https://doi.org/10.1051/ro/2020017 www.rairo-ro.org
TREES WITH UNIQUE MINIMUM GLOBAL OFFENSIVE ALLIANCE
Mohamed Bouzefrane
1,∗, Isma Bouchemakh
2, Mohamed Zamime
3and Noureddine Ikhlef-Eschouf
4Abstract.Let G= (V, E) be a simple graph. A non-empty set D ⊆ V is called a global offensive alliance ifDis a dominating set and for every vertexvinV−D,|NG[v]∩D| ≥ |NG[v]−D|. The global offensive alliance number is the minimum cardinality of a global offensive alliance inG. In this paper, we give a constructive characterization of trees having a unique minimum global offensive alliance.
Mathematics Subject Classification. 05C69, 05C70.
Received July 18, 2018. Accepted February 19, 2020.
1. Introduction
In this paper,G = (V, E) is a simple graph with vertex set V =V(G) and edge setE =E(G). Theopen neighborhood of a vertex v ∈ V is NG(v) = {u∈V |uv∈E} and the closed neighborhood of v is NG[v] = NG(v)∪ {v}. The degree of a vertex v ∈ V, denoted by dG(v), is the cardinality of its open neighborhood.
A vertex of degree one is called a leaf and its neighbor is called asupport vertex. If v is a support vertex of a treeT, then LT(v) will denote the set of the leaves adjacent tov, and let lT(v) =|LT(v)|.The set of leaves and support vertices ofT is denoted byL(T) andS(T), respectively. Further, let l(T) =|L(T)|.As usual, the path of ordern is denoted byPn,and thestar of order nbyK1,n−1. Adouble star, denoted bySp,q, is a tree obtained fromP2 by joiningppendent edges to one end andqpendent edges to other end ofP2. Asubdivision of an edgeuv is obtained by introducing a new vertex w and replacing the edgeuv by the edges uw and wv.
A subdivided star, denoted bySSk, is a starK1,k such that each edge is subdivided exactly once. Awounded spider is the tree formed by subdividingredges ofK1,k,where 1≤r≤k−1.For a vertexv,in a rooted tree T,letC(v) andD(v) denote the set of children and descendants ofv,respectively. LetD[v] =D(v)∪ {v}.The maximal subtree atv,denoted byTv,is the subtree of T induced by D[v].
Keywords.Tree, domination, global offensive alliance, characterization.
1 Department of Mathematics, B.P. 270, Lamda-RO Laboratory, University of Blida, Blida, Algeria.
2 Faculty of Mathematics, Laboratory L’IFORCE, University of Sciences and Technology Houari Boumediene (USTHB), B.P.
32 El-Alia, Bab-Ezzouar 16111, Algiers, Algeria.
3 Faculty of Technology, University of M´ed´ea, M´ed´ea, Algeria.
4 Faculty of Sciences, Department of Mathematics and Computer Science, University of M´ed´ea, M´ed´ea, Algeria.
∗Corresponding author:[email protected]
Article published by EDP Sciences c EDP Sciences, ROADEF, SMAI 2021
If G andH are two vertex-disjoint graphs, the union of G and H is the graph G∪H whose vertex-set is V(G)∪V(H) and edge-set isE(G)∪E(H).For an integerk≥2, the union ofk copies of graphGis denoted kGand the union of graphsG1, G2, . . . , Gk is denoted ∪ki=1Gi.
LetD⊆V be a non-empty set of vertices of graphG. Dis adominating set of Gif every vertex inV −D has at least one neighbor inD.Thedomination number ofG,denoted byγ(G),is the minimum cardinality of a dominating set ofG. Dis an offensive alliance ofGif for everyv∈V −D,we have
|NG[v]∩D| ≥ |NG[v]−D|. (1.1) D is aglobal offensive alliance, or briefly a goa, of GifD is both an offensive alliance and a dominating set of G. Theglobal offensive alliance number, denoted byγo(G),is the minimum cardinality of a goa ofG.Note that γo is defined for every graph. For a graphG,the vertex set V ofG is a goa, asV −V =∅, and the condition forV to be a goa is (vacuously) true. A goa ofGof cardinalityγo(G) is called aγo(G)-set.Gis aunique global offensive alliance graph, or briefly anugoa-graph, ifGhas a uniqueγo(G)-set. A unique global offensive alliance on a graphGallows vertices to have a unique assignment. This can be used for an extension of the graph while keeping the same layout which existed.
The literature on the concept of domination and its variants has been surveyed and detailed in the books of Hayneset al.[25,26]. Among these variants, there are the alliances in graphs. The study of alliances in graphs is first investigated by Kristiansenet al.[31]. Favaronet al.[11] initiated and made a thorough study of offensive alliances, they derived several bounds on the offensive alliance number. Several other bounds on (global) offensive alliance number are obtained by Rodr´ıguez-Vel´azquez and Sigarreta [34,35], Rad [33], Harutyunyan [22,23] and, Yero and Rodr´ıguez-Vel´a zquez [38]. Concerning the computational complexity, the NP-completeness is proved by Balakrishnanet al.[1] for the problem of global offensive alliance and by Fernauet al.[12] for both problems (offensive and global offensive alliances) in general graphs. More results can be found in [3–6,10,32].
Graphs with a unique minimumµ-set, whereµ is some domination parameter, has attracted the interest of several researchers over the last decades. In particular, graphs with a unique minimum dominating sets have been studied, by Guntheret al.[21]. These authors initiated this related concept to uniqueness in the domination theory and investigated some of the structural properties of graphs having this property. Some works have been established on different classes of graphs, for example, by Fischermann and Volkmann [15] for trees and in [16]
for cactus graphs, by Fischermann [13] for block graphs, by Hedetniemi [27] for some cartesian product graphs and in [28] for some repeated cartesian products graphs. Regarding the uniqueness related to other invariants of domination, studies have been undertaken by Blidia et al.[2] for the locating-domination number, by Chellali and Haynes for paired-domination number in [7] and for double domination number in [8], Chellali and Rad [9]
for roman domination number, and by Haynes and Henning [24] for total domination number. Further works on this topic can be found in [14,17–20,29,30,36,37].
In this paper, we are interested in determining the structure of trees that are ugoa-trees. The paper is organized as follows. Section 2 deals with the preservation of the ugoa-tree property after having performed some operation on it. In Section3, we provide a constructive characterization of ugoa-tree. Finally, we conclude by setting some problems that we seem interesting.
2. Preliminary results
We first present some elementary facts concerning the ugoa-trees. Some results are straightforward and so their proofs are omitted.
Observation 2.1. LetT be a tree of order at least three and letu∈S(T). Then, (i) there is aγo(T)-set that contains all support vertices ofT,
(ii) ifD is a uniqueγo(T)-set, thenD contains all support vertices ofT but no leaf, (iii) iflT(u)≥2,thenubelongs to anyγo(T)-set.
Proof. (i) and (ii) are obvious. LetSbe anyγo(T)-set and assume that item (iii) is not satisfied. In this case, all leaves adjacent touwould be contained inS and the set (S−LT(u))∪ {u}would be a goa ofT of cardinality
less thanγo(T),a contradiction.
Observation 2.2. LetT be a tree obtained from a nontrivial treeT0 by joining a new vertex v at a support vertexuofT0.LetD andD0 beγo(T) -sets ofT andT0,respectively. Then,
(i) |D0|=|D|,
(ii) D∩V(T0) is aγo(T0)-set,
(iii) ifT is an ugoa-tree such thatuis in anyγo(T0)-set, thenT0 is an ugoa-tree.
Proof. According to Observation2.1(iii), and sincelT(u)≥2,umust be inD.
(i) D is clearly a goa ofT0, and then|D0| ≤ |D|.By Observation 2.1(i) we can assume thatu∈ D0. Hence, D0 can be extended to a goa of T, which leads to|D| ≤ |D0|.Thus, equality holds.
(ii) SinceD∩V(T0) =D is a goa ofT0 of cardinality|D|=|D0|, we deduce thatD∩V(T0) is aγo(T0)-set.
(iii) Item (i) together with the fact thatubelongs to anyγo(T0) imply thatD0can be extended to aγo(T)-set.
Therefore, the uniqueness ofD as aγo(T)-set leads toD0=D. Hence,D0 is the uniqueγo(T0).
Observation 2.3. LetT be a tree obtained from a nontrivial treeT0, different fromP2, by joining the center vertexyof the pathP3=x–y–zat a support vertexvofT0. LetDandD0beγo(T)-sets ofT andT0,respectively such that each of them contains all support vertices. Then,
(i) |D0|=|D| −1,
(ii) D∩V(T0) is aγo(T0)-set,
(iii) ifT is an ugoa-tree, thenT0 is an ugoa-tree.
Proof. (i) Sincey∈Dandv∈D∩D0,the setD− {y}is a goa ofT0and then|D0| ≤ |D| −1. Moreover, since v∈D0,D0 can be extended to a goa ofT by addingy. It follows,|D| ≤ |D0∪ {y}|=|D0|+ 1 and equality holds.
(ii) SinceD∩V(T0) =D− {y}is a goa ofT0 of cardinality|D| −1 =|D0|, D∩V(T0) is aγo(T0)-set.
(iii) Let B ={y}. In view of item (i), D0 can be extended to aγo(T)-set by adding the unique vertex of B.
This and item (ii) together with the uniqueness ofD imply thatD0=D∩V(T0) is the uniqueγo(T0)-set.
Observation 2.4. LetT be a tree obtained from a nontrivial treeT0 by adding a subdivided starSSk withk leaves,k≥1, and then identifying its center to some leafv ofT0.Letwbe the support vertex adjacent tovin T0, and letD andD0 beγo(T)-sets ofT andT0,respectively. Ifw∈D∩D0,then the following three properties are satisfied.
(i) |D0|=|D| −k,
(ii) D∩V(T0) is aγo(T0)-set,
(iii) ifT is an ugoa-tree, thenT0 is an ugoa-tree.
Proof. LetB={y1, y2, . . . , yk}be the set of support vertices of SSk.Definition ofT implies that every vertex inB is adjacent tov.
(i) Obviously, each vertex inB remains a support vertices in T.Hence, in view of Observation2.1(i), we can assume that D contains all vertices of B. Therefore, since w∈D andv /∈D, D−B is a goa ofT0, and consequently,|D0| ≤ |D−B|=|D| −k.We now prove the opposite inequality.w∈D0 implies thatD0can be extended to a goa ofT by adding all vertices ofB.Hence,|D| ≤ |D0∪B|=|D0|+kand equality holds.
(ii) The proof is similar to that of Observation2.3(ii), by takingD∩V(T0) =D−B.
(iii) The proof is similar to that of Observation2.3(iii).
Observation 2.5. Let V(T0) be the vertex set of a nontrivial tree T0, and let D0 be a γo(T0)-set such that V(T0)−D0 has a vertexw of degreeq≥2 inT0 and |NT0(w)∩(V(T0)−D0)| ≤1. Letpbe a positive integer such that
p≤q−1, if |NT0(w)∩(V(T0)−D0)|= 0, or
p≤q−3, if |NT0(w)∩(V(T0)−D0)|= 1.
(2.1)
LetT be a tree obtained fromT0by addingpsubdivided starsSSk1, . . . , SSkp,ki ≥2 for 1≤i≤p, and joining each centerxi ofSSki, 1 ≤i≤p, at w. LetD be a γo(T)-set. Ifw and x1, x2, . . . , xp are not in D,then the following three properties are satisfied.
(i) |D0|=|D| −Pp i=1ki, (ii) D∩V(T0) is aγo(T0)-set,
(iii) ifT is an ugoa-tree, thenT0 is also an ugoa-tree.
Proof. Fori∈ {1, . . . , p}, letBi the set of the support vertices of SSki.
(i) Since the verticesw,x1, x2, . . . , xpare not inD,all vertices of∪pi=1Bimust be inD.Therefore,D− ∪pi=1Bi is a goa of T0, giving that |D0| ≤ |D| −Pp
i=1ki. For the opposite inequality, we first prove that the set A=∪pi=1Bi∪D0 is a goa ofT, that is, |NT[z]∩A| ≥ |NT[z]−A|holds, for each z∈ {w, x1, x2, . . . , xp}.
For this, we consider two cases depending on the value ofz.
Case 1.z=xi,for somei∈ {1, . . . , p}.
We then have
|NT[z]∩A|=|NT[z]∩(∪pi=1Bi)|=ki≥2, and
|NT[z]−A|=|{z, w}|= 2.
Thus,|NT[z]∩A| ≥ |NT[z]−A|, for eachz∈ {x1, x2, . . . , xp}.
Case 2.z=w.
We have
|NT[z]∩A|=
q, if |NT0(w)∩(V(T0)−D0)|= 0, q−1, if |NT0(w)∩(V(T0)−D0)|= 1, and
|NT[z]−A|=
p+ 1, if |NT0(w)∩(V(T0)−D0)|= 0, p+ 2, if |NT0(w)∩(V(T0)−D0)|= 1.
According to (2.1), we have in this case also,|NT[z]∩A| ≥ |NT[z]−A|, forz =w. Therefore,A is a goa ofT, and
|D| ≤ |A|=|D0|+
p
X
i=1
ki.
Hence, the equality holds, as required.
(ii) Using the fact thatD∩V(T0) =D− ∪pi=1Bi, the proof is similar to that of Observation2.3(ii).
(iii) The proof is similar to that of Observation2.3(iii), by takingB=∪pi=1Bi.
3. The main result
In order to characterize the trees with a unique minimum global offensive alliance, we define a familyF of all treesT that can be obtained from a sequenceT1, T2, . . . , Tr,r≥1, of trees, whereT1is the pathP3centered at a vertexy, T =Tr,and, if r≥2, the treeTi+1 is obtained fromTi, 1≤i≤r−1, by one of the following operations listed below. LetA(T1) ={y}.
– OperationO1: Attach a vertex by joining it to any support vertex ofTi.LetA(Ti+1) =A(Ti).
– OperationO2: Attach a pathP3=u-v-wby joiningvto any support vertex ofTi.LetA(Ti+1) =A(Ti)∪{v}.
– OperationO3: Adding a subdivided star SSk withk leaves,k≥1, and then identifying the center of it to the leaf ofTi,where its support vertexwinTi satisfies the following conditionC0.
(C0) : If|NTi[w]∩A(Ti)| ≥ |NTi(w)∩(V(Ti)−A(Ti))|+ 1, then
• either lTi(w) = 2 and NTi(w)−A(Ti) has a least one vertex wt, such that |NTi(wt)∩A(Ti)| ≤
|NTi[wt]∩(V(Ti)−A(Ti))|+ 1,
• or lTi(w) = 1 and NTi(w) −A(Ti) has two vertices wp and wq, such that, |NTi(wl)∩A(Ti)| ≤
|NTi[wl]∩(V(Ti)−A(Ti))|+ 1, forl=p, q.
LetA(Ti+1) =A(Ti)∪B,whereB is the set of support vertices ofSSk.
– Operation O4: Let w ∈ V(Ti) − A(Ti) be a vertex of degree q ≥ 2 in Ti, such that
|NTi(w)∩(V (Ti)−A(Ti))| ≤ 1. Attach p ≥ 1 subdivided stars SSki, ki ≥ 2 for 1 ≤ i ≤ p, by joining each center xi ofSSki, atw, such that
p≤
q−1 if |NTi(w)∩(V (Ti)−A(Ti))|= 0, q−3 if |NTi(w)∩(V (Ti)−A(Ti))|= 1.
LetA(Ti+1) =A(Ti)∪(∪pi=1Bi,whereBi is the set of support vertices ofSSki. Before stating our main result, we need to prove an additional lemma.
Lemma 3.1. If T ∈ F, thenA(T)is the uniqueγo(T)-set.
Proof. LetT ∈ F. We proceed by induction on the number of operations, sayr, required to constructT. The property is true ifr= 1, since T =T1 is the pathP3 centered at y andA(T) ={y} is the unique γo(T)-set.
This establishes the base case.
Assume that for any treeT0 ∈ F that can be constructed withr−1 operations fromT1,A(T0) is the unique γo(T0)-set. LetT =Tr, with r≥2, andT0=Tr−1. We consider four cases, depending on the used operation.
Case 1.T is obtained fromT0 by using OperationO1.
Assume that T is obtained from T0 by attaching an extra vertex at a support vertex u of T0. In view of Observation2.1(ii), u∈A(T0). Hence, A(T0) can be extended to a goa ofT. By Observation 2.2(i), γo(T) = γo(T0),and thenA(T0) is aγo(T)-set. SinceA(T0) is the uniqueγo(T0)-set, we infer thatA(T) =A(T0) is the uniqueγo(T)-set.
Case 2.T is obtained fromT0 by using OperationO2.
A(T0)∪ {v}is a goa ofT. By Observation2.3(i),γo(T) =γo(T0) + 1,and thenA(T0)∪ {v} is aγo(T)-set.
SinceA(T0) is the uniqueγo(T0)-set, we infer that A(T) =A(T0)∪ {v}is the unique γo(T)-set.
Case 3.T is obtained fromT0 by using OperationO3.
A(T0)∪B is a goa ofT. By Observation 2.4(i), γo(T) = γo(T0) +k, and then A(T0)∪B is a γo(T)-set.
SinceA(T0) is the uniqueγo(T0)-set, we infer that, A(T) =A(T0)∪B is the uniqueγo(T)-set.
Case 4.T is obtained fromT0 by using OperationO4.
A(T0)∪(∪pi=1Bi) is a goa ofT. By Observation2.5(i),γo(T) =γo(T0) +Pp
i=1ki,and thenA(T0)∪(∪pi=1Bi) is a γo(T)-set. SinceA(T0) is the unique γo(T0)-set, we infer that A(T) = A(T0)∪(∪pi=1Bi) is the unique
γo(T)-set.
Note that in each one of the previously four cases, A(Ti+1) is obtained from A(Ti) by adding all support vertices inTi+1−Ti. Hence, the following corollary is immediate.
Corollary 3.2. LetT ∈ F andS(T)be a set of support vertices inT. Then, γo(T)>|S(T)|.
We are now ready to prove our main result.
Theorem 3.3. A tree T is an ugoa-tree if and only ifT =K1 orT ∈ F.
Proof. It is obvious thatT =K1is an ugoa-tree. Also, Lemma3.1states that any member ofF is an ugoa-tree.
We prove the converse by induction on the numbernof vertices of T. For n= 1 orn= 3, the result trivially holds, but not forn= 2, sinceP2is not an ugoa-tree. Forn= 4, T ∈ {K1,3, P4}. ClearlyP4is not an ugoa-tree, whilst K1,3 is an ugoa-tree that can be obtained from aP3 using OperationO1. Hence,K1,3∈ F. For n= 5, T ∈ {S1,2, K1,4, P5}. The double star S1,2 is not an ugoa-tree, while K1,4 and P5 are ugoa-trees, since K1,4
can be obtained fromK1,3 by using Operation O1, andP5can be obtained from aP3 by using OperationO3. Therefore,K1,4 andP5 are in F. This establishes the base case.
Assume nown≥6, and any treeT0 of ordern0, 3≤n0 < n, has a uniqueγo(T0)-set inF. LetT be a tree of ordernwith the uniqueγo(T)-setDand lets∈S(T).By Observation2.1(ii), we haves∈D.IflT(s)≥3,then letT0 be the tree obtained fromT by removing a leaf adjacent tos and letD0 be a γo(T0)-set. Then, clearly n0 =|V(T0)|=n−1≥5, lT0(s)≥2 and so,s∈D0 by Observation2.1(iii). According to Observation2.2(iii), T0 is an ugoa-tree. Applying the inductive hypothesis toT0, we get T0 ∈ F. Thus, T is obtained fromT0 by OperationO1,implying thatT ∈ F. Assume now that
for eachx∈S(T), lT(x)≤2. (3.1)
RootT at a vertexrof maximum eccentricity. Letube a support vertex of maximum distance fromrand let u0 be a leaf adjacent tou. Letv andwbe the parents ofuandv,respectively, in the rooted tree. We consider two cases.
Case 1.v∈D.
If lT(u) = 1,then D∪ {u0} − {u} is a γo(T)-set, contradicting the uniqueness ofD as a γo(T)-set. Hence, by (3.1),lT(u) = 2.We claim thatv∈S(T). Assume to the contrary thatv6∈S(T).Then, either w∈D and so, D− {v} is a goa ofT of cardinality less than |D|, contradicting the minimality of D, or w /∈ D and so, D− {v} ∪ {w} is a γo(T)-set, contradicting the uniqueness ofD as a γo(T)-set. This completes the proof of the claim. Let T0 =T −Tu and D0 be a γo(T0)-set. By Observation 2.1(i), we can assume that D0 contains all support vertices in T0. Since |V(Tu)| = 3, it follows that n0 = |V(T0)| = n−3 ≥3 and so, T0 6= P2. By Observation 2.3(iii), T0 is an ugoa-tree. Applying our inductive hypothesis, we get T0 ∈ F. Thus, T can be obtained fromT0 by OperationO2 and so,T ∈ F.
Case 2.v /∈D.
According to Observation2.1(ii),v /∈S(T) and then,lT(v) = 0.Letk=|NT(v)−{w}|.We havedT(v) =k+1 and sinceu∈NT(v)− {w}, we infer thatk≥1. Fori∈ {1, . . . , k},letui∈NT(v)− {w}such thatu1=u. The choice ofv sets that
ui∈S(T), lT(ui)≥1 and so,ui∈D for alli. (3.2) Hence, by (3.1), we have 1≤lT(ui)≤2 for alli.Assume first thatlT(uj) = 2 for somej∈ {1, . . . , k}. Without loss of generality, let j = 1. Then, uhas another neighbor u00 6= u0 in T. Let T0 = T − {u00} and D0 be any γo(T0)-set. Clearlyu0 is the unique leaf ofuin T0.We claim thatu∈D0.Suppose to the contrary thatu6∈D0. Then, u0 andv must be in D0. It follows that D00 = (D0− {u0})∪ {u} is a γo(T)-set which is different from D (since v belongs to D00 and not to D), a contradiction. This completes the proof of the claim. We have n0 =n−1 ≥ 5. By Observation 2.2(iii), T0 is an ugoa-tree. Applying our inductive hypothesis toT0, we get T0∈ F.Hence,T is obtained fromT0 by OperationO1, implying that T∈ F.Assume now that
lT(ui) = 1 and hence,dT(ui) = 2 for alli. (3.3)
For all i ∈ {1, . . . , k}, let u0i be the unique leaf adjacent to ui (with u01 = u0). We distinguish two subcases, depending on whetherwbelongs toD or not.
Case 2.1.w∈D.
In view of (3.3),Tv is a subdivided starSSk centered atv.LetT0 =T−(Tv− {v}).Clearlyv∈L(T0) and w∈S(T0). Ifn0 =|V(T0)|= 2, thenT is a wounded spider with exactly one non-subdivided edge and in this case, it is not difficult to see that such a graph is not an ugoa-tree. Hence, assume that n0 ≥3. We claim the following.
If|NT[w]∩D| ≥ |NT(w)∩(V(T)−D)|+ 1 withlT(w)∈ {0,1}, then – whenlT(w) = 1, NT(w)−D has at least one vertexwtsuch that
|NT(wt)∩D| ≤ |NT[wt]∩(V(T)−D)|+ 1. (3.4) – WhenlT(w) = 0, NT(w)−Dhas at least two vertices wp andwq for which (3.4) is fulfilled.
Indeed, assume first thatlT(w) = 1 and suppose there is no vertex inNT(w)−Dfor which (3.4) is fulfilled. Let w0 be the leaf adjacent towinT. In this case (D− {w})∪ {w0}is aγo(T)-set different fromD, a contradiction.
Now, assume that lT(w) = 0. Suppose first thatNT(w)−D has exactly one vertex, sayw00 for which (3.4) is fulfilled. Then (D− {w})∪ {w00}is aγo(T)-set different fromD,a contradiction. Suppose now there is no vertex inNT(w)−Dfor which (3.4) is fulfilled. Then (D− {w})∪ {v}is aγo(T)-set different fromD, a contradiction.
This complete the proof of the claim.
Observe that whenlT0(w)∈ {1,2},the previous claim remains true by replacingDbyD0 andT byT0.Thus, according to Observation 2.4(iii), T0 is an ugoa-tree. By induction onT0, we getT0 ∈ F. SinceT is obtained fromT0 by using OperationO3,we directly obtainT ∈ F.
Case 2.2.w /∈D.
By Observation2.1(ii), w /∈S(T) and so, lT(w) = 0.Since v and ware in V(T)−D, v must have at least two neighbors inD. Hence,dT(v) =k+ 1≥3. Lett be the parent ofw,and let X, Y and Z be the following sets
Y =C(w)∩S(T), X=C(w)−Y andZ=D(w)∩(S(T)−Y).
Observe thatv ∈X, u∈Z, NT(w) ={t} ∪X∪Y and every vertex in Z plays the same role as u.Therefore, by (3.2), we haveZ ⊂D sinceZ⊂S(T). We also have, by (3.3), every vertex inZ has exactly two neighbors such that one of them is a leaf and the other one is in X. Furthermore, asv∈X, ui ∈Z for alli∈ {1, . . . , k}
and so,|Z| ≥k≥2. Notice also that|X| ≥1 since v∈X.Likewise |Y| ≥1 since D is aγo(T)-set. It is clear thatY ⊆S(T) and then, by Observation2.1(ii),Y ⊆D. Put
X={x1, x2, . . . xp}, p≥1, withx1=vand |Y|=q−1, q≥2.
The fact that every vertex inX plays the same role asv, xi∈V(T)−Dfor alli∈ {1, . . . , p}, let pi=|NT(xi)− {w}| fori= 1, . . . , p.
We havep1=kand since for alli∈ {1, . . . , p},xi andware inV(T)−D, ximust have at least two neighbors inZ. Hence,dT(xi) =pi+ 1≥3. This means that for all i∈ {1, . . . , p}, V(Txi) induces a subdivided starSSpi
of orderpi+ 1 centered atxi. Becausew∈V(T)−D,inequality (1.1) is valid by replacingv byw.This gives
p≤q−1 ift∈D, or p≤q−3 otherwise. (3.5)
LetT0=T−(∪pi=1Txi) andD0 be aγo(T0)-set. Observe thatT0 contains at least oneP3as an induced subgraph, which means that n0 =|V(T0)| ≥3. For all i∈ {1, . . . , p},let Bi be the support vertex set ofSSpi. We have
∪pi=1Bi =Z andNT0(w) =Y ∪ {t}, so
dT0(w) =q≥2.
According to Observation2.1(i), we can assume that Y ⊂D0 sinceY ⊂S(T0).Then, tis the only neighbor of win T0 that may not be inD0, that is
|NT0(w)∩(V(T0)−D0)| ≤1.
Ift∈D0,then the minimality ofD0leads tow∈V(T0)−D0,because otherwise, we could replacewbytinD0. By Observation2.5(ii) and (iii), we haveD0=D∩V(T0).Hence,t∈Dif and only ift∈D0. Notice that ift∈D0, thenNT0(w)∩(V(T0)−D0) is an empty-set, otherwise,twould be the unique vertex ofNT0(w)∩(V(T0)−D0).
Thus, (3.5) can be rewritten as follows:
If |NT0(w)∩(V(T0)−D0)|= 0, thenp≤q−1, and
if |NT0(w)∩(V(T0)−D0)|= 1, thenp≤q−3.
By Observation2.5(iii), T0 is an ugoa-tree. Applying the inductive hypothesis to T0, we deduce thatT0 ∈ F. Since T can be obtained from T0 by Operation O4, we finally conclude that T ∈ F. This completes the
proof.
Theorem3.3suggests a polynomial algorithm for testing if a given tree T withnvertices belongs toF and, if so, the algorithm returns the set A(T). The steps of this algorithm can be summarized as follows. If T is a path on 2 or 4 vertices, answerT /∈ F and stop. Else ifn≤5,answerT ∈ F, return the obvious setA(T),and stop.Supposen≥6.IfT has a strong support vertex, sayssuch thatlT(s)≥3, call the algorithm recursively on the treeT0 obtained from T by removing a leaf neighbor ofs; if the answer to the recursive call isT0 ∈ F, then answer T ∈ F, return A(T) = A(T0) and stop, else answerT /∈ F and stop. Assume now that for each x∈S(T),lT(x)≤2.Pick a vertexr, root the treeT atr, pick a vertex u0 at maximum distance fromr. Letu be the parent of u0 in the rooted tree andv be the parent ofu.If v ∈S(T) and lT(u) = 2, call the algorithm recursively on the treeT0 =T −Tu; if the answer to the recursive call is T0 ∈ F, then answer T ∈ F, return A(T) =A(T0)∪ {u} and stop, else answerT /∈ F and stop. Ifv ∈S(T) andlT(u) = 1, then return the answer T /∈ F and stop. Assume now thatv /∈S(T).If some child ofv has two leavesu0, u00, then call the algorithm recursively on the treeT0 =T− {u00}; if the answer to the recursive call isT0∈ F, then answerT ∈ F, return A(T) =A(T0) and stop, else answerT /∈ F and stop. Now, assume that every child ofv is a support vertex of degree two. Letwthe parent ofvand further letX ={x1, x2, . . . xp}andY be two subset ofV(T) defined as in the proof of Theorem3.3. Assume first thatw∈S(T) or|Y|<|X|.Call the algorithm recursively on the tree T0=T−(Tv− {v}); if the answer to the recursive call isT0∈ F andwfulfills the conditionC0(see operation O3),then answerT ∈ Fand returnA(T) =A(T0)∪B and stop, else answerT /∈ F and stop. Assume now that w /∈S(T) and |Y| ≥ |X|. Call the algorithm recursively on the treeT0 =T −(∪pi=1Txi); if the answer to the recursive call is T0 ∈ F, andw /∈A(T0), then answer T ∈ F and return A(T) =A(T0)∪(∪pi=1Bi) and stop, else answerT /∈ F and stop.
4. Conclusion
In this paper, we have provided a constructive characterization of all ugoa-trees. After this work, several problems could be raised. The following seem of particular interest:
(1) Although the case of trees was solved in this paper, it is still interesting to find a characterization of all ugoa-graphs in the class of unicyclic graphs, or more generally, cactus graphs.
(2) The characterization problem of all ugoa-graphs does not seem easy, nevertheless it is an attractive point to explore.
(3) Apart from the constructive characterization of ugoa-trees, it is interesting to investigate other types of characterizations.
(4) Knowing that any graph is not necessarily an ugoa-graph, it would therefore be interesting to characterize all graphsGwith the property that each vertex ofGbelongs to at least oneγo(G)-set.
(5) We can prospect another type of alliance, that is defensive alliance. A non-empty set of verticesD⊆V, of a graphG, is a defensive allianceofG, if for everyv∈D,|NG[v]∩D| ≥ |NG[v]−D|. Dis aglobal defensive alliance, or briefly agda, ofG, ifDis a defensive alliance and a dominating set ofG. An interesting problem consists to investigating all trees with a unique minimum global defensive alliance.
Acknowledgements. The authors would like to thank the anonymous referees for their valuable comments and suggestions leading to improvement of this paper.
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