Distributed algorithms for finding the unique minimum distance dominating set in directed split-stars

ARTICLE IN PRESS

J. Parallel Distrib. Comput. 63 (2003) 481–487

Distributed algorithms for finding the unique minimum distance dominating set in directed split-stars Fu-Hsing Wang,a Jou-Ming Chang,b Yue-Li Wang,a,
and Sun-Jen Huanga a

Department of Information Management, National Taiwan University of Science and Technology, 43, Section 4, Kee-Lung Road, Taipei 106, Taiwan, Republic of China b Department of Information Management, National Taipei College of Business, Taipei, Taiwan, Republic of China Received 1 March 2002; revised 25 February 2003; accepted 20 March 2003

Abstract A distance-k dominating set D of a directed graph G is a set of vertices such that for every vertex v of G; there is a vertex uAD and the distance between u and v is at most k: Minimum distance-k dominating set is especially important in communication networks for distributed data structures and for server placement. In this paper, we show that there is a unique minimum distance-k dominating set for k ¼ 1; 2 in a directed split-star, which has recently been developed as a new model of the interconnection network for parallel and distributed computing systems. Moreover, we shall present simple distributed algorithms for finding such sets. r 2003 Elsevier Science (USA). All rights reserved. Keywords: Distributed algorithms; Distance-k dominating set; Interconnection networks; Directed split-stars

1. Introduction Researches on domination and related topics have attracted graph theorists for their strong practical applications and theoretical interest. The motivation of studying distance domination in a graph arises mainly from communication network design in a real world. Over the decades, people have realized the significance of data communication. As usual, a communication network is modeled by an undirected graph that nodes and edges in a graph correspond to the communication sites and links, respectively. Distance k-dominating set has important applications in several fields. For example, Slater [19] showed that the distance domination problem can be applied to communication networks. A set of distance k-dominating sets can be selected to be servers [3] or copies of a distributed directory [18] so that every node is within distance k of some server or directory copy, respectively. Dominating sets have also been proposed for storing location information of the network nodes in the mobile adhoc network routing [17]. Recently, Jia et al. gave an efficient distributed algorithm for constructing small


Corresponding author. Fax: +886-02-27376777. E-mail address: [email protected] (Y.-L. Wang).

dominating sets [16]. However, the minimum dominating set problem is NP-complete in general graphs [12]. Fortunately, polynomial time solutions exist in many special graphs. A thorough treatment of domination in graphs and digraphs can be found in [13,14]. Star graphs provide an attractive interconnection scheme for parallel computers and distributed networks. Hence, characterizations of this architecture have been widely investigated [1,2,10]. In [8], Cheng et al. gave a variant architecture of star graphs, known as split-stars, which can be viewed as companion graphs of alternating group graphs. The split-stars act as bridges between the star graphs and the alternating group graphs [9]. Some nice properties of split-stars are given in [6,7,9,5,4]. Later on, Cheng and Lipman [6] proposed an assignment of directions to the edges of split-stars as a new class of digraph called the directed split-stars. They also showed that the resulting digraphs are not only strongly connected, but, in fact, they have maximal arc-fault tolerance for connectivity and a small diameter. In this paper, we shall show that the directed split-stars have the unique minimum distance-k dominating set for k ¼ 1; 2: Furthermore, we shall present distributed algorithms for finding such sets. The remaining part of this paper is organized as follows. In Section 2, we give the definition of directed

0743-7315/03/$ - see front matter r 2003 Elsevier Science (USA). All rights reserved. doi:10.1016/S0743-7315(03)00040-6

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split-stars and introduce some basic terminologies and notations. In Section 3, we study the problem of finding the minimum domination sets on directed split-stars. A result shows that n-dimensional directed split-star has the unique minimum dominating set of size ðn  1Þ!: In Section 4, we further investigate the distance-2 domination in directed split-stars and obtain a similar result to that in the previous section. Section 5 presents two distributed algorithms for finding the unique minimum distance-k dominating set for k ¼ 1; 2; and shows the correctness of the proposed algorithms. Finally, we give concluding remarks and address our future research in the last section.

2. Preliminaries All graphs considered in this paper are finite and simple (i.e., without loops and multiple edges). Let G ¼ ðV ; AÞ be a directed graph (abbreviated digraph) with vertex set V and arc set A; where ADV  V : An arc /u; vS is said to be directed from u to v; in which case we say that u dominates v; and v is dominated by u: The outset of a vertex u in G is the set OðuÞ ¼ fv j vAV and /u; vSAAg; while the closed outset is O½u ¼ OðuÞ,fug: For a subset SDV ; we also define S OðSÞ ¼ uAS OðuÞ and O½S ¼ OðSÞ,S: The distance of two vertices u and v in a digraph G; denoted by dðu; vÞ; is the number of arcs along a shortest path from u to v; and dðu; vÞ ¼ N if there is no path from u to v in G: Let k be a positive integer. A set Dk of vertices in a digraph G ¼ ðV ; AÞ is called a distance-k dominating set if every vertex in V \Dk is within distance k from some vertex of Dk : A distance-k dominating set with the minimum number of vertices is called a minimum distance-k dominating set, and its cardinality is denoted by gk ðGÞ: The distance-k outset of a vertex u in G is the set Ok ðuÞ ¼ fv j vAV and dðu; vÞ ¼ kg; while the distance-k inset is Ik ðuÞ ¼ fv j vAV and dðv; uÞ ¼ kg: For simplicity, we write gðGÞ; OðuÞ and IðuÞ for k ¼ 1: Let odðuÞ and idðuÞ denote the outdegree and the indegree, respectively, of a vertex uAV ; i.e., odðuÞ ¼ jOðuÞj and idðuÞ ¼ jIðuÞj: Let DðGÞ be the maximum outdegree among all vertices in the digraph G: For terms of digraphs not defined here refer to [15]. ~2 is a digraph The n-dimensional directed split-star S n whose vertices are in a one-to-one correspondence with n! permutations ½p1 ; p2 ; y; pn  of the set N ¼ ~2 are connected f1; 2; y; ng; and two vertices u; v of S n by an arc /u; vS if and only if the permutation of v can be obtained from u by either the 2-exchange or a 3rotation. Let u ¼ ½p1 ; p2 ; y; pn : A 2-exchange interchanges the first symbol p1 with the second symbol p2 whenever p1 4p2 ; in which case v ¼ ½p2 ; p1 ; y; pn  is called a 2-exchange neighbor of u: A 3-rotation rotates the symbols in positions 1, 2 and i from left to

right for some iAf3; 4; y; ng; and in this case v ¼ ½pi ; p1 ; p3 ; y; pi1 ; p2 ; piþ1 ; y; pn  is said to be a ~2 is denoted 3-rotation neighbor of u: The vertex set of S n 2 ~ by V ðSn Þ: For simplicity, we also use V to represent ~2 Þ if not confusing. Fig. 1 depicts an example of S ~2 V ðS n n for n ¼ 4: 3. The unique minimum dominating set Let u ¼ ½p1 ; p2 ; y; pn : The 2-exchange neighbor of u is EðuÞ ¼ ½p2 ; p1 ; y; pn ; and the set of 3-rotation neighbors of u is RðuÞ ¼ fv j v ¼ ½pi ; p1 ; p3 ; y; pi1 ; p2 ; ~2 Þ; 3pipng: We use Ri ðuÞ to denote piþ1 ; y; pn AV ðS n the 3-rotation neighbor by rotating p1 ; p2 and pi : ~2 : If v is the 2Proposition 1. Let u be a vertex of S n exchange neighbor of u; then odðuÞ ¼ n  1 and odðvÞ ¼ n  2: Proof. By definition, u has n  2 3-rotation neighbors and a 2-exchange neighbor. Thus, odðuÞ ¼ n  1: In contrast with u; v has no 2-exchange neighbor and odðvÞ ¼ n  2: & ~2 Þ ¼ n  1; where DðS ~2 Þ is the In particular, DðS n n 2 ~ maximum outdegree of Sn : Clearly, there are n!2 vertices ~2 : Let D1 ¼ fv j v ¼ with the maximum outdegree in S n ~ 2 ½p1 ; 1; p3 ; y; pn AVðSn Þg: Þ It is obvious that D1 is an ~2 Þ ¼ n  1 ~2 and odðvÞ ¼ DðS independent set of S n n for every vertex vAD1 : ~2 Þ ¼ ðn  1Þ!: Lemma 2. gðS n ~2 can dominate at most Proof. Since any vertex in S n 2 ~ DðSn Þ þ 1 vertices including itself, the number of ~2 Þ þ 1Þ ¼ vertices dominated by D1 is at most jD1 j ðDðS n ~2 ðn  1Þ! n ¼ n!: Note that the number of vertices in S n 2 ~ is n!: If we can prove that no vertex in Sn is dominated by more than one vertex in D1 ; then D1 is a minimum ~2 and the lemma holds. dominating set of S n Suppose to the contrary that there exists a vertex rAV \D1 which is dominated by at least two vertices of D1 : Let u ¼ ½u1 ; 1; u3 ; y; un  and v ¼ ½v1 ; 1; v3 ; y; vn  be two distinct vertices in D1 and both of them dominate the vertex r: Since r is dominated by u; r is either the 2exchange neighbor or a 3-rotation neighbor of u: Assume that r is the 2-exchange neighbor of u; i.e., r ¼ ½1; u1 ; u3 ; y; un : Thus, r cannot be the 2-exchange neighbor of v: Since the symbol 1 is at the second position of v; r cannot also be a 3-rotation neighbor of v: This contradicts that r is a neighbor of vertex v: Therefore, r must be a 3-rotation neighbor of u: Let r ¼ Ri ðuÞ ¼ ½ui ; u1 ; u3 ; y; ui1 ; 1; uiþ1 ; y; un  for some iAf3; y; ng: Since the symbol 1 is not at the first position of r; r cannot be the 2-exchange neighbor of v:

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Fig. 1. Four-dimensional directed split-star.

It means that r must be a 3-rotation neighbor of v: Since the symbol 1 is at the ith position of r; it implies r ¼ Ri ðvÞ ¼ ½vi ; v1 ; v3 ; y; vi1 ; 1; viþ1 ; y; vn : It leads to u ¼ v; which is a contradiction. &

implies that there is a vertex in jD01 j which dominates ~2 Þ ¼ more than n  1 vertices. This contradicts that DðS n n  1: We conclude that D1 is the unique minimum ~2 : & dominating set of S n

~2 Þg: Theorem 3. Let D1 ¼ fv j v ¼ ½p1 ; 1; p3 ; y; pn AV ðS n ~ Then, D1 is the unique minimum dominating set of Sn2 :

4. The unique minimum distance-2 dominating set

Proof. Suppose that D01 is a minimum dominating set of ~2 and D0 aD1 : Then there exists a vertex u ¼ S n 1 ½u1 ; 1; u3 ; y; un  in D1 and ueD01 : Since D01 is a dominating set, u must be either the 2-exchange neighbor or a 3-rotation neighbor of some vertex vAD01 : However, u cannot be the 2-exchange neighbor of vertex v since the symbol 1 is at the second position of u and u1 41: Vertex u must be a 3-rotation neighbor of vertex v: Let u ¼ Ri ðvÞ where vAD01 and iAf3; y; ng: Then v ¼ ½1; ui ; u3 ; y; ui1 ; u1 ; uiþ1 ; y; un  which is the 2-exchange neighbor of vertex ½ui ; 1; u3 ; y; ui1 ; u1 ; uiþ1 ; y; un : By Proposition 1, odðvÞ ¼ n  2: It

The distance-2 outset of a vertex uAV is given by S O ðuÞ ¼ OðwÞ: For a set D2 DV ; O2 ðD2 Þ ¼ wAOðuÞ S2 O ðuÞ: Obviously, a set D2 is a distance-2 uAD2 2 dominating set if and only if O½D2 ,O2 ðD2 Þ ¼ V : We denote by EðD2 Þ and RðD2 Þ the sets of the 2-exchange neighbors of D2 and the 3-rotation neighbors of D2 ; respectively. That is, EðD2 Þ ¼ fEðvÞ j vAD2 g and RðD2 Þ ¼ fRi ðvÞ j vAD2 ; 3pipng: In order to give a simple representation of O2 ðD2 Þ; we use the following notation: REðD2 Þ ¼ RðEðD2 ÞÞ; ERðD2 Þ ¼ EðRðD2 ÞÞ; and RRðD2 Þ ¼ RðRðD2 ÞÞ: Note that REðD2 Þ, ERðD2 Þ,RRðD2 Þ ¼ O2 ðD2 Þ:

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~2 Þg: By deLet D2 ¼ fv j v ¼ ½2; 1; p3 ; y; pn AV ðS n ~2 Þg and finition, EðD2 Þ ¼ fv j v ¼ ½1; 2; p3 ; y; pn AV ðS n ~2 Þg: RðD2 Þ ¼ fv j v ¼ ½pi ; 2; p3 ; y; pi1 ; 1; piþ1 ; y; pn AV ðS n Also we have REðD2 Þ ¼ fv j v ¼ ½pi ; 1; p3 ; y; pi1 ; 2; ~2 Þg and ERðD2 Þ ¼ fv j v ¼ ½2; pi ; p3 ; y; piþ1 ; y; pn AV ðS n ~2 Þg: In addition, RRðD2 Þ can be pi1 ; 1; piþ1 ; y; pn AV ðS n partitioned into the following three sets in considering the positions affected by the second 3-rotation. Let RR1 ðD2 Þ ¼ fv j v ¼ ½pj ; pi ; p3 ; y; pi1 ; 1; piþ1 ; y; pj1 ; ~2 Þg; 2; pjþ1 ; y; pn AV ðS n RR2 ðD2 Þ ¼ fv j v ¼ ½pj ; pi ; p3 ; y; pj1 ; 2; pjþ1 ; y; pi1 ; ~2 Þg; and 1; piþ1 ; y; pn AV ðS n RR3 ðD2 Þ ¼ fv j v ¼ ½1; pi ; p3 ; y; pi1 ; 2; piþ1 ; y; pn A ~2 Þg: Thus, O2 ðD2 Þ ¼ ERðD2 Þ,REðD2 Þ,RR1 ðD2 Þ, V ðS n RR2 ðD2 Þ,RR3 ðD2 Þ: ~2 Þg: Lemma 4. Let D2 ¼ fv j v ¼ ½2; 1; p3 ; y; pn AV ðS n The following statements are true: (i) The sets REðD2 Þ; ERðD2 Þ; and RRðD2 Þ are pairwise disjoint. (ii) The sets D2 ; OðD2 Þ; and O2 ðD2 Þ are pairwise disjoint. (iii) For any two vertices u; vAOðD2 Þ; OðuÞ-OðvÞ ¼ |: Proof. By definition, O2 ðD2 Þ is divided into five disjoint subsets ERðD2 Þ; REðD2 Þ; RR1 ðD2 Þ; RR2 ðD2 Þ and RR3 ðD2 Þ: This shows that statement (i) holds. Since OðD2 Þ ¼ EðD2 Þ , RðD2 Þ ¼ fv j v ¼ ½p1 ; 2; p3 ; y; pn A ~2 Þg; OðD2 Þ contains all vertices with the symbol 2 at V ðS n the second position. So, OðD2 Þ-D2 ¼ |: Furthermore, there are no vertices of O2 ðD2 Þ having the symbols 2 and 1 in this order at the leftmost two positions, O2 ðD2 Þ-D2 ¼ |: Similarly, O2 ðD2 Þ-OðD2 Þ ¼ | since O2 ðD2 Þ contains no vertices with the symbol 2 at the second position. Therefore, statement (ii) holds. We now prove statement (iii) by contradiction. Let u and v be any two distinct vertices in OðD2 Þ and assume wAOðuÞ-OðvÞ: The following five cases need to be considered. Case 1: wAREðD2 Þ: In this case, u and v are 2-exchange neighbors of some vertices in D2 : Let u ¼ ½1; 2; u3 ; y; un  and v ¼ ½1; 2; v3 ; y; vn : Assume that w ¼ Ri ðuÞ ¼ Rj ðvÞ for some i; jAf3; y; ng: Since Ri ðuÞ ¼ ½ui ; 1; u3 ; y; ui1 ; 2; uiþ1 ; y; un  and Rj ðvÞ ¼ ½vj ; 1; v3 ; y; vj1 ; 2; vjþ1 ; y; vn ; it implies that i ¼ j and uk ¼ vk for 3pkpn: This contradicts that u and v are two distinct vertices in OðD2 Þ: The following cases consider that u and v are 3-rotation neighbors of some vertices in D2 ; and let u ¼ ½ui ; 2; u3 ; y; ui1 ; 1; uiþ1 ; y; un  and v ¼ ½vj ; 2; v3 ; y; vj1 ; 1; vjþ1 ; y; vn : Case 2: wAERðD2 Þ: In this case, w ¼ EðuÞ ¼ EðvÞ: Thus, ½2; ui ; u3 ; y; ui1 ; 1; uiþ1 ; y; un  ¼ ½2; vj ; v3 ; y; vj1 ; 1; vjþ1 ; y; vn :

This implies i ¼ j and uk ¼ vk for 3pkpn: It reaches a contradiction again that u and v are two distinct vertices. Case 3: wARR1 ðD2 Þ: In this case, there exist i0 and j 0 with ioi0 and joj 0 such that w ¼ Ri0 ðuÞ ¼ Rj0 ðvÞ: Since Ri0 ðuÞ ¼ ½ui0 ; ui ; u3 ; y; ui1 ; 1; uiþ1 ; y; ui0 1 ; 2; ui0 þ1 ; y; un  and Rj 0 ðvÞ ¼ ½vj0 ; vj ; v3 ; y; vj1 ; 1; vjþ1 ; y; vj0 1 ; 2; vj 0 þ1 ; y; vn ; it implies that i ¼ j; i0 ¼ j 0 ; and uk ¼ vk for 3pkpn; which is a contradiction. Case 4: wARR2 ðD2 Þ: By considering i0 oi and j 0 oj; a proof similar to that of case 3 gives a contradiction. Case 5: wARR3 ðD2 Þ: In this case, w ¼ Ri ðuÞ ¼ Rj ðvÞ: Thus ½1; ui ; u3 ; y; ui1 ; 2; uiþ1 ; y; un  ¼ ½1; vj ; v3 ; y; vj1 ; 2; vjþ1 ; y; vn : This implies i ¼ j and uk ¼ vk for 3pkpn; a contradiction. & Let uAD2 : Then odðEðuÞÞ ¼ n  2 and each vertex v in RðuÞ has outdegree n  1: Since odðuÞ ¼ n  1; by Lemma 4, jO2 ðuÞj ¼ ðn  2Þ þ ðn  1Þ ðn  2Þ ¼ n2  2n: ~2 Þ ¼ ðn  2Þ!: Lemma 5. g2 ðS n Proof. Clearly, jD2 j ¼ ðn  2Þ!: Recall that EðD2 ÞRðD2 Þ ¼ |; and OðD2 Þ ¼ EðD2 Þ,RðD2 Þ contains all vertices with the symbol 2 at the second position. Thus, jOðD2 Þj ¼ ðn  1Þ!; jEðD2 Þj ¼ jD2 j ¼ ðn  2Þ!; and jRðD2 Þj ¼ ðn  1Þ!  ðn  2Þ!: By Lemma 4, the sets D2 ; OðD2 Þ; and O2 ðD2 Þ are pairwise disjoint. Hence, for proving D2 is a distance-2 dominating set, it suffices to show that O2 ðD2 Þ contains exactly n!  ðn  1Þ!  ðn  2Þ! vertices. For each vertex vAEðD2 Þ; it has the form ½1; 2; p3 ; y; pn : Thus v contains n  2 3-rotation neighbors. For each vertex vARðD2 Þ; it has the form ½pi ; 2; p3 ; y; pi1 ; 1; piþ1 ; y; pn : Since pi 42; v has n  2 3-rotation neighbors and one 2-exchange neighbor. Also we have shown in Lemma 4 that for any two vertices u; vAOðD2 Þ; OðuÞ-OðvÞ ¼ |: Thus, jREðD2 Þj ¼ ðn  2Þ ðn  2Þ!; jERðD2 Þj ¼ jRðD2 Þj ¼ ðn  1Þ!  ðn  2Þ! ¼ ðn  2Þ ðn  2Þ!; and jRRðD2 Þj ¼ ðn  2Þ2 ðn  2Þ!: In particular, REðD2 Þ; ERðD2 Þ; and RRðD2 Þ are pairwise disjoint. We can derive jO2 ðD2 Þj ¼ jREðD2 Þj þ jERðD2 Þj þ jRRðD2 Þj ¼ ðn  2Þ ðn  2Þ! þ ðn  2Þ ðn  2Þ! þ ðn  2Þ2 ðn  2Þ! ¼ n!  ðn  1Þ!  ðn  2Þ!: Thus, D2 is a minimum distance-2 dominating set of ~2 and the lemma follows. & S n

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~2 Þg: Theorem 6. Let D2 ¼ fv j v ¼ ½2; 1; p3 ; y; pn AV ðS n Then, D2 is the unique minimum distance-2 dominating set ~2 : of S n Proof. Suppose that D02 is a minimum distance-2 ~2 and D0 aD2 : Clearly, jD0 j ¼ dominating set of S n 2 2 P jD2 j ¼ ðn  2Þ! and jOðD02 Þjp wAD0 odðwÞpjD02 j 2 ðn  1Þ ¼ ðn  1Þ!: Thus, V \O½D02  contains at least n!  ðn  1Þ!  ðn  2Þ! vertices. Also, we have shown that for any vertex wAV the number of vertices contained in O2 ðwÞ is at most n2  2n: An upper bound of the total outdegree of vertices in OðD02 Þ can be computed as follows: X odðwÞpjD02 j ðn2  2nÞ ¼ n!  ðn  1Þ!  ðn  2Þ!: wAOðD02 Þ

Since D02 is a distance-2 dominating set, the two sides of the above inequality are equal. In particular, the outdegree of every vertex contained in O½D02  is indeed n  1: Since D02 aD2 ; there exists a vertex u ¼ ½2; 1; u3 ; y; un  of D2 such that ueD02 : Since D02 is a distance-2 dominating set, u must be contained in one of the sets EðD02 Þ; RðD02 Þ; REðD02 Þ; ERðD02 Þ; and RRðD02 Þ: However, u cannot be a 2-exchange neighbor of some vertex since the symbol 2 comes before symbol 1 in u: Hence, ueEðD02 Þ and ueERðD02 Þ: This implies that there exists a vertex vAO½D02  such that u ¼ Ri ðvÞ where iAf3; y; ng: Thus, v ¼ ½1; ui ; u3 ; y; ui1 ; 2; uiþ1 ; y; un  ¼ Eð½ui ; 1; u3 ; y; ui1 ; 2; uiþ1 ; y; un Þ: By Proposition 1, odðvÞ ¼ n  2; which is a contradiction to the above argument. &

5. The distributed algorithms A distributed network can be viewed as a system consisting of two types of components: processors and interconnections between processors. Usually, such a system is modeled by a digraph where each vertex represents a processor, and a directed arc between processors represents message channel for communications. A distributed algorithm is defined to be a collection of local algorithms from the same copy one for each processor. Every processor independently executes its local algorithm and cooperates with the other processors to achieve a certain objective. To communicate among themselves, the processors share information through transmitting messages. That is, the processors can communicate only by sending and receiving messages over the communication links. In this section, we assume that individual processor operates at different speeds and the timing model used in the communications network is asynchronous, i.e., the transmitting message is placed in a queue waiting for the receiver to accept it and the sending processor can proceed

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immediately. We also assume that each vertex (processor) u knows the indegree idðuÞ and the outdegree odðuÞ; while the identities of all vertices in the graph are unknown. Since the graph dimension and the larger value of idðuÞ and odðuÞ differ by 1, each processor knows the network size. Note that each processor has a local state and all variables are locally maintained by each processor. By Theorems 3 and 6, D1 ¼ fv j v ¼ ½p1 ; 1; p3 ; y; pn  ~2 Þg and D2 ¼ fv j v ¼ ½2; 1; p3 ; y; pn AV ðS ~2 Þg are AV ðS n n ~2 the unique minimum distance-k dominating set of S n for k ¼ 1; 2; respectively. Before introducing our algorithm for computing the distance-1 dominating set in a directed split-star, we first introduce a property which will be used for proving the correctness of the algorithm. Lemma 7. Vertex x is in D1 if and only if odðyÞoidðyÞ for every yAIðxÞ: Proof. Let x ¼ ½p1 ; 1; p3 ; y; pn : It is clear that x is a 3-rotation neighbor of y: Then y ¼ ½1; pi ; p3 ; y; pi1 ; p1 ; piþ1 ; y; pn  for some iAf3; y; ng: Since the symbol 1 is at the first position of y; odðyÞ ¼ n  2oidðyÞ ¼ n  1: To prove the necessary part, we assume to the contrary that xeD1 and idðyÞoodðyÞ for every vertex yAIðxÞ: Consider the following two cases. Case 1: The symbol 1 is at the first position of x: Let x ¼ ½1; p2 ; p3 ; y; pn : There exists a yAIðxÞ which is ½p2 ; 1; p3 ; y; pn : Clearly, x is the 2-exchange neighbor of y: Since the symbol 1 is at the second position of y; idðyÞ ¼ n  2oodðyÞ ¼ n  1: Thus, this case is impossible. Case 2: The symbol 1 is at the ith position of x where 3pipn: Let x ¼ ½p1 ; p2 ; p3 ; y; pi1 ; 1; piþ1 ; y; pn : Then, there is a yAIðxÞ and y ¼ ½p2 ; 1; p3 ; y; pi1 ; p1 ; piþ1 ; y; pn : Clearly, x is a 3-rotation neighbor of y: Since the symbol 1 is at the second position of y; idðyÞoodðyÞ: This case is also impossible. We conclude that xAD1 if and only if odðyÞoidðyÞ for every yAIðxÞ: & According to Lemma 7, we design an algorithm for computing the distance-1 dominating set by two stages. In the first stage, each processor sets up the messages and sends them to its neighbors. In the second stage, the received messages are compared with the indegree for determining the membership of dominating set. Below we present the details of the algorithm in which x is a processor for running the algorithm. Algorithm Distance-1 Domination Stage 1. Set up the transmitting message to be 1 if x’s outdegree is less than its indegree, and to be 0 otherwise. Send message to each neighbor yAOðxÞ:

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Stage 2.

Receive the incoming messages and count the number of 1 to the sum until all messages are received. If the sum accumulated by x is equal to x’s indegree, then xAD1 ; otherwise xeD1 :

Theorem 8. Algorithm Distance-1 Domination correctly determines the unique minimum distance-1 dom~2 : inating set of S n

Lemma 9. Let x ¼ ½p1 ; p2 ; p3 ; y; pn  and xAD1 : Then the following statements are true: (i) If yAI2 ðxÞ and y ¼ ½y1 ; y2 ; y; yn ; then y1 ef1; p1 g: (ii) If Y ¼ fy j yAI2 ðxÞ and odðyÞoidðyÞg; then jY j ¼ ½1 þ 2 þ ? þ ðn  2Þ  ðn  p1 Þ: Proof. Since p1 4p2 ¼ 1; x cannot be a 2-exchange neighbor of any vertex. Suppose yAI2 ðxÞ and let w be a vertex in IðxÞ such that yAIðwÞ: Since x is a 3-rotation neighbor of w; we have w ¼ ½p2 ¼ 1; pi ; p3 ; y; pi1 ; p1 ; piþ1 ; y; pn 

Proof. From the algorithm, it is easy to see that the sum accumulated by x indicates the number of vertices directed to x with the outdegree less than the indegree. By Lemma 7, x is a member of the unique minimum distance-1 dominating set if and only if each vertex directed to x has the outdegree less than the indegree, and therefore it can be determined by the last statement of Stage 2. & We now give the complexity analysis as follows. Since the message transferred once only occurs in the first stage of the algorithm, the communication complexity is evidently proportional to the number of directed arcs in the digraph. ~2 ; For computing the distance-2 dominating set of S n we design an algorithm having three stages where the first two stages are the same as the previous one, except that the sum accumulated by x is sent out to x’s neighbors again. Finally, in the third stage, the secondly received messages are compared with certain condition for determining the membership of distance-2 dominating set. Since each message is received twice in each processor and these incoming messages must be discriminated for controlling the sequence of stages, we may assume that every message provides with a tag for indicating the different type of messages. Algorithm Distance-2 Domination Stages 1 and 2. Perform the same two stages as that in the algorithm Distance-1 Domination, and send the sum as a transmitting message to each neighbor yAOðxÞ: Stage 3. Receive the incoming messages and count them to a total sum until all messages are received. If xAS and the total sum equals to ðn2Þðn3Þ ; then xAD2 ; 2 otherwise xeD2 : To show the correctness of the above algorithm, we need the following lemma.

for some iAf3; 4; y; ng:

ð1Þ

No matter what w is either the 2-exchange or a 3-rotation neighbor of y; it is easy to see that pi must be at the first position of y: Since pi a1 and pi ap1 for i ¼ 3; y; n; this shows that statement (i) holds. To show statement (ii), we first define Yw ¼ fy j yAI2 ðxÞ and odðyÞoidðyÞg for each vertex wAIðxÞ: We now consider a vertex yAYw : Since odðyÞoidðyÞ; it implies that y is a 2-exchange neighbor of some vertex. Thus, if y1 and y2 are the symbols at the first position and the second position, respectively, of y; then y1 oy2 : In particular, w is a 3-rotation neighbor of y in this case. According to (1), we can obtain y ¼ ½pi ; pj ; p3 ; y; pj1 ; p2 ¼ 1; pjþ1 ; y; pn  where jAf3; 4; y; i  1; i þ 2; i þ 3; y; ng: By the fact pi opj ; we have jYw j ¼ n  pi : Indeed, the above argument shows that all vertices in Yw have the same symbol, say pi ; at the first position. Thus Yw -Yw0 ¼ | for any two vertices w; w0 AIðxÞ: Note that Y ¼ fy j yAI2 ðxÞ and odðyÞoidðyÞg [ fy j yAIðwÞ and odðyÞoidðyÞg ¼ wAIðxÞ

[

¼

Yw :

wAIðxÞ

Therefore, we have the following consequence: X X jY j ¼ jYw j ¼ n  pi wAIðxÞ

" ¼

n X

pi AN1 \fp1 g

#

ðn  kÞ  ðn  p1 Þ:

k¼2

This completes the proof.

&

The following corollary follows directly from the fact D2 CD1 and statement (ii) of Lemma 9. Corollary 10. Suppose xAD1 and let Y ¼ fy j yA2 ðxÞ and odðyÞoidðyÞg: Then xAD2 if and only if jY j ¼ 1 þ 2 þ ? þ ðn  3Þ:

ARTICLE IN PRESS F.-H. Wang et al. / J. Parallel Distrib. Comput. 63 (2003) 481–487

Theorem 11. The Algorithm Distance-2 Domination determines the unique minimum distance-2 dominating set ~2 correctly. of S n Proof. Since D2 CD1 ; the vertices of D2 are filtered through D1 by the algorithm. Let x be a vertex in D1 : Since the total sum accumulated by x indicates the number of vertices with the distance 2 directed to x and satisfying the property that the outdegree is less than the indegree, the correctness of the algorithm follows directly from Corollary 10 and the last statement of Stage 3. &

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6. Concluding remarks Cheng et al. [6] gave an orientation to the split-stars ~2 are maximally and showed that the oriented graphs S n arc-connected and arc-fault tolerant. In this paper, based on the structural properties of directed split-stars, we show that the minimum distance-k dominating set of ~2 is unique for k ¼ 1; 2: An immediate consequence is S n that the bondage number is 1 for any directed split-star. The bondage number of a nonempty undirected graph G to be the minimum cardinality among all sets of edges E such that gðG  EÞ4gðGÞ [11,20,21]. However, finding the minimum distance-k dominating set of a directed split-star is still unknown when kX3: Our recent work has tended towards such a research. For the algorithmic purpose, we demonstrate two efficient distributed algorithms, Algorithm Distance-1 Domination and Algorithm Distance-2 Domination, which are suitable to the applications in interconnection networks. Since stabilized algorithms are close related to distributed algorithms for some senses, a natural question to ask is what characterizations are necessary for designing an efficient stabilized algorithm on this problem?

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