The Unique Minimum Dominating Set of Directed Split-Stars

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(1)²}’rÕæ˜íñø|üXºÕ¯ The Unique Minimum Dominating Set of Directed Split-Stars Ùr "p Ù r×à Fu–Hsing Wang1 1. Jou–Ming Chang2. Yue–Li Wang1,∗. Cheng–Ju Hsu3. Department of Information Management, National Taiwan University of Science and Technology, Taipei, Taiwan, Republic of China.. 2. Department of Information Management, National Taipei College of Business, Taipei, Taiwan, Republic of China.. 3. Department of Computer Science, National Taiwan Ocean University, Keelung, Taiwan, Republic of China.. Abstract. 1. A dominating set S of a directed graph D is a set of vertices such that every vertex not in S is dominated by at least one vertex of S. In this paper, we show that there is a unique minimum distance-k dominating set, for k = 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.. Let D = (V, A) be a directed graph (digraph) with vertex set V and arc set A, where A ⊆ V × V . An arc < u, v > 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 D is the set O(u) = {v ∈ V | < u, v >∈ A}, while the closed outset is O[v] = O(v) ∪ {v}. For a subS set S ⊆ V , we also define O(S) = v∈S O(v) and O[S] = O(S) ∪ S. Let od(v) = |O(v)| be the outdegree of a vertex v ∈ V , and ∆(D) the maximum outdegree among all vertices in the digraph D. Undirected graphs form in a sense a special class of digraphs, and an undirected edge (u, v) is a pair of arcs < u, v > and < v, u >.. ¿b Ê…¹d2, ø„pÊ²}’rÕæ˜2, × Ñø£ùí|üXºÕ¯, uñøíø Õ¯. Keyword: Dominating set, Interconnection network, Directed split-star.. ∗ All correspondence should be addressed to Professor Yue–Li Wang, Department of Information Management, National Taiwan University of Science and Technology, 43, Section 4, Kee–Lung Road, Taipei, Taiwan, Republic of China. (Phone: 886– 02–27376768, Fax: 886–02–27376777, Email: ylwang@cs.ntust.edu.tw).. Introduction. An independent set S of a digraph D is a set of vertices such that no two vertices of S are joined by an arc. A set S ⊆ V is a dominating set of D if every vertex v ∈ V \S is dominated by at least one vertex of S, where V \S = {v ∈ V | v 6∈ S}. A dominating set with the minimum number of vertices is called a minimum dominating set, and its cardinal-.

(2) ity, denoted by γ(D), is termed the domination number of D. Researches on domination number and related parameters have much been attracted by graph theorists for their strongly practical applications and theoretical interesting. For instance, consider an interconnection network modeled by a graph, for which vertices represent processors and edges represent direct communication links between pairs of processor. Assume that from time to time we need to collect information from all processors, and this work must be done relatively often and fast. Thus we cannot route this information over too long a path. This suggests to identify a small set of processors (a dominating set) which are close to all other processors. It is well-know that the problem of finding a minimum dominating set is NP-complete for general undirected graphs and specifically for many restricted classes of graphs [3]. For a thorough treatment of domination in graphs, we refer the reader to [4, 5].. 4231. 2431. 4321. 3241. a. b. 3421. 2341. c. d. 1234. 4132. 1432. 2134. 1324. 4312. 3214. 3124. 3142. 1342. 3412. 2314. 4213. 2413. 4123 c. 1243 d 2143. 1423 a. b. Figure 1: 4-dimensional directed split-star.. Although domination and related topics have been extensively studied, the respective analogs on digraphs have not received much attention. In this paper, we present results concerning domination in a special class of digraphs called directed split-stars, which has recently been developed as a new model of the interconnection network for parallel and distributed computing systems. Cheng et al. [2] gave a variant distributed processor architecture of the star graphs which is known as the split-stars. Cheng and Lipman [1] proposed an assignment of orientation to the split-stars and showed that the resulting digraphs are not only strongly connected, but, in fact, they have maximal arc-fault tolerance and a small diameter. − → The n-dimensional directed split-star Sn2 is a directed graph whose vertices are in a oneto-one correspondence with n! permutations [p1 , p2 , . . . , pn ] of the set N = {1, 2, . . . , n}, − → and two vertices u, v of Sn2 are connected by an arc < u, v > if and only if the permutation of v can be obtained from u by either a 2-exchange or a 3-rotation. Let u = [p1 , p2 , . . . , pn ]. A. 2-exchange interchanges the first symbol p1 with the second symbol p2 whenever p1 > p2 , i.e., v = [p2 , p1 , . . . , pn ]. A 3-rotation counterclockwise rotates the symbols in positions 1, 2 and i for some i ∈ {3, 4, . . . , n}, i.e., v = [pi , p1 , p3 , . . . , pi−1 , p2 , pi+1 , . . . , pn ]. Fig− → ure 1 depicts an example of Sn2 for n = 4.. The remaining part of this paper is organized as follows. In Section 2, we study the problem of finding minimum domination set on directed split-stars. A result shows that − → Sn2 has a unique minimum dominating set of size (n − 1)!. In Section 3, we further investigate the distance-2 domination in directed split-stars and obtain a similar result as the previous section. Finally, a concluding remark is given in the last section. 2.

(3) 2. The Unique Minimum Dominating Set. least two vertices of S. This can easily be seen from the fact that the total outdegree of the vertices in S is equal to the number of vertices in V \S and there are no arcs between any two vertices in S.. Let v = [p1 , p2 , . . . , pn ]. The 2-exchange neighbor of v is E(v) = [p2 , p1 , . . . , pn ], and the 3-rotation neighbors of v are F i (v) = [pi , p1 , p3 , . . . , pi−1 , p2 , pi+1 , . . . , pn ] for 3 ≤ i ≤ n.. Let u = [u1 , 1, u3 , . . . , un ] and v = [v1 , 1, v3 , . . . , vn ] be any two distinct vertices in S and r ∈ V \S such that r ∈ O(u) ∩ O(v). Since r is dominated by u, either a 2-exchange or a 3-rotation exists from u to r. Assume that r is the 2-exchange neighbor of u, i.e., r = [1, u1 , u3 , . . . , un ]. Since the symbol 1 is on the second position of v, r cannot be a 3-rotation neighbor of v. This implies that r must be the 2-exchange neighbor of v and u = v, which contradicts that u and v are two distinct vertices of S. On the other hand, if r is a 3-rotation neighbor of u, then r = F i (u) = [ui , u1 , u3 , . . . , ui−1 , 1, ui+1 , . . . , un ] for some i ∈ {3, . . . , n}. Since the symbol 1 is not on the first position of r, r cannot be the 2-exchange neighbor of v. So r must be also a 3-rotation neighbor of v. Since the symbol 1 is on the ith position of r, it implies r = F i (v) = [vi , v1 , v3 , . . . , vi−1 , 1, vi+1 , . . . , vn ]. Thus u = v, which leads to a contradiction. Q. E. D.. − → Proposition 1 Let u be a vertex of Sn2 . If v is the 2-exchange neighbor of u, then od(u) = n − 1 and od(v) = n − 2. Proof. By definition, u has n − 2 3rotation neighbors and one 2-exchange neighbor. Thus, od(u) = n − 1. Since v has no 2-exchange neighbor, od(v) = n − 2. Q. E. D. − →2 Since each vertex of Sn is either a 2exchange neighbor of some vertex or a vertex having a 2-exchange neighbor, there are n! − →2 vertices with the maximum outdegree in Sn2 . − → In particular, ∆(Sn2 ) = n − 1. To simplify the description of our results, we define the following sets. Let Ni = N \{i} for i = 1, 2, and let N12 = N \{1, 2}, where N = {1, 2. . . . , n}.. Theorem 3 Let S = {[p1 , 1, p3 , . . . , pn ] | pi ∈ N1 , i ∈ N2 }. Then, S is the unique minimum − → dominating set of Sn2 .. Lemma 2 Let S = {[p1 , 1, p3 , . . . , pn ] | pi ∈ N1 , i ∈ N2 }. Then, S is a minimum domi− → − → nating set of Sn2 , and γ(Sn2 ) = (n − 1)!. In particular, every vertex not in S is dominated by exactly one vertex of S.. Proof. Suppose that S 0 is a minimum dom− → inating set of Sn2 and S 0 6= S. Then there exists a vertex u = [u1 , 1, u3 , . . . , un ] of S such that u ∈ / S 0 . Since S 0 is a dominating set, u must be either a 2-exchange neighbor or a 3rotation neighbor of some vertex v ∈ S 0 . However, u cannot be the 2-exchange neighbor of some vertex in S 0 since the symbol 1 is on the second position of u and u1 > 1. Let u = F i (v) where v ∈ S 0 and i ∈ {3, . . . , n}. Then v = [1, ui , u3 , . . . , ui−1 , u1 , ui+1 , . . . , un ] = E([ui , 1, u3 , . . . , ui−1 , u1 , ui+1 , . . . , un ]). By Proposition 1, od(v) = n − 2. Recall that P w∈S od(w) = |S| · (n − 1) = n! − (n − 1)!. Since |S 0 | = |S| and v ∈ S 0 , the total outdegree of the vertices in S 0 is at most n! − (n −. Proof. It is clear that S is an independent − → − → set of Sn2 and od(v) = ∆(Sn2 ) = n − 1 for every vertex v ∈ S. Thus, the total outdegree of the P vertices in S is v∈S od(v) = n! − (n − 1)!. − → Since any vertex in Sn2 can dominate at most − →2 ∆(Sn ) vertices, we have − → γ(Sn2 ) ≥. n! = (n − 1)! = |S|. − →2 ∆(Sn ) + 1. Thus we only need to show that S is a dom− → inating set of Sn2 . Suppose that S is not a dominating set. We claim that there exists a vertex r ∈ V \S which is dominated by at 3.

(4) 1)!−1 and |V \S 0 | = n!−(n−1)!. This contradicts to the assumption that S 0 is a dominating set. Therefore, the result directly follows from Lemma 2. Q. E. D.. the symbol 2 on the second position. O(S) ∩ S = ∅. Also we have. So. RX(S) = {[pi , 1, p3 , . . . , pi−1 , 2, pi+1 , . . . , pn ] | p3 , . . . , pn ∈ N12 , 3 ≤ i ≤ n} and. 3. XR(S) = {[2, pi , p3 , . . . , pi−1 , 1, pi+1 , . . . , pn ] | p3 , . . . , pn ∈ N12 , 3 ≤ i ≤ n}.. The Unique Minimum Distance-2 Dominating Set. In addition, RR(S) can be partitioned into the following three sets (where the position of the second 3-rotation is considered):. A set S of vertices in a directed graph D = (V, A) is called a distance-2 dominating set if every vertex in V \S is within distance 2 from some vertex of S. A distance-2 dominating set with the minimum number of vertices is called a minimum distance-2 dominating set, and its cardinality is denoted by γ2 (D). The distance-2 outset of a vertex v ∈ V is given S by O2 (v) = w∈O(v) O(w). If S ⊆ V then S O2 (S) = v∈S O2 (v). Obviously, a set S is a distance-2 dominating set if and only if O[S]∪ O2 (S) = V .. RR1 (S) = {[pj , pi , p3 ,. . ., pi−1 , 1, pi+1 ,. . ., pj−1 , 2, pj+1 ,. . ., pn ]|p3 ,. . ., pn ∈ N12 , 3 ≤ i < j ≤ n}, RR2 (S) = {[pj , pi , p3 ,. . ., pj−1 , 2, pj+1 ,. . ., pi−1 , 1, pi+1 ,. . ., pn ] | p3 ,. . ., pn ∈ N12 , 3 ≤ j < i ≤ n}, and RR3 (S) = {[1, pi , p3 , . . . , pi−1 , 2, pi+1 , . . . , pn ] | p3 , . . . , pn ∈ N12 , 3 ≤ i ≤ n}. Furthermore, O2 (S) is divided into five disjoint subsets XR(S), RX(S), RR1 (S), RR2 (S) and RR3 (S). This shows that the statement (1) holds. Since there are no vertices of O2 (S) having the symbols 2 and 1 in the foremost two positions, O2 (S) ∩ S = ∅. Similarly, we can verify O2 (S) ∩ O(S) = ∅ since O2 (S) contains no vertices with the symbol 2 on the second position. Therefore, statement (2) holds. We now prove statement (3) by contradiction as follows. Let u and v be any two distinct vertices in O(S) and assume w ∈ O(u) ∩ O(v). We consider the following cases.. Let S be a set of vertices in a directed split− → star Sn2 . We denote by X(S) and R(S) the sets consisting of the 2-exchange neighbors of S and the 3-rotation neighbors of S, respectively. That is, X(S) = {E(v) | v ∈ S} and R(S) = {F i (v) | v ∈ S, 3 ≤ i ≤ n}. In order to give a simple representation of O2 (S), we use the following notation: RX(S) = R(X(S)), XR(S) = X(R(S)), and RR(S) = R(R(S)). Note that RX(S) ∪ XR(S) ∪ RR(S) = O2 (S). Lemma 4 Let S = {[2, 1, p3 , . . . , pn ] | p3 , . . . , pn ∈ N12 }. The following statements are true:. Case 1: w ∈ RX(S). In this case, u and v are 2-exchange neighbors of some vertices in S. Let u = [1, 2, u3 , . . . , un ] and v = [1, 2, v3 , . . . , vn ]. Assume that w = F i (u) = F j (v) for some i, j ∈ {3, . . . , n}. Since F i (u) = [ui , 1, u3 , . . . , ui−1 , 2, ui+1 , . . . , un ] and F j (v) = [vj , 1, v3 , . . . , vj−1 , 2, vj+1 , . . . , vn ], it implies that i = j and uk = vk for 3 ≤ k ≤ n. This contradicts that u and v are two distinct vertices in O(S).. Proof. By definition, X(S) = {[1, 2, p3 , . . . , pn ] | p3 , . . . , pn ∈ N12 } and R(S) = {[pi , 2, p3 , . . . , pi−1 , 1, pi+1 , . . . , pn ] | p3 , . . . , pn ∈ N12 , 3 ≤ i ≤ n}. Clearly, X(S) ∩ R(S) = ∅. Moreover, O(S) = X(S) ∪ R(S) = {[p1 , 2, p3 , . . . , pn ] | pi ∈ N2 , i ∈ N2 }, i.e., O(S) contains all vertices with. The following cases consider that u and v. 4.

(5) bors and one 2-exchange neighbor. Also we have shown in Lemma 4 that for any two vertices u, v ∈ O(S), O(u) ∩ O(v) = ∅. Thus, |RX(S)| = (n − 2) · (n − 2)!, |XR(S)| = |R(S)| = (n − 1)! − (n − 2)! = (n − 2) · (n − 2)!, and |RR(S)| = (n − 2)2 · (n − 2)!. In particular, RX(S), XR(S), and RR(S) are pairwise disjoint. Therefore,. are 3-rotation neighbors of some vertices in S, and let u = [ui , 2, u3 , . . . , ui−1 , 1, ui+1 , . . . , un ] and v = [vj , 2, v3 , . . . , vj−1 , 1, vj+1 , . . . , vn ]. Case 2: w ∈ XR(S). In this case, w = E(u) = E(v). Thus, [2, ui , u3 , . . . , ui−1 , 1, ui+1 ,. . . ,un ] = [2,vj ,v3 ,. . . ,vj−1 ,1,vj+1 ,. . . ,vn ]. This implies i = j and uk = vk for 3 ≤ k ≤ n. It again reaches a contradiction that u and v are two distinct vertices.. |O2 (S)| = |RX(S)| + |XR(S)| + |RR(S)| =. Case 3: w ∈ RR1 (S). In this case, there exist i0 and j 0 with i < i0 and j < j 0 0 0 such that w = F i (u) = F j (v). Since 0 F i (u) = [ui0 , ui , u3 , . . . , ui−1 , 1, ui+1 , . . . , 0 ui0 −1 , 2, ui0 +1 , . . . , un ] and F j (v) = [vj 0 , vj , v3 , . . . , vj−1 , 1, vj+1 , . . . , vj 0 −1 , 2, vj 0 +1 , . . . , vn ], it implies that i = j, i0 = j 0 , and uk = vk for 3 ≤ k ≤ n, which is a contradiction.. (n − 2) · (n − 2)! + (n − 2) · (n − 2)! + (n − 2)2 · (n − 2)!. = n! − (n − 1)! − (n − 2)! − → Let ∆2 (Sn2 ) = maxv∈V |O(v)| + |O2 (v)|. By Proposition 1, it is easy to verify that for any vertex v ∈ V , O2 (v) contains at most (n − 2) · (n − 1) + 1 · (n − 2) = n2 − 2n vertices. Thus − → ∆2 (Sn2 ) = (n − 1) + (n2 − 2n) = n2 − n − 1. Consequently,. Case 4: w ∈ RR2 (S). By considering i0 < i and j 0 < j, a proof similar to that of case 3 gives a contradiction.. − → γ2 (Sn2 ) ≥. Case 5: w ∈ RR3 (S). In this case, w = F (u) = F j (v). Thus, [1, ui , u3 , . . . , ui−1 , 2, ui+1 ,. . ., un ] = [1,vj ,v3 ,. . . ,vj−1 ,2,vj+1 ,. . . ,vn ]. This implies i = j and uk = vk for 3 ≤ k ≤ n, a contradiction. Q. E. D. i. n! = (n − 2)! = |S|. − →2 ∆2 (Sn ) + 1. This shows that S is a minimum distance-2 − → dominating set of Sn2 . Q. E. D. Now, we can state our main result as follows.. Lemma 5 Let S = {[2, 1, p3 , . . . , pn ] | p3 , . . . , pn ∈ N12 }. Then, S is a minimum − → distance-2 dominating set of Sn2 .. Theorem 6 Let S = {[2, 1, p3 , . . . , pn ] | p3 , . . . , pn ∈ N12 }. Then, S is the unique min− → imum distance-2 dominating set of Sn2 .. Proof. Clearly, |S| = (n − 2)!. Recall that X(S) ∩ R(S) = ∅, and O(S) = X(S) ∪ R(S) contains all vertices with the symbol 2 on the second position. Thus, |O(S)| = (n − 1)!, |X(S)| = |S| = (n − 2)!, and |R(S)| = (n − 1)! − (n − 2)!. By Lemma 4, the sets S, O(S), and O2 (S) are pairwise disjoint. Hence, for proving S is a distance-2 dominating set, it suffices to show that O2 (S) contains exactly n! − (n − 1)! − (n − 2)! vertices.. Proof. Suppose that S 0 is a minimum − → distance-2 dominating set of Sn2 and S 0 6= S. Clearly, |S 0 | = |S| = (n − 2)! and |O(S 0 )| ≤ P 0 w∈S 0 od(w) ≤ |S | · (n − 1) = (n − 1)!. Thus, V \O[S 0 ] contains at least n!−(n−1)!−(n−2)! vertices. Also, we have shown that for any vertex w ∈ V the number of vertices contained in O2 (w) is at most n2 − 2n. An upper bound of the total outdegree of vertices in O(S 0 ) can be computed as follows: X od(w) ≤ |S 0 | · (n2 − 2n). For each vertex v ∈ X(S), it has the form [1, 2, p3 , . . . , pn ]. Thus v contains n − 2 3rotation neighbors. For each vertex v ∈ R(S), it has the form [pi ,2,p3 ,. . ., pi−1 ,1,pi+1 ,. . ., pn ]. Since pi > 2, v has n − 2 3-rotation neigh-. w∈O(S 0 ). = n! − (n − 1)! − (n − 2)!. 5.

(6) Since S 0 is a distance-2 dominating set, the two hand sides of the above inequality are equal. In particular, the outdegree of every vertex contained in O[S 0 ] is indeed n − 1.. [3] M.R. Garey and D.S. Johnson, Computers and Intractability: A Guide to the Theory of NP-completeness, Freeman, San Francisco, 1979.. Since S 0 6= S, there exists a vertex u = [2, 1, u3 , . . . , un ] of S such that u ∈ / S 0 . Since S 0 is a distance-2 dominating set, u must be contained in one of the sets X(S 0 ), R(S 0 ), RX(S 0 ), XR(S 0 ), and RR(S 0 ). However, u cannot be a 2-exchange neighbor of some vertex since the symbol 2 comes before symbol 1 in u. Hence, u ∈ / / XR(S 0 ). This implies that X(S 0 ) and u ∈ there exists a vertex v ∈ O[S 0 ] such that u = F i (v) where i ∈ {3, . . . , n}. Thus, v = [1, ui , u3 , . . . , ui−1 , 2, ui+1 , . . . , un ] = E([ui , 1, u3 , . . . , ui−1 , 2, ui+1 , . . . , un ]). By Proposition 1, od(v) = n − 2, which is a contradiction to the above argument. Q. E. D.. [4] T.W. Haynes, S.T. Hedetniemi and P.J. Slater, Fundamentals of Domination in Graphs, Marcel Dekker, New York, 1998.. 4. [5] T.W. Haynes, S.T. Hedetniemi and P.J. Slater, Domination in Graphs: Advanced Topics, Marcel Dekker, New York, 1998. [6] J.B. Jensen and G. Gutin, Digraphs: Theory, Algorithms and Applications, SpringerVerlag, London, 2001.. Concluding Remark. Cheng et al.[1] gave an orientation to the splitstars, and showed that the oriented graphs − → Sn2 are maximally arc-connected and arc-fault tolerant. In this paper, we show that there is a unique minimum distance-k dominating set, for k = 1, 2, in a directed split-star. However, the problem of finding a minimum distance-k dominating set for k ≥ 3 on directed splitstars is still unsolvable. Further, a natural question to ask is whether the distance-k dominating set for k ≥ 3 is unique.. References [1] Eddie Cheng and Marc J. Lipman, Orienting split-stars and alternating group graphs, Networks, Vol. 35, pp. 139-144, 2000. [2] Eddie Cheng, Marc J. Lipman and H.A. Park, An attractive variation of the star graphs: split-stars, Technical Report No. 98-3, Oakland University, 1998. 6.

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