Global Strong Defensive Alliances of Sierpinski-Like´...

Theory Comput Syst (2013) 53:365–385DOI 10.1007/s00224-012-9423-2

Global Strong Defensive Alliances of Sierpinski-LikeGraphs

Chien-Hung Lin · Jia-Jie Liu · Yue-Li Wang

Published online: 29 August 2012© Springer Science+Business Media, LLC 2012

Abstract A strong alliance in a graph G = (V ,E) is a set of vertices S ⊆ V satisfy-ing the condition that, for each v ∈ S, the number of its neighbors, including itself,in S is greater than the number of those neighbors not in S. A strong alliance S isglobal if S forms a dominating set of G. In this paper, we shall propose a way forfinding a minimum global strong alliance for each of those Sierpinski-like graphs.Furthermore, we also derive the exact values of those global strong alliance numbers.

Keywords Global defensive alliances · Dominating sets · Sierpinski graphs ·Extended Sierpinski graphs · Sierpinski-like graphs

1 Introduction

Let G = (V ,E) be an undirected graph, where V (G) and E(G) are vertex and edgesets of G respectively. For simplicity, we also use V and E to represent V (G) andE(G), respectively, when only one graph is mentioned. All graphs consider in thispaper are finite, undirected, without loops and multiple edges. For any vertex v ∈ V

and a set S ⊆ V , the open neighborhood of v in S is the set NS(v) = {u ∈ S|uv ∈ E}.The closed neighborhood of v in S is NS[v] = NS(v)∪{v}. If S = V , then we simplywrite N(v) and N [v] rather than NV (v) and NV [v], respectively. A nonempty set S ⊆

This work was supported in part by the National Science Council of Republic of China undercontracts NSC 100-2221-E-011-067-MY3, NSC 100-2221-E-011-068-, andNSC 100-2221-E-128-003-.

C.-H. Lin · Y.-L. Wang (�)Department of Information Management, National Taiwan University of Science and Technology,Taipei, Taiwane-mail: [email protected]

J.-J. LiuDepartment of Information Management, Shih Hsin University, Taipei, Taiwan

mailto:[email protected]

366 Theory Comput Syst (2013) 53:365–385

Fig. 1 Sierpinski graphs

V of vertices is called a defensive alliance (respectively, strong defensive alliance) ifand only if, for every v ∈ S, |NS[v]| � |NV \S(v)| (respectively, |NS[v]| > |NV \S(v)|)[21]. An alliance S is called global if it forms a dominating set (i.e., every vertex inV \ S is adjacent to at least one vertex in S). A generalization of defensive allianceswas presented by Shafique and Dutton in [28, 29]. They defined a defensive k-allianceas a set S of vertices in G with the property that every vertex in S has at least k

more neighbors in S than it has outside of S. Thus, a strong defensive alliance isalso a defensive 0-alliance. In the following, a global strong defensive alliance isabbreviated as d-alliance. The global strong defensive alliance number of G, denotedγd(G), is the cardinality of a minimum d-alliance.In [8], the authors proposed several bounds on the cardinality of the minimum

global defensive alliance set for some classes of graphs, such as cycles, paths, trees,complete graphs, etc. Later on, various alliance problems have been defined in recentdecade, such as (strong) defensive alliances [5, 7, 30], (strong) offensive alliances [6,24, 25], and powerful alliances [1]. For more discussions, please refer to [2, 5, 8, 13,26, 30].

In this paper, we are concerned with the global strong defensive alliance numbersof Sierpinski-like graphs. The Sierpinski graph S(n, k) consists of k copies of S(n −1, k) for n > 1, where S(1, k) is the complete graph of k vertices [16]. Sierpinskigraphs are also called WK-recursive networks in [33]. For example, S(1,3), S(2,3),and S(3,3) are shown in Figs. 1(a), 1(b), and 1(c), respectively. Formal definitions ofSierpinski-like graphs including: Sierpinski graphs, extended Sierpinski graphs, andSierpinski gasket graphs will be introduced in Sect. 2.

The organization of this paper is as follows. In Sect. 2, we introduce Sierpinski-like graphs in detail. The global strong defensive alliance of Sierpinski graphs andextended Sierpinski graphs are discussed and constructed in Sects. 3 and 4, respec-tively. In Sect. 5, we construct a d-alliance from a Sierpinski gasket graph Sn. Wealso prove that those constructed sets are minimum. Finally, concluding remarks aregiven in Sect. 6.

2 Sierpinski-Like Graphs

The definitions of Sierpinski-like graphs are described as follows. The reader is re-ferred to [3, 4, 11, 12, 16, 27, 33] for the details. The vertex set of S(n, k) con-sists of all n-tuples of integers 1,2, . . . , k, for integers n � 1 and k � 3, namely

Theory Comput Syst (2013) 53:365–385 367

Fig. 2 Labeled Sierpinski graphs

V (S(n, k)) = {1,2, . . . , k}n. Thus, we use 〈v1, v2, . . . , vn〉 to represent the label ofvertex v, denoted �(v), or in the regular expression form v1v2 . . . vn for short whenno confusion arises. By using a convention on representing regular expressions, wealways use w,x, y, and z to denote a substring of v1v2 . . . vn and a, b, c, and d to de-note a number in v1v2 . . . vn, i.e., a, b, c, d ∈ {1,2, . . . , k}. The length of a substringw is denoted by |w|. For example, �(v) = wabn−h, for 1 � h � n, means that the la-bel of v begins with prefix w, then concatenates with number a, and finally ends withn − h b′s, where bh is the Kleene closure in regular expression. Thus |w| = h − 1.For convenience, we also say that v1v2 . . . vn is a vertex if �(v) = v1v2 . . . vn.

Two different vertices u and v are adjacent in S(n, k) if and only if �(u) = wabn−h

and �(v) = wban−h with a �= b, for some 1 � h � n. Note that if h = 1, then w = ε

which is a null string. Further, if h = n, then both bn−h and an−h are empty. Weuse an unordered pair (u, v) (or, in the label form, (�(u), �(v))) to denote the edgeincident to both u and v. By the above definition, the subgraph of S(n, k) induced bythe set of vertices whose labels begin with a is a Sierpinski subgraph S(n − 1, k) andis denoted by Sa(n−1, k). Similarly, we also use Sw(n−h, k) to denote the subgraphinduced by the vertices with prefix w in their labels. When n−h = 1, it is obvious thatSw(1, k) is a complete graph and we call it a terminal clique. Vertex v ∈ V (S(n, k))

is an extreme vertex if �(v) = an. Therefore, there are exactly k extreme vertices inS(n, k). Since the label of an extreme vertex v is an, by definition, v has exactly k −1neighbors whose labels are of the form an−1b with b �= a. Every non-extreme vertexv with �(v) = wabn−h has exactly k neighbors whose labels are of the form wban−h

and wabn−h−1c with 1 � c � k and c �= b. Thus, the degree of every extreme vertex,say v, is |N(v)| = k−1 while all other vertices have degree k. Figure 2 depicts S(3,3)

and S(3,4). An interesting connection is that S(n,3), for n � 1, is isomorphic to thegraphs of the Tower of Hanoi puzzle with n disks [10, 16] and has been extensivelystudied (see [11] for an overview and the references therein for the details).

The extended Sierpinski graphs S+(n, k) and S++(n, k) were introduced byKlavzar and Mohar [18]. The graph S+(n, k) is obtained from S(n, k) by adding


Fig. 3 Extended Sierpinski graphs: S+(3,3) and S++(3,3)

a special vertex, say s, and edges joining s to all extreme vertices of S(n, k) (seeFig. 3(a)). The graph S++(n, k) is obtained from S(n, k) by adding a new copyof S(n − 1, k) which is denoted by Sk+1(n − 1, k), and joining extreme vertex an

in S(n, k) to vertex ban−1 in the added Sk+1(n − 1, k), for a = 1,2, . . . , k, whereb = k+1 (see Fig. 3(b)). The vertex-, edge-, and total-colorings on S(n, k), S+(n, k),and S++(n, k) have been studied by Jakovac and Klavzar [14]. In [22], Lin, Liu, andWang investigated the hub numbers of Sierpinski-like graphs.

The Sierpinski gasket graph Sn is a variant of Sierpinski graph S(n,3). Sn can beobtained from S(n,3) by contracting every edge of S(n,3) that lies in no triangle. Forexample, see Figs. 2(a) and 4. Vertices 〈1,1,2〉 and 〈1,2,1〉 in S(3,3) are contractedto be a vertex in S3 in which we use the regular expression 1(12|21) to denote thelabel of the resulting vertex, where “|” is the union operation in regular expression.According to the definition of extreme vertices in S(n, k), the vertices with labels 1n,2n, and 3n in Sn are also called extreme vertices. The labels of other vertices are of theform w(abh|bah) where 1 � h � n − 1, w ∈ {1,2, . . . , k}n−h−1, and a and b are oneof the pairs: 1 and 2, 1 and 3, or 2 and 3. For convenience, we also use wabh or wbah

to represent the contracted vertex w(abh|bah). The vertices with labels 12n−1|21n−1,13n−1|31n−1, and 23n−1|32n−1 are called the waist vertices of Sn. The neighbors ofthe extreme vertex an are of the form an−2(ab|ba) with a �= b. The neighbors ofvertex v with label w(abh|bah) are of the form: wabh−2(bc|cb) and wbah−2(ad|da)

for c �= b and d �= a. The Sierpinski gasket graph Sn also contains 3x copies of Sn−x

which are denoted by Sn−x,a , for a ∈ {1,2,3}x , where Sn−x,a contains all verticeswhose labels begin with a.

Many properties of Sierpinski gasket graphs have been studied such as the hamil-tonicity [19, 32], the pancyclicity [32], the domination number [16, 17], the chromaticnumber [19, 23, 32], the total chromatic number [14, 32], and the hub number [22].


Fig. 4 Sierpinski gasketgraph S3

Moreover, Sierpinski gasket graphs play an important role in dynamic systems andprobability [9, 15] as well as in psychology [20, 31].

3 Computing γd(S(n,k))

For a vertex v in Sw(1, k), vertex u ∈ N(v) is called an outer vertex of v if u is notin Sw(1, k). Note that every vertex v ∈ S(n, k) has exactly one outer vertex exceptthat extreme vertices have no outer vertex. Furthermore, if u is the outer vertex ofv, then v is also the outer vertex of u. Let S be a d-alliance in S(n, k). A terminalclique Sw(1, k) is called a free clique with respect to S if S ∩ V (Sw(1, k)) = ∅;otherwise, Sw(1, k) is called a nonfree clique with respect to S . A d-alliance S inS(n, k) is called a dn-alliance if there exists a free clique with respect to S , namelyS ∩ V (Sw(1, k)) = ∅ for some Sw(1, k); otherwise, it is called a dn-alliance. A dn-alliance (respectively, dn-alliance) S in S(n, k) is denoted by Sf (respectively, Sn).

Lemma 3.1 If there exists a vertex v ∈ S in Sw(1, k), then |S ∩ V (Sw(1, k))| � k+12

if k is odd; otherwise, |S ∩ V (Sw(1, k))| � k2 , where S is a d-alliance in S(n, k).

Proof If v ∈ S is an extreme vertex, then |N [v]| = k. This implies that |S ∩V (Sw(1, k))| = |S ∩ N [v]| � k+1

2 if k is odd; otherwise, |S ∩ V (Sw(1, k))| > k2 . If

v is a non-extreme vertex, then |N [v]| = k + 1. Thus, if k is odd, then |S ∩ N [v]| �k+1

2 + 1; otherwise, |S ∩N [v]| � k2 + 1. Since the outer vertex of v may also be in S ,

this results in |S ∩ V (Sw(1, k))| � k+12 when k is odd and |S ∩ Sw(1, k)| � k

2 whenk is even. �

Corollary 3.2 For odd k � 3, |Sn| � k+12 · kn−1, and, for even k � 4, |Sn| � k

2 · kn−1,where n � 2.

To prove that γd(S(n, k)) is equal to the bound described in Corollary 3.2, we shall

show that there exists a strong defensive alliance S whose cardinality is exactly equal


to that bound. Note that, by the proof of Lemma 3.1, we have to ensure that, for eachnon-extreme vertex in S , its outer vertex is also in S .

In the following, the plus operation on computing the label of a vertex is alwaystaken modulo k. However, if the resulting value is 0, then we use k to replace it.Based on the possible values of k, we define the set DS(n, k) of vertices for differentSierpinski graphs S(n, k) as follows:

DS(n, k) =

⎧⎪⎪⎨

⎪⎪⎩

{wab|ab ∈ (X × Y) ∪ (Y × X)} if k is even,

{wab|b = a + i,− k−14 � i � k−1

4 } if k is odd and k−12 is even,

{wab|b = a + i, k+14 � i � 3k−1

4 } if both k and k−12 are odd,

where X = {1,2, . . . , k2 } and Y = { k

2 + 1, k2 + 2, . . . , k}.

For example, see Figs. 5(a)–5(c). In Fig. 5(a), since k is even, X = {1,2} andY = {3,4}. Thus X ×Y = {13,14,23,24}, Y ×X = {31,32,41,42}, and DS(2,4) ={13,14,23,24,31,32,41,42}. Figure 5(b) shows the set DS(n, k) when k is oddand k−1

2 is even while Fig. 5(c) is for the case where both k and k−12 are odd. We

can find that DS(2,5) and DS(2,7) contain all black vertices in Figs. 5(b) and 5(c),respectively.

Proposition 3.3 For every terminal clique Sw(1, k) in S(n, k), |V (Sw(1, k)) ∩DS(n, k)| = � k+1

2 �.

Proof By counting the number of elements in Y and the possible values of i in thedefinition of DS(n, k), the proposition follows. �

Proposition 3.4 |DS(n, k)| = k+12 ·kn−1 if k is odd; otherwise, |DS(n, k)| = k

2 ·kn−1.

Proof By Proposition 3.3, |V (Sw(1, k)) ∩ DS(n, k)| = � k+12 �. If k is even, then

|DS(n, k)| = kn−1 · � k+12 � = kn−1 · k

2 . If k is odd and k−12 is even, then |DS(n, k)| =

kn−2 · k · � k+12 � = k+1

2 · kn−1. If k is odd and k−12 is odd, then |DS(n, k)| =

kn−2 · k · � k+12 � = k+1

2 · kn−1. �

Proposition 3.5 If v ∈ DS(n, k) is an extreme vertex, then k is odd and k−12 is even.

Proof If v is an extreme vertex, then v = bn. By inspection on the definition ofDS(n, k), k cannot be an even number. Furthermore, for the case where both k andk−1

2 are odd, i cannot be equal to 0 in the equation b = a + i. This establishes theproof. �

Lemma 3.6 If v ∈ DS(n, k) is not an extreme vertex, then the outer vertex of v mustbe in DS(n, k).

Proof Since v is not an extreme vertex, we may assume that �(v) = wb, where w =w1w2 . . .wn−1. According to the construction of DS(n, k), we consider the followingcases.


Fig. 5 Illustrations of DS(n, k)

Case 1. k is even.By the definition of DS(n, k) for this case, wn−1 �= b, and if b is in X, then wn−1

must be in Y and vice versa, where X = {2,3, . . . , k2 + 1} and Y = {1, k

2 + 2, k2 +

3, . . . , k}. Thus w1w2 . . .wn−2bwn−1 is the outer vertex of v and is in DS(n, k).Case 2. k is odd and k−1

2 is even.


In this case, b = wn−1 + i, for some − k−14 � i � k−1

4 . If i = 0, then b = wn−1 andthe outer vertex of v is w1w2 . . .wj−1b(wj )

n−j , where j is the maximum index suchthat wj �= b in w. By definition, w1w2 . . .wj−1b(wj )

n−j is in DS(n, k). If i �= 0,then b �= wn−1 and the outer vertex of v is u = w1w2 . . .wn−2bwn−1. To prove that u

is in DS(n, k), we have to prove that there exists j such that wn−1 = b + j for some− k−1

4 � j � k−14 . By comparing b = wn−1 + i and wn−1 = b+j , this yields i = −j .

Since − k−14 � i � k−1

4 implies − k−14 � j � k−1

4 , u is also in DS(n, k).Case 3. Both k and k−1

2 are odd.In this case, b = wn−1 + i, for some k+1

4 � i � 3k−14 . Note that, in this case, i

cannot be equal to 0. We only need to consider the case where i �= 0. By a similarargument as in Case 2, the outer vertex of v is u = w1w2 . . .wn−2bwn−1. It remainsto prove that there exists j such that wn−1 = b + j for some k+1

4 � j � 3k−14 . Since

wn−1 = b + j is taken modulo k, it is equivalent to proving that k+14 � k + j �

3k−14 . Clearly, i = −j . By replacing i with −j in the inequality k+1

4 � i � 3k−14 ,

this yields − k+14 � j � − 3k−1

4 . Adding k in each term of the above inequality, thisresults in k+1

4 � k + j � 3k−14 . Therefore, u ∈ DS(n, k). This concludes the proof of

this lemma. �

Lemma 3.7 DS(n, k) is a dn-alliance in S(n, k).

Proof By the definition of DS(n, k), DS(n, k) ∩ V (Sw(1, k)) �= ∅ for every terminalclique Sw(1, k). This implies that every vertex in S(n, k) is dominated by some ver-tex in DS(n, k). It remains to show that, for every v ∈ DS(n, k), |NDS(n,k)[v]| >

|NV \DS(n,k)(v)|. By Proposition 3.3, |V (Sw(1, k)) ∩ DS(n, k)| = � k+12 �. Let v ∈

DS(n, k) be a vertex in Sw(1, k). If v is an extreme vertex, then, by Proposition 3.5,k is odd and |N [v]| = k. Thus, |V (Sw(1, k)) ∩ DS(n, k)| = k+1

2 and |NDS(n,k)[v]| >

|NV \DS(n,k)(v)|. If v is not an extreme vertex, then |N [v]| = k + 1. By Lemma 3.6,the outer vertex of v is also in DS(n, k). Therefore, |NDS(n,k)[v]| = |V (Sw(1, k)) ∩DS(n, k)| + 1 = � k+1

2 � + 1 > k − � k+12 � = k + 1 − (� k+1

2 � + 1) = |NV \DS(n,k)(v)|.This completes the proof. �

Lemma 3.8 For n � 2, |Sf| > |Sn| except that |Sf| � |Sn| when k = 3.

Proof Assume that Sx(1, k) is a free clique with respect to Sf. Thus, for every vertexin Sx(1, k), its outer vertex must be in Sf. Let Y be the set of vertices whose outervertices are in Sx(1, k). Note that if Sx(1, k) is a free clique with respect to S , thenSx(1, k) cannot contain an extreme vertex; otherwise, the extreme vertex in Sx(1, k)

cannot be dominated by any vertex in Sf. Moreover, there are k − 1 vertices in Y

which are in the same Sw(2, k) as Sx(1, k) located. Let Sy(1, k) be a terminal cliquecontaining a vertex, say v, in Y . By Lemma 3.1 and considering the outer vertex of v,|V (Sy(1, k)) ∩ S| � � k+1

2 � + 1. Assume that there are h free cliques with respect toSf. Then, |Sf| � (� k+1

2 �+ 1) · (k − 1) ·h+ (kn−2 −h) · k · k+12 = k+1

2 · kn−1 + k−32 ·h

if k is odd; and |Sf| � (� k+12 �+1) · (k −1) ·h+ (kn−2 −h) · k · k

2 = k2 · kn−1 + k−2

2 ·hif k is even. By inspection, |Sf| > |Sn| except that |Sf| might be equal to |Sn| whenk = 3. �


Fig. 6 |Sn| = 32 and |Sf| = 36 in S(3,4)

Fig. 7 |Sn| = 18 and |Sf| = 18 in S(3,3)

For example, see Figs. 6 and 7. Figure 6(a) depicts DS(3,4) which is an Sn. Fig-ure 6(b) shows an Sf with 3 free cliques in S(3,4). We can find out the fact that ifa terminal clique is a free clique in some S(2, k) for k even, then each of the otherterminal cliques in the same S(2, k) has k

2 + 1 vertices in Sf. Thus, it is easy to ver-ify that |Sf| > |Sn| when k � 4 is an even integer. Figures 7(a) and 7(b) are used toillustrate the case where k is odd. We can find that |Sf| = |Sn| when k = 3.

Theorem 3.9 For S(n, k) with n � 2,

γd

(S(n, k)

) ={

k+12 · kn−1 if k is odd,

k2 · kn−1 otherwise.

Proof By Corollary 3.2, Proposition 3.4, and Lemmas 3.7 and 3.8, this theoremholds. �


4 Computing γd(S+(n,k))

Since S+(n, k) is obtained from S(n, k) by adding a new vertex s and edges joining s

with all extreme vertices in S(n, k), S+(n, k) also consists Si(n−1, k) as a subgraph,for i = 1,2, . . . , k. Let S be the minimum d-alliance of S+(n, k). In the following,we investigate γ

d(S+(n, k)) for k > 3 and k = 3, respectively.

Lemma 4.1 γd(S+(n, k)) � γ

d(S(n, k)) + 1.

Proof By the construction of DS(n, k), if all extreme vertices are in DS(n, k), thenDS(n, k) ∪ {s} is a d-alliance of S+(n, k); otherwise, DS(n, k) ∪ {an} is a d-allianceof S+(n, k), where vertex an is not in DS(n, k). This completes the proof. �

Lemma 4.2 For any d-alliance S of S+(n, k), |S| � γd(S(n, k))+1 when k > 3 and

n � 2.

Proof Since S(n, k) is a subgraph of S+(n, k), by Lemma 3.8, |S| > γd(S(n, k))

when k > 3 if there exists a free clique with respect to S in subgraph S(n, k). Thus,in this case, |S| � γ

d(S(n, k)) + 1.

Now we consider the case where there is no free clique with respect to S in sub-graph S(n, k). Note that S+(n, k) is a regular graph and |N [v]| = k + 1 for eachvertex v in S+(n, k). We may assume that s is not in S and there is an extremevertex an in S dominating s. Assume also that an is in terminal clique Sx(1, k). Itis clear that |S ∩ V (Sx(1, k))| > k+1

2 if k is odd; otherwise, |S ∩ V (Sw(1, k))| � k+1

2 � = k2 + 1. For every vertex v ∈ S ∩ V (Sw(1, k)) with w �= x, by Lemma 3.1,

|S ∩ V (Sw(1, k))| � k+12 if k is odd; otherwise, |S ∩ V (Sw(1, k))| � k

2 . Therefore,

|S| � ∑kn−1−1i=1 |S ∩ V (Sw(1, k))| + |S ∩ V (Sx(1, k))|. After replacing and simplify-

ing, |S| � k+12 · kn−1 + 1 if k is odd and |S| � k

2 · kn−1 + 1 if k is even. Comparingthe inequality of |S| with the value of γ

d(S(n, k)) in Theorem 3.9, this results in

|S| � γd(S(n, k)) + 1. �

Define the set DS+(n,3) = {wabc|c ∈ {1,2,3}, ab /∈ {13,21,32}} for S+(n,3)

with n � 3. See Fig. 8 for an example in which D+S (3,3) contains all black vertices.

Lemma 4.3 DS+(n,3) is a d-alliance in S+(n,3).

Proof By the definition of DS+(n,3), all vertices are in DS+(n,3) except s and allvertices in terminal cliques Sw(1, k) with w ended by 13,21, and 32. Thus if a vertexv ∈ DS+(n,3) (respectively, v /∈ DS+(n,3)) is in terminal clique Sw(1,3), then allother vertices in Sw(1,3) are also in (respectively, not in) DS+(n,3). This meansthat, for every v ∈ DS+(n,3), |NDS+ (n,3)[v]| > |NV \DS+ (n,3)(v)|, and DS+(n,3) is astrong alliance. It is clear that all extreme vertices of S+(n,3) are in DS+(n,3). Thuss is dominated by some vertex in DS+(n,3). If v(�= s) is not in DS+(n,3), then, byinspection, its outer vertex must be in DS+(n,3). Therefore, DS+(n,3) is a d-alliancein S+(n, k). �


Fig. 8 Global strong defensivealliance of S+(3,3)

Lemma 4.4 For S+(n,3) with n > 2, γd(S+(n,3)) = γ

d(S(n,3)).

Proof Assume that S is a d-alliance of S+(n, k). If S is a dn-alliance in S(n, k), thenby Corollary 3.2, |S| � 2 · 3n−1. For the case where S is a df-alliance in S(n, k), weassume that Sx(1,3) is a free clique with respect to S and v ∈ V (Sx(1,3)). Then, bya similar argument as in Lemma 3.8, we can derive that |S| � k+1

2 · kn−1 + k−32 · h =

2 · 3n−1, where h is the number of free cliques with respect to S in S+(n, k). Con-sequently, |S| � 2 · 3n−1 no matter what S is a dn- or df-alliance of S+(n, k). ByLemma 4.3, DS+(n,3) is a d-alliance in S+(n, k) and |DS+(n,3)| = 3n−2 · 6 =2 · 3n−1. This results in γ

d(S+(n,3)) = 2 · 3n−1. Comparing with the value of

γd(S(n,3)) in Theorem 3.9, this yields γ

d(S+(n,3)) = γ

d(S(n,3)) and the lemma

follows. �

Theorem 4.5 For S+(n, k) with n � 2,

γd

(S+(n, k)

) =

⎧⎪⎪⎨

⎪⎪⎩

7 if n = 2 and k = 3,

γd(S(n,3)) if n > 2 and k = 3,

γd(S(n, k)) + 1 if n � 2 and k > 3.

Proof It is easy to verify that γd(S+(2,3)) = 7. Then, by Lemmas 4.1, 4.2 and 4.4,

this theorem follows immediately. �

5 Computing γd(S++(n,k))

A vertex v is an extreme vertex in S++(n, k) if it is either an extreme vertex in S(n, k)

or its label �(v) = (k + 1)an−1, for 1 � a � k, in Sk+1(n − 1, k). For S++(n, k) withn > 2, denote by DS++(n, k) = DS(n, k) ∪ D

S(n − 1, k), where we use D

S(n − 1, k)

to denote the set DS(n − 1, k) for Sk+1(n − 1, k) in S++(n, k). In the following, weinvestigate γ

d(S++(n, k)) for n = 2 and n > 2 separately.


Lemma 5.1 For S++(n, k) with n > 2, DS++(n, k) is a d-alliance of S++(n, k).

Proof Since DS(n, k) and DS(n− 1, k) are d-alliances of S(n, k) and Sk+1(n− 1, k)

respectively, DS++(n, k) is a dominating set of S++(n, k). All we have to prove isthat, for every v ∈ DS++(n, k), |NDS++ (n,k)[v]| > |NV \DS++ (n,k)(v)|. Let E standfor the set of extreme vertices in S++(n, k). By the definition of DS(n, k), E ⊂DS++(n, k) only when k is odd and k−1

2 is even. Note that, in S++(n, k), the outervertex of any extreme vertex is also an extreme vertex. By a similar argument asin Lemma 3.7, for every v ∈ DS++(n, k), |NDS++ (n,k)[v]| > |NV \DS++ (n,k)(v)|. Thiscompletes the proof. �

Corollary 5.2 For n > 2, γd(S++(n, k)) � γ

d(S(n, k)) + γ

d(S(n − 1, k)).

Lemma 5.3 For any d-alliance S of S++(n, k) with n > 2, |S| � γd(S(n, k)) +

γd(S(n − 1, k)).

Proof Suppose to the contrary that there exists a d-alliance S of S++(n, k) with|S| < γ

d(S(n, k)) + γ

d(S(n − 1, k)). Then either |S ∩ V (S(n, k))| < γ

d(S(n, k)) or

|S ∩ V (Sk+1(n − 1, k))| < γd(S(n − 1, k)) must hold. We only consider the case that

|S ∩ V (S(n, k))| < γd(S(n, k)). The case that |S ∩ V (Sk+1(n − 1, k))| < γ

d(S(n −

1, k)) can be handled similarly.Let S ′ = S ∩ V (S(n, k)). By Proposition 3.4, if k is odd, then |S ′| < k+1

2 · kn−1;

otherwise, |S ′| < k2 · kn−1. By Theorem 3.9, S ′ is not a d-alliance in the subgraph

S(n, k). This means that it is either the case that there exists an extreme vertex v ∈V (S(n, k)) \ S ′ which is not dominated by any vertex in S ′ or the case that there isan extreme vertex v ∈ S ′ with |NS ′ [v]| = |NV (S(n,k))\S ′(v)| in S(n, k).

If it is the former case, then assume that v = an ∈ V (S(n, k))\ S ′ is not dominatedby any vertex in S ′. This implies that San−1(1, k) is a free clique with respect to S ′.By using a similar argument as in Lemma 3.8, |S ′| � k+1

2 ·kn−1 if k is odd, otherwise,|S ′| � k

2 · kn−1, a contradiction.For the latter case, we assume that extreme vertex v ∈ S ′ ∩ V (S(n, k)) is with

|NS ′ [v]| = |NV (S(n,k))\S ′(v)| in S(n, k). This means that k must be even and eachterminal clique in S(n, k) has at least k

2 vertices in S ′. This implies that there are atleast k

2 · kn−1 vertices in S ′ which is a contradiction. This concludes the proof of thislemma. �

Now we consider the case where n = 2, i.e., S++(2, k). To construct a d-allianceof S++(2, k), we define the following sets:

A1 ={

v|�(v) = a2 or (k + 1)a,1 � a �⌈

k

4

⌉

ork

2+ 1 � a �

⌈3k

4

⌉}

,

R1 ={

ab|ab = i

(k

2+ i

)

or

(k

2+ i

)

i,1 � i �⌈

k

4

⌉}

,

A2 ={

v|�(v) = ab or ba, b = a − k + 3

4, a ∈ X

}

∪ {(k + 1)b|b �∈ X ∪ Y

},


R2 ={

a2|2 � a � k + 3

4or

3k + 5

4� a � k

}

,

A3 ={

v|�(v) = a2 or (k + 1)a,1 � a � k + 1

2

}

, and

R3 ={

v|�(v) = ab or ba, b = a + k + 1

4,1 � a � k + 1

4

}

,

where X = {2,3, . . . , k+34 } and Y = { 3k+5

4 , 3k+54 + 1, . . . , k}. The set DS++(2, k) is

defined as follows:

DS++(2, k) =

⎧⎪⎪⎨

⎪⎪⎩

(DS(2, k) ∪ A1) \ R1 if k is even,

(DS(2, k) ∪ A2) \ R2 if k is odd and k−12 is even,

(DS(2, k) ∪ A3) \ R3 otherwise.

For example, see Fig. 9. In Fig. 9(a), A1 = {11,51,33,53} and R1 = {13,31}.Thus, after removing the vertices in R1 from and adding the vertices in A1 toDS(2, k), this yields DS++(2,5) containing all black vertices in Fig. 9(a). Similarly,we can find that A2 = {25,52,61,63,64}, R2 = {22,55}, A3 = {11,22,33,44,81,

82,83,84} and R3 = {13,31,24,42} in Figs. 9(b) and 9(c), respectively. It is easy toverify DS++(2,6) and DS++(2,7) contain the black vertices in Figs. 9(b) and 9(c),respectively.

Lemma 5.4 If v is in DS++(2, k), then its outer vertex is also in DS++(2, k).

Proof By inspection, if vertex v is in Ai (respectively, Ri ), for i = 1 or 3, then itsouter vertex is also in Ai (respectively, Ri ). By Lemma 3.6, if v ∈ DS(n, k) is not anextreme vertex, then the outer vertex of v must be in DS(n, k). By Proposition 3.5, ifv ∈ DS(n, k) is an extreme vertex, then k is odd and k−1

2 is even. Thus all we have toconsider is the case that k is odd and k−1

2 is even. By the definition of DS++(2, k) =(DS(2, k) ∪ A2) \ R2, it is easy to verify that if v is in DS++(2, k), then its outervertex is also in DS++(2, k) no matter whether v is an extreme vertex or not. �

Lemma 5.5 DS++(2, k) is a d-alliance of S++(2, k).

Proof Since there exists at least one vertex v ∈ DS++(2, k) in each terminal clique,every vertex u �∈ DS++(2, k) is dominated by a vertex in DS++(2, k). It remains toprove that |NDS++ (2,k)[v]| > |NV \DS++ (2,k)(v)|. According to the construction ofDS++(2, k), we consider the following three cases.

Case 1. k is even.In this case, DS++(2, k) = (DS(2, k) ∪ A1) \ R1. We can find the fact that if

a vertex in R1, say v, which is in terminal clique Sw with �(v) = i( k2 + i) (re-

spectively, ( k2 + i)i) is removed from DS(2, k), then an extreme vertex i2 ∈ A1

(respectively, ( k2 + i)2 ∈ A1) in Sw is added into DS++(2, k). Furthermore, an

extreme vertex in Sk+1 with label (k + 1)i (respectively, (k + 1)( k2 + i) is also


Fig. 9 DS++ (2,5) and DS++ (2,7)

added into DS++(2, k) which is the outer vertex of i2 (respectively, ( k2 + i)2). Thus

DS(2, k) ∩ V (Sw(1, k)) = DS++(2, k) ∩ V (Sw(1, k)) for each terminal clique insubgraph S(2, k) of S++(2, k). By Proposition 3.3 and Lemma 5.4, for each v ∈DS(2, k)∩V (Sw(1, k)), |NDS++ (2,k)[v]| = � k+1

2 �+1 = k2 +1 which is clearly greater

than |NV \DS++ (2,k)(v)|. The number of vertices in DS++(2, k) ∩ V (Sk+1(1, k)) is

equal to k4� + 3k

4 � − k2 which is equal to k

2 (respectively, k2 + 1) when k

2 is


even (respectively, odd). This also results in |NDS++ (2,k)[v]| > |NV \DS++ (2,k)(v)| forv ∈ DS(2, k) ∩ V (Sw(1, k)). Thus this case holds.

Case 2. k is odd and k−12 is even.

In this case, DS++(2, k) = (DS(2, k) ∪ A2) \ R2. We can find that the number ofextreme vertices in {(k + 1)b|b �∈ X ∪ Y } is equal to k − ( k+3

4 − 1 + k − 3k+54 +

1) = k+12 . Furthermore, the number of vertices in DS++(2, k) ∩ V (Sw(1, k)) remains

unchanged on the construction of DS++(2, k) for every terminal clique Sw(1, k) inDS(2, k). By Proposition 3.3 and Lemma 5.4, for each v ∈ DS(2, k) ∩ V (Sw(1, k)),|DS++(2, k) ∩ V (Sw(1, k))| = � k+1

2 � + 1 = k+32 and |NV \DS++ (2,k)(v)| = k − k+1

2 =k−1

2 , where Sw(1, k) is a terminal clique in S++(2, k). This case is also holds.Case 3. k is odd and k−1

2 is odd.By using a similar argument as in Case 2, this case holds. This completes the

proof. �

Corollary 5.6 For S++(2, k) with k � 3,

γd

(S++(2, k)

)�

⎧⎪⎪⎨

⎪⎪⎩

γd(S(2, k)) + k

2 if 0 ∼= k(mod 4),

γd(S(2, k)) + k

2 + 1 if 2 ∼= k(mod 4),

γd(S(2, k)) + k+1

2 otherwise.

Lemma 5.7 For any d-alliance S in S++(2, k) with k � 3,

|S| �

⎧⎪⎪⎨

⎪⎪⎩

γd(S(2, k)) + k

2 if 0 ∼= k(mod 4),

γd(S(2, k)) + k

2 + 1 if 2 ∼= k(mod 4),

γd(S(2, k)) + k+1

2 otherwise.

Proof According to whether Sk+1(1, k) is a free clique or not with respect to S , wehave two cases to consider:

Case 1. |S ∩ V (Sk+1(1, k))| = ∅.In this case, every vertex in Sk+1(1, k) has to be dominated by its outer vertex.

Thus all extreme vertices i2, for 1 � i � k, must be in S . Clearly, |N [v]| = k + 1 foreach v ∈ S and |NS(2,k)[v] ∩ S| � k+1

2 + 1 if k is odd; otherwise, |NS(2,k)[v] ∩ S| �k2 + 1. Thus, if k is odd, then |S| � ( k+1

2 + 1) · k = ( k+12 · k) + k = γ

d(S(2, k)) + k;

otherwise, |S| � ( k2 + 1) · k = k

2 · k + k = γd(S(2, k)) + k.

Case 2. |S ∩ V (Sk+1(1, k))| �= ∅.Let v ∈ S ∩V (Sk+1(1, k)). Then |N [v]∩ S| � k+1

2 +1 if k is odd and |N [v]∩ S| �k2 + 1 otherwise. It is obvious that the smallest value of |S ∩ V (Sw(1, k))| is equal tok+1

2 for odd k and k2 for even k. Furthermore, if k is odd, then |S| � k+1

2 · (k + 1) =γd(S(2, k)) + k+1

2 ; otherwise, |S| � k2 · (k + 1) = γ

d(S(2, k)) + � k+1

2 �.However, for the case that 2 ∼= k(mod 4), |S ∩ V (Sw(1, k))| cannot be equal to k

2 .The reason is that if |S ∩ V (Sw(1, k))| is equal to the smallest value, then, for everyv ∈ S , its outer vertex must be also in S . This means that if S contains only thosepairs of vertices, then |S| is an even number. Unfortunately, when 2 ∼= k(mod 4),


k2 · (k + 1) is not an even number, and the minimum lower bound for |S| is greaterthan k

2 · (k+1). Accordingly, there is at least one vertex in S whose outer vertex is notin S when 2 ∼= k(mod 4). This implies that, when 2 ∼= k(mod 4), |S| � k

2 ·k+ k2 +1 =

γd(S(2, k)) + k

2 + 1. This establishes the proof. �

By Theorem 3.9, Corollaries 5.2 and 5.6, and Lemmas 5.3 and 5.7, we have thefollowing theorem.

Theorem 5.8 For S++(n, k) with n � 2,

γd

(S++(n, k)

) ={

γd(S(2, k)) + k

2 + 1 if 2 ∼= k(mod 4),

γd(S(n, k)) + γ

d(S(n − 1, k)) otherwise.

6 Computing γd(Sn)

Recall that, all vertices in Sierpinski gasket graphs Sn are contracted from the ver-tices in S(n, k) except extreme vertices. The label of every contracted vertex can beexpressed as w(abh|bah) for some 1 � h � n− 1 where the possible value-pairs of a

and b are: 1 and 2, 1 and 3, or 2 and 3. By using an exhaustive search algorithm, wecan find that γ

d(S2) = 3 and γ

d(S3) = 8. In the rest of this section, we only investigate

γd(Sn), for n � 4.First, we introduce a relaxation of the global strong defensive alliance problem

on Sierpinski gasket graphs Sn. A nonempty set S ⊆ V (Sn) is called an improper d-alliance, denoted di-alliance, if and only if S is a dominating set of V \ {1n,2n,3n}and, for every v ∈ S \ {1n,2n,3n}, |NS[v]| > |NV \S(v)|. A di-alliance S of Sn iscalled an e-element di-alliance if |S ∩{1n,2n,3n}| = e. The cardinality of a minimume-element di-alliance is denoted by γ e

di(Sn). The improper global strong defensive

alliance number of Sn, denoted by γdi(Sn), is equal to min0�e�3{γ e

di(Sn)}.

Proposition 6.1 γd(Sn) � γ

di(Sn).

By using an exhaustive search, we can find that the values of γ e

di(S3), for 0 � e � 3,

as shown in Proposition 6.2 (see Figs. 10(a)–10(d), respectively).

Proposition 6.2 For S3, γ 0di(S3) = 6, γ 1

di(S3) = γ 2

di(S3) = 7, and γ 3

di(S3) = 8.

Now we prove the following lemma.

Lemma 6.3 For Sn with n � 3, γ e

di(Sn) � 2 · 3n−2 + e

2�, for 0 � e � 3.

Proof We prove this lemma by induction on k. By Proposition 6.2, the basis holdsimmediately. In the induction step, we shall prove that the statement holds for n = k,for k > 3. Recall that Sk contains three copies of Sk−1 which are Sk−1,a , for 1 �


Fig. 10 Minimum e-elementdi-alliances in S3

a � 3. Let Se be an e-element di-alliance in Sk , for 0 � e � 3, and let S ae be a di-

alliance in Sk−1,a , for 1 � a � 3. By taking into consideration the possible e-elementdi-alliances in Sk , we have four cases to consider.

Case 1: e = 0.In this case, all extreme vertices of Sk are not in S0. We consider all possible

numbers of waist vertices in S0. If none of waist vertices of Sk is in S0, then all S a0 ,

for 1 � a � 3, can only be 0-element di-alliance in Sk−1,a (see Fig. 11(a)). By theinduction hypothesis, this yields |S0| = ∑3

a=1 |S0 ∩V (Sk−1,a)| = ∑3a=1 |S a

0 | � 3 ·2 ·3(k−1)−2 = 2 ·3k−2. If exact one waist vertex of Sk is in S0, then one of Sk−1,a , for 1 �a � 3, contains a 0-element di-alliance and each of the other two Sk−1,a contains a1-element di-alliance (see Fig. 11(b)). We may assume without loss of generality thatthe former is Sk−1,3 and the latter contains Sk−1,1 and Sk−1,2. Thus, by the inductionhypothesis, |S0| = |S0 ∩ V (Sk−1,1)| + |S0 ∩ V (Sk−1,2)| + |S0 ∩ V (Sk−1,3)| − 1 =|S 1

1 | + |S 21 | + |S 3

0 | − 1 � 2(2 · 3(k−1)−2 + 1) + (2 · 3(k−1)−2) − 1 = 2 · 3k−2 + 1. Ifexact two waist vertices of Sk are in S0, then one of Sk−1,a , for 1 � a � 3, containsa 2-element di-alliance and each of the other two Sk−1,a contains a 1-element di-alliance (see Fig. 11(c)). We may assume without loss of generality that the former isSk−1,1 and the latter contains Sk−1,2 and Sk−1,3. Thus, by the induction hypothesis,|S0| = |S0 ∩ V (Sk−1,1)| + |S0 ∩ V (Sk−1,2)| + |S0 ∩ V (Sk−1,3)| − 2 = |S 1

2 | + |S 21 | +

|S 31 | − 2 � 3(2 · 3(k−1)−2 + 1) − 2 = 2 · 3k−2 + 1. If all waist vertices of Sk are in S0,

then |S0| = |S 12 |+|S 2

2 |+|S 32 |−3 � 3(2 ·3(k−1)−2 +1)−3 = 2 ·3k−2 (see Fig. 11(d)).

Therefore, in this case, |S0| � 2 · 3k−2.Case 2: e = 1.We may assume without loss of generality that 1k ∈ S1. If none of waist vertices

of Sk is in S1, then, by the induction hypothesis, |S1| = |S1 ∩ V (Sk−1,1)| + |S1 ∩V (Sk−1,2)| + |S1 ∩ V (Sk−1,3)| = |S 1

1 | + |S 20 | + |S 3

0 | � (2 · 3(k−1)−2 + 1) + 2 · 2 ·3(k−1)−2 = 2 · 3k−2 + 1 (see Fig. 12(a)).


Fig. 11 All possible numbersof waist vertices in S0

Fig. 12 All possible numbers of waist vertices in S1

If exact one waist vertex of Sk is in S1, then we have two subcases to consider.If this waist vertex is 12k−1|21k−1 (respectively, 13k−1|31k−1), then |S1| = |S1 ∩V (Sk−1,1)| + |S1 ∩ V (Sk−1,2)| + |S1 ∩ V (Sk−1,3)| − 1 = |S 1

2 | + |S 21 | + |S 3

0 | − 1 �2(2 ·3(k−1)−2 +1)+(2 ·3(k−1)−2)−1 = 2 ·3k−2 +1 (see Fig. 12(b)). If 23k−1|32k−1 isthe mentioned waist vertex in S1, then |S1| = |S 1

1 |+|S 21 |+|S 3

1 |−1 � 3(2 ·3(k−1)−2 +1) − 1 = 2 · 3k−2 + 2 (see Fig. 12(c)).

If exact two waist vertices of Sk are in S1, then we also have two subcases toconsider. If these two waist vertices are 12k−1|21k−1 and 13k−1|31k−1, then |S1| =|S 1

3 |+|S 21 |+|S 3

1 |−2 � (2 ·3(k−1)−2 +2)+2(2 ·3(k−1)−2 +1)−2 = 2 ·3k−2 +2 (seeFig. 12(d)). If these two waist vertices are 13k−1|31k−1 and 23k−1|32k−1, then |S1| =


Fig. 13 DSg (4) in S4

|S 12 | + |S 2

1 | + |S 32 | − 2 � 2(2 · 3(k−1)−2 + 1) + (2 · 3(k−1)−2 + 1) − 2 = 2 · 3k−2 + 1

(see Fig. 12(e)).If all waist vertices of Sk are in S1, then |S1| = |S 1

3 | + |S 22 | + |S 3

2 | − 3 � (2 ·3(k−1)−2 + 2) + 2(2 · 3(k−1)−2 + 1) − 3 = 2 · 3k−2 + 1 (see Fig. 12(f)). Therefore, inthis case, |S1| � 6 · 3k−3 + 1.

Case 3: e = 2.Analogously, by considering all possible numbers of waist vertices in S2, we can

obtain that |S2| � 2 · 3k−2 + 1 in this case.Case 4: e = 3.By using a similar argument as in the above case, we can obtain that |S3| � 2 ·

3k−2 + 2 in this case. This completes the proof. �

Corollary 6.4 For Sn with n � 4, γd(Sn) � 2 · 3n−2.

Proof By definition, γdi(Sn) = min0�e�3{γ e

di(Sn)}. This implies that γ

d(Sn) �

γdi(Sn) = min0�e�3{γ e

di(Sn)} � 2 · 3n−2. Then, by Proposition 6.1, this corollary fol-

lows directly. �

Define DSg (n) = {v|vn−3vn−2vn−1vn ∈ {(1|2)(133|311), (1|3)(233|322), (2|3)

(122|211), (11|12|22|23)(13|31), (11|12|31|33)(23|32), (22|23|31|33)(12|21)}}. Forexample, DSg (4) is depicted in Fig. 13.

Lemma 6.5 For Sn with n � 4, DSg (n) is a d-alliance in Sn with |DSg (n)| = 2 ·3n−2.

Proof It is easy to verify that DSg (4) is a d-alliance in S4. For Sn with n > 4, it is

clear that every DSg(4) of S4,a , for a ∈ {1,2,3}n−4, is also a d-alliance of that S4,a ,

and thus DSg (n) is the union of all those DSg (4)’s which form a d-alliance of Sn.Accordingly, |DSg (n)| = 3n−4 · 2 · 32 = 6 · 3n−3. This completes the proof. �

We summarize the results in this section as the following theorem.


Theorem 6.6 For Sn with n � 1,

γd(Sn) =

⎧⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎩

2 if n = 1,

3 if n = 2,

8 if n = 3,

2 · 3n−2 if n � 4.

7 Concluding Remarks

In this paper, we derive exact values of d-alliance numbers for various Sierpinski-likegraphs. Furthermore, we also find a way to construct a d-alliance for each of them.In our proposed constructions, every vertex only needs to examine its own label todetermine whether it is in an alliance or not.

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http://dx.doi.org/10.1007/s00373-011-1076-4

http://dx.doi.org/10.1007/s00373-011-1076-4


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