Power domination polynomials of graphs
Boris Brimkov, Rutvik Patel, Varun Suriyanarayana, Alexander Teich
aa r X i v : . [ m a t h . C O ] M a y Power domination polynomials of graphs
Boris Brimkov ∗ , Rutvik Patel ∗ , Varun Suriyanarayana ∗ , Alexander Teich ∗ Abstract
A power dominating set of a graph is a set of vertices that observes every vertex in thegraph by combining classical domination with an iterative propagation process arising fromelectrical circuit theory. In this paper, we study the power domination polynomial of a graph G of order n , defined as P ( G ; x ) = P ni =1 p ( G ; i ) x i , where p ( G ; i ) is the number of powerdominating sets of G of size i . We relate the power domination polynomial to other graphpolynomials, present structural and extremal results about its roots and coefficients, and identifysome graph parameters it contains. We also derive decomposition formulas for the powerdomination polynomial, compute it explicitly for several families of graphs, and explore graphswhich can be uniquely identified by their power domination polynomials. Keywords:
Power domination polynomial, power dominating set, domination, zero forcing
Let G = ( V, E ) be a graph, let S ⊂ V be a set of initially colored vertices, and consider the followingcolor change rules:1) Every neighbor of an initially colored vertex becomes colored.2) Whenever there is a colored vertex with exactly one uncolored neighbor, that neighbor becomescolored. S is a power dominating set of G if all vertices in G become colored after applying rule 1) once, andrule 2) as many times as possible (i.e. until no more vertices can change color). The application ofrule 1) is called the domination step , and each application of rule 2) is called a forcing step . The power domination number of G , denoted γ P ( G ), is the cardinality of a minimum power dominatingset. S is a zero forcing set of G if all vertices in G become colored after applying rule 2) as manytimes as possible (and not applying rule 1) at all). The zero forcing number of G , denoted Z ( G ),is the cardinality of a minimum zero forcing set. S is a dominating set of G if all vertices in G become colored after applying rule 1) once; the cardinality of a minimum dominating set is calledthe domination number , denoted γ ( G ). Thus, the power domination process can be regarded as acombination of classical graph domination and zero forcing.Power domination arises from a graph theoretic model of the Phase Measurement Unit (PMU)placement problem from electrical engineering: in order to monitor their networks, electrical powercompanies place PMUs at select locations in the power network; the physical laws by which PMUs ∗ Department of Computational and Applied Mathematics, Rice University, Houston, TX, 77005, USA([email protected], [email protected], [email protected], [email protected]) power propagation time [1, 26, 31, 40], where the objective is to find the smallest number of timesteps it takes for thegraph to be colored by a minimum power dominating set. Sets of vertices related to complements ofpower dominating sets (of arbitrary size) have also been used in integer programming approaches forcomputing the power domination number (cf. [12]). The PMU placement literature also considerspower dominating sets with various additional properties, such as redundancy, controlled islanding,and connectedness, and optimizes over them in addition to the cardinality of the set (see, e.g.,[5, 15, 41, 49]). In order to study the collection of power dominating sets of a graph in a moregeneral framework, we introduce the power domination polynomial , which counts the number ofdistinct power dominating sets of a given size.
Definition 1.
Let G be a graph on n vertices and p ( G ; i ) be the number of power dominating setsof G with cardinality i . The power domination polynomial of G is defined as P ( G ; x ) = n X i =1 p ( G ; i ) x i . We study the basic algebraic and graph theoretic properties of the power domination polyno-mial, present structural and extremal results about its roots and coefficients, compute it explic-itly for several families of graphs, and identify some graphs which can be recognized by theirpower domination polynomials. We also relate the power domination polynomial to the zero forc-ing polynomial and the domination polynomial of a graph G , which respectively count the num-ber of zero forcing sets and dominating sets of G . More precisely, the zero forcing polynomialof G is defined as Z ( G ; x ) = P ni =1 z ( G ; i ) x i and the domination polynomial of G is defined as D ( G ; x ) = P ni =1 d ( G ; i ) x i , where z ( G ; i ) is the number of zero forcing sets of G of size i , and d ( G ; i )is the number of dominating sets of G of size i .Work on the domination polynomial and zero forcing polynomial includes derivations of recur-rence relations [38], analysis of the roots [16], and characterizations for specific graphs [3, 6, 11, 36].Similar results have been obtained for the connected domination polynomial [24], independencepolynomial [35], clique polynomial [32], vertex cover polynomial [25], and edge cover polynomial [4],which are defined as generating functions of their eponymous sets. In general, graph polynomialscontain important information about the structure and properties of graphs that can be extractedby algebraic methods. In particular, the values of graph polynomials at specific points, as well astheir coefficients, roots, and derivatives, often have meaningful interpretations. Such informationand other unexpected connections between graph theory and algebra are sometimes discovered longafter a graph polynomial is originally introduced (see, e.g., [19, 48]). For more definitions, results,and applications of graph polynomials, see the survey of Ellis-Monaghan and Merino [29] and thebibliography therein. 2his paper is organized as follows. In the next section, we recall some graph theoretic notionsand notation. In Section 3, we present a variety of structural and extremal results about thecoefficients of the power domination polynomial, relate the power domination polynomial to othergraph polynomials, and give several decomposition formulas. In Section 4, we give closed-formexpressions for the power domination polynomials of several families of graphs, and explore graphswhich can be uniquely identified by their power domination polynomials. In Section 5, we studythe roots of the power domination polynomial. We conclude with some final remarks and openquestions in Section 6. A graph G = ( V, E ) consists of a vertex set V = V ( G ) and an edge set E = E ( G ) of two-elementsubsets of V . The order of G is denoted by n ( G ) = | V | . We will assume that the order of G isnonzero, and when there is no scope for confusion, dependence on G will be omitted. Two vertices v, w ∈ V are adjacent , or neighbors , if { v, w } ∈ E ; we will sometimes write vw to denote an edge { v, w } . The neighborhood of v ∈ V is the set of all vertices which are adjacent to v , denoted N ( v ); the degree of v ∈ V is defined as d ( v ) = | N ( v ) | . The closed neighborhood of v is the set N [ v ] = N ( v ) ∪ { v } . Given S ⊂ V , N [ S ] = S v ∈ S N [ v ], and N ( S ) = (cid:0)S v ∈ S N ( v ) (cid:1) \ S .Given S ⊂ V , the induced subgraph G [ S ] is the subgraph of G whose vertex set is S and whoseedge set consists of all edges of G which have both endpoints in S . An isomorphism between graphs G and G will be denoted by G ≃ G . An isolated vertex , or isolate , is a vertex of degree 0. A dominating vertex is a vertex which is adjacent to all other vertices. A cut vertex is a vertex which,when removed, increases the number of connected components in G . A block of G is a maximal(with respect to inclusion) subgraph which does not contain cut vertices. The path, cycle, completegraph, and empty graph on n vertices will respectively be denoted P n , C n , K n , K n . An endpoint of P n is a degree 1 vertex if n >
1, and a degree 0 vertex if n = 1.Given two graphs G and G , the disjoint union G ˙ ∪ G is the graph with vertex set V ( G ) ˙ ∪ V ( G ) and edge set E ( G ) ˙ ∪ E ( G ). The join of G and G , denoted G ∨ G , is thegraph obtained from G ˙ ∪ G by adding an edge between each vertex of G and each vertex of G .The corona of two graphs G and G , denoted G ◦ G , is the graph obtained by taking one copyof G and n ( G ) copies of G , and adding an edge between the i th vertex of G and each vertex ofthe i th copy of G , 1 ≤ i ≤ n ( G ). The complete bipartite graph with parts of size a and b , denoted K a,b , is the graph K a ∨ K b . The graph K n − , , n ≥
3, will be called a star and denoted S n , andthe graph C n − ∨ K , n ≥
4, will be called a wheel and denoted W n . For other graph theoreticterminology and definitions, we refer the reader to [10].Given integers a and b with 0 ≤ a < b , we adopt the convention that (cid:0) ab (cid:1) = 0. We adopt also theconventions that P i ∈∅ s i = 0, Q i ∈∅ s i = 1, and S i ∈∅ S i = ∅ . For any positive integer n , [ n ] denotesthe set { , . . . , n } . N , Z , Q , R , and C respectively denote the set of positive integers, integers,rational numbers, real numbers, and complex numbers. In this section, we give several structural and extremal results about the coefficients of powerdomination polynomials. We begin with the following basic observation.3 bservation 1.
Let G = ( V, E ) be a graph and R ⊂ V . If S is a power dominating set of G ,then S ∪ R is a power dominating set of G . Equivalently, any superset of a power dominating set ispower dominating, and any subset of a non-power dominating set is not power dominating. Proposition 2.
Let G = ( V, E ) be a graph. If p ( G ; i ) = (cid:0) ni (cid:1) for some i ∈ [ n ] , then p ( G ; j ) = (cid:0) nj (cid:1) for j ∈ { i, . . . , n } .Proof. Note that p ( G ; j ) ≤ (cid:0) nj (cid:1) for all j ∈ [ n ]. Suppose there exists a j ∈ { i, . . . , n } such that p ( G ; j ) < (cid:0) nj (cid:1) . Then there exists a set S ⊂ V of size j which is not power dominating. ByObservation 1, no subset of S is power dominating; in particular, the subsets of S of size i arenot power dominating, which contradicts the assumption that every subset of V of size i is powerdominating.A matching of G = ( V, E ) is a set M ⊂ E such that no two edges in M have a common endpoint.A matching M saturates a vertex v , if v is an endpoint of some edge in M . Theorem 3 (Hall’s Theorem [33]) . Let G be a bipartite graph with parts X and Y . G has amatching that saturates every vertex in X if and only if for all S ⊂ X , | S | ≤ | N ( S ) | . Proposition 4.
Let G = ( V, E ) be a graph. Then, p ( G ; i ) ≤ p ( G ; i + 1) for ≤ i < n .Proof. Let X be the set of all power dominating sets of G of size i and Y be the set of all subsetsof V of size i + 1. Let H be the bipartite graph with parts X and Y , where a vertex x ∈ X isadjacent to a vertex y ∈ Y whenever x ⊂ y . Each x ∈ X is adjacent to n − i sets in Y , each ofwhich consists of x and a vertex of G not in x . Each y ∈ Y is adjacent to at most i + 1 vertices in X , since a set of size i + 1 has i + 1 subsets of size i . Now, suppose for contradiction that for someset S ⊂ X , | S | > | N ( S ) | . Since d ( v ) = n − i for each v ∈ S and | S | > | N ( S ) | , there must be somevertex w ∈ N ( S ) with d ( w ) > n − i . Thus, i + 1 ≥ d ( w ) > n − i , which contradicts the assumptionthat i < n . It follows that for each S ⊂ X , | S | ≤ | N ( S ) | , so by Theorem 3, H has a matchingthat saturates all vertices of X . Thus, for 1 ≤ i < n , to each power dominating set of size i , therecorresponds a distinct superset of size i + 1, which by Observation 1 is also a power dominatingset. Theorem 5.
Let G = ( V, E ) be a graph. Then, for k ∈ { , . . . , n − } , p ( G ; n − k ) = (cid:12)(cid:12) { S ⊂ V : | S | = k, S ∩ N ( V \ S ) is a zero forcing set of G [ S ] } (cid:12)(cid:12) . Proof.
We will show that for any S ⊂ V , S ∩ N ( V \ S ) is a zero forcing set of G [ S ] if and onlyif V \ S is a power dominating set of G . Suppose first that S ∩ N ( V \ S ) is a zero forcing set of G [ S ]. If the vertices in V \ S are initially colored, the vertices in N ( V \ S ) will become colored in thedomination step, and by assumption, any uncolored vertices in S can be forced by the vertices in S ∩ N ( V \ S ). Thus V \ S is a power dominating set of G . Now suppose V \ S is a power dominatingset of G . Then, after the domination step, the vertices in N ( V \ S ) become colored, and the onlyuncolored vertices are in G [ S ]. Thus, in order for V \ S to be a power dominating set, the vertices in S ∩ N ( V \ S ) must be able to force any uncolored vertices in G [ S ], so S ∩ N ( V \ S ) is a zero forcingset of G [ S ]. Thus, for any k ∈ { , . . . , n − } and S ⊂ V with | S | = k , V \ S is a power dominatingset of size n − k if and only if S ∩ N ( V \ S ) is a zero forcing set of G [ S ].4ote that it can be determined whether a set is a zero forcing set of a graph G in linear time; thus,while computing all the coefficients of the power domination polynomial of G is in general NP-hard(since, e.g., its smallest nonzero coefficient corresponds to γ P ( G ) which is NP-hard to find [34]),computing p ( G ; n − k ) is in the complexity class XP, for the parameter k . The following are simpleconsequences of Theorem 5. Corollary 6.
Let G be a graph with I isolates and k K -components. Then, p ( G ; n ) = 1 , p ( G ; n −
1) = n − I p ( G ; n −
2) = (cid:0) n (cid:1) − I ( n − I ) − (cid:0) I (cid:1) − k . If G is connected and has at least 3 vertices, then p ( G ; n − k ) = (cid:0) nk (cid:1) for k ∈ { , , } .Proof.
1) and 2) are trivial. For 3), two vertices can be excluded from a power dominating setwhenever neither of them is an isolate and they do not form a K component in G . 4) follows from1), 2), and 3).Corollary 6 shows that the power domination polynomial counts the number of isolates of a graphand the number of K -components; however we will show in Proposition 26 that in general, it cannotcount the number of components.A fort of a graph G = ( V, E ), as defined in [30], is a non-empty set F ⊂ V such that no vertexoutside F is adjacent to exactly one vertex in F . The following facts are well-known in the literature. Proposition 7 ([12, 30, 46]) . Let G = ( V, E ) be a graph and let S ⊂ V .1. S is a zero forcing set of G if and only if S intersects every fort F of G .2. S is a power dominating set of G if and only if S intersects N [ F ] for every fort F of G . In [12, 46], it was shown that the following integer program can be used to compute the powerdomination number of a graph G = ( V, E ), where N ( G ) = { N [ F ] : F is a fort of G } .min X v ∈ V s v s.t.: X v ∈ N s v ≥ ∀ N ∈ N ( G ) s v ∈ { , } ∀ v ∈ V We now give a way to bound the number of constraints in this model using the power dominationpolynomial.
Proposition 8.
Let G be a graph. Then, |N ( G ) | ≤ n − P ( G ; 1) . roof. P ( G ; 1) = P ni =1 p ( G ; i ) equals the number of distinct power dominating sets of G , andhence also the number of complements of power dominating sets of G . By Proposition 7, a powerdominating set must intersect every element of N ( G ). Thus, the complement of a power dominatingset cannot be an element of N ( G ), and so the number of complements of power dominating sets of G is at most the number of sets of G which are not neighborhoods of forts, i.e., 2 n − |N ( G ) | . Thus, P ( G ; 1) ≤ n − |N ( G ) | . In this section, we characterize the graphs whose power domination polynomials are identical totheir zero forcing polynomials or their domination polynomials. We begin with the following basicobservation, which follows from the fact that every zero forcing set and every dominating set of agraph is also a power dominating set.
Observation 9.
Let G be a graph. Then, z ( G ; i ) ≤ p ( G ; i ) , and d ( G ; i ) ≤ p ( G ; i ) for all ≤ i ≤ n .Also, γ P ( G ) ≤ Z ( G ) and γ P ( G ) ≤ γ ( G ) . Note that in general, Z ( G ) and γ ( G ) (as well as d ( G ; i ) and z ( G ; i )) are incomparable. Recall alsothe following result relating zero forcing sets and power dominating sets. Theorem 10 ([23]) . S is a power dominating set of G if and only if N [ S ] is a zero forcing set of G . Theorem 11.
Let G be a graph. Z ( G ; x ) = P ( G ; x ) if and only if G ≃ P a ˙ ∪ · · · ˙ ∪ P a k , where a i ≤ for ≤ i ≤ k .Proof. Let G be a graph such that Z ( G ; x ) = P ( G ; x ). Suppose G has a power dominating set S which is not a zero forcing set. Let S , . . . , S k be all zero forcing sets of G of size | S | . Since P ( G ; x ) = Z ( G ; x ), there are the same number of zero forcing sets of size | S | as power dominatingsets of size | S | . However, since S is different from S , . . . , S k and since every zero forcing set is alsoa power dominating set, it follows that p ( G ; | S | ) > z ( G ; | S | ), a contradiction. Thus, S is a zeroforcing set of G if and only if S is a power dominating set of G . From this and from Theorem 10,it follows that S is a zero forcing set of G if and only if N [ S ] is a zero forcing set of G . In otherwords, a dominating set of a zero forcing set of G is also zero forcing.Now, suppose some component G ′ of G has zero forcing number greater than one. Let Z bea minimum zero forcing set of G ′ ; G ′ [ Z ] has no edges, since otherwise a dominating set of G ′ [ Z ]would be a smaller zero forcing set. Let v and v p be two vertices in Z such that a shortest path v , v , . . . , v p between v and v p in G ′ contains no other vertices of Z (clearly such vertices exist, e.g.if they are a closest pair among all pairs of vertices in Z ). The set N [ Z \{ v } ∪ { v } ] is also a zeroforcing set of G ′ since it contains Z ; thus, Z \{ v } ∪ { v } is also a zero forcing set. Similarly, the set N [ Z \{ v } ∪ { v } ] is a zero forcing set of G ′ since it contains Z \{ v } ∪ { v } ; hence, Z \{ v } ∪ { v } is also zero forcing. By the same reasoning, the sets Z \{ v } ∪ { v } , . . . , Z \{ v } ∪ { v p − } are allzero forcing. However, G ′ [ Z \{ v } ∪ { v p − } ] contains the edge v p v p − , and hence G ′ has a smallerzero forcing set than Z , a contradiction. Thus, Z ( G ′ ) = 1 and so G ′ is a path. If G ′ is a pathof length greater than 1, then there are subsets of V ( G ′ ) of size 1 which are not zero forcing sets,but all subsets of size 1 are power dominating sets. Since P ( G ; x ) = Z ( G ; x ), it follows that everycomponent of G is a path of length at most 1. 6onversely, if every component of G is a path of length at most 1, then every power dominatingset is clearly a zero forcing set (and vice versa), so P ( G ; x ) = Z ( G ; x ). Theorem 12.
Let G = ( V, E ) be a graph. D ( G ; x ) = P ( G ; x ) if and only if for each non-isolate u ∈ V , there exists a v ∈ N ( u ) such that N [ v ] ⊂ N [ u ] .Proof. Suppose that u is a non-isolate of G such that for each v ∈ N ( u ), there is a q ∈ N [ v ] with q / ∈ N [ u ]. Let S = V \ N [ u ] be a set of initially colored vertices. Then, each vertex in N ( u ) has aneighbor in S , so all vertices in N ( u ) will be colored in the domination step. Since u is the onlyuncolored vertex left after the domination step, and since u is not an isolate, u will be colored in theforcing step, so S is a power dominating set of G . On the other hand, clearly S is not a dominatingset since u has no neighbor in S . Since all dominating sets of G (and in particular those of size | S | )are also power dominating sets, it follows that p ( G ; | S | ) > d ( G ; | S | ) and so D ( G ; x ) = P ( G ; x ).Conversely, suppose that for each non-isolate u ∈ V , there exists a v ∈ N ( u ) such that N [ v ] ⊂ N [ u ], but D ( G ; x ) = P ( G ; x ). Since every dominating set of G is a power dominating set, theremust be some power dominating set S that is not a dominating set. Since all isolates must be inevery power dominating set, there is some non-isolate u ∈ V that is the last vertex to be forced by S . By our assumption, there is a v ∈ N ( u ) such that N [ v ] ⊂ N [ u ]. Since u is the last vertex tobe colored, v must have already been colored at the timestep when u gets forced. If v was initiallycolored, then u would have been colored in the domination step, contradicting the fact that u shouldbe forced rather than dominated. If v was colored in the domination step, then since every neighborof v is also a neighbor of u , u would again have been dominated rather than forced, by the samevertex that dominates v . If v was colored in some forcing step, then since every neighbor of v isalso a neighbor of u , u would have had to be colored in order for the forcing to occur, contradictingthe assumption that u is the last vertex to be colored. Thus, in all cases, it follows that S cannotexist, so D ( G ; x ) = P ( G ; x ).Note that the condition in Theorem 12 can be verified in polynomial time. Graphs satisfying thiscondition include, e.g., graphs where each block is a clique and contains at least two non-cut vertices. Theorem 13.
For any connected graph G = ( V, E ) , z ( G ; i ) = p ( G ; i ) if and only if z ( G ; i ) = p ( G ; i ) = (cid:0) ni (cid:1) or z ( G ; i ) = p ( G ; i ) = 0 .Proof. One direction is trivial. For the other direction, let G be a connected graph such that z ( G ; i ) = p ( G ; i ) for some i ∈ [ n ]. Let F be a fort of G with minimum cardinality, and let f = | F | .Suppose first that i > n − f . Then for every S, S ′ ⊂ V with | S | = i and | S ′ | ≥ f , S ∩ S ′ = ∅ .Since the size of every fort is at least f , every set of size i will intersect every fort, and hence byProposition 7 will be zero forcing and power dominating. Thus z ( G ; i ) = p ( G ; i ) = (cid:0) ni (cid:1) .Now suppose i ≤ n − f . If z ( G ; i ) = p ( G ; i ) = 0, we are done. Thus suppose that z ( G ; i ) = p ( G ; i ) = 0. Let Z be a zero forcing set of size i and let k := | Z ∩ F | ; note that since Z is zeroforcing, k >
0. Then, | V \ ( Z ∪ F ) | = | V | − ( | Z | + | F | − | Z ∩ F | ) = n − i − f + k ≥ k . If F = V ,every set of size 1 is zero forcing and hence z ( G ; i ) = p ( G ; i ) = (cid:0) ni (cid:1) . Otherwise, if F = V , since G isconnected, it follows that N ( F ) = ∅ . Let Z ′ be a set obtained by starting from Z , removing all k vertices in Z ∩ F , and adding k vertices from V \ ( Z ∪ F ) in such a way that Z ′ ∩ N ( F ) = ∅ . Notethat this is always possible, since | V \ ( Z ∪ F ) | ≥ k and N ( F ) = ∅ . Then, by Proposition 7, Z ′ is apower dominating set of size i , since it intersects every fort of G except F (because if Z ′ does notintersect some fort F ′ = F , then F ′ ( F , contradicting that F is minimum) and Z ′ intersects N [ F ].7owever, Z ′ is not zero forcing since it does not intersect F . This contradicts the assumption that z ( G ; i ) = p ( G ; i ).Note that the condition “ G is connected” in Theorem 13 is necessary, since, e.g., 0 = p ( K ˙ ∪ K ; 3) = 3 = z ( K ˙ ∪ K ; 3) = (cid:0) (cid:1) . Moreover, the analogous statement for the dominationpolynomial does not hold, since, e.g., 0 = p ( S ; 1) = 1 = d ( S ; 1) = (cid:0) (cid:1) . This fact, along withTheorems 11, 12, and 13 shows that in general, the power domination polynomial of a graph coin-cides more often (both partially and completely) with its domination polynomial than with its zeroforcing polynomial. In this section, we present several results about computing the power domination polynomial of agraph in terms of the power domination polynomials of smaller graphs.
Proposition 14. If G is a graph such that G ≃ G ˙ ∪ G , then P ( G ; x ) = P ( G ; x ) P ( G ; x ) .Proof. A power dominating set of size i in G consists of a power dominating set of size i in G and a power dominating set of size i = i − i in G . Since power dominating sets of size i and i can be chosen independently in G and G for each i ≥ γ P ( G ), i ≥ γ P ( G ), andsince p ( G ; i ) p ( G ; i ) = 0 for each i < γ P ( G ) or i < γ P ( G ), it follows that p ( G ; i ) = P i + i = i p ( G ; i ) p ( G ; i ). The left-hand-side of this equation is the coefficient of x i in P ( G ; x ),and since P ( G ; x ) = P n ( G ) j = γ P ( G ) p ( G ; j ) x j and P ( G ; x ) = P n ( G ) j = γ P ( G ) p ( G ; j ) x j , the right-hand-side of the equation is the coefficient of x i in P ( G ; x ) P ( G ; x ). Thus, P ( G ; x ) P ( G ; x ) and P ( G ; x )have the same coefficients and the same degree, so they are identical. Corollary 15.
Let G be a graph on n vertices. Then p ( G ˙ ∪ K ; i ) = p ( G ; i − for each i ∈ [ n + 1] ,and P ( G ˙ ∪ K k ; x ) = x k P ( G ; x ) for all k ∈ N . Lemma 16.
For any n , n ∈ N , n + n X i =1 X i ,i ∈ N i + i = i (cid:18) n i (cid:19)(cid:18) n i (cid:19) x i = (( x + 1) n − x + 1) n − . Proof.
Let A and B be disjoint sets with | A | = n and | B | = n . The number of ways to choose asubset of A ∪ B of size i which intersects both A and B is X i ,i ∈ N i + i = i (cid:18) n i (cid:19)(cid:18) n i (cid:19) . Counting another way, this quantity is equal to (cid:18) n + n i (cid:19) − (cid:18) n i (cid:19) − (cid:18) n i (cid:19) , (cid:0) n i (cid:1) of the (cid:0) n + n i (cid:1) subsets of A ∪ B of size i contain vertices only from A , and (cid:0) n i (cid:1) containvertices only from B . Then, n + n X i =1 X i ,i ∈ N i + i = i (cid:18) n i (cid:19)(cid:18) n i (cid:19) x i = n + n X i =1 (cid:18)(cid:18) n + n i (cid:19) − (cid:18) n i (cid:19) − (cid:18) n i (cid:19)(cid:19) x i = n + n X i =1 (cid:18) n + n i (cid:19) x i − n X i =1 (cid:18) n i (cid:19) x i − n X i =1 (cid:18) n i (cid:19) x i = (( x + 1) n + n − − (( x + 1) n − − (( x + 1) n − x + 1) n − x + 1) n − , where the second and third equalities follow from the convention that (cid:0) ab (cid:1) = 0 if a < b and from thebinomial theorem. Theorem 17.
Let G and G be graphs on n and n vertices, respectively. For j ∈ { , } , let I j equal the number of isolates of G j if n j > , and equal zero if n j = 1 . Let G = G ∨ G . Then, P ( G ; x ) = (1 + I /x ) P ( G ; x ) + (1 + I /x ) P ( G ; x ) + (( x + 1) n − x + 1) n − . Proof.
The power dominating sets of G can be partitioned into those which intersect both V ( G )and V ( G ), and those which intersect only one of V ( G ) or V ( G ). Since each vertex of G isadjacent to each vertex of G and vice versa, any set of vertices which intersects both V ( G ) and V ( G ) is a power dominating set of G . Moreover, there are X i ,i ∈ N i + i = i (cid:18) n i (cid:19)(cid:18) n i (cid:19) such sets of size i , for any i ∈ [ n + n ]. For j ∈ { , } , a set S ⊂ V ( G j ) is a power dominating setof G if and only if one of the following conditions holds:1. S is a power dominating set of G j .2. I j ≥ S is a power dominating set of G j − v , where v is an isolate of G j .Note that in the second case, S is a power dominating set because V ( G ) \ V ( G j ) will be coloredin the domination step by some vertex other than v , then N [ S ] ∩ V ( G j ) will force any uncoloredvertices of G j − v , and finally v can be forced by any vertex in V ( G ) \ V ( G j ). For any i ∈ [ n + n ],there are p ( G j ; i ) power dominating sets of G j of size i , and p ( G j ; i + 1) power dominating sets of G j − v of size i for each isolate v of G j (recall that p ( G j ; k ) = 0 if k > n j ). Therefore, P ( G ; x ) = n + n X i =1 X i ,i ∈ N i + i = i (cid:18) n i (cid:19)(cid:18) n i (cid:19) + p ( G ; i ) + I p ( G ; i + 1) + p ( G ; i ) + I p ( G ; i + 1) x i = (( x + 1) n − x + 1) n −
1) + P ( G ; x ) + P ( G ; x ) ++ I ( P ( G ; x ) /x − p ( G ; 1)) + I ( P ( G ; x ) /x − p ( G ; 1))= (1 + I /x ) P ( G ; x ) + (1 + I /x ) P ( G ; x ) + (( x + 1) n − x + 1) n − , j ∈ { , } , I j p ( G j ; 1) = 0 since if I j >
0, then then G j has at least two connected components,so p ( G j ; 1) = 0. Corollary 18.
Let G be a graph on n vertices, and I equal the number of isolates of G if n > ,and equal zero if n = 1 . Let G ′ be the graph obtained from G by adding a dominating vertex. Then P ( G ′ ; x ) = (1 + I/x ) P ( G ; x ) + x ( x + 1) n . Theorem 19.
Let H be a graph with vertex set { v , . . . , v n } . For ≤ i ≤ n , let G i be a graph oforder n i with p ( G i ; 1) = n i , and let u i be any vertex of G i , except, if G i is a path, u i is not anendpoint of the path. Let G be the graph obtained by identifying u i and v i for ≤ i ≤ n . Then, P ( G ; x ) = n Y i =1 (( x + 1) n i − . Proof.
Let G ′ = ˙ S ni =1 G i . We will show that S is a power dominating set of G ′ if and onlyif S is a power dominating set of G . By Propositions 2 and 14, it will follow that P ( G ; x ) = Q ni =1 (cid:16)P n i j =1 (cid:0) n i j (cid:1) x j (cid:17) = Q ni =1 (( x + 1) n i − S be a power dominating set of G ′ . Then S contains at least one vertex of V ( G i ), 1 ≤ i ≤ n .Moreover, for 1 ≤ i ≤ n , since u i is the only vertex of G [ V ( G i )] which may have neighbors outside G [ V ( G i )] in G , S ∩ V ( G i ) will power dominate in G [ V ( G i )] as it does in G ′ [ V ( G i )] until u i iscolored, at which point the forcing could possibly stop. However, since forcing will proceed in each G [ V ( G i )], all vertices u i in G will eventually be colored by the respective S ∩ V ( G i ). Then, allneighbors of each u i outside G [ V ( G i )] will be colored, and forcing will resume in each G [ V ( G i )] asin G ′ [ V ( G i )]. Thus, S will power dominate all of G .Now let S be a power dominating set of G . Suppose there is some j ∈ [ n ] for which S ∩ V ( G j ) = ∅ .Since S ⊂ V ( G ) \ V ( G j ), by Observation 1, V ( G ) \ V ( G j ) is a power dominating set of G . Then, byTheorem 10, N [ V ( G ) \ V ( G j )] ∩ V ( G j ) is a zero forcing set of G [ V ( G j )]. If v j is an isolate in H ,then N [ V ( G ) \ V ( G j )] ∩ V ( G j ) = ∅ , so S cannot be a power dominating set of G , a contradiction.Otherwise, N [ V ( G ) \ V ( G j )] ∩ V ( G j ) = { u j } . However, the only graph for which a single vertexis a zero forcing set is a path, and this happens only when the vertex is an endpoint of the path.This contradicts our assumption that if G i is a path, u i is not an endpoint of the path. Thus, S ∩ V ( G j ) = ∅ , so S contains at least one vertex x i of V ( G i ) for all i ∈ [ n ]. Then, since p ( G i ; 1) = n i , { x i } is a power dominating set of G ′ [ V ( G i )] for all i , so S is a power dominating set of G ′ . P ( G ; x ) for specific graphs In this section, we give closed-form expressions and algorithms to compute the power dominationpolynomials of several families of graphs, as well as some characterizations of uniqueness. We havealso implemented a brute force algorithm for computing the power domination polynomials ofarbitrary graphs (cf. https://github.com/rsp7/Power-Domination-Polynomial ), and used it tocompute the power domination polynomials of all graphs on fewer than 10 vertices.10 roposition 20. P ( K n ; x ) = P ( P n ; x ) = P ( C n ; x ) = P ( W n ; x ) = ( x + 1) n − .2. P ( K n ; x ) = x n .3. P ( S n ; x ) = x ( x + 1) n − + x n − + ( n − x n − , for n ≥ .4. P ( H ◦ K k ; x ) = (( x + 1) k +1 − n for any graph H of order n , and k > .Proof.
1. It is easy to see that any vertex of K n , P n , C n , or W n forms a power dominating set of size1. Then, by Observation 1, every set of vertices of these graphs is a power dominating set. Itfollows that P ( K n ; x ) = P ( P n ; x ) = P ( C n ; x ) = P ( W n ; x ) = P ni =1 (cid:0) ni (cid:1) x i = ( x + 1) n − K n is the set containing all of its vertices, so P ( K n ; x ) = x n .3. Since S n ≃ K n − ∨ K , by Corollary 18 it follows that P ( S n ; x ) = (1 + ( n − /x ) P ( K n − ; x ) + x ( x + 1) n − = x n − + ( n − x n − + x ( x + 1) n − .4. This follows by Theorem 19, where G i ≃ K k +1 for 1 ≤ i ≤ n . Note that since k >
1, each( k + 1)-clique whose vertex is identified with a vertex of H is different from a path (i.e., it isnot K or K ).A graph G = ( V, E ) is a threshold graph if there exists a real number t and a function w : V → R such that uv ∈ E if and only if w ( u ) + w ( v ) ≥ t . For a binary string B , the threshold graphgenerated by B , written T ( B ), is the graph whose vertices are the symbols in B , and which has anedge between a pair of symbols x and y with x to the left of y if and only if y = 1. It was shown byChv´atal and Hammer [22] that every threshold graph is generated by some binary string, and thatfor a particular threshold graph, this string is unique apart from the first symbol (since changingthe first symbol from 0 to 1, or from 1 to 0, would not affect T ( B )). Thus we will refer to B and T ( B ) interchangeably, where symbols of B correspond to vertices in T ( B ).A block of a binary string B is a maximal contiguous substring consisting either of only 0s oronly 1s. The partition of B into its blocks is called the block partition of B , and when the blockpartition of B has ω blocks, we label the blocks B i , 1 ≤ i ≤ ω , and write B = B B . . . B ω . Each B i consisting of 0s is called a 0-block, and each B i consisting of 1s is a 1-block. Similarly, we willcall a vertex in a 0-block a 0-vertex, and a vertex in a 1-block a 1-vertex. We will refer to | B i | as b i . We will assume that T ( B ) is connected, i.e. that B ω is a 1-block. Note that if B ω is a1-block, then by Corollary 15, P ( T ( B B . . . B ω B ω +1 ); x ) = x b ω +1 P ( T ( B B . . . B ω ); x ). We willalso assume without loss of generality that B has at least two symbols, and that the first symbol of B is the same as the second symbol, so that b ≥
2. We now give a characterization of the powerdominating sets of a threshold graph, and an algorithm to efficiently compute the coefficients of thepower domination polynomial. 11 heorem 21.
Let T = T ( B ) be a threshold graph generated by a binary string B . A set S ⊂ V ( T ) is a power dominating set of T if and only if there exists v ∈ S such that a ) All 1-vertices are in N [ v ] . b ) All 0-blocks following or including the block containing v have at most one vertex not in S .Proof. Suppose conditions a ) and b ) are true. By condition a ), all 1-vertices are in N [ v ] and willbe colored in the domination step. Let B i be the block with the smallest index which contains anuncolored vertex u after the domination step. Since all 1-vertices are colored, B i is a 0-block; bycondition b ), u is the only uncolored vertex in B i . Then, any 1-vertex in B i +1 can force u . Thisprocess can be repeated until all vertices are colored; thus, S is a power dominating set.Now suppose condition a ) is false. Since all 1-vertices are adjacent to each other, S contains no1-vertices; moreover, since for each vertex v in B , all 1-vertices are in N [ v ], S does not contain anyvertices from B . The vertices in B are therefore not in S and not adjacent to any vertex in S , sothey can only be colored in a forcing step. However, since any vertex not in B is adjacent to all ornone of the vertices in B , and since | B | ≥
2, these vertices can never be forced.Finally, suppose condition a ) is true and b ) is false. Then, for any v ∈ S satisfying condition a ),some 0-block following or including the block containing v has at least two vertices not in S . Let B i be the block with the largest index which contains a vertex of S satisfying condition a ). B i = B ω ,since then condition b ) would be true. Thus, some 0-block B j with j ≥ i has at least two verticesnot in S , and S does not contain vertices of any 1-block B k with k > j . Then, the vertices in B j which are not in S cannot be colored by the domination step; they also cannot be forced, since anyvertex which is adjacent to one of them is adjacent to all of them. Theorem 22.
Let T = T ( B ) be a threshold graph generated by a binary string B . The coefficientsof P ( T ; x ) can be computed in O ( n ) time with Algorithm 1.Proof. Let B = B B . . . B ω and for i ∈ [ ω ], let T i = T ( B B . . . B i ); thus, T = T ω . For 2 ≤ i ≤ ω ,if B i is a 0-block, then T i = T i − ˙ ∪ K b i , so by Proposition 14, P ( T i ; x ) = x b i P ( T i − ; x ) . (1)For 2 ≤ i ≤ ω , if B i is a 1-block, then T i = T i − ∨ K b i , so by Theorem 17 (and since by assumption, b > P ( T i ; x ) = P ( T i − ; x ) + b i − x P ( T i − ; x ) + ( x + 1) b + ... + b i − ( x + 1) b + ... + b i − . (2)Algorithm 1 begins with an all-zero array of coefficients for P ( T ; x ). In the first if-statement, if B is a 1-block, the coefficients of P ( T ; x ) are computed by the binomial expansion of ( x + 1) b . In thesecond if-statement, if B is a 0-block, by our assumption T has at least one 1-block; thus, P ( T ; x )is computed by first setting “ a b ←
1” for P ( T ; x ) = x b , and then setting “ a b − ← b ” for b x P ( T ; x ), adding the binomial expansion of ( x + 1) b + b , and subtracting the binomial expansionof ( x + 1) b as in (2).After the two initial if-statements, the while-loop evaluates P ( T i ; x ) and P ( T i +1 ; x ) for i ≤ ω − i equal to either 2 or 3 depending on whether B was a 1-block or a 0-block; in eithercase, the first block (if any) to be evaluated by the while-loop is a 0-block, followed by a 1-block.12 lgorithm 1: Finding coefficients of P ( T ( B ); x ) Input:
Binary string B = B . . . B ω ; Output:
Values a , . . . , a n such that P ( T ( B ); x ) = P ni =1 a i x i ; for i = 1 to n do a i ← if B is a 1-block thenfor j = 1 to b do a j ← (cid:0) b j (cid:1) ; i ← if B is a 0-block then a b ← a b − ← b ; for j = 1 to b + b do a j ← a j + (cid:0) b + b j (cid:1) ; for j = 1 to b do a j ← a j − (cid:0) b j (cid:1) ; i ← while i ≤ ω − do s ← b + . . . + b i − ; for j = s + b i to b i + 1 do a j ← a j − b i ; for j = 1 to b i do a j ← i ← i + 1; s ← b + . . . + b i − ; for j = 1 to s − do a j ← a j + a j +1 b i − ; for j = 1 to s + b i do a j ← a j + (cid:0) s + b i j (cid:1) ; for j = 1 to s do a j ← a j − (cid:0) sj (cid:1) ; i ← i + 1; return a , . . . , a n ;The first two for-loops in the while-loop shift the indices of the coefficients of P ( T i − ; x ) to the rightby b i , which is equivalent to multiplying P ( T i − ; x ) by x b i to obtain P ( T i ; x ) as in (1). Then, i isincremented as a 1-block is to be added. The third for-loop adds to each coefficient a j the coefficientto the right of a j multiplied by b i − ; this is equivalent to adding b i − x P ( T i − ; x ) to P ( T i − ; x ) as in(2). The last two for-loops respectively add the binomial expansion of ( x + 1) b + ... + b i and subtract13he binomial expansion of ( x + 1) b + ... + b i − , as in (2). Thus, by the end of the last for-loop, P ( T i ; x )is computed from P ( T i − ; x ) as in (2), and then i is incremented. When i = ω + 1, the while-loopterminates, and the last computed polynomial is P ( T ω ; x ) = P ( T ; x ); its coefficients a , . . . , a n arereturned by the algorithm.To verify the runtime, note that the first for-loop and the two if-statements in Algorithm 1 caneach be evaluated in O ( n ) time; the while-loop executes O ( ω ) = O ( n ) times, with each for-loopinside it taking O ( n ) time. Thus, the total runtime is O ( n ). P -unique graphs In this section, we identify several families of graphs which can be recognized by their power domina-tion polynomials; more precisely, we identify families F such that if G ∈ F , then P ( H ; x ) = P ( G ; x )implies H ≃ G . Let us call families of graphs satisfying this property P -unique . We also identifyarbitrarily large sets of graphs which all have the same power domination polynomial and show thatthe power domination polynomial is generally not effective at measuring vertex connectivity. Theorem 23.
Let G be a graph.1. G is P -unique if and only if G ˙ ∪ K is P -unique.2. G is P -unique if and only if G ˙ ∪ K is P -unique.3. For any k ≥ , G ˙ ∪ K k is not P -unique.Proof. Note that for any graphs G , G , and G , G ≃ G if and only if G ˙ ∪ G ≃ G ˙ ∪ G . Let K be either the graph K or K .Suppose that G is P -unique, and let H be a graph such that P ( G ˙ ∪ K ; x ) = P ( H ; x ). By Corollary6, G ˙ ∪ K and H have the same number of K -components; in particular, H = H ′ ˙ ∪ K for some H ′ .Using Proposition 14, we have that P ( G ; x ) = P ( G ˙ ∪ K ; x ) / P ( K ; x ) = P ( H ; x ) / P ( K ; x ) = P ( H ′ ; x ).Since G is P -unique, this implies that G ≃ H ′ and hence that G ˙ ∪ K ≃ H , which means that G ˙ ∪ K is also P -unique.Suppose G ˙ ∪ K is P -unique and let H be a graph such that P ( G ; x ) = P ( H ; x ). By Proposition14, we have that P ( G ˙ ∪ K ; x ) = P ( G ; x ) P ( K ; x ) = P ( H ; x ) P ( K ; x ) = P ( H ˙ ∪ K ; x ). Since G ˙ ∪ K is P -unique, this implies that G ˙ ∪ K ≃ H ˙ ∪ K and hence that G ≃ H , which means that G is also P -unique.By Propositions 14 and 20, for k ≥ P ( G ˙ ∪ K k ) = P ( G ˙ ∪ P k ) but G ˙ ∪ K k G ˙ ∪ P k . Corollary 24. K n is P -unique.2. ˙ S ki =1 K is P -unique, for all k ∈ N .3. (cid:16) ˙ S ki =1 K (cid:17) ˙ ∪ K ℓ is P -unique, for all k, ℓ ∈ N . heorem 25. S n is P -unique for n ≥ .Proof. Since γ P ( S n ) = 1, it suffices to show that if H is a graph on n vertices such that γ P ( H ) = 1and H S n , then P ( H ; x ) differs from P ( S n ; x ) in at least one coefficient. Let { v } be a powerdominating set of H . Since any power dominating set of H must contain at least one vertex fromeach connected component, H is connected and hence v has at least one neighbor. If v has exactlyone neighbor, then the fact that { v } is power dominating implies that H ≃ P n , in which case p ( H ; 1) = n > p ( S n ; 1) by Proposition 20.Hence, suppose that v has at least two neighbors. If some neighbor u of v is adjacent to a vertex w that v is not adjacent to, then V ( H ) \ { u, v, w } is a power dominating set of H , since some otherneighbor of v will dominate v , and then v can force u and u can force w . Otherwise, since n ≥ H is connected but not a star, v has at least three neighbors, two of which, say u and w , areadjacent. If x is a third neighbor of v , then V ( H ) \ { u, v, x } is a power dominating set of H , since w can dominate v and u , and v can force x . In each of these two cases, each of the (cid:0) n − n − (cid:1) subsetsof V ( H ) including v and excluding any three vertices is power dominating by Observation 1. Notethat p ( S n ; n −
3) = (cid:0) n − n − (cid:1) by Proposition 20, so p ( H ; n − > p ( S n ; n − P ( H ; x ) = P ( S n ; x ).Note that the condition “ n ≥
4” in Theorem 25 is necessary, since for the degenerate case n = 3, P ( S ; x ) = P ( K ; x ) but S K . Proposition 26.
1. There exist arbitrarily large sets of graphs that all have the same power domination polynomial.2. A connected and a disconnected graph can have the same power domination polynomial.3. Graphs with vertex connectivity , , and n − can have the same power domination polyno-mial.Proof.
1. Adding a single chord to C n , n ≥
4, produces a graph in which any set of size 1 is a powerdominating set. Thus, by Observation 1 and Proposition 20, all such graphs have the samepower domination polynomial as C n . Since the number of distinct (up to isomorphism) waysto add a chord to C n is ⌊ n/ ⌋ −
1, this yields arbitrarily large sets of graphs with the samepower domination polynomial.2. Let G be the graph obtained by connecting two cycles by a single edge e . S is a powerdominating set of G if and only if it contains at least one vertex from each cycle; the same istrue for G − e . Thus, P ( G ; x ) = P ( G − e ; x ).3. By Proposition 20, P n , C n , and K n all have the same power domination polynomial. We define a power domination root of a graph G to be a root of P ( G ; x ). In this section, we studyvarious properties of power domination roots. We begin with the following basic facts.15 roposition 27. Let G be a graph. Then:1. Zero is a power domination root of G of multiplicity γ P ( G ) .2. G cannot have positive power domination roots.3. G may have complex power domination roots.4. If r is a real rational power domination root of G , then r is an integer.Proof.
1. This follows from the fact that the first nonzero coefficient of P ( G ; x ) corresponds to x γ P ( G ) .2. P ( G ; x ) is a non-constant polynomial with nonnegative coefficients, and hence the derivativeof P ( G ; x ) is a nonzero polynomial with nonnegative coefficients. Thus P ( G ; x ) is strictlyincreasing on (0 , ∞ ), and P ( G ; 0) = 0, so it cannot have positive roots.3. By Proposition 20, K n , n ≥
3, has complex power domination roots, since there exist complex n th roots of unity for n ≥
3. For example, P ( K ; x ) = x + 4 x + 6 x + 4 x has complex roots − ± i .4. This follows from the Rational Root Theorem and the fact that p ( G ; n ) = 1.We will now characterize graphs having a small number of distinct power domination roots. Thenext observation follows immediately from Propositions 20 and 27. Observation 28.
A graph G has exactly one distinct power domination root if and only if G ≃ K n . Definition 2 ([51]) . Let F be the family of graphs obtained from connected graphs H by addingtwo new vertices v ′ and v ′′ , two new edges vv ′ and vv ′′ , and possibly the edge v ′ v ′′ , for each vertex v of H . Theorem 29 ([51]) . If G = ( V, E ) is a connected graph of order n ≥ , then γ P ( G ) ≤ n/ withequality if and only if G ∈ F ∪ { K , } . Theorem 30.
A graph G has exactly two distinct power domination roots if and only if G ≃ G ˙ ∪ . . . ˙ ∪ G k ˙ ∪ K r , where k ≥ , r ≥ , and G i ≃ P for ≤ i ≤ k . Moreover, if G has exactly twopower domination roots, these roots are and − .Proof. If G ≃ G ˙ ∪ . . . ˙ ∪ G k ˙ ∪ K r , where k ≥ r ≥
0, and G i ≃ P for 1 ≤ i ≤ k , then by Proposition14 and Proposition 20, P ( G ; x ) = x r ( x + 2 x ) k ; thus, G has exactly two distinct power dominationroots: 0 and − G has exactly two distinct power domination roots. Let G ′ be the largestcomponent of G (by number of vertices), and let n ′ = | V ( G ′ ) | . Clearly n ′ >
1, since otherwise G ≃ K n and G would have only one power domination root by Observation 28.Suppose for contradiction that n ′ ≥
3. Since 0 is a power domination root of G ′ of multiplicity γ P ( G ′ ), we have that P ( G ′ ; x ) = x p ( x + a ) n ′ − p for some p ∈ [ n ′ −
1] and a > x n ′ − in x p ( x + a ) n ′ − p is ( n ′ − p ) a . Since G ′ is connected and hasmore than one vertex, it has no isolates, so it follows from Corollary 6 that ( n ′ − p ) a = n ′ . Since16 ′ and p are integers, a is rational, and by Proposition 27, a is an integer. Moreover, since p > a ≥
2. Since the lowest-order term in P ( G ′ ; x ) involves x p , we have that γ P ( G ′ ) = p .By Theorem 29, p = n ′ ( a − a ≤ n ′ , which implies that a ≤ /
2, a contradiction. Thus it cannothold that n ′ ≥
3, so n ′ = 2. Then, G ≃ G ˙ ∪ . . . ˙ ∪ G k ˙ ∪ K r , for some k ≥ r ≥
0, and G i ≃ P for1 ≤ i ≤ k . Theorem 31.
A graph G has exactly three distinct power domination roots if and only if G ≃ G ˙ ∪ . . . ˙ ∪ G k ˙ ∪ K r , where k ≥ , r ≥ , and G i ∈ F for ≤ i ≤ k . Moreover, if G has exactly threepower domination roots, these roots are , − √ i , and − −√ i .Proof. If G ≃ G ˙ ∪ . . . ˙ ∪ G k ˙ ∪ K r , where k ≥ r ≥
0, and G i ∈ F for 1 ≤ i ≤ k , then G ˙ ∪ . . . ˙ ∪ G k can be viewed as a graph obtained from a graph H by identifying a vertex of K or a non-endpointvertex of P to each vertex of H . Thus, G ˙ ∪ . . . ˙ ∪ G k satisfies the conditions of Theorem 19; then,by Proposition 14 and Proposition 20, P ( G ; x ) = x r ( x + 3 x + 3 x ) k , and therefore G has exactlythree distinct power domination roots: 0, − √ i , and − −√ i .Now suppose that G has exactly three distinct power domination roots. If all components of G have at most 2 vertices, then by Theorem 30, G has at most two distinct power domination roots.Thus, let G ′ be a component of G with n ′ ≥ G ′ cannot have fewer than three distinct power domination roots. Thus, G ′ must have exactly 3 distinctpower domination roots, so P ( G ′ ; x ) = x j ( x − a ) k ( x − b ) l for some j, k, l ∈ N with j + k + l = n ′ and some a, b ∈ C \{ } with a = b . Using the coefficient of x n ′ − in P ( G ′ ; x ) and Corollary 6, wehave that (cid:18) k (cid:19) a + (cid:18) l (cid:19) b + klab = p ( G ′ ; n ′ −
2) = (cid:18) n ′ (cid:19) . (3)Using the coefficient of x n ′ − in P ( G ′ ; x ), Corollary 6, and the fact that G ′ has no isolates, we alsohave that − ( ka + lb ) = p ( G ′ ; n ′ −
1) = n ′ . (4)Because P ( G ′ ; x ) is a monic polynomial with integer coefficients, a is an algebraic integer and henceits minimal polynomial A ( x ) over Q is a monic irreducible polynomial with integer coefficients. Notethat by minimality, A ( x ) cannot have 0 as a root. Since A ( x ) divides P ( G ′ ; x ), since irreduciblepolynomials with rational coefficients are separable, and since P ( G ′ ; x ) has three distinct roots, wehave that either A ( x ) = x − a or A ( x ) = x + rx + s for some r, s ∈ Z . Thus, we consider two cases:1. Suppose that A ( x ) = x + rx + s . Since a , b , and 0 are roots of P ( G ′ ; x ), and A ( x ) divides P ( G ′ ; x ), and 0 is not a root of A ( x ), b must be a root of A ( x ). Thus, A ( x ) must also bethe minimal polynomial of b over Q . Thus P ( G ′ ; x ) = x j ( x + rx + s ) k , which implies that j + 2 k = n ′ . By Theorem 29, we have that j = γ P ( G ′ ) ≤ n ′ /
3, which implies that k ≥ n ′ / a + b ∈ Z , since A ( x ) is monic and r ∈ Z . Since − k ( a + b ) = n ′ > − ( a + b ) ∈ { , , } . We consider three subcases:(a) If a + b = −
1, then k = n ′ and hence j = − n ′ , a contradiction.(b) If a + b = −
2, then k = n ′ / j = 0, a contradiction.(c) If a + b = −
3, then γ P ( G ) = j = k = n ′ /
3. By Theorem 29, G ′ ∈ F , since P ( K , ; x ) = x + 6 x + 15 x + 20 x + 15 x , which has five distinct roots.17. Suppose that A ( x ) = x − a ; then, the minimal polynomial of b over Q must be x − b , and − a, − b ∈ N by Proposition 27. Since n ′ ≥
3, by Theorem 29 we have j ≤ n ′ /
3, which impliesthat k + l ≥ n ′ / > n ′ /
2. If a ≤ − b ≤ −
2, then − ( ka + lb ) > n ′ , a contradiction to (4).Therefore without loss of generality, we can assume that a = −
1. Then by (4), k − lb = n ′ .On the other hand, by (3), (cid:0) k (cid:1) + (cid:0) l (cid:1) b − klb = (cid:0) n ′ (cid:1) . Thus k − k + ( l − l ) b − klb = n ′ − n ′ .This implies that ( k − lb ) − k − lb = n ′ − n ′ . Substituting n ′ for k − lb , we get k + lb = n ′ ,but since k − lb = n ′ , we have that b = 0 or b = −
1. But then P ( G ′ ; x ) would only have twodistinct roots: − G ′ of G has at least 3 vertices, G ′ ∈ F . Suppose G also has a K -component. Then G would have − G must have a component G ′ withat least 3 vertices, G ′ must be in F , and hence by Theorem 19, G ′ has power domination roots0, − √ i , and − −√ i ; by Proposition 14, it follows that G would have at least 4 distinct powerdomination roots, a contradiction. Thus, all components of G are either isolates or graphs in F .Note that it can be verified in polynomial time whether a graph G is isomorphic to G ˙ ∪ . . . ˙ ∪ G k ˙ ∪ K r for some k ≥ r ≥
0, and G i ∈ F for 1 ≤ i ≤ k .We now identify regions of the complex plane in which no power domination roots can exist. Theorem 32 (Rouch´e’s Theorem) . Let f , g , and h be analytic functions on a region Ω , and let C be a simple closed connected curve (i.e., a closed curve that does not intersect itself ) in Ω . If f ( z ) = g ( z ) + h ( z ) in Ω and | h ( z ) | > | g ( z ) | on C , then on the set enclosed by C , f and h have thesame number of roots (counting repeated roots multiple times). Corollary 33.
Let f , g , and h be polynomials such that f ( z ) = g ( z ) + h ( z ) , and on a circle ofradius r centered at in the complex plane, | h ( z ) | > | g ( z ) | . Then, on the disk of radius r centeredat , f and h have the same number of roots. Note that in the case that h ( z ) is a non-zero constant, f ( z ) has no roots on the disk in question. Theorem 34.
Let G be a graph. Let a be a positive real number, and let f ( G ; a ) = P ni =1 p ( G ; i ) a i P ni =1 P nk = i p ( G ; k ) (cid:0) ki (cid:1) a k − i . If a + bi is a root of P ( G ; x ) , then | b | ≥ min n f ( G ; a ) , ( f ( G ; a )) n o .Proof. Define the polynomial P ( G ; y ) = n X j =1 p ( G ; j )( a + y ) j , which is obtained by shifting P ( G ; x ) to the left by a . If P ( G ; x ) has a complex root with realpart a > P ( G ; y ) has a purely imaginary root bi (which is non-zero since the power dominationpolynomial has no real positive roots). Then, P ( G ; y ) = n X j =1 p ( G ; j ) j X k =0 (cid:18) jk (cid:19) a j − k y k = n X k =0 n X j = k p ( G ; j ) (cid:18) jk (cid:19) a j − k y k , a + y ) j and the second equalityfollows from moving p ( G ; j ) into the sum, interchanging the order of summation, and from the factthat p ( G ; 0) = 0. Hence, for 0 ≤ k ≤ n , the coefficient of y k in P ( G ; y ) is c k = n X j = k p ( G ; j ) (cid:18) jk (cid:19) a j − k . By Corollary 33 (with f = P ( G ; y ), g = P nk =1 c k y k , and h = c ), if c > P nk =1 c k y k on | y | = r , then P ( G ; y ) has no roots inside a disk of radius less than r . Let r ∗ = sup { r : c > P nk =1 c k r k } . Then,the root bi of P ( G ; y ) satisfies | b | ≥ r ∗ . We will now show that r ∗ ≥ min { f ( G ; a ) , ( f ( G ; a )) n } .Define A ( r ) := c P nk =1 c k r k for r >
0, and note that r ∗ = sup { r : A ( r ) > } . Note also that A ( r )is strictly monotonically decreasing as r increases (since r > A (1) <
1, then r ∗ < A (1) >
1, then r ∗ >
1; if A (1) = 1 then r ∗ = 1. Finally, note that A (1) = f ( G ; a ). We will nowconsider 3 cases. Case 1: A (1) = 1. Then f ( G ; a ) = ( f ( G ; a )) n = 1 = r ∗ . Case 2: A (1) <
1. Since r ∗ <
1, for any r ≤ r ∗ , if c P nk =1 c k r >
1, then c P nk =1 c k r k >
1. Thus, f ( G ; a ) = c P nk =1 c k = sup { r : c P nk =1 c k r > } ≤ sup { r : c P nk =1 c k r k > } = r ∗ . Case 3: A (1) >
1. If r ≥
1, then c P nk =1 c k r n > c P nk =1 c k r k >
1. If r <
1, since A (1) > A is strictly monotonically decreasing, A ( r ) >
1. So, for 0 < r < c P c k r n > c P c k r k > r > f ( G ; a )) n = (cid:18) c P nk =1 c k (cid:19) n = sup { r : c P nk =1 c k r n > } ≤ sup { r : c P nk =1 c k r k > } = r ∗ . We now characterize complex power domination roots in an expression that is independent of thegraph.
Corollary 35.
Let G be a connected graph. Let a be a positive real number, and let f ( a ) = P ni = ⌈ n ⌉ (cid:0) n −⌈ n/ ⌉ i −⌈ n/ ⌉ (cid:1) a i P ni =1 P nk = i (cid:0) nk (cid:1)(cid:0) ki (cid:1) a k − i . If a + bi is a root of P ( G ; x ) , then | b | > min { f ( a ) , ( f ( a )) n } .Proof. If G has fewer than 3 vertices, P ( G ; x ) could only have the real roots 0 and −
2. Thus, n ≥ γ P ( G ) ≤ n . Moreover any superset of a power dominating set is also powerdominating, and p ( G ; k ) ≤ (cid:0) nk (cid:1) . Thus, f ( a ) ≤ f ( G ; a ) from Theorem 34, and the result follows. In this paper, we introduced the power domination polynomial of a graph in order to study theenumeration problem associated with power domination. We explored various structural proper-ties of P ( G ; x ), related it to other graph polynomials, characterized P ( G ; x ) for specific familiesof graphs, and analyzed some properties of power domination roots. We now offer several openquestions about the power domination polynomial.19 uestion 1. For any graph G , is P ( G ; x ) unimodal?There is some evidence the answer to Question 1 is affirmative, e.g. as seen in Propo-sition 4. We have also computationally verified that the power domination polynomialsof all graphs on fewer than 10 vertices are unimodal; computer code can be found at https://github.com/rsp7/Power-Domination-Polynomial .Another direction for future work is to derive conditions which guarantee that a polynomial P is or is not the power domination polynomial of some graph. For instance, Corollary 35 gave anecessary condition for this using the complex roots of the polynomial. In addition, it would beinteresting to find other families of graphs which are uniquely identified by their power dominationpolynomials. The following question related to P -unique graphs (and inspired by Theorem 23) couldalso be investigated. Question 2.
Given two P -unique graphs G and G , when is G ˙ ∪ G P -unique? What other graphoperations preserve P -uniqueness?It would also be interesting to characterize or count all power dominating sets (or at least allminimum power dominating sets) of some other nontrivial families of graphs such as trees andgrids. In Proposition 20, the power domination polynomial of the corona of any graph with K k , k > k = 1 appears to be more difficult as Theorem19 does not apply directly; this motivates the following question: Question 3.
For any graph H , is there an efficient way to compute P ( H ◦ K ; x )?A graph polynomial f ( G ; x ) satisfies a linear recurrence relation if f ( G ; x ) = P ki =1 g i ( x ) f ( G i ; x ),where the G i ’s are obtained from G using vertex or edge elimination operations, and the g i ’s are fixedrational functions. For example, the chromatic polynomial P ( G ; x ) satisfies the deletion-contractionrecurrence P ( G ; x ) = P ( G − e ; x ) − P ( G/e ; x ). Similarly, a splitting formula for a graph polynomial f ( G ; x ) is an expression for f ( G ; x ) in terms of the polynomials of certain subgraphs of G ; severalsuch formulas were derived in Section 3.2. In view of this, it would be interesting to investigate thefollowing question: Question 4.
Are there linear recurrence relations for P ( G ; x ), or splitting formulas for P ( G ; x )based on cut vertices or separating sets?Answering these questions would be useful for computational approaches to the problem; in par-ticular, a linear recurrence relation would allow the power domination polynomial of a graph to becomputed recursively. References [1] A. Aazami. Domination in graphs with bounded propagation: algorithms, formulations andhardness results.
Journal of Combinatorial Optimization , 19(4): 429–456, 2010.[2] AIM Special Work Group. Zero forcing sets and the minimum rank of graphs.
Linear Algebraand its Applications , 428(7):1628–1648, 2008.[3] S. Akbari, S. Alikhani, and Y.-H. Peng. Characterization of graphs using domination polyno-mials.
European Journal of Combinatorics , 31(7):1714–1724, 2010.204] S. Akbari and M. R. Oboudi. On the edge cover polynomial of a graph.
European Journal ofCombinatorics , 34(2):297–321, 2013.[5] S. Akhlaghi, N. Zhou, and N. E. Wu. PMU placement for state estimation considering mea-surement redundancy and controlled islanding. In
Power and Energy Society General Meeting(PESGM) , pp. 1–5, 2016.[6] S. Alikhani and Y.-H. Peng. Introduction to domination polynomial of a graph.
Ars Combi-natoria , 114:257–266, 2014.[7] F. Aminifar, M. Fotuhi-Firuzabad, M. Shahidehpour, and A. Khodaei. Probabilistic multistagePMU placement in electric power systems.
IEEE Transactions on Power Delivery , 26(2):841–849, 2011.[8] T. Baldwin, L. Mili, M. Boisen, and R. Adapa. Power system observability with minimal phasormeasurement placement.
IEEE Transactions on Power Systems , 8(2):707–715, 1993.[9] K. F. Benson, D. Ferrero, M. Flagg, V. Furst, L. Hogben, V. Vasilevska, and B. Wissman.Zero forcing and power domination for graph products.
Australasian Journal of Combinatorics ,70(2): 221, 2018.[10] J. A. Bondy and U. S. R. Murty.
Graph Theory with Applications , volume 290. MacmillanLondon, 1976.[11] K. Boyer, B. Brimkov, S. English, D. Ferrero, A. Keller, R. Kirsch, M. Phillips, and C. Reinhart.The zero forcing polynomial of a graph. arXiv:1801.08910 , 2018.[12] C. Bozeman, B. Brimkov, C. Erickson, D. Ferrero, M. Flagg, and L. Hogben. Restricted powerdomination and zero forcing problems. arXiv:1711.05190 , 2017.[13] B. Brimkov, C. C. Fast, and I. V. Hicks. Computational approaches for zero forcing and relatedproblems. arXiv:1704.02065 , 2017.[14] B. Brimkov and I. V. Hicks. Complexity and computation of connected zero forcing.
DiscreteApplied Mathematics , 229: 31–45, 2017.[15] B. Brimkov, D. Mikesell, and L. Smith. Connected power domination in graphs. arXiv:1712.02388 , 2017.[16] J. I. Brown and J. Tufts. On the roots of domination polynomials.
Graphs and Combinatorics ,30(3):527–547, 2014.[17] D. J. Brueni.
Minimal PMU placement for graph observability: a decomposition approach . PhDthesis, Virginia Tech, 1993.[18] D. J. Brueni and L. S. Heath. The PMU placement problem.
SIAM Journal on DiscreteMathematics , 19(3):744–761, 2005.[19] T. H. Brylawski. A decomposition for combinatorial geometries.
Transactions of the AmericanMathematical Society , 171:235–282, 1972. 2120] D. Burgarth and V. Giovannetti. Full control by locally induced relaxation.
Physical ReviewLetters , 99(10):100501, 2007.[21] Y. Caro and R. Pepper. Dynamic approach to k-forcing.
Theory and Applications of Graphs ,2(2): 2015.[22] V. Chv´atal and P. L. Hammer. Aggregation of inequalities in integer programming.
Annals ofDiscrete Mathematics , 1:145–162, 1977.[23] N. Dean, A. Ilic, I. Ramirez, J. Shen, and K. Tian. On the power dominating sets of hypercubes.In , pp. 488–491. IEEE, 2011.[24] G. Deepak and N. Soner. Connected domination polynomial of a graph.
International Journalof Mathematical Archive , 4(11):90–96, 2014.[25] F. M. Dong, M. D. Hendy, K. L. Teo, and C. H. Little. The vertex-cover polynomial of a graph.
Discrete Mathematics , 250(1-3):71–78, 2002.[26] P. Dorbec and S. Klavˇzar. Generalized power domination: propagation radius and Sierpi´nskigraphs.
Acta Applicandae Mathematicae
SIAM Journal on Discrete Mathematics , 22(2): 554–567, 2008.[28] M. Dorfling and M. A. Henning, A note on power domination in grid graphs.
Discrete AppliedMathematics , 154(6): 1023–1027, 2006.[29] J. A. Ellis-Monaghan and C. Merino. Graph polynomials and their applications II: Interrelationsand interpretations. In
Structural Analysis of Complex Networks , pp. 257–292, 2011.[30] C. C. Fast.
Novel Techniques for the Zero-Forcing and p-Median Graph Location Problems .PhD thesis, Rice University, 2017.[31] D. Ferrero, L. Hogben, F. H. J. Kenter, and M. Young. Note on power propagation time andlower bounds for the power domination number.
Journal of Combinatorial Optimization , 34(3):736–741, 2017.[32] H. Hajiabolhassan and M. Mehrabadi. On clique polynomials.
Australasian Journal of Com-binatorics , 18:313–316, 1998.[33] P. Hall. On representatives of subsets.
Journal of the London Mathematical Society , 10(1):26–30, 1935.[34] T. W. Haynes, S. M. Hedetniemi, S. T. Hedetniemi, and M. A. Henning. Domination in graphsapplied to electric power networks.
SIAM Journal on Discrete Mathematics , 15(4):519–529,2002.[35] C. Hoede and X. Li. Clique polynomials and independent set polynomials of graphs.
DiscreteMathematics , 125(1-3):219–228, 1994. 2236] S. Jahari and S. Alikhani. Domination polynomials of k-tree related graphs.
InternationalJournal of Combinatorics , 2014, 2014.[37] J. Kneis, D. M¨olle, S. Richter, and P. Rossmanith. Parameterized power domination complexity.
Information Processing Letters , 98(4): 145–149, 2006.[38] T. Kotek, J. Preen, F. Simon, P. Tittmann, and M. Trinks. Recurrence relations and splittingformulas for the domination polynomial.
Electronic Journal of Combinatorics , 19(3), 2012.[39] Q. Li, T. Cui, Y. Weng, R. Negi, F. Franchetti, and M. D. Ilic. An information-theoreticapproach to PMU placement in electric power systems.
IEEE Transactions on Smart Grid ,4(1):446–456, 2013.[40] C. S. Liao. Power domination with bounded time constraints.
Journal of Combinatorial Opti-mization , 31(2): 725–742, 2016.[41] A. Mahari and H. Seyedi. Optimal PMU placement for power system observability using BICA,considering measurement redundancy.
Electric Power Systems Research , 103: 78–85, 2013.[42] N. M. Manousakis, G. N. Korres, and P. S. Georgilakis. Taxonomy of PMU placement method-ologies.
IEEE Transactions on Power Systems , 27(2):1070–1077, 2012.[43] L. Mili, T. Baldwin, and A. Phadke. Phasor measurements for voltage and transient stabil-ity monitoring and control. In
Workshop on Application of advanced mathematics to PowerSystems, San Francisco , 1991.[44] J. Peng, Y. Sun, and H. Wang. Optimal PMU placement for full network observability usingTabu search algorithm.
International Journal of Electrical Power & Energy Systems , 28(4):223–231, 2006.[45] D. D. Row. A technique for computing the zero forcing number of a graph with a cut-vertex.
Linear Algebra and its Applications , 436(12): 4423–4432, 2012.[46] L. Smith.
Fort Neighborhoods: A Set Covering Formulation for Power Domination in Graphs .Masters thesis, Rice University, 2018.[47] R. Sodhi, S. Srivastava, and S. Singh. Multi-criteria decision-making approach for multi-stageoptimal placement of phasor measurement units.
IET Generation, Transmission & Distribution ,5(2):181–190, 2011.[48] R. P. Stanley. Acyclic orientations of graphs. In
Classic Papers in Combinatorics , pp. 453–460.Springer, 1987.[49] N. Xia, H. B. Gooi, S. X. Chen, and M. Q. Wang. Redundancy based PMU placement in stateestimation.
Sustainable Energy, Grids and Networks , 2: 23–31, 2015.[50] G. Xu, L. Kang, E. Shan, and M. Zhao, Power domination in block graphs.
Theoretical Com-puter Science , 359: 299–305, 2006.[51] M. Zhao, L. Kang, and G. J. Chang. Power domination in graphs.