open k-monopolies in graphs: complexity and related concepts · open k-monopolies in graphs:...
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Open k-monopolies in graphs: complexity and relatedconcepts
Dorota KuziakDepartament d’Enginyeria Informatica i Matematiques
Universitat Rovira i VirgiliAv. Paısos Catalans 26, 43007 Tarragona, Spain
Iztok Peterin∗
University of Maribor, FEECSSmetanova 17, 2000 Maribor, Slovenia
Ismael G. Yero†
Departamento de MatematicasEscuela Politecnica Superior, Universidad de Cadiz
Av. Ramon Puyol s/n, 11202 Algeciras, [email protected]
Abstract
Closed monopolies in graphs have a quite long range of applications in several problemsrelated to overcoming failures, since they frequently have some common approaches aroundthe notion of majorities, for instance to consensus problems, diagnosis problems or votingsystems. We introduce here open k-monopolies in graphs which are closely related to differentparameters in graphs. Given a graph G = (V,E) and X ⊆ V , if δX(v) is the number ofneighbors v has in X, k is an integer and t is a positive integer, then we establish in thisarticle a connection between the following three concepts:
• Given a nonempty set M ⊆ V a vertex v of G is said to be k-controlled by M ifδM (v) ≥ δV (v)
2 + k. The set M is called an open k-monopoly for G if it k-controls everyvertex v of G.
∗Supported by the Ministry of Science of Slovenia under the grant P1-0297. The author is also with IMFM,Jadranska 19, 1000 Ljubljana, Slovenia.†The research was partially done while the author was at University of Maribor, Slovenia, supported by “Ministe-
rio de Educacion, Cultura y Deporte”, Spain, under the “Jose Castillejo” program for young researchers. Referencenumber: CAS12/00267.
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• A function f : V → {−1, 1} is called a signed total t-dominating function for G iff(N(v)) =
∑v∈N(v) f(v) ≥ t for all v ∈ V .
• A nonempty set S ⊆ V is a global (defensive and offensive) k-alliance in G if δS(v) ≥δV−S(v) + k holds for every v ∈ V .
In this article we prove that the problem of computing the minimum cardinality of an open0-monopoly in a graph is NP-complete even restricted to bipartite or chordal graphs. Nev-ertheless, for the case of trees the problem becomes polynomial and we give an algorithm forcomputing the open 0-monopoly number of a tree in polynomial time. In addition we presentsome general bounds for the minimum cardinality of open k-monopolies and we derive someexact values.
Key words: open k-monopolies; k-signed total domination; global defensive k-alliance; globaloffensive k-alliance
2010 Mathematics Subject Classification: 05C85; 05C69; 05C07
1 Introduction
We begin stating some terminology and notation which we will use. Throughout this article, Gdenotes a simple graph with vertex set V (G) and edge set E(G) (we will use only V and E if thegraph is clear from the context). The order of G is n = |V (G)| and the size is m = |E(G)|. Wedenote two adjacent vertices u and v by u ∼ v. Given a vertex v ∈ V, the set N(v) = {u ∈ V :u ∼ v} is the open neighborhood of v, and the set N [v] = N(v)∪ {v} is the closed neighborhood ofv. So, the degree of a vertex v ∈ V is δ(v) = |N(v)|. Given a set S ⊂ V , the open neighborhood ofS is N(S) =
⋃v∈S N(v) and the closed neighborhood of S is N [S] = N(S)∪S. The minimum and
maximum degree of G are denoted by δ(G) and ∆(G), respectively (again we use δ and ∆ for shortif G is clear from the context). For a nonempty set S ⊆ V and a vertex v ∈ V , NS(v) denotesthe set of neighbors v has in S, i.e., NS(v) = S ∩N(v). The degree of v in S will be denoted byδS(v) = |NS(v)|. Also, S = V − S is the complement of a set S in V and ∂S = N [S] − S is theboundary of a set S. The subgraph of G induced by a set S is denoted by 〈S〉.
In the first article, [19], on closed monopolies in graphs (called as monopolies there) thefollowing terminology was used. A vertex v in G is said to be controlled by a set M ⊂ V if at leasthalf of its closed neighborhood is in M . The set M is called a closed monopoly if it controls everyvertex v of G. Equivalently, the set M is a closed monopoly in G, if for any vertex v ∈ V (G) it
follows that |N [v] ∩M | ≥⌈|N [v]|
2
⌉. In this article, we introduce open k-monopolies in a natural
way, by replacing closed neighborhoods with open neighborhoods. Hence, we can use the degreeof vertices instead of cardinalities of closed neighborhoods. Given some integer k, a vertex v of Gis said to be k-controlled by a set M if δM(v) ≥ δ(v)
2+ k. Analogously, the set M is called an open
k-monopoly if it k-controls every vertex v of G. Notice that not for every value of k there existsan open k-monopoly in G (further on we give some suitable interval for such k). Also, note that,close and open monopolies cannot be exactly compared, since in a closed monopoly a vertex v alsocounts itself in controlling v, which is not the case in any open monopoly. The smallest exampleis already K2, where is only one vertex in a closed monopoly, but both vertices are necessary in
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an open 0-monopoly. Differently, there are only two vertices in a minimum open 0-monopoly ofP5, while we need at least three vertices in every closed monopoly of P5. In this article, we arefocused only in open k-monopolies. In this sense, from now on we omit the term “open” andjust use the terminology of k-monopolies. On the other hand, we remain using the term closedmonopoly whenever referring to some previous work on this topic.
According to the article [1], several problems related to overcoming failures have some com-mon approaches around the notion of majorities. Their ideas are directed toward decreasing, asmuch as possible, the damage caused due to failed vertices; by maintaining copies of the mostimportant data and performing a voting process among the participating processors in situationthat failures occur; and by adopting as true those data stored at the majority of the not failedprocessors. This idea is also commonly used in some fault tolerant algorithms including agree-ment and consensus problems [3], diagnosis problems [27] or voting systems [10], among otherapplications and references.
The authors of [1] were interested into locality based on the following facts. Frequently,processors running in a system are better aware of whatever happens in their neighborhood thanoutside of it. Moreover, some distributed network models allow only for computations developedwith local processors, which means that, a processor can only obtain a data from other processorshaving a “relative” close distance from itself. Therefore, it is more efficient to store data as locallyas possible.
Nevertheless, there could exists also a risk in this way. If the voting is restricted to localneighborhoods, we could produce a sufficiently large set of failures which will probably constitutethe majority in some of these neighborhoods. In this sense, in [1] the authors assert the following:once the voting is performed over subsets of vertices, the ability of failed vertices to influence theoutcome of the votes becomes not only a function of their number but also a function of theirlocation in the network: well situated vertices can acquire greater influence. This simple fact ledthem to study the problem of characterizing the potential power of a set of failures in a networkof processors, and as a consequence, the study of (closed) monopolies in graphs.
Notions of closed monopolies in graphs were introduced first in [19], where several ideasregarding voting systems were described. Once such article appeared, a high number of researcheswere devoted to such parameter and its relationship with other similar structures like (defensiveand offensive) alliances [15] or signed dominating functions [2], among other works. An interestingarticle, where several of these connections are dealt with, is [6]. Moreover, such article presentsa possible generalization of all these (closed) monopolies-related structures which comprise themaltogether. The complexity of closed monopolies in graphs is also well studied. The NP-hardnessof finding the minimum cardinality of a closed monopoly in a graph is easy to observe as statedin [19]. There was also posted a conjecture concerning the inapproximation of such problem. Aweaker version of such conjecture has been proved in [22]. In addition some other inapproximationresults of this problem have appeared in [20, 21]. Particularly, in [21], the inapproximation resultsare centered in regular graphs. Moreover, there is also proved that for the case of tree graphs, aclosed monopoly of minimum cardinality can be computed in linear time. On the other hand, in[16], some relationships and bounds for the minimum cardinality of closed monopolies in graphs arestated in terms of matchings and/or girths. Also, dynamic closed monopolies has been introducedin connection with modeling some problems of spreading the influence in social networks (see
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[1, 23]). Other studies in dynamic closed monopolies can be found in [9, 28].
2 Concepts related to monopolies
Many times mathematical concepts are defined independently in two or even more papers. Whenthis occurs, the equivalence sometimes is obvious (mostly when papers occur in the same timeperiod), but sometimes we need more effort to find the connection (mostly when there is a longertime period between publications). This may yield not sufficient effort of later authors with thehistory, but we rather present it as an enough important concept to start to investigate it fromdifferent point of view.
The above holds (at least partial) for signed (total) domination (introduced first in 1995 in[11] (2001 in [29])) and for different types of alliances (introduced first in 2004 in [15]). We addmonopolies to this list and present these connections in this section.
2.1 Alliances
Alliances in graphs were introduced first in [15] and generalized to k-alliances in [25, 26]. After thatseveral works have been developed in this topic. Remarkable cases are [4, 12]. Relationships withdifferent parameters of the graphs have been obtained and the alliances of several families of graphshave been studied. A nonempty set S ⊆ V is a defensive k-alliance in G for k ∈ {−∆, . . . ,∆} iffor every v ∈ S
δS(v) ≥ δS(v) + k. (1)
Moreover, for k ∈ {2 − ∆, . . . ,∆}, a nonempty set S ⊆ V is an offensive k-alliance in G if forevery v ∈ ∂S
δS(v) ≥ δS(v) + k. (2)
A nonempty set S ⊆ V is a powerful k-alliance if S is a defensive k-alliance and an offensive(k + 2)-alliance. A set D is a dominating set in G if every vertex outside of D is adjacent to atleast one vertex of G.
A (defensive, offensive or powerful) k-alliance is called global if it is a dominating set. Theglobal defensive (offensive) k-alliance number of G, denoted by γdk(G) (γok(G)), is defined as theminimum cardinality of a global defensive (offensive) k-alliance in G. For k ∈ {−∆, ...,∆−2}, theglobal powerful k-alliance number of G, denoted by γpk(G), is defined as the minimum cardinality ofa global powerful k-alliance in G. A global powerful alliance of minimum cardinality in G is calleda γpk(G)-set of G. Notice that there exist graphs not containing any global powerful k-alliancefor some specific values of k. In this sense, in this work we are interested in those graphs havingglobal powerful k-alliances. It means that whenever we study such an alliances we are supposingthat the graph contains it.
Notice that the terminology used for alliances provides a very useful tool which can be usedwhile proving several results, i.e., a set of vertices M is a k-monopoly in G if and only if forevery vertex v of G, δM(v) ≥ δM(v) + 2k (from now we will call this expression the k-monopolycondition) and we will say that M is a k-monopoly in G if and only if every v of G satisfies thek-monopoly condition for M .
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An interesting possible generalization of alliances in graphs (and some other related param-eters) is given in [6]. In this work is proposed a new framework, which the authors call (D,O)-alliances. The main idea of this allows not only to characterize several known variants of al-liances, but also suggest a unifying framework for its study. In this sense, a (D,O)-alliance, withD,O ⊆ Z in a graph G = (V,E) is a set S such that for any v ∈ S, δS(v) − δS(v) ∈ D and forany v ∈ N(S) \ S, δS(v) − δS(v) ∈ O. According to this, it is clear to observe that a defensivek-alliance can be understood as a ({z ∈ Z : z ≥ k},Z)-alliance, and an offensive k-alliance as a(Z, {z ∈ Z : z ≥ k})-alliance.
2.2 Signed (total) domination
Given a graph G = (V,E) and a function f : V → {−1, 1} we consider the following for f :
• f is a signed dominating function for G if f(N [v]) =∑
u∈N [v] f(u) ≥ 1, for all v ∈ V .
• f is a signed total dominating function for G if f(N(v)) =∑
u∈N(v) f(u) ≥ 1, for all v ∈ V .
• f is a signed k-dominating function for G if f(N [v]) ≥ k for all v ∈ V .
• f is a signed total k-dominating function for G if f(N(v)) ≥ k for all v ∈ V .
The minimum weight∑
v∈V f(v) of a signed (total) (k-dominating) dominating function f is thesigned (total) (k-domination) number of G and they are denoted in the following way.
signed domination signed total domination signed k-domination signed total k-dominationγs(G) γst(G) γks (G) γkst(G)
Notice that, if k = 1, then a signed (total) 1-dominating function is a standard signed (total)dominating function for G. Also, any kind of signed (total) (k-dominating) dominating function fof G induces two disjoint sets of vertices B1 and B−1, such that for every vertex v ∈ Bi, f(v) = iwith i ∈ {−1, 1}. Hereby we will represent such a function f by the sets B1 and B−1 inducedby f and we will write f = (B1, B−1). A signed (total) (k-dominating) dominating function f ofminimum weight is called a γ-function with γ ∈ {γs(G), γst(G), γks (G), γkst(G)}, respectively.
2.3 Connections between concepts
Observing the definitions of monopoly and alliance we see that both concepts are closely related.That is, let M be a 0-monopoly in G = (V,E) and let v ∈ V . Hence, v has at least half of its
neighbors in M , i.e., δM(v) ≥ δ(v)2
, which leads to δM(v) ≥ δM(v). Since this is satisfied for everyvertex of G we obtain that M is a global defensive 0-alliance and also a global offensive 0-alliance.On the contrary, let A be a global defensive 0-alliance which is also a global offensive 0-alliancein G. Hence, for every vertex u ∈ V we have that δA(v) ≥ δA(v), which leads to δA(v) ≥ δ(v)
2.
Therefore, A is a 0-monopoly.In [25] was defined the concept of global powerful k-alliances. Nevertheless, it was not taken
into account the possibility of studying the cases in which a set is a global defensive k-alliance and
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also a global offensive k-alliance. According to the concept of monopoly we observe the importanceof such a case, which is one of our motivations to develop the present investigation.
We continue with a relationship between signed total domination, alliances and monopolies.
Theorem 1. Let G = (V,E) be a graph and let k ∈{
1, . . . ,⌊δ(G)
2
⌋}be an integer. The following
statements are equivalent:
(i) M ⊂ V is a k-monopoly in G;
(ii) M is a global defensive (2k)-alliance and a global offensive (2k)-alliance in G;
(iii) f = (B1 = M,B−1 = M) is a signed total (2k)-dominating function for G.
Moreover, if k = 0, then (i) and (ii) are also equivalent.
Proof. The equivalence between (i) and (ii) is straightforward since for every set of vertices M
and every vertex v of G, the conditions δM(v) ≥ δ(v)2
+ k and δM(v) ≥ δM(v) + 2k are equivalent
for every k ∈{
0, . . . ,⌊δ(G)
2
⌋}.
Let M be a global defensive (2k)-alliance and a global offensive (2k)-alliance in G. Let thefunction f : V → {−1, 1} such that for any v ∈ V , it follows f(v) = 1 if v ∈ M and, f(v) = −1otherwise. If v ∈M , then since M is a global defensive (2k)-alliance in G, we have that
f(N(v)) = f(NM(v)) + f(NM(v))
= δM(v)− δM(v)
≥ δM(v) + 2k − δM(v)
= 2k.
Now, if v ∈M , then by using that M is a global offensive (2k)-alliance in G, the same computationas above gives that f = (B1 = M,B−1 = M) is a signed total (2k)-dominating function for G.
On the other hand, let f ′ = (B′1, B′−1) be a signed total (2k)-dominating function for G. Let
M ′ = B′1 and let the vertex u ∈ V . If u ∈ M ′, then since f ′ is a signed total (2k)-dominatingfunction in G, we have that
δM ′(u) = f ′(NM ′(u))
= f ′(N(u))− f ′(NM ′(u))
≥ 2k − f ′(NM ′(u))
= δM ′(u) + 2k.
Thus, M ′ is a global defensive (2k)-alliance in G. Finally, since f ′ is a signed total (2k)-dominatingfunction for G, if u ∈M ′, then as above we deduce that M ′ is a global offensive (2k)-alliance.
The following corollary is a direct consequence of Theorem 1 (ii) and (iii). We omit the proof.
Corollary 2. Let G = (V,E) be a graph and let k ∈{
1, . . . ,⌊δ(G)
2
⌋}be an integer. A set
M ⊂ V is a global defensive k-alliance and a global offensive k-alliance in G if and only iff = (B1 = M,B−1 = M) is a signed total k-dominating function for G.
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Now we prove a connection between signed domination and powerful alliances.
Theorem 3. Let G = (V,E) be a graph and let k ∈ {0, . . . , δ(G)}. Then S ⊂ V is a globalpowerful k-alliance in G if and only if f = (B1 = S,B−1 = S) is a signed (k + 1)-dominatingfunction for G.
Proof. Let S be a global powerful k-alliance in G. So, S is a global defensive k-alliance and aglobal offensive (k + 2)-alliance in G. Let f = (B1 = S,B−1 = S) be a function in G and letv ∈ V . We consider the following cases.
Case 1: v ∈ S. Since S is a global defensive k-alliance in G, we have that
f(N [v]) = f(NS(v)) + f(NS(v)) + 1
= δS(v)− δS(v) + 1
≥ δS(v) + k − δS(v) + 1
= k + 1.
Case 2: v ∈ S. Since S is a global offensive (k + 2)-alliance in G, we have that
f(N [v]) = f(NS(v)) + f(NS(v))− 1
= δS(v)− δS(v)− 1
≥ δS(v) + k + 2− δS(v)− 1
= k + 1.
Thus, f = (B1 = S,B−1 = S) is a signed (k + 1)-dominating function for G.On the other hand, let f ′ = (B′1, B
′−1) be a signed (k + 1)-dominating function in G. We will
show that A = B′1 is a global powerful k-alliance in G. Let u ∈ V . We consider the following.
Case 3: u ∈ A. Since f ′ is a signed (k + 1)-dominating function for G, we have that
δA(u) = f ′(NA(u))
= f ′(N [u])− f ′(NA(u))− 1
≥ k + 1− f ′(NA(u))− 1
= δA(u) + k.
Thus, A is a global defensive k-alliance in G.
Case 4: u ∈ A. Since A is a signed (k + 1)-dominating function in G, we have that
δA(u) = f ′(NA(u))
= f ′(N [u])− f ′(NA(u)) + 1
≥ k + 1− f ′(NA(u)) + 1
= δA(u) + k + 2
Thus, A is a global offensive (k+2)-alliance and, as a consequence, A is a global powerful k-alliancein G. Therefore, the proof is complete.
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Corollary 4. For any graph G of order n and any integer k ∈ {0, . . . , δ(G)},
γk+1s (G) = 2γpk(G)− n.
Proof. Let S be a γpk(G)-set. By Theorem 3, f = (B1 = S,B−1 = S) is a signed total (k + 1)-dominating function of minimum weight in G. Thus γk+1
s (G) = |S| − |S|. Since |S|+ |S| = n andγpk(G) = |S|, the result follows by adding these two equalities above.
According to the above ideas we can resume the relationships which motivated our work inthe following table.
k-monopoly (k ≥ 0) ⇔ Global defensive (2k)-alliance andglobal offensive (2k)-alliance
k-monopoly (k ≥ 1) ⇔ Signed total (2k)-domination
Signed total k-domination (k ≥ 1) ⇔ Global defensive k-alliance andglobal offensive k-alliance
Signed (k + 1)-domination (k ≥ 0) ⇔Global defensive k-alliance andglobal offensive (k + 2)-alliance(A global powerful k-alliance)
Notice that the definition of signed (total) k-dominating function is restricted to k ≥ 1 whilek-alliances are defined for any k ∈ {−∆(G), . . . ,∆(G)} and k-monopolies can be defined for someinteger k whose limits are presented further. In this sense, these concepts being quite similarbetween them could be generalized for k being zero or negative. To obtain a meaningful negativelower bound for k-monopolies we involve another well-known concept: total domination1. Namely,every k-monopoly, k ≥ 0 is also a total dominating set for G. To remain this property also fork < 0, we need to demand δM(v) ≥ 1 for every v ∈ V (G). Therefore in this work we propose thefollowing definition of monopolies and we study some of its mathematical properties.
Given a integer k ∈{
1−⌈δ(G)
2
⌉, . . . ,
⌊δ(G)
2
⌋}and a set M , a vertex v of G is said to be
k-controlled by M if δM(v) ≥ δ(v)2
+ k. The set M is called a k-monopoly if it k-controls everyvertex v of G. The minimum cardinality of any k-monopoly is the k-monopoly number and it isdenoted byMk(G). A monopoly of cardinalityMk(G) is called aMk(G)-set. In particular noticethat for a graph with a leaf (vertex of degree one), there exist only 0-monopolies and the neighborof every leaf is in each M0-set.
Notice that every non trivial graph G contains at least one k-monopoly, with k ∈{
1−⌈δ(G)
2
⌉,
. . . ,⌊δ(G)
2
⌋}, since every vertex of G satisfies the k-monopoly condition for the whole vertex set
V (G). Also, if G has an isolated vertex,Mk(G) does not exists. But if G has no isolated vertices,then, sinceMk(G)-set is also a total dominating set, we haveMk(G) ≥ 2. Thus, we can say thatin general for any graph G of order n, 2 ≤Mk(G) ≤ n.
The last result of this section reveals a connection between M1−d δ(G)
2 e(G) and γt(G).
1A set D is a total dominating set in a graph G if every vertex of G is adjacent to a vertex of D. The minimumcardinality of a total dominating set is the total domination number, denoted by γt(G).
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Theorem 5. For any r-regular graph G,
M1−d r2e(G) = γt(G).
Proof. Let q = r2
+ 1 −⌈r2
⌉and let M be a M1−d r2e(G)-set. If r is even, then q = 1 and if r
is odd, then q = 12. In both cases, for any vertex v of G, δM(v) ≥ 1, since δM(v) is an integer.
Hence M is a total dominating set and M1−d r2e(G) ≥ γt(G). If A is a γt(G)-set, then for every
vertex v ∈ V we obtain δA(v) ≥ 1 ≥ q, since q ∈ {12, 1}. Thus, A is also a (1 −
⌈r2
⌉)-monopoly
and M1−d r2e(G) ≤ γt(G), which yields the equality.
3 Complexity
Studies about complexity of signed domination were first presented by Hattingh et al. in [11].After that Henning has shown in [14] that signed total domination problem is NP-complete evenrestricted to bipartite or chordal graphs. This last work was continued in [18], where the NP-completeness of signed (total) k-domination problem was shown for k ≥ 2. Consequently, byTheorem 1 the k-monopoly problem is also NP-complete for every k ≥ 1. Hence, it remains
to investigate the complexity of k-monopolies for 1 −⌈δ(G)
2
⌉≤ k ≤ 0. As mentioned in the
introduction, the complexity and also several inapproximation results [20, 21, 22, 23] are knownfor a closed monopoly.
On the other hand, also the global defensive k-alliance problem is NP-complete (unpublishedmanuscript [5]) as well as global offensive k-alliance problem (see [8]), but not both together.Notice that global powerful k-alliance problem is NP-complete as shown in [7] but, as we mentionbefore, a global powerful k-alliance is a global defensive k-alliance and a global offensive (k + 2)-alliance. Here we follow a similar approach as in [11] to show that 0-monopoly problem is NP-complete. We will show the polynomial time reduction on the total domination set problem:
Problem: TOTAL DOMINATION SET (TDS)INSTANCE: A graph G and a positive integer k ≤ |V (G)|.QUESTION: Is γt(G) ≤ k?
Problem: 0-MONOPOLYINSTANCE: A graph G and a positive integer k ≤ |V (G)|.QUESTION: Is M0(G) ≤ k?
Recall that the total domination set problem is NP-complete even when restricted to bipartitegraphs (see [17]) or to chordal graphs (see [24]).
Theorem 6. Problem 0-MONOPOLY is NP-complete, even when restricted to bipartite or chordalgraphs.
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Proof. It is obvious that 0-monopoly is a member of NP since for a given set M with |M | ≤ k wecan check in polynomial time for each vertex v of a graph G if v is controlled by M .
Let G be a graph of order n and size m. We construct a graph H from G as follows. For everyvertex v add δG(v) − 1 paths on five vertices and connect v with an edge to every middle vertexof these paths. Hence to obtain H from G we added 5
∑v∈V (G)(δG(v) − 1) = 10m − 5n vertices
and the same amount of edges. (Notice that we have added exactly 4m− 2n leaves.) Clearly thiscan be done in polynomial time. Also, if G is bipartite or chordal graph, so is H. Next we claimM0(H) = 6m− 3n+ γt(G).
To prove this, let M be a 0-monopoly of H. Let v1v2v3v4v5 be an arbitrary path added to G.Clearly v2, v4 ∈M , since they are unique neighbors of v1 and v5, respectively. Moreover, if v3 is notin M , then both v1 and v5 must be in M to control v2 and v4, respectively. Since M has minimumcardinality, this implies that v3 ∈ M . Let v ∈ V (G). By the above, v has δG(v)− 1 neighbors inM outside of G. Since δH(v) = 2δG(v)− 1, v needs an additional neighbor in M ∩ V (G) = P tobe controlled by M . Hence, P forms a total dominating set of G and so γt(G) ≤ |P |. Altogether
M0(H) = |M | = |P |+ 3∑
v∈V (G)
(δG(v)− 1) ≥ γt(G) + 6m− 3n.
On the other hand, suppose S is a γt(G)-set of G. We will show that M = S ∪ {v ∈V (H)− V (G) : δH(v) > 1} is a 0-monopoly for H. Every vertex v ∈ V (H) with δH(v) = 1 has aneighbor of degree two which is in M . Without loss of generality, every vertex v ∈ V (H)− V (G)with δH(v) = 2 has one neighbor of degree 1 and the other neighbor which is in M and we have1 = δM(v) ≥ δM(v) = 1. Every other vertex v ∈ V (H) − V (G) has degree three, and two of itsneighbors are in V (H) − V (G) with degree two and thus they are in M . Hence 2 = δM(v) ≥δM(v) = 1. It remains to check vertices from V (G). Let v be a vertex with degH(v) = 2 degG(v)−1.Since S is a γt(G) set, v has at least one neighbor in S and additional δG(v) − 1 vertices in Min V (H)− V (G). Altogether v has at least δG(v) neighbors in M , which is more than half of itsneighbors. The next calculation ends the proof of the claim:
M0(H) ≤ |M | = |S|+ |{v ∈ V (H)− V (G) : δH(v) > 1}|= γt(G) + 3
∑v∈V (G)
(δG(v)− 1)
= γt(G) + 6m− 3n.
Therefore, we have that if j = 6m − 3n + k, then γt(G) ≤ k if and only if M0(H) ≤ j andthe proof is completed.
Once studied the complexity of finding a 0-monopoly in a graph, it remains to investigate the
complexity for 1−⌈δ(G)
2
⌉≤ k ≤ −1, which we leave as an open problem.
4 0-monopolies in trees
According to the section above the problem of computing the 0-monopoly number is an NP-complete problem. Nevertheless, when restricted to trees, it becomes polynomial as we show at
10
next. We must recall that finding a closed monopoly of minimum cardinality in a tree graph can bedone in linear time, as proved in [21]. In this sense, it is perhaps not surprising to think that com-puting a k-monopoly in a tree is also of linear or polynomial order. However, this is actually not atrivial task and we next give a polynomial time algorithm to compute the k-monopoly number of
a tree T . Note that the suitable interval for k in the case of a tree T is{
1−⌈δ(T )
2
⌉, . . . ,
⌊δ(T )
2
⌋},
which gives only one possibility, that is k = 0, since δ(T ) = 1. In the algorithm we use thefollowing sets:
M - vertices belonging to a 0-monopoly of T ,O - vertices not in M ,S - supports of T (vertices adjacent to a leaf).
We use a data structure called parent array. By choosing the root r as an arbitrary vertexthat is not a leaf, the function p(v) yields the parent of v with respect to r (that is the neighborof v on the path to r). Functions n(v) and l(v) give lists of vertices that are neighbors of v andneighbors of v which are leaves, respectively. First we order V (T ) by a BFS-algorithm startingin r and we create a list L with respect to a BFS-ordering. Clearly all supports must be in M ,so that all leaves are controlled by M . Hence we delete all leaves from L. At each step of thealgorithm we use the last vertex v of L, update his neighborhood’s location with respect to theabove sets, and then delete l(v) from T and v from L.
Operation set v ∈ M at the same time removes v from the set O and puts it in M . Thefunction mon(X) gives list of vertices of the set X belonging to M .
Next result shows the correctness of Algorithm 1.
Theorem 7. Set M produced by Algorithm 1 is an optimal 0-monopoly of a tree T . The timecomplexity of Algorithm 1 is O(n∆(T )).
Proof. The time complexity of Algorithm 1 is clearly O(n∆(T )), since we use the function mon(v)for every vertex (that is not a leaf) and in this function we check which neighbors of v are in M .
We will proceed by induction on n. For n ≤ 4 the result can be checked by some simplecalculations. Also, notice that Algorithm 1 produces an optimal 0-monopoly for all paths. Letn ≥ 5 and let T be a tree that is not a path. Let V be the vertex set of T . We represent T asa rooted tree with root r. According to distances between vertices of V and the root r, we willpartition the set V into level sets `1, . . . , `k, for some integer k, where for every vertex x ∈ `i,d(r, x) = i. A vertex v ∈ `i is called a child of its neighbor p(v) in `i−1 and on the contraryv is called the parent p(u) of its neighbors u ∈ `i+1, if they exist. Let M be a set obtained byAlgorithm 1 for the tree T .
Let v ∈ `k−1 be the support in T with which we start Algorithm 1 and let u ∈ `k be a neighborof v. We distinguish three possibilities with respect to the degree of v.
Case 1: δT (v) = 2p, p > 1. Let G = T−{u}. If u /∈M , then M can also be obtained by Algorithm1 for G which is aM0-set for G by induction hypothesis. Otherwise, if u ∈M , then we can obtainM −{u}∪ {w} by Algorithm 1 for G, where w is some degree one neighbor of v not in M . Hence
11
Algorithm 1 0-monopoly of a tree T
Input: a tree T with at least three vertices, its root r with δ(r) > 1, the list of supports SOutput: a 0-monopoly M of T
O = T − S,M = Sorder V (T ) by BFS algorithmL is the list of vertices ordered with respect to BFS-orderingdelete all leaves of T from Lwhile |L| ≥ 2 do
take last vertex v ∈ Lif |mon(n(v))| <
⌈δ(v)
2
⌉then
if p(v) ∈ O thenset p(v) ∈M
end ifwhile |mon(n(v))| <
⌈δ(v)
2
⌉do
take (u ∈ l(v)) && (u ∈ O)set u ∈M
end whileend ifdelete l(v) from T and v from L
end whiletake rwhile |mon(l(r))| <
⌈δ(r)
2
⌉do
take (u ∈ l(r)) && (u ∈ O)set u ∈M
end while
12
M − {u} ∪ {w} is also a M0-set for T .
Case 2: δT (v) = 2p− 1, p ≥ 2. By Algorithm 1 for G = T − {u} we can obtain set M − {w} as aM0-set (by induction hypothesis), where δ(w) = 1, w ∈ N(v) and, in addition, w = u wheneveru ∈ M . There are exactly p − 1 neighbors of v in M − {w} since δG(v) = 2p − 2 and v is thestarting vertex of Algorithm 1. Hence M −{w} is not aM0-set for T since v is not controlled byM − {w}. Therefore M is a M0-set for T .
Case 3: δT (v) = 2. Without loss of generality, we may assume that there is a vertex x withδT (x) > 2 between v and r, otherwise we make the same analysis as follows in a branch withsuch a vertex. We divide this case additionally with respect to the distance from v to the closestvertex x with δT (x) > 2.
Case 3.1: d(v, x) = 1. Let B be the set of neighbors of x from `k−1 which have neighbors in `k.We consider the following subcases.
Case 3.1.1: |B| 6=⌈δT (x)
2
⌉. If |B| >
⌈δT (x)
2
⌉, then M − {v} is one of the possible outcomes of Al-
gorithm 1 for G = T − {u}, which is a M0-set by induction hypothesis. Clearly M − {v} is nota M0-set for T , since u is not controlled by M − {v}. Hence M is a M0-set for T , since v is the
only neighbor of u. Otherwise |B| <⌈δT (x)
2
⌉. Then M is a possible outcome of Algorithm 1 for G
and v is in M to control x. By the induction hypothesis is M aM0-set for G and hence also for T .
Case 3.1.2: |B| =⌈δT (x)
2
⌉. Then M − B − {x} is one of the possible outcomes of Algorithm 1 for
G = (T − NT [NT [x] − p(x)]) ∪ p(x), which is a M0-set by induction hypothesis. Thus, M is aM0-set for T , since B ∪ {x} are necessary to control NT [NT [x]− {p(x)}] ∪ {p(x)}.
Case 3.2: d(v, x) = 2. Let w be the common neighbor of v and x and G = T − {u, v, w}. Clearlyv, w ∈ M since we started our algorithm in v. If δT (x) is an odd number 2p + 1, p ≥ 1, thenM ′ = M − {v, w} is a possible outcome of the Algorithm 1 for G, which is optimal by inductionhypothesis. Hence x has at least p+ 1 neighbors in M in T and is controlled by M . Hence M isa M0-set for T since we need at least two vertices of {u, v, w} in M . Let now δT (x) be an evennumber 2p, p ≥ 2. Note that every neighbor of x different than p(x) that is not a leaf is in M . Ifx has at least p leaves as neighbors, then M ′ is not a M0-set for G, since x is not controlled byM ′. Hence if x has less than p leaves in NT (x), then M ′ is aM0-set for G and can be obtained byAlgorithm 1. This yields that M is aM0-set for T , since we need at least two vertices of {u, v, w}in M . If x has at least p leaves in NT (x), then M ′∪{t} is aM0-set for G that can be obtained byAlgorithm 1, where t is any leaf adjacent to x not contained in M . Since M ′ ∪{t} is aM0-set forG by induction hypothesis, M is a M0-set for T . Namely x has exactly p neighbors in M in T .
Case 3.3: d(v, x) = 3. Let uvwzx be the path from u to x and G = T − {u, v, w, z}. Clearlyv, w ∈ M since we started the algorithm in v. Let A be the set of all neighbors of x in `k−3
which are not leaves, and also have not any descendant in `k. Notice that in particular z /∈ A. By
13
Algorithm 1, A ⊂M . We consider the following subcases.
Case 3.3.1: |A| ≥ δT (x)2
. Hence x is already controlled by A in T and thus, also by M ′ = M−{v, w}in G. Again M ′ is aM0-set for G since M ′ is a possible outcome of Algorithm 1 for G. ThereforeM is a M0-set for T since we need at least two vertices out of {u, v, w} in a 0-monopoly of T .
Case 3.3.2: δT (x)2− 1 ≤ |A| < δT (x)
2. Hence, p(x) ∈ M by Algorithm 1, and the proof is analogous
to Case 3.3.1.
Case 3.3.3: |A| < δT (x)2− 1. By Algorithm 1 M contains some vertices from B = NT (x) − (A ∪
{p(x)}). Let t be a vertex from M∩B. Let first z /∈M . If δT (x) = 2p, p ≥ 2, then M ′ = M−{v, w}is a possible outcome of Algorithm 1 for G, which is optimal by induction hypothesis. Again Mis a M0-set for T . If δT (x) = 2p + 1, p ≥ 1, then M ′ contains p + 1 neighbors of x which is too
much since δG(x) = 2p and |A| < δT (x)2− 1. Hence M ′ cannot be produced by Algorithm 1 for
G, but M ′ − {t} is a possible outcome of Algorithm 1 for G. Clearly M − {t} does not controlx in T and we need also t in a 0-monopoly. Thus M is a M0-set for T . Let now z ∈ M . IfδT (x) = 2p + 1, p ≥ 1, then M ′ = M − {v, w, z} is a possible result of Algorithm 1 for G, whichis optimal by induction hypothesis. Clearly M is a M0-set for T . If δT (x) = 2p, p ≥ 2, then M ′
contains p− 1 neighbors of x which is not enough since δG(x) = 2p− 1 and x is not controlled byM ′ in G. For s ∈ B −M is M ′ ∪ {s} a possible result of Algorithm 1 for G, which is optimal byinduction hypothesis. Again, by the same reasons we get that M is a M0-set for T .
Case 3.4: d(v, x) ≥ 4. Let uvwzy . . . x be the path from u to x and G = T − {u, v, w, z}. Noticethat x can be adjacent to y. Clearly v, w ∈ M and z /∈ M , since we started our algorithm in v.It is easy to see that M ′ = M − {v, w} is a possible outcome of the Algorithm 1 for G, which isaM0-set for G by induction hypothesis. Hence M is an optimal 0-monopoly for T since we needat least two vertices of {u, v, w} in any M0-set of T .
5 Bounding Mk(G)
In this section we present bounds forMk(G) with respect to the minimum and maximum degreesof G and with respect to the order and size. First notice that the k-monopoly condition δM(v) ≥δ(v)
2+ k is equivalent to the following expressions:
δM(v) ≤ δ(v)
2− k. (3)
Theorem 8. Let G be a graph of order n, minimum degree δ and maximum degree ∆. Then for
any integer k ∈{
1−⌈δ(G)
2
⌉, . . . ,
⌊δ(G)
2
⌋},⌈
∆ + 2k + 2
2
⌉≤Mk(G) ≤ n−
⌊δ − 2k
2
⌋.
14
Proof. Let A be a set of vertices of G such that |A| =⌊δ−2k
2
⌋and let v be a vertex of G. Hence
δA(v) ≤⌊δ−2k
2
⌋≤ δ−2k
2. So,
δA(v) ≥ δ(v)− δ − 2k
2≥ δ(v)− δ(v)− 2k
2=δ(v) + 2k
2.
Thus we have 2δA(v) ≥ δ(v) + 2k = δA(v) + δA(v) + 2k, which leads to δA(v) ≥ δA(v) + 2k.Therefore A is a k-monopoly in G and the upper bound follows.
On the other hand, let M be a Mk(G)-set and let u be a vertex of maximum degree in G.By (3) we have that
∆ = δM(u) + δM(u) ≤ δM(u) +δ(u)
2− k = δM(u) +
∆
2− k,
which leads to ∆2
+ k ≤ δM(u). Now, if u ∈ M , then we obtain that ∆2
+ k ≤ δM(u) ≤ |M | − 1and, as a consequence, ∆+2k+2
2≤ |M |. Conversely, if u /∈ M , then ∆
2+ k ≤ δM(u) ≤ |M | which
leads to ∆+2k2≤ |M |. Therefore, |M | ≥ max
{∆ + 2k
2,∆ + 2k + 2
2
}=
∆ + 2k + 2
2and the lower
bound follows.
As the following corollary shows the above bounds are tight.
Corollary 9. For every complete graph Kn and every k ∈{
1−⌈δ(G)
2
⌉, . . . ,
⌊δ(G)
2
⌋},
Mk(Kn) =
⌈n+ 2k + 1
2
⌉.
Proof. From Theorem 8 we have that⌈n+2k+1
2
⌉≤Mk(Kn) ≤ n−
⌊n−2k−1
2
⌋. If n− 2k− 1 is even,
then n+ 2k + 1 is even and we obtain that⌈n+ 2k + 1
2
⌉≤Mk(Kn) ≤ n−
⌊n− 2k − 1
2
⌋= n− n− 2k − 1
2=n+ 2k + 1
2=
⌈n+ 2k + 1
2
⌉.
On the other hand, if n− 2k − 1 is odd, then n+ 2k + 1 is odd and we have that⌈n+ 2k + 1
2
⌉≤Mk(Kn) ≤ n−
⌊n− 2k − 1
2
⌋= n− n− 2k − 2
2=n+ 2k + 2
2=
⌈n+ 2k + 1
2
⌉.
Next we obtain a lower bound for Mk(G) in terms of order and size of G.
Theorem 10. For any graph G of order n and size m and for every k ∈{
1−⌈δ(G)
2
⌉, . . . ,
⌊δ(G)
2
⌋}−
{0},
Mk(G) ≥⌈
3kn−m2k
⌉.
15
Proof. Let M be aMk(G)-set. Since every vertex v ∈M satisfies that δM(v) ≥ δM(v) + 2k ≥ 2k,we have that c(M,M) ≥ 2k|M | = 2k(n − |M |), where c(M,M) is the edge cut set between Mand M . Since δM(v) ≥ δM(v) + 2k holds for every vertex v ∈M , we have
2k|M | ≤ c(M,M)
=∑v∈M
δM(v)
≤∑v∈M
(δM(v)− 2k)
= 2|E(〈M〉)| − 2k|M |,
which leads to |E(〈M〉)| ≥ kn. Since m ≥ |E(〈M〉)|+ c(M,M), we obtain that m ≥ kn+ 2k(n−|M |) = 3kn− 2k|M | and the result follows.
To see the tightness of the above bound we consider the following family F of graphs. Webegin with a complete graph Kt with set of vertices V = {v0, v1, . . . , vt−1} and t− 1 ≡ 0 (mod 4)and t isolated vertices U = {u0, u1, . . . , ut−1}. From now on all the operations with subindexes ofvi or ui are done modulo t. To obtain a graph G ∈ F , for every i ∈ {0, . . . , t − 1}, we add theedges uivi, uivi+1, uivi+2, . . . , uivi+(t−3)/2. Notice that G has order 2t and size t(t− 1) and everyvertex vi ∈ V has t−1
2neighbors in U and vice versa. Hence δ(G) = t−1
2. Suppose k =
⌊t−1
4
⌋.
If v ∈ V , then δV (v) = t − 1 = t−12
+ t−12
= δU(v) + t−12
= δU(v) + 2k. Also if v ∈ U , thenδV (v) = t−1
2= δU(v) + t−1
2= δU(v) + 2k. Thus V is a k-monopoly in G. By Theorem 10 we have
Mk(G) = t and the bound is achieved.
Theorem 11. For any r-regular graph G of order n and for every k ∈{
1−⌈δ(G)
2
⌉, . . . ,
⌊δ(G)
2
⌋},
Mk(G) ≥⌈n(2k + r)
2r
⌉.
Proof. Let V be the vertex set of G and let M be a Mk(G)-set. For any vertex v ∈ V and anyM ⊂ V we have that δ(v) = δM(v) + δM(v). By subtracting 2δM(v) in both sides of the equalitywe obtain δ(v)− 2δM(v) = δM(v)− δM(v). Making a sum for every vertex of G and using the factthat G is r-regular, it follows∑
v∈V
(δM(v)− δM(v)) =∑v∈V
(δ(v)− 2δM(v)) = nr − 2∑v∈V
δM(v) = nr − 2r|M | = r|M | − r|M |.
Thus,∑
v∈V (δM(v)−δM(v)) = r|M |−r|M |. Since every vertex v ∈ V satisfies δM(v) ≥ δM(v)+2k,we have
2kn =∑v∈V
2k ≤∑v∈V
(δM(v)− δM(v)) = r|M | − r|M | = 2r|M | − rn
and the result follows.
As we will see in Proposition 16, the above bound is tight. For instance, it is achieved for thecase of cycles C4t for k = 0.
16
6 Exact values for Mk(G)
As already mentioned, for any graph G of order n, 2 ≤ Mk(G) ≤ n. We first characterize theclasses of graphs achieving the limit cases for these bounds.
Proposition 12. Let G be a graph of order n. Then Mk(G) = 2 if and only if G is isomorphicto P2, P3, P4, C3 or C4. Moreover, k is either 0 or 1.
Proof. If G is isomorphic to P2, P3, P4, C3 or C4, then δ(G) ≤ 2, k ∈ {0, 1} and Mk(G) = 2. Onthe contrary, suppose thatMk(G) = 2. Let S = {u, v} be aMk(G)-set. Notice that u and v mustbe adjacent. So, δS(u) ≤ 1 and δS(v) ≤ 1 and G must contain at most four vertices. Moreover, forevery vertex x /∈ {u, v} it follows δS(x) ≤ 1. Thus, δ(G) ≤ 2, k ∈ {0, 1} and we have the followingcases. If δS(u) = 0 and δS(v) = 0, then G is isomorphic to P2. If δS(u) = 1 and δS(v) = 0 (or viceversa), then G is isomorphic to P3. If δS(u) = 1 and δS(v) = 1, then G is isomorphic either to P4,C3 or C4, which completes the proof.
Proposition 13. Let G be a graph of order n and minimum degree δ. Then Mk(G) = n if andonly if k =
⌊δ2
⌋and either
(i) δ is even and every vertex of G is adjacent to a vertex of degree δ or δ + 1, or
(ii) δ is odd and every vertex of G is adjacent to a vertex of degree δ.
Proof. Suppose Mk(G) = n. Hence, for any vertex v ∈ V (G), M = V (G) − {v} is not k-monopoly in G. Thus the vertex v or some vertex u ∈ N(v) does not satisfy the monopolycondition. If δM(v) < δM(v) + 2k, then we have that δ(v) < 2k ≤ δ, a contradiction. ThusδM(u) < δM(u) + 2k, which leads to δ(u)− 1 < 1 + 2k. So δ(u) ≤ 2k + 1. As a consequence, weobtain that k ≥ δ−1
2(or equivalently k ≥
⌈δ−1
2
⌉). Since k ≤
⌊δ2
⌋, we obtain that k =
⌊δ2
⌋=⌈δ−1
2
⌉.
Thus, δ(u) ≤ 2k+ 1 = 2⌈δ−1
2
⌉+ 1. Hence, if δ is even, then we have that δ(u) ≤ δ+ 1, and if δ is
odd, then we have that δ(u) ≤ δ. Therefore, (i) and (ii) follow.On the other hand, suppose k =
⌊δ2
⌋. Assume δ is even and every vertex of G is adjacent to
a vertex of degree δ or δ + 1. Hence, let M ⊂ V (G), let x /∈M and let u ∈ N(x) having degree δor δ + 1. So we have,
δM(u) ≤ δ < 2
⌊δ
2
⌋+ 1 = 2k + 1 ≤ δM(u) + 2k.
Thus, M is not a k-monopoly.Now, suppose δ is odd and every vertex of G is adjacent to a vertex of degree δ. As above let
M ′ ⊂ V (G), let x′ /∈M ′ and let u′ ∈ N(x′) having degree δ. So we have,
δM ′(u′) < δ = 2
⌊δ
2
⌋+ 1 = 2k + 1 ≤ δM ′(u′) + 2k.
Thus, M ′ is not a k-monopoly.Therefore, any proper subset of V (G) is not a k-monopoly and we have thatMk(G) = n.
17
The wheel graph of order n is defined as W1,n−1 = K1 +Cn−1, where + represents the join ofmentioned graphs. The fan graph F1,n−1 of order n is defined as the graph K1 + Pn−1.
Corollary 14.
(i) For any r-regular graph G of order n, Mb r2c(G) = n.
(ii) For any wheel graph W1,n−1, M1(W1,n−1) = n.
(iii) For any fan graph F1,n−1, M1(F1,n−1) = n.
(iv) For any bipartite graph Kr,r+1, r even, Mb r2c(Kr,r+1) = 2r + 1.
We continue this section by obtaining exact values for some graph classes. Recall that, by
Corollary 9, for k ∈{
1−⌈δ(G)
2
⌉, . . . ,
⌊δ(G)
2
⌋}we have Mk(Kn) =
⌈n+2k+1
2
⌉. We continue with
complete bipartite graphs.
Proposition 15. For every complete bipartite graph Kr,t and every k ∈{
1−⌈δ(G)
2
⌉, . . . ,
⌊δ(G)
2
⌋},
Mk(Kr,t) =
⌈r + 2k
2
⌉+
⌈t+ 2k
2
⌉.
Proof. Let X and Y be the partition sets of Kr,t such that |X| = r and |Y | = t and let S be asubset of vertices of Kr,t such that |S ∩X| =
⌈r+2k
2
⌉and |S ∩ Y | =
⌈t+2k
2
⌉. Let v be a vertex of
Kr,t. If v ∈ X, then
δS(v) =
⌈t+ 2k
2
⌉≥ t+ 2k
2= t+ 2k − t+ 2k
2≥ t−
⌈t+ 2k
2
⌉+ 2k = δS(v) + 2k.
Analogously, if v ∈ Y , then we obtain that δS(v) ≥ δS(v) + 2k. Thus, S is a k-monopoly in Kr,t
and we have that Mk(Kr,t) ≤⌈r+2k
2
⌉+⌈t+2k
2
⌉.
Now, let M be a Mk(Kr,t)-set and let u be a vertex of Kr,t. If u ∈ X, then we have thatδM(u) ≥ δM(u) + 2k = t − δM(u) + 2k, which leads to δM(u) ≥ t+2k
2and, as a consequence,
|Y ∩M | = δM(u) ≥⌈t+2k
2
⌉. Analogously, if u ∈ Y , then we obtain that |X ∩M | ≥
⌈r+2k
2
⌉. Thus,
Mk(Kr,t) = |M ∩X|+ |M ∩ Y | ≥⌈r+2k
2
⌉+⌈t+2k
2
⌉and the proof is complete.
Next we study k-monopolies of cycles and paths. First notice that the case k = 1 for cyclesfollows directly from Corollary 14 (i), that is, M1(Cn) = n.
Proposition 16. For every integer n ≥ 3,
M0(Cn) =M0(Pn) =
n2
if n ≡ 0 mod 4,
n+22
if n ≡ 2 mod 4,
n+12
if n ≡ 1 mod 4 or n ≡ 3 mod 4.
18
Proof. By Theorem 5,M0(Cn) = γt(Cn) and it is known from [13] that γt(Cn) =⌊n2
⌋+⌈n4
⌉−⌊n4
⌋.
Hence we are done with cycles.Let V (Pn) = {v0, . . . , vn−1}. We proceed by induction on k ≥ 1 where n = 4k + i and
i ∈ {−1, 0, 1, 2}. Let Mn be a subset of V (Pn) defined as follows.
• If n ≡ 0 (mod 4), then Mn = {v1, v2, v5, v6, . . . , vn−3, vn−2}.
• If n ≡ 1 (mod 4), then Mn = {v1, v2, v3, v6, v7, v10, v11, . . . , vn−3, vn−2}.
• If n ≡ 2 (mod 4), then Mn = {v0, v1, v3, v4, v7, v8, v11, v12, . . . , vn−3, vn−2}.
• If n ≡ 3 (mod 4), then Mn = {v0, v1, v4, v5, . . . , vn−3, vn−2}.
It is straightforward to check that Mn is aM0(Pn)-set for k = 1. Notice that M4 is the uniqueM0(P4)-set. Let k > 1. Set M4(k−1)+i is aM0(P4(k−1)+i)-set by induction hypothesis. Clearly, any0-monopoly M ′ of Pn contains at least two vertices of the last three vertices vn−3, vn−2, vn−1. Hence,these two vertices have no influence on the vertices of M ′ from the first 4(k− 1) + i vertices of thepath P4k+i. Therefore |M ′∩{v0, . . . , v4(k−1)+i}| ≥ |M4(k−1)+i| and M4k+i = M4(k−1)+i∪{vn−3, vn−2}is a M0(P4k+i)-set. It is easy to see that |M4k+i| gives the desired values.
7 Partitions into k-monopolies
In this section we present some results about partitioning a graphs into monopolies. To this end,we say that a graph G = (V,E) is k-monopoly partitionable if there exists a vertex partitionΠ = {S1, . . . , Sr} of V , r ≥ 2, such that for every i ∈ {1, . . . , r}, Si is a k-monopoly in G.
Theorem 17. If a graph G is k-monopoly partitionable, for some k ∈ {1 −⌈δ(G)
2
⌉, . . . ,
⌊δ(G)
2
⌋},
then r ≤ 2− 2k and k ≤ 0.
Proof. Let Si, Sj ∈ Π and let v be a vertex of G. Then we have that
δSi(v) ≥ δSi(v) + 2k
= 2k +r∑
`=1,` 6=i
δS`(v)
= δSj(v) + 2k +r∑
`=1,` 6=i,j
δS`(v)
≥ δSj(v) + 4k +r∑
`=1,`6=i,j
δS`(v)
Since for every u of G, δS`(u) ≥ 1 for every ` ∈ {1, . . . , r}, we obtain thatr∑
`=1,`6=i,j
δS`(u) ≥ r − 2.
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So,
δSi(v) ≥ δSj(v) + 4k + r − 2
= 4k + r − 2 +r∑
`=1,` 6=j
δS`(v)
= δSi(v) + 4k + r − 2 +r∑
`=1,` 6=i,j
δS`(v)
≥ δSi(v) + 4k + 2r − 4.
Thus 2k + r − 2 ≤ 0, which leads to r ≤ 2 − 2k and k ≤ 1 − r/2. Since r ≥ 2, we have thatk ≤ 0.
From the above result we have that G can be only partitioned into at most 2−2k k-monopoliesfor k ≤ 0. The particular case k = 0 is next studied. Notice that for instance, cycles of order 4tand hypercubes Q2t with t ≥ 1 are examples of graphs having a partition into two 0-monopolies.
Proposition 18. Let G be a graph having a vertex partition into two 0-monopolies {X, Y }. Thenthe following assertions are satisfied.
(i) For every vertex v of G, δX(v) = δY (v).
(ii) For every vertex v of G, δ(v) is an even number.
(iii) The size mX of 〈X〉 equals the size mY of 〈Y 〉.
(iv) The cardinality of the edge cut set c(X, Y ) produced by the vertex partition {X, Y } equals thesize m of G minus two times the size of 〈X〉.
Proof. For every vertex v of G we have that δX(v) ≥ δY (v) and δY (v) ≥ δX(v). Thus, (i) follows.Now, (ii) follows from the fact that δ(v) = δX(v) + δY (v) = 2δX(v) = 2δY (v). To prove (iii) weconsider the following ∑
v∈X
δX(v) +∑v∈Y
δX(v) =∑v∈X
δY (v) +∑v∈Y
δY (v).
Since,∑
v∈Y δX(v) =∑
v∈X δY (v) we have the result. As a consequence, m = c(X, Y ) +mX +mY
and by (iii) we obtain (iv).
A natural question which now arises concerning the computational complexity on the existenceof such partitions mentioned above. That is for instance, given a graph G, can we decide whetherG is k-monopoly partitionable? Moreover, if the answer is positive, can we find such partitionsby using some efficient algorithm?
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