP,L(G) = γP,L′(G′), where a′y is redefined to be R.
(5) Otherwise, if ax = F, bx > 0 and ay ̸= R, then γP,L(G) = γP,L′(G′), where a′y is redefined to be F.
An example of application of these rules for 2-power domination is de- picted in Figure 1: the labels obtained from deleting the leaves bottom-up, from left to right are given.
(B,2) (B,2) (B,2) (B,2)
(B,2) (B,2)
(B,2) (B,2) (B,2)
(B,2) (B,0)
(R,0)
(F,1)
(F,1) (R,0)
Figure 1: Example of labels after application of the algorithm.
Proof: (1) Suppose S is a γP,L(G)-set. Since the label axis R, x is necessarily in S. Let S′ = S \ {x}. We prove by induction on i that for any step i,
PGi′,L′(S′) = PG,Li (S) \ {x}. Clearly, since a′y was redefined not to be B, this is true initially, for i = 0. Suppose the property is true until some step i ≥ 0. Let v ∈ PG,Li (S)\ {x}. By induction hypothesis, v ∈ PGi′,L′(S′) and NG[v]\ PG,Li (S) = NG′[v]\ PGi′,L′(S′). Added to the fact that the bv are unchanged, we obtain that if v satisfies the L-PC in G, it also satisfies the L′-PC in G′, and vice versa. Thus,PGi′,L′(S′) =PG,Li (S)\ {x} for all i and so S′ is an L′-power dominating set of G′. This gives γP,L(G)≥ γP,L′(G′) + 1.
On the other hand, suppose S′ is a γP,L′(G′)-set. Let S = S′∪ {x}. Simi- larly, PG,Li (S) =PGi′,L′(S′)∪ {x} for all i and so S is an L-power dominating set of G. This gives γP,L(G)≤ γP,L′(G′) + 1.
In the following, ax ̸= R. Suppose S is a γP,L(G)-set. We may assume that x /∈ S, for otherwise the set S† = S \ {x} ∪ {y} satisfies that PG,L0 (S) ⊆ PG,L0 (S†) and so S† is a γP,L(G)-set which can be used to replace S.
(2) First remark that if ay = R, then the property clearly holds. Suppose now that ax = F and bx= 0.
Consider a subset of vertices S not containing x. We claim thatPGi′,L′(S)∪ {x} = PG,Li (S) for all i. Clearly, this is true for i = 0. Suppose it is true until some step i ≥ 0. We check that any vertex satisfying the L′-PC in G′ at step i also satisfies the L-PC in G, and vice versa. The only vertex that may satisfy the L′-PC in G′ but not the L-PC in G is y. But since ax = F , x ∈ PG,L0 (S)⊆ PG,Li (S), and therefore NG[y]\ PG,Li (S) ⊆ NG′[y]\ PGi′,L′(S).
By induction hypothesis, any vertex (different from x) satisfying the L-PC in G satisfies the L′-PC in G′. In the case of x, since bx = 0, the satisfaction of the L-PC is not relevant for the computation of the monitored set at the next step.
(3) Consider a γP,L(G)-set S not containing x. We claim that S is a L′-power dominating set of G′. To prove it, we show that for any i ≥ 0, PG,Li (S)\ {x} ⊆ PGi′,L′(S). Clearly, this is true when i = 0. Suppose it is true until step i. Let v ̸= x satisfy the L-PC condition in G. If v ̸= y, then bv = b′v and v satisfies the L′-PC in G′ by induction hypothesis. Suppose now that y satisfies the L-PC.
First assume that x∈ PG,Li (S). Then, either x∈ PG,L0 (S), which implies that y ∈ S and N[y] ⊆ PG,L0 (S), or x was added later, meaning that y satisfied the L-PC in G at some earlier step i∗ < i. In both cases, NG[y]⊆ PG,Li (S) and by induction hypothesis, NG′[y] ⊆ PGi′,L′(S′), thus y also satisfies the L′-PC condition in G′ (b′y ≥ 0).
Suppose now that x /∈ PG,Li (S). Then NG[y] \ PG,Li (S) = NG′[y] \ PGi′,L′(S) ˙∪{x}. Therefore, the hypothesis that NG[y]\PG,Li (S) ≤by implies that NG′[y]\ PGi′,L′(S) ≤by − 1 = b′y, and y satisfies the L′-PC in G′.
Finally, remark that if x satisfies the L-PC in G, then x ∈ PG,Li (S) and as remarked above, y is also in PG,Li (S). Therefore, this is not relevant for the computation of the monitored set at the next step. This concludes the proof of our claim.
Now, consider a γP,L′(G′)-set S′. First remark that for any step i ≥ 0, PGi′,L′(S′) ⊆ PG,Li (S′). Indeed, the only reason for this assertion not to be true would be that at some step i, y satisfies the L′-PC in G′ but not the L-PC in G. However, since NG[y]\ NG′[y] = {x} and by = b′y+ 1, this may not happen. Moreover, since S′ is a L′-power dominating set of G′, y has to satisfy the L′-PC in G′ at some step, thus the L-PC in G, and x gets monitored at the next step.
(4) Any L′-power dominating set S′ of G′ contains y since a′y = R, so S′ is also an L-power dominating set of G.
From our earlier remark, we may consider a γP,L(G)-set S that does not contain x. Since by = 0, y may not satisfy the L-PC unless x is in PG,L0 (S).
Yet, ax = B, so for x to be in PG,L0 (S), it need to be in N [S]. Therefore, y is in S, and S is a L′-power dominating set of G′.
(5) Suppose now that ax = F, bx > 0 and by ̸= R. For any set S not containing x and for any step i ≥ 0, we claim that PG,Li (S) ⊆ PGi′,L′(S)∪ {x} ⊆ PG,Li+1(S), thus implying that S is a L-power dominating set of G if and only if S is a L′-power dominating set of G′.
Since a′y = F, the only neighbour of x (namely y) is in PG,L0 (S) and the first inclusion is easy to check.
For the second inclusion, first remark that PG0′,L′(S)\ {y} ⊆ PG,L0 (S).
Moreover, x is in PG,L0 (S) and satisfies the L-PC in G, so y ∈ PG1′,L′(S).
Thus, the inclusion is true for i = 0. Now, one can check that every vertex satisfying the L′-PC in G′ at step i also satisfies the L-PC in G at step i + 1.
This concludes the proof. 2
3. Bounds for k-power domination
Let G be a connected graph. We first establish that the natural relation on the k-power domination numbers of a graph when k varies is best possible.
We also prove that if n = |V (G)| ≥ k + 2, then γP,k(G) ≤ k+2n . A better bound is also proposed when G is claw-free and (k + 2)-regular.
3.1. Relation between k- and (k + 1)-power domination
Comparing k-power domination for different values of k, we note that if S is a k-power dominating set, then it is also a k′-power dominating set for any larger k′. Thus, we clearly get for any graph G the following inequality chain:
γ(G) = γP,0(G)≥ γP,1(G)≥ . . . ≥ γP,k−1(G)≥ γP,k(G)≥ γP,k+1(G)≥ . . . . Actually, this inequality chain cannot be improved in general, as shows the following observation:
Observation 6. If (xk)0≤k≤n is a finite non-increasing sequence of positive integers, then there exists a graph G such that γP,k(G) = xk for 0≤ k ≤ n.
Proof: For 0 ≤ k ≤ n, take xk − xk+1 copies of the star K1,k+1, where xn+1 is set as 0, and form a complete subgraph on the centres of all these stars. An example of such a graph for the sequence (7, 5, 5, 3, 2) is depicted by Figure 2. One can check easily that for a given k, a k-power dominating set must contain one vertex from each of the xk stars K1,j where j ≥ k + 1, and that the centres of these stars form a γP,k(G)-set. 2
K
7Figure 2: The graph for the k-power domination number sequence (7, 5, 5, 3, 2).
3.2. General bounds
In this part, we mainly prove Theorem 11, which states that any graph G of order n satisfies γP,k(G)≤ k+2n . We first need to prove the following few lemmas.
Lemma 7. If G is connected and ∆(G)≤ k + 1, then γP,k(G) = 1.
Proof: For any vertex v, we claim that S = {v} is a k-power dominating set of G. Let w ∈ Pi(S), i≥ 0. At least one neighbour of w is in Pi(S), so
|N(w) \ Pi(S)| ≤ d(w) − 1 ≤ k and then N(w) ⊆ Pi+1(S). This being true for any vertex w and for any i≥ 0, Pi+1(S) = N [Pi(S)]. By connectivity of
the graph, P∞(S) = V (G). 2
In the following we consider graphs with maximum degree at least k + 2.
Lemma 8. If S is a k-power dominating set of G containing a vertex v of degree at most k + 1, then (S\ {v}) ∪ {u} is also a k-power dominating set of G for any u ∈ N(v).
Proof: Let S′ = (S \ {v}) ∪ {u}. The set P0(S′) contains both u and v. Therefore, N (v)\ P0(S′) ≤ d(v)− 1 ≤ k, and N[v] ⊆ P1(S′). Thus, P0(S) ⊆ P1(S′) and by Observation 3, S′ is a k-power dominating set of G.
2 Lemma 9. If G is a connected graph with maximum degree at least k + 2, then there exists a γP,k(G)-set containing only vertices of degree at least k +2.
Proof: The proof is based on Lemma 8. Let S be a γP,k(G)-set containing as many vertices of degree at least k +2 as possible. Suppose S contains a vertex v of degree at most k + 1. Let w be a vertex of degree at least k + 2 closest to v, that is such that j = d(v, w) is minimum. Consider a shortest path µ = (x0, x1, . . . , xj) from x0 = v to xj = w. Iteratively applying Lemma 8, we obtain that for any 1 ≤ i ≤ j, Si = (S \ {v}) ∪ {xi} is also a k-power dominating set of G. In particular, this is true for xj = w, thus contradicting the assumption that S contains the maximum number of degree at least k + 2
vertices, or the minimality of S. 2
For any vertex v of a subset S of V (G), the S-private neighbourhood of v, denoted epn(v, S), is the set of neighbours of v which are not neighbours of any vertex of S\ {v}.
Lemma 10. If G is a connected graph with ∆(G) ≥ k + 2, then there exists a γP,k(G)-set S such that every vertex in S has at least k + 1 S-private neighbours, i.e., |epn(x, S)| ≥ k + 1 for x ∈ S.
Proof: Let S be a γP,k(G)-set having only vertices of degree at least k + 2 (by Lemma 9) such that G[S] has the minimum number of components. If
every vertex x of S has|epn(x, S)| ≥ k + 1, then we are done. Suppose there exists v in S such that |epn(v, S)| ≤ k. We consider two cases:
Case 1. There exists w ∈ S adjacent to v. We claim that S′ = S \ {v}
is a k-power dominating set of G, contradicting the minimality of S. Indeed, since v ∈ P0(S′) and|N(v)\P0(S′)| = |epn(v, S)| ≤ k, N[v] ⊆ P1(S′). Thus, P0(S) ⊆ P1(S′) and S′ is a k-power dominating set of G, contradicting the minimality of S.
Case 2. Vertex v is an isolated vertex of G[S]. Since d(v)≥ k + 2 and
|epn(v, S)| ≤ k, there exist w ∈ N(v) and x ∈ S such that w ∈ N(x) (w is not a S-private neighbour of x). Set S′ = (S\ {v}) ∪ {w} if d(w) ≥ k + 2 and S′ = S \ {v} otherwise. We claim that S′ is a k-power dominating set of G. Indeed, since v ∈ P0(S′) and |N(v) \ P0(S′)| ⊆ |epn(v, S)| ≤ k, N [v]⊆ P1(S′). Thus,P0(S)⊆ P1(S′) and S′ is a k-power dominating set of G. Moreover, S′ has one less component than S, contradicting the choice of
S. 2
As a consequence of Lemmas 7 and 10, we obtain the following theorem.
Theorem 11. If G is a connected graph of order n≥ k + 2, then γP,k(G)≤ n
k + 2 and this bound is best possible.
Proof: That k+2n is a upper bound for the k-power domination number of G is a direct consequence of Lemmas 7 and 10.
Let us describe graphs for which this bound is attained. Let G be any connected graph on n vertices, denoted v1, v2, . . . , vn and let H1, H2, . . . , Hn
be a family of graphs on k + 1 vertices. We form a new graph G′ by taking the disjoint union of all these graphs and adding edges linking vi to every vertices of Hi.
We prove by contradiction that a k-power dominating set of such a graph G′ must contain at least one vertex in each V (Hi)∪{vi}. Suppose there exists some k-power dominating set S not containing any vertex in V (H1)∪ {v1}, without loss of generality. At the beginning, no vertex of H1 is monitored, P0(S) ∩ V (H1) = ∅. Let i be the smallest integer such that there exists x ∈ V (H1)∩ Pi(S). It exists since S is a k-power dominating set, and i ≥ 1. For x to be added into Pi(S), it means that some vertex from N (x) was in Pi−1(S) and had at most k neighbours not in Pi−1(S). Since
N (x) ⊆ V (H1)∪ {v1} and V (H1)∩ Pi−1(S) = ∅, this vertex is necessarily v1. Yet N [v1]\ Pi−1(S)⊇ V (H1) which contain k + 1 vertices, contradicting our assumption.
Therefore, a k-power dominating set of G′must contain at least one vertex in each V (Hi)∪{vi}, and this is at least n vertices among the n(k+2) vertices of G′. Note that V (G) for example is a k-power dominating set of G′. 2 3.3. Regular claw-free graphs
This subsection is dedicated to claw-free graphs. A claw-free graph is a graph that does not contain a claw, i.e. K1,3, as an induced subgraph.
For positive integers k and r, let Dk,r be the graph obtained from the disjoint union of r copies of Di ∼= Kk+3 − xiyi, a complete graph on k + 3 vertices minus one edge xiyi, for 1 ≤ i ≤ r by adding r edges yixi+1 (1≤ i ≤ r) where xr+1 = x1; see Figure 3 for Dk,6.
Kk+3 K
k+3
x Kk+3
x x Kk+3
Kk+3
x
Kk+3
x1
3
4 5 6
y
y y y
y
1
3 4 5
6
x2
y2
Figure 3: The graph Dk,6.
Theorem 12. For a positive integer k, if G is a connected claw-free (k + 2)- regular graph of n vertices, then γP,k(G)≤ n/(k +3) with equality if and only if G is isomorphic to Dk,r for some r ≥ 1.
Proof: Let S be a γP,k(G)-set such that (1) G[S] has as few edges as possible and under this condition, (2) |N[S]| is as large as possible.
Claim 1. G[S] is an independent set.
Proof: Suppose to the contrary that there exist two adjacent vertices in G[S], say u and v. Observe first that |epn(v, S)| = k + 1; otherwise, we have |epn(v, S)| ≤ k and S′ = S \ {v} is a k-power dominating set with less vertices than S, contradicting the minimality of S. Now let w be any S-private neighbour of v and consider S′′= (S\ {v}) ∪ {w}. The set S′′ is a γP,k(G)-set, but G[S′′] contains less edges than G[S], contradicting (1). This
completes the proof of Claim 1. 2
Claim 2. If u and v are two distinct vertices of S, then |N(u) ∩ N(v)| ≤ 1.
Proof: Suppose to the contrary that |N(u) ∩ N(v)| ≥ 2. Let x and y be two distinct vertices in N (u)∩ N(v). As the graph is claw-free, G contains either the edge xy or the edge xz or yz where z is a third neighbour of v. In the two cases, S′ = S\ {u} is a k-power dominating set of G with one vertex less than S, contradicting the minimality of S. This completes the proof of
Claim 2. 2
Claim 3. Let u and v be two distinct vertices of S and x be a vertex in N (u)∩ N(v). Moreover assume that u and x are adjacent to a vertex y.
Then, the k +1 other neighbours z1, z2, . . . , zk+1 of v are S-private neighbours of v, form a clique of size k + 1, and N [zi] ⊆ N[S] for 1 ≤ i ≤ k + 1, see Figure 4.
clique
u ∈ S v ∈ S
x
y z1 zk+1
Figure 4: Claim 3.
Proof: First, observe that there is no edge xzi; otherwise S′ = S \ {v}
is a k-power dominating set with less vertices than S, contradicting the minimality of S. Second, since the graph is claw-free, it follows that the vertices zi’s form a clique of size k + 1. If some zi belongs to N [S\ {v}], then as previously, S′ = S \ {v} is a k-power dominating set with less vertices than S, contradicting the minimality of S. To complete the proof, assume that N [zi]̸⊆ N[S] for some i. Then S′′ = S\ {v} ∪ {zi} is a γP,k(G)-set with
|N[S′′]| > |N[S]|, contradicting (2). This completes the proof of Claim 3. 2
Claim 4. Let v be a vertex of S with neighbourhood x, z1, z2, . . . , zk+1 such that x is adjacent to u ∈ S \ {v} and to y with uy ∈ E(G), and such that the vertices z1, z2, . . . , zk+1 form a clique of size k + 1. Moreover assume that z1 is linked to a vertex t which is linked to a vertex w ∈ S \ {v}. Then, t is also linked to the vertices z2, z3, . . . , zk+1. The k + 1 other neighbours s1, s2, . . . , sk+1 of w are S-private neighbours of w, form a clique of size k + 1, and N [si]⊆ N[S] for 1 ≤ i ≤ k + 1, see Figure 5.
t
u ∈ S v ∈ S
x
y z1 zk+1
clique s1 sk+1
w ∈ S
Figure 5: Claim 4.
Proof: First observe that there is no edge tsi; otherwise S′ = S\ {v} is a k- power dominating set with less vertices than S, contradicting the minimality of S. It follows that, since the graph is claw-free, t is linked to all zi (1≤ i ≤ k + 1), and that the vertices s1, s2, . . . , sk+1 form a clique of size k + 1. Now, every si is a S-private neighbour of w; otherwise, S′′ = S\ {w} is a k-power dominating set with less vertices than S, contradicting the minimality of S.
Finally, N [si] ⊆ N[S] for all i; otherwise, S′′′ = S \ {v, w} ∪ {z1, si} is a γP,k(G)-set with |N[S′′′]| > |N[S]|, contradicting (2). This completes the
proof of Claim 4. 2
By using similar arguments, one can extend the previous claim as follows:
Claim 5. Let u, v1, v2, . . . , vi be distinct vertices of S. Every vj with 1 ≤ j ≤ i has neighbourhood ti−1, zj1, z2j, . . . , zk+1j . Assume for all 1 ≤ j ≤ i, the vertices zj1, z2j, . . . , zk+1j form a clique of size k + 1 and are linked to tj. The vertex u is linked to two vertices y and t0, and yt0 ∈ E(G). Finally suppose that z1i is linked to a vertex t which is linked to a vertex vi+1 ∈ S\ {v1, v2, . . . , vi}. Then, t is also linked to the vertices zi2, zi3, . . . , zk+1i . The k + 1 other neighbours z1i+1, zi+12 , . . . , zk+1i+1 of vi+1 are S-private neighbours of vi+1, form a clique of size k + 1, and N [zi+1j ]⊆ N[S] for 1 ≤ j ≤ k + 1, see Figure 6.
u ∈ S
y
v1∈S
t0 z11 z1k+1 z12 zk+12
v2∈S
t1 zk+1i+1
vi−1∈S
ti−2 zi−11 z1i zk+1i
vi∈S
zi−1k+1 ti−1
vi+1∈S
zi+11
t
clique
Figure 6: Claim 5.
Claim 6. If u and v are two distinct vertices of S, then N (u)∩ N(v) = ∅.
Proof: Assume that two vertices u and v of S have a common neighbour x. Since x must not be in a claw and by symmetry, there exists a vertex y linked to u and x. We apply now Claims 3, 4 and repetitively Claim 5. Since S is finite, we obtain with Claim 5 that vi+1 = u and zji+1 = t0 for some j which is not a S-private neighbour of vi+1, a contradiction. 2 It follows that|V (G)| ≥ γP,k(G)(k + 3) and so γP,k(G)≤ |V (G)|/(k + 3).
Now observe that γP,k(Dk,r) = r = |V (Dk,r)|/(k + 3). Suppose that γP,k(G) = |V (G)|/(k + 3). By Claim 6, one can choose S such that the N [v]’s with v ∈ S are a partition of V (G). A vertex z ∈ N(S) is said to be special if there are two distinct vertices u and v in S such that z is linked to v and N (z)⊆ N(u) ∪ {v}.
Claim 7. Let u and v be two distinct vertices of S. Assume that u is adjacent to two adjacent vertices x and y. Finally, let z be a vertex adjacent to x and v. Then, z is special. Moreover G[N (v)\ {z}] is a clique of size k + 1 and so it is for G[N (z)∩ N(u)].
Proof: Suppose that z is not special. Then, N (z) contains at most k vertices of N (u). Let r be a vertex in N (z)\(N(u)∪{v}). By the choice of S, r belongs to N (S) (like every vertex not in S). It follows that S′ = S\{u} is a k-power dominating set having less vertices than S, contradicting the minimality of S. Hence, z is special. Since the graph is claw-free and N (u)∩ N(v) = ∅, G[N (z)∩ N(u)] is a clique of size k + 1. Finally, the graph being claw-free, v, G[N (v)\ {z}] is also a clique of size k + 1. 2 If |S| = 1, then G = Kk+3 = Dk,1. So assume that |S| ≥ 2. Let A(v) = {x ∈ N(v): N(x) \ N[v] ̸= ∅}. First observe that for all v ∈ S,
|A(v)| ≥ 2. If every v ∈ S has |A(v)| = 2, then by connectedness, G is Dk,|S|
as desired. Let v1 be a vertex of S with|A(v1)| ≥ 3. Since the graph is claw- free v1 is adjacent to two adjacent vertices x1 and y1 with x1, y1 ∈ A(v1).
Moreover we can assume that x1 is linked to a vertex z2, which is linked to a vertex v2 ∈ S\{v1}. By Claim 7, z2 is special, and both G[N (z2)∩N(v1)] and G[N (v2)\{z2}] are cliques of size k+1. Now, there exist two adjacent vertices x2, y2 in N (v2)\ {z2} with x2 ∈ A(v2). By repeating the same argument, we have S = {v1, v2, . . . , vr}; each N(vi) contains a special vertex zi such that G[N (vi)\ {zi}] is a clique of size k + 1 and N(zi) ={vi} ∪ N(vi−1)\ {zi−1} for 1 ≤ i ≤ r, where v0 = vr and z0 = zr. This implies that G = Dk,r. This
completes the proof of Theorem 12. 2
3.4. Graphs of diameter two
In [16], the authors propose some bounds on the power domination num- ber of planar graphs when the diameter is bounded. A natural question is then whether one can propose a general (constant) bound on the k-power domination number of graphs with bounded diameter. The answer to this question is negative, as the following proposition shows.
Proposition 13. For any k, there exist graphs of diameter 2 with arbitrarily large k-power domination number.
Proof: Suppose (P, L) is a finite projective plane of order n, where the set P contains exactly n2+ n + 1 points, the set L contains exactly n2+ n + 1 lines, each point is in exactly n + 1 lines and each line contains exactly n + 1 points. Consider the graph G with vertex set V (G) = P ∪ L and edge set E(G) ={xℓ : x ∈ P, ℓ ∈ L, x ∈ ℓ} ∪ {ℓℓ′ : ℓ ̸= ℓ′ in L}.
Since every two points are in exactly one line, G is of diameter 2.
If S is a γP,k(G)-set, then without loss of generality we may assume that S ⊆ L. Since N[S] = L ∪∪
{x ∈ ℓ : ℓ ∈ S}, any line not in S is adjacent to n + 1− |S| points which are not in N[S]. In order for S to be a k-power dominating set of G, it must be the case that n + 1− |S| ≤ k or equivalently
|S| ≥ n + 1 − k. This gives that γP,k(G)≥ n + 1 − k.
On the other hand, consider a point x and a subset S ⊆ L of size n+1−k, such that any line in S contains x. The set N [S] contains all lines. Moreover, any line ℓ that do not contain x share a point with every line in S, and these points are all distinct. Indeed, if two lines in S shared the same point with ℓ, then they would share both x and that point, but by any two points goes only one line. Therefore, any line not containing x contain n−k+1 points in N[S].
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