Bounds and closed formulaes for the (k-)monopoly number of lexicographic product. graphs.

Bounds and closed formulaes for the (k-)monopoly number of lexicographic product graphs Ismael González Yero a, Dorota Kuziak b and Iztok Peterin c,d a Depament of Mathematics, EPS, University of Cádiz Av Ramón Puyol s/n, 110 Algeciras, Spain b Depament of Computer Engineering and Mathematics, Rovira i Virgili University Av Països Catalans 6, 43007 Tarragona, Spain c University of Maribor, FEECS, Smetanova 17, 000 Maribor, Slovenia d IMFM, Jadranska 19, 1000 Ljubljana, Slovenia E-mail addresses: ismaelgonzalez@ucaes, dorotakuziak@urvcat, iztokpeterin@umsi Abstract We consider a simple graph G = (V, E) without isolated veices and minimum degree δ(g) Let the integer k {1 δ(g)/,, δ(g)/ } A veex v of G is said to be k-controlled by a set M V, if δ M (v) δ(v) +k where δ M (v) represents the number of neighbors v has in M and δ(v) the degree of v The set M is called a k-monopoly if it k-controls every veex v of G The minimum cardinality of any k-monopoly in G is the k-monopoly number of G In this aicle we study the k-monopolies of the lexicographic product of graphs Specifically we obtain several relationships between the k-monopoly number of this product graph and the k-monopoly numbers and/or order of its factors Moreover, we bound (or compute the exact value) of the k-monopoly number of several families of lexicographic product graphs Keywords: Monopolies; lexicographic product graphs AMS Subject Classification (010): 05C69, 05C76 1 Introduction and preliminaries Monopolies in graphs have been related to several problems regarding overcoming failures, according to the fact that monopolies have very frequently some common approaches in the field of majorities For instance, some connections with consensus problems [3], diagnosis problems [17] or voting systems [5], among other 0 The research was paially done while the third author was visiting the University of Maribor, Slovenia, suppoed by Ministerio de Educación, Cultura y Depoe, Spain, under the Jose Castillejo program for young researchers Reference number: CAS1/0067 1

applications and references are already described Monopolies in graphs are also closely related to different parameters in graphs like global alliances and signed domination (see [11]) On the other hand, it is also known that the complexity of computing the k-monopoly number of a graph is an NP-complete problem for k 0, see [11] In this sense, it is desirable the reduction of the problem of computing the k-monopoly number of some complex graph classes to other ones much simpler classes, in order to compute the value of such a parameter An interesting case of that is precisely the product graphs, which has been initiated in the aicles [1] and [13] for the direct and strong product of graphs, respectively To continue pursuing this goal, in this aicle we obtain several relationships between the k-monopoly number of the lexicographic product graphs and the k-monopoly number of its factors In this aicle we consider a simple graph G = (V, E) Given a set S V and a veex v V, we denote by δ S (v) the number of neighbors v has in S If S = V, then δ V (v) is the degree of v and we just write δ(v) The minimum degree of G is denoted by δ(g) and the maximum degree by (G) Given an integer k { 1 δ(g),, δ(g) } and a set M, a veex v of G is said to be k-controlled by M if δ M (v) δ(v) + k The set M is called a k-monopoly if it k-controls every veex v of G The minimum cardinality of any k-monopoly is the k-monopoly number and it is denoted by M k (G) A monopoly of cardinality M k (G) is called a M k (G)-set In paicular notice that for a graph with a leaf (veex of degree one), there exist only 0-monopolies and the neighbor of every leaf is in each M 0 -set Monopolies in graphs were defined first in [14] and they were generalized to k-monopolies recently in [11] Other studies about monopolies in graphs and some of its applications can be found in [4, 15, 16, 19, 9] If M represents the complement of the set M, then we can use the following equivalent definition for a k-monopoly in G A set of veices M is a k-monopoly in G if and only if for every veex v of G, δ M (v) δ M (v) + k (from now we will call this expression the k-monopoly condition) and we will say that M is a k-monopoly in G if and only if every v of G satisfies the k-monopoly condition for M We use for a graph G standard notations N G (g) for the open neighborhood {g : gg E(G)} and N G [g] for the closed neighborhood N G (g) {g} Let S V (G) Neighborhoods over S form a subpaition of a graph G if N G (u) N G (v) = for every different u, v S A set S forms a maximum subpaition if S forms a subpaition where V (G) v S N G(v) has the minimum cardinality among all possible sets S V (G) In the extreme case where v S N G(v) = V (G) we call G an efficient open domination graph and the set S an efficient open dominating set of G Efficient open domination graphs were first studied in [6] The work was continued in [7], where all efficient open domination trees have been described

inductively and it was shown that it is an NP-complete problem to show whether G is an efficient open domination graph or not Recently, in [18], the efficient open dominating sets of Cayley graphs were considered and a discussion with respect to product graphs in general can be found in [10] Clearly not all graphs are efficient open domination graphs The lexicographic product G H (also sometimes denoted by G[H] and called composition) of graphs G and H is a graph with V (G H) = V (G) V (H) Two veices (g, h) and (g, h ) are adjacent in G H whenever gg E(G) or g = g and hh E(H) For a fix h V (H) we call G h = {(g, h) V (G H) : g V (G)} a G-layer in G H H-layers g H for a fix g V (G) are defined symmetrically Notice that the subgraph induced by G h or g H is isomorphic to G or H, respectively The lexicographic product is clearly not commutative, while it is associative [8] Upper bounds We begin with the following general result in which the k-monopoly number of the second factor of the lexicographic product graph contains at least half its order Theorem 1 Let G and H be graphs and let k M k (H) V (H), then we have { 1 M k (G H) M k (H) V (G) δ(h),, } δ(h) If V (H) Proof Let M be a M k (H)-set where M Notice that H has no isolated veex, since M k (H) exists We claim that M = V (G) M is a k-monopoly set of G H For (g, h) V (G H) we have Therefore the result follows δ M (g, h) = δ G (g) M + δ M (h) V (G) δ G (g) + δ H(h) = δ G H(g, h) + k + k V (H) The expectation, that the condition M k (H) from the above theorem is fulfilled, is greater when k is big Hence, this theorem is more useful for positive k For instance, if H is a graph with even number of veices with an V (H) universal veex, then M k (H) for k 0 3

V (H) If M k (H) >, then we can add (at least one) an isolated veex to H to obtain graph H + and the upper bound from Theorem 1 still holds for G H + Next we continue with a general result for non negative values of k Theorem Let G and H be graphs and let l = min{δ(g), δ(h)} For every k {0,, l }, M k (G H) M k (G) V (H) Proof Let M G be a M k (G)-set We will show that M = (M G V (H)) is a k -monopoly set of G H Let (g, h) V (G H) We consider the following cases Case 1: g M G Hence, Case : g / M G Hence, δ M (g, h) = δ MG (g) V (H) + δ H (h) δ MG (g) V (H) + k V (H) + δ H (h) = δ M (g, h) + k V (H) + δ H (h) δ M (g, h) + kδ H (h) + δ H (h) δ M (g, h) + 4k + k δ M (g, h) = δ MG (g) V (H) δ MG (g) V (H) + k V (H) = δ MG (g) V (H) + δ H (h) + k V (H) δ H (h) = δ M (g, h) + k V (H) δ H (h) δ M (g, h) + k(δ H (h) + 1) δ H (h) δ M (g, h) + δ H (h)(k 1) + k δ M (g, h) + k(k 1) + k = δ M (g, h) + 4k Thus, we have that M is a k -monopoly in G H, which completes the proof 3 Lower bounds The following observation and its corollary presented in [1] are useful to present a lower bound for M k (G H) 4

Observation 31 [1] Let G be a graph If S V (G) forms a subpaition of G, then M k (G) k S + δ(v) v S Corollary 3 [1] Let G be a graph If S V (G) forms a subpaition of G, then M k (G) S (k + δ(g)/) In [10] all efficient open domination graphs among lexicographic products have been characterized It was shown that G H is an efficient open domination graph if and only if one of the following conditions is fulfilled (i) G is a graph without edges and H an efficient open domination graph, or (ii) G is an efficient open domination graph and H contains an isolated veex Notice that (ii) can be generalized to every graph for maximum paitions The following proposition describes this case Proposition 33 Let G and H be graphs, where H contains an isolated veex h and G is without them If S G is a maximum subpaition of G, then S = S G {h} is a maximum subpaition of G H Proof Let S G be a maximum subpaition of G and h an isolated veex of H Since N G (g) N G (g ) = holds for different veices g and g of S G, we have N G H (g, h) N G H (g, h) = since h has no neighbor in H If S G {h} is not a maximum paition of G H, then there exists (g 1, h 1 ) V (G H) with N G H (g, h) N G H (g 1, h 1 ) = for every g S G Clearly g / S G since G contains no isolated veices Hence S G {g } is a paition of G, which yields a contradiction with S G being a maximal paition of G The following theorem follows directly from Observation 31, Proposition 33 and the fact that δ G H (g, h) = δ G (g) V (H) + δ H (h) (Notice that h is an isolated veex of H and hence δ H (h) = 0) Theorem 34 Let G and H be graphs, H with an isolated veex h and G without them and let l = δ(g) V (H) + δ(h) If S G is a maximum subpaition of G, then for k {1 l,, l } we have δ(g) V (H) M k (G H) k S G + (g,h) S G {h} 5

Finally we consider the case where H has no isolated veices To do so we need to introduce some notation Given a graph G and a set S V (G), we say that the closed neighborhoods of the veices of S form a closed subpaition for G if it is satisfied N G [u] N G [v] = for every two different veices u, v S A closed subpaition for G, formed by a set S, is a maximum closed subpaition, if V (G) v S N G[v] has the minimum possible cardinality among all closed subpaitions for G formed by sets S V (G) It is already known that, if S forms a maximum closed subpaition which is also a paition of V (G), then the set S is called a perfect code [] or an efficient dominating set [1] in G The result, regarding the maximum closed subpaitions of graphs, yields an interesting consequence for our purposes Proposition 35 Let G and H be graphs, both without isolated veices If S G is a maximum closed subpaition of G and h a veex of maximum degree in H, then S = S G {h} is a maximum subpaition of G H Proof Let S G be a maximum closed subpaition of G and h a veex of maximum degree in H Since G h is isomorphic to G and S G is a closed subpaition of G, we have N G H (g, h) N G H (g, h) = for every different veices g and g from S G Hence S G {h} is a subpaition of G H If S G {h} is not a maximum subpaition of G H, then there exists (g 1, h 1 ) for which N G H (g, h) N G H (g 1, h 1 ) = for every g S G Clearly N G (g 1 ) N G (g) = for every g S G Also g 1 g for any g S G since G has no isolated veices Similar g 1 is not adjacent to any g S G since H has no isolated veices Hence, S G {g 1 } is a closed subpaition of G, which is a contradiction with the maximality of S G Therefore the proof is complete Again, the following theorem is obtained directly from the Observation 31, Proposition 35 and the fact that δ G H (g, h) = δ G (g) V (H) +δ H (h) (Notice that we can choose the veex of maximum degree (H) in H) Theorem 36 Let G and H be graphs, both without isolated veices and let l = δ(g) V (H) + δ(h) If S G is a maximum closed subpaition of G and h is a veex of maximum degree in H, then for k {1 l,, l } we have δg (g) V (H) + (H) M k (G H) k S G + (g,h) S G {h} The next result presented [1] is also useful to present a bound for the k- monopoly number of lexicographic product graphs 6

{ Proposition 37 [1] If G is a graph order n, then, for any k 1 we have ( n δ(g) M k (G) (G) ) + k δ(g),, Corollary 38 Let G and H be two { graphs without isolated veices } of order n and m, respectively For any k 1 mδ(g)+δ(h),, mδ(g)+δ(h) we have ( ) nm mδ(g) + δ(h) M k (G H) + k m (G) + (H) 4 Bounds and exact values for some specific families of lexicographic product graphs The following result from [11] will be useful in this section Lemma 41 [11] For every integer r 3, r, if r 0 (mod 4), r+ M 0 (C r ) = M 0 (P r ) =, if r (mod 4), r+1, if r x (mod 4), x {1, 3} It is clear from the previous section that lower bounds for M k (G H) depends on (non) existence of an isolated veex in H This is the reason that we sta with an empty graph E n as second factor By E n we mean a graph on n veices without any edges Proposition 4 Let r, t 3 be integers (i) If r 0 (mod 4), then M 0 (C r E t ) = (ii) If r x (mod 4), x {1, 3}, then t M 0(C r E t ) + t (iii) If r (mod 4), then t M 0(C r E t ) + t Proof The upper bound follows directly from Theorem and Lemma 41 For the lower bound we use Theorem 34 For this we need the cardinality of a maximum subpaition S of C r It is easy to see that S = r, if r 0 (mod 4), r, if r (mod 4), r 1, if r x (mod 4), x {1, 3} 7 } δ(g)

and the result follows Proposition 43 Let r, t 3 be integers (i) If r 0 (mod 4), then M 0 (P r E t ) = (ii) If r 1 (mod 4), then 3t + t M0 (P r E t ) + t (iii) If r (mod 4), then t + t (iv) If r 3 (mod 4), then t + t M0 (P r E t ) + t M0 (P r E t ) + t Proof Again the upper bound follows directly from Theorem and Lemma 41 For the lower bound we need the cardinality and the structure of a maximum subpaition S of C r Notice that there exists S which contains no veices of degree 1 and r veices of degree if r 0 (mod 4); two veices of degree 1 and r 3 veices of degree if r 1 (mod 4); two veices of degree 1 and r veices of degree if r (mod 4); one veex of degree 1 and r 1 veices of degree if r 3 (mod 4) From this the lower bound follows directly from Theorem 34 Next results are given for the lexicographic products of cycles and/or paths Proposition 44 Let r, t 3 be integers (i) If t 0 (mod 4), then M 0 (C r C t ) = (ii) If t 1 (mod 4), then (a) if r 0 (mod 4), then M 0 (C r C t ) =, (b) if r is odd, then M 0 (C r C t ) = +1, (c) if r (mod 4), then M 0(C r C t ) (iii) If t (mod 4), then M0 (C r C t ) r + M 0(C r ) Moreover, + 1 M0 (C r C t ) + r (iv) If t 3 (mod 4), then M0 (C r C t ) r(t+1) 8

Proof First notice that, since C r C t is a (t+)-regular graph, from Corollary 38 it follows that M 0 (C r C t ) and all the lower bounds follow Now, let V 1 = {u 0, u 1,, u r 1 } and V = {v 0, v 1,, v t 1 } be the veex sets of C r and C t, respectively (operations with the subscripts of u i and v j will be done modulo r and t, respectively) We consider the following cases Case 1: t 0 (mod 4) From Lemma 41 we have that M 0 (C t ) = t Hence, by Theorem 1, we obtain that M 0 (C r C t ) rm 0 (C t ) = and (i) follows Case : t 1 (mod 4) By Lemma 41 we have that M 0 (C t ) = t+1 Thus, there exists a M 0 (C t )-set A such that v 0, v 1, v A and let X be a M 0 (C r )- set We claim that the set M = (V 1 A) (X {v 0 }) (recall that X is the complement of X) is a 0-monopoly in C r C t To see this, notice that if u i X, then the corresponding u i C t -layer contains t+1 veices of M and, if u i / X, then the u i C t -layer contains t+1 1 veices of M Also, notice that every veex of (u i, v j ) M has at least one neighbor in M u i C t and at least one neighbor in v M C j r Hence, for every veex (u i, v j ) we have that δ M (u i, v j ) = δ M u i 1 Ct (u i, v j ) + δ M u ict (u i, v j ) + δ M u i+1 Ct (u i, v j ) t + 1 + 1 + t + 1 1 = t + 1 = δ(u i, v j ) As a consequence, we obtain that every veex of C r C t satisfies the 0-monopoly condition Thus, the upper bound of (ii) is obtained from the fact that M = r(t+1) r + M 0 (C r ) = r + M 0(C r ) Now, by making some simple calculations, from the bounds of (ii) and Lemma 41, the items (a), (b) and (c) are obtained Case 3: t (mod 4) We proceed similarly as in the Case By Lemma 41 we have that M 0 (C t ) = t+ Hence, we consider a M 0(C t )-set B such that v 0, v 1, v B Let Y V 1 such that Y = {u 0, u, u 4,, u r/ } We shall show that M = (V 1 B) (Y {v 0 }) is a 0-monopoly in C r C t Notice that, if u i Y, then the corresponding u i C t -layer contains t+ 1 veices of M and, if u i / Y, then the u i C t -layer contains t+ veices of M Also, notice that every veex of (u i, v j ) M has at least one neighbor in M u i C t Hence, for every 9

veex (u i, v j ) we have that δ M (u i, v j ) = δ M u i 1 C t (u i, v j ) + δ M u ic t (u i, v j ) + δ M u i+1 C t (u i, v j ) t + 1 + 1 + t + 1 = t + 1 = δ(u i, v j ) Thus, every veex of C r C t satisfies the 0-monopoly condition Since M = r(t+) r = + r, the upper bound of (iii) is proved Case 4: t 3 (mod 4) The upper bound of (iv) follows directly from Theorem 1 and Lemma 41 Proposition 45 Let r, t 3 be integers If H is either a path P t or a cycle C t, then r (t + 1), if r 0 (mod 3), 3 ( M 0 (P r H) r ) 3 (t + 1) + t +, if r 1 (mod 3), and ( r 3 M 0 (P r H) ) 1 (t + 1) + t+, if r (mod 3),, if r 0 (mod 4), r(t+), if r (mod 4), r(t+1), if r x (mod 4), x {1, 3} Proof Notice that the graph P r has always a maximum closed subpaition formed by a set S of cardinality r 3 which has the following shape If r 0 (mod 3), then every veex of S has degree two If r 1 (mod 3), then there are two veices in S having degree one and the other veices of S have degree two If r (mod 3), then there is only one veex in S having degree one and the other veices of S have degree two 10

Hence, by using Theorem 36 we have the lower bounds bounds follow from Theorem 1 and Lemma 41 Finally, the upper The above result can be compared with Proposition 43 and we can observe that the gap between upper and lower bound is bigger The reason of this is related of using Theorems 34 and 36 for the lower bounds Proposition 46 Let r, t 3 be integers (i) If t 0 (mod 4), then M 0 (C r P t ) = (ii) If t x (mod 4), x {1, 3}, then (iii) If t (mod 4), then M 0(C r P t ) r(t+) M0 (C r P t ) r(t+1) Proof Notice that the minimum degree of P t is one Since t + 1 is always an odd number, Corollary 38 leads to the following t + 1 t + M 0 (C r P t ) = = t + t + Therefore, the lower bounds follow and the upper bounds follow from Theorem 1 and Lemma 41 The following results were obtained in [13] for the strong product of graphs We recall that the strong product G H of graphs G and H is a graph with veex set V (G H) = V (G) V (H) Two veices (g, h) and (g, h ) are adjacent in G H whenever (gg E(G) and h = h ) or (g = g and hh E(H)) or (gg E(G) and hh E(H)) Since C m K n = Cm K n and P m K n = Pm K n, the next two results are direct consequences of Proposition 35 and Proposition 36,respectively, from [13] Proposition 47 For integers r, t 3 we have + r t + 1 6 M 0(C r K t ) r r 3 Moreover, if r 0(mod 3) and t is odd, then M 0 (C r K t ) = + r 6 Proposition 48 For integers r, t 3 we have r 3 3t+1, if r 0 (mod 3), tr + r 6 t 3 M 0(P r K t ) r 3 3t+1 + 1, if r 1 (mod 3), r 3 3t+1 + t, if r (mod 3) 11

Neveheless, since the lexicographic product is not commutative, we can exchange the factors to obtain K r C t and K r P t, which we do at next Proposition 49 Let r, t 3 be integers (i) If t 0 (mod 4), then M 0 (K r C t ) = (ii) If t x (mod 4), x {1, 3}, then M0 (K r C t ) r(t+1) (iii) If t (mod 4), then M 0(K r C t ) r(t+) Proof We can use Theorem 1 and the values of M 0 (C t ) given in Lemma 41 to obtain upper bounds On the other hand, by using the Corollary 38 it follows that t(r 1) + M 0 (K r C t ) t(r 1) + t(r 1) + t(r 1) + = Therefore, the lower bounds are obtained and the proof is complete A similar approach can be used for the case of the graph K r P t Notice that the expression t(r 1) + 1 is an even number if and only if t is odd and r is even So, for t even or (t odd and r odd), since the minimum degree of P t is one, Corollary 38 leads to the following: t(r 1) + 1 M 0 (K r P t ) = t(r 1) + According to that, we have the next result Proposition 410 Let r, t 3 be integers t(r 1) + t(r 1) + = (i) If t 0 (mod 4), then M 0 (K r P t ) = (ii) If t x (mod 4), x {1, 3} and r is odd, then M0 (K r P t ) r(t+1) (iii) If t (mod 4), then M 0(K r P t ) r(t+) (iv) If t x (mod 4), x {1, 3} and r is even, then M 0 (K r P t ) r(t+1) t(r 1)+ t(r 1)+1 1

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