On the domination numbers of generalized de Bruijn digraphs and generalized Kautz digraphs

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1 Information Processing Letters 86 (2003) On the domination numbers of generalized de Bruijn digraphs and generalized Kautz digraphs Yosuke Kikuchi, Yukio Shibata Department of Computer Science, Gunma University, Kiryu, Gunma , Japan Received 11 March 2002; received in revised form 22 October 2002 Communicated by M. Yamashita Abstract This work deals with the domination numbers of generalized de Bruijn digraphs and generalized Kautz digraphs. Dominating sets for digraphs are not familiar compared with dominating sets for undirected graphs. Whereas dominating sets for digraphs have more applications than those for undirected graphs. We construct dominating sets of generalized de Bruijn digraphs where obtained dominating sets have some qualifications. For generalized Kautz digraphs, there is a minimum dominating set in those constructed dominating sets Elsevier Science B.V. All rights reserved. Keywords: Generalized de Bruijn digraphs; Generalized Kautz digraphs; Dominating set; Domination number; Absorbant; Kernel; Interconnection networks 1. Introduction It is one of major areas in theoretical and algorithmic observation to study domination and its related topics for undirected graphs [1,10]. For an undirected graph, domination of a vertex is the neighborhood of the vertex intuitively. Besides domination has been defined on digraphs, concepts of solution and kernel which are peculiar to digraphs have been defined. Solution is the vertex set whose vertices are non-adjacent and other vertices are adjacent from some vertex in this set intuitively. Kernel is also the vertex set whose vertices are non-adjacent and other vertices are adjacent to some vertex in this set intuitively. The origin * Corresponding author. addresses: kikuchi@msc.cs.gunma-u.ac.jp (Y. Kikuchi), shibata@msc.cs.gunma-u.ac.jp (Y. Shibata). of solution and kernel was given in game theory [9]. Solution and kernel have applications for logic and facility location [4,6,8,17]. Considering applications of domination and its related topics for game theory and logic, digraphs are more natural and have more applications than undirected graphs. Although digraphs have many applications, there are only a few studies for domination on digraphs. Barkauskas and Host, and Bar-Yehuda and Vishkin [2,3] have studied algorithms for domination on digraphs. Barkauskas and Host showed that the problem of determining whether or not an arbitrary digraph has an efficient dominating set is NP-complete and gave a linear time algorithm for finding efficient dominating sets in directed trees [2]. There are some researches for kernel in the digraph with no odd circuits [16] and for domination in tournaments that is a class of digraphs [7,14] /03/$ see front matter 2003 Elsevier Science B.V. All rights reserved. doi: /s (02)

2 80 Y. Kikuchi, Y. Shibata / Information Processing Letters 86 (2003) Each edge has orientation in digraphs and these edges are called arcs. The purpose of this study is to investigate the minimum dominating sets of generalized de Bruijn digraphs and generalized Kautz digraphs. These digraphs have good properties as interconnection network topologies. The resource location problem in an interconnection network is one of the facility location problems. Constructing the dominating set corresponds to solve a kind of resource location problem. We introduce the notion of consecutive minimum dominating set. The consecutive minimum dominating set can be considered important to determine the domination number of generalized de Bruijn digraphs and generalized Kautz digraphs, due to the adjacency of vertices in those digraphs. The rest of the paper is organized as follows. Section 2 gives definition and terminology. In Section 3 we give simple lower and upper bounds for the domination number of generalized de Bruijn digraphs. We also investigate the condition when the domination number of generalized de Bruijn digraphs is equal to the lower bound of the domination number. In this investigation, the consecutive minimum dominating set is used. Section 4 deals with the domination number of generalized Kautz digraphs. Determining the domination number of generalized Kautz digraphs is easier than that of generalized de Bruijn digraphs. The difference between definition of generalized de Bruijn digraphs and that of generalized Kautz digraphs causes this easiness. Section 5 concludes with discussions and an open problem. An early version of the paper has been presented in [13]. 2. Definition and terminology V(G) and A(G) are the vertex set and the arc set of a digraph G(V, A), respectively. There is an arc from x to y if (x, y) A(G). Thevertexx is called a predecessor of y and y is called a successor of x. ThesetsO(v) ={w (v, w) A(G)} and I(v) = {u (u, v) A(G)} are called the outset and the inset of the vertex v, respectively. Similarly, we define O[v] =O(v) {v}, I[v] =I(v) {v}. IfS V(G) then O(S) = s S O(s), I(S)= s S I(s), O[S]= s S O[s],andI[S]= s S I[s]. AsetS V(G)is a dominating set of a digraph G if v is a successor of some vertex u S for all v/ S. AsetS V(G) is an absorbant of a digraph G if there is a vertex u S such that u is a successor of v for each v/ S. AsetS V(G) is independent if for any x,y S, (x, y) / A(G). IfS V(G) is independent and a dominating set of a digraph G, S is called a solution of G. IfS V(G) is independent and an absorbant of a digraph G, S is called a kernel of G [9]. X denotes the cardinality of a set X. Foravertexu V(G), O(u) is called the out degree of u. Weuse the notation γ(g), called domination number, forthe minimum cardinality of a set S V(G) which is a dominating set. If S is a dominating set of G whose cardinality is equal to γ(g), then S is called the minimum dominating set of G. Letm, n be positive integers. (m, n) denotes the greatest common divisor of m and n. m n and m n mean that m divides n and m does not divide n, respectively. In this paper, we will describe domination numbers of generalized de Bruijn digraphs and generalized Kautz digraphs. Furthermore we construct minimum dominating sets of generalized de Bruijn digraphs and generalized Kautz digraphs. Definitions of generalized de Bruijn digraphs and generalized Kautz digraphs are as follows. The generalized de Bruijn digraph G B (n, d) is defined by congruence equations. V(G B (n, d)) ={0, 1, 2,...,n 1}, A(G B (n, d)) ={(x, y) y dx + i(mod n), 0 i<d}. If n = d D, G B (n, d) is the de Bruijn digraph B(d,D). Whereas definition of the generalized Kautz digraph G K (n, d) is given by V(G K (n, d)) ={0, 1, 2,...,n 1}, A(G K (n, d)) ={(x, y) y dx i(mod n), 0 <i d}. If n = d(d 1) D 1, G K (n, d) is the Kautz digraph K(d 1,D). Generalized de Bruijn digraphs and generalized Kautz digraphs are introduced by Imase and Itoh [11,12] and Reddy et al. [15], so-called Imase Itoh digraphs. It is well known that de Bruijn digraphs and Kautz digraphs have good properties as network topologies for interconnection networks compared with hypercubes [5]. These generalizations remove the restriction on the cardinality of vertex

3 Y. Kikuchi, Y. Shibata / Information Processing Letters 86 (2003) sets of these digraphs and make these digraphs more valuable as network models. For generalized de Bruijn digraphs and generalized Kautz digraphs, dominating sets can be investigated on number systems. If S = {x,x + 1,...,x + k} (mod n) is a dominating set of G B (n, d), thens is called a consecutive dominating set. Similar definition is given on G K (n, d). 3. Domination numbers of generalized de Bruijn digraphs There is a simple lemma for determining domination number of a generalized de Bruijn digraph. Lemma 3.1. γ(g B (n, d)) n/(). Proof. Let S be a minimum dominating set of G B (n, d). We obtain S +d S n from the definition of G B (n, d). The above lemma gives a lower bound for the domination number of G B (n, d). We also obtain the following lemma with respect to the upper bound of the domination number of G B (n, d). Lemma 3.2. γ(g B (n, d)) n/d. Proof. For G B (n, d), lets ={1, 2,..., n/d }. Then O(S) ={d,,d + 2,..., n/d d + d 1} with duplications. O(S) = n/d d n. Thus S is a dominating set of G B (n, d). From the proof of Lemma 3.2, S ={1, 2,..., n/d } is a dominating set, but not minimum. For example, {1, 2, 3, 4, 5, 6} is a dominating set of G B (12, 2). Furthermore {1, 2, 3, 4} is not a dominating set of G B (12, 2), since the outset of a subset {1, 2, 3, 4} of G B (12, 2) is {2, 3, 4, 5, 6, 7, 8, 9}. Whereas γ(g B (12, 2)) = 4 because the vertex subset {4, 5, 6, 7} is a minimum dominating set of G B (12, 2). Thus {1, 2, 3, 4, 5, 6} is a dominating set but not minimum. The following theorem shows a sufficient condition for the domination number of G B (n, d) to be n/() and a method for determining a dominating set of G B (n, d). Theorem 3.3. Let n and d be positive integers with d 2. If there is a vertex x V(G B (n, d)) satisfying ( ) n (d 1)x l 0 (mod n) (1) for some l(0 l () n/() n), then γ ( G B (n, d) ) n =. Proof. From Lemma 3.1, we show the correctness of this statement by constructing a dominating set of G B (n, d) that contains x satisfying Eq. (1). Let D = {x,x + 1,...,x + n/() 1} V(G B (n, d)). Then out sets of vertices in D are given as follows. O(x)={dx,dx + 1,...,dx+ d 1}, O(x + 1) = { d(x + 1), d(x + 1) + 1,..., d(x + 1) + d 1 },. O ( x + n/() 1 ) = { d ( x + n/() 1 ), d ( x + n/() 1 ) + 1,..., d ( x + n/() 1 ) + d 1 }. Thus O(D) n = d and Since x satisfies Eq. (1), dx x + n/() l(mod n) D = n for some l (0 l () n/() n). If n/() l<0, D O(x) = D = n/() <l.if n/() l 0, then D O(x) =l. Thus we obtain D O(D) = D + O(D) D O(x). n/() + d n/() l n/() + d n/() { () n/() n } = n. Hence D is a minimum dominating set of G(n, d).

4 82 Y. Kikuchi, Y. Shibata / Information Processing Letters 86 (2003) One can see the following statement as a special case of Theorem 3.3. Corollary 3.4. For G B (n, d), γ(g B (n, d)) = 1 if and only if d n or n is odd and d = n 1. Proof. We only show the necessary condition because sufficiency is obvious. Since γ(g B (n, d)) = 1, there is a vertex x V(G B (n, d)) such that O[x] n. If O(x) n, thend n because of the definition of G B (n, d). If O(x) =n 1, then d = n 1andx V(G B (n, d)) has no loop. Hence, we obtain x + 1 (n 1)x (mod n). Therefore 2x (mod n),and n is odd. From Theorem 3.3, one can see the following corollary for the solution of G B (n, d). Corollary 3.5. Let n. If there is a vertex x V(G B (n, d)) such that (d 1)x n 0 (mod n), then there is a solution S of G B (n, d) such that S = n. We prepare the reversal of a digraph G, denoted by G 1. The reversal of a digraph G is defined by V(G 1 ) = V(G), (x, y) A(G 1 ) if and only if (y, x) A(G). For generalized de Bruijn digraphs, G B (n, d) is not necessarily isomorphic to G B (n, d) 1. For de Bruijn digraph B(d,D), one can see that B(d,D) = B(d,D) 1. Then there is one to one correspondence between the dominating sets and the absorbants in B(d,D).ForB(d,D), the minimum cardinality of an absorbant is equal to the domination number. Therefore, the following statement is derived. Corollary 3.6. For B(d,D), both the minimum cardinalities of the absorbants and the dominating sets are d D /(). Proof. We first assume that D is even. Then the value l in Theorem 3.3 takes an integer between 0 to d and from the definition of B(d,D), wehaved 2. The equations x d D 2 + d D 4 + +d D 2k + l = 1, + d (mod d D ), lead x to satisfy Eq. (1) in Theorem 3.3. Next let us assume that D is odd, then the value l in Theorem 3.3 takes an integer between 0 and 1. We put x d D 2 + d D 4 + +d D 2k + l = 1, + d 3 + d(mod d D ), then x satisfies Eq. (1) in Theorem 3.3. The minimum dominating set is not necessarily a set of vertices whose labels are consecutive. For G B (10, 5), both {1, 2} and {3, 6} are minimum dominating sets of G B (10, 5). Theset{1, 2} is constituted of consecutive vertices and the set {3, 6} is not constituted of consecutive vertices. Thus Theorem 3.3 gives a sufficient condition but not a necessary condition. It is not easy to find all minimum dominating sets of G B (n, d). It is not enough to determine that domination number of a given G B (n, d) is equal to n/(d +1) only using Theorem 3.3 because G B (n, d) may have a minimum dominating set which is constituted of non-consecutive vertices and whose cardinality is equal to n/(). Then we give a necessary and sufficient condition for G B (n, d) to have a minimum dominating set S with S = n/() and constituted of consecutive vertices. Theorem 3.7. G B (n, d) satisfies one of the following four conditions if and only if there is a vertex x V(G B (n, d)) such that {x,x + 1,...,x + n/(d + 1) 1} is a consecutive minimum dominating set. Conditions: (i) () n and n/() takes the value between 0 and n/() (n + 1) n under the modulus (d 1,n), (ii) () n and d is even, (iii) 2() n, d is odd and 4 (), (iv) 2 k () n, d is odd and contains only one 2 as a factor (d 1 = 2 k s and s is odd).

5 Y. Kikuchi, Y. Shibata / Information Processing Letters 86 (2003) Proof. It is obvious that γ(g B (n, d)) = 1, when d n. Suppose that d<n. We show that there is a vertex x in V(G B (n, d)) such that x + n/() dx + i(mod n), (2) 0 i min { d 1, n/() () n }. (3) From Eq. (2), we obtain (d 1)x n/() i(mod n). (4) If is not a divisor of n and n/() () n>d 1, the value i takes an integer between 0 and d 1 in Eq. (3). One can note that if n, then i = 0. G B (n, d) has a consecutive dominating set S with S = n/() if and only if there is a vertex x V(G B (n, d)) satisfying Eq. (4). If d +1 is not a divisor of n,let(d 1,n)= s.then there are q and r(0 r<s)such that n/() = qs+ r. One can put r = i from the ranges of r and i,if r n/() () n. Otherwise, there does not exist x that satisfies Eq. (4) in G B (n, d). Thus, when () n, r n/() () n if and only if there are x and i satisfying Eqs. (2) and (3). Next, if is a divisor of n,wehave (d 1)x n/() (mod n) (5) by Eq. (4). Eq. (5) has a solution if and only if ( ) (d 1), n n/(). (6) Hence if d is even, there is an integer x satisfying Eq. (5) in Z n. Whereas if d is odd, d 1, andn are even integers. One can put = 2 h α (h is an integer, α is an odd integer). If h 2, then d 1 = 2(2 h 1 α 1). It holds that 2 (d 1) and 4 (d 1),since2 h 1 α 1 is odd. Therefore n is divisible by 2 h+1.thenn is divisible by 2(). For = 2 h α, let us consider the case h = 1, then one can put d 1 = 2(α 1) = 2 k β (β is odd). 2 h+1 divides n by (6). Therefore n is divisible by 2 h (). Conversely if G B (n, d) has a consecutive minimum dominating set {x,x + 1,...,x + n/() 1}, then x satisfies Eq. (4) for some i (0 i min{d 1, n/() () n}). Therefore n and d satisfy one of above four conditions. Consider whether there exists G B (n, d) that does not have a minimum consecutive dominating set and γ(g B (n, d)) = n/(). We show that if n/(d + 1) =1, 2, or 3, then there exists G B (n, d) that does not have a minimum dominating set D with D = n/(). Theorem 3.8. For domination number of generalized de Bruijn digraphs, we obtain γ ( G B (2s,2s 1) ) = 2, (7) γ ( G B (8s 4, 4s 3) ) = 3, (8) and γ ( G B (6s,2s 1) ) = 4, (9) where s is a natural number. Proof. For Eq. (7), each vertex of G B (2s,2s 1) has a loop. Then γ(g B (2s,2s 1)) 1. For Eq. (8), let us contrary assume that γ(g B (8s 4, 4s 3)) = 2 and a vertex x is contained in a dominating set of G B (8s 4, 4s 3). Thenx + 1 or x + 4s 2 is contained in the dominating set of G B (8s 4, 4s 3). If the dominating set is {x,x+ 1}, then (4s 3)x x + 2 (mod 8s 4), 4(s 1)x 2 (mod 8s 4). Therefore we have a contradiction. If the dominating set is {x,x + 4s 2}, thenwe obtain (4s 3)x x + 1 (mod 8s 4) (10) or (4s 3)(x + 4s 2) x + 1 (mod 8s 4). (11) If Eq. (10) holds, then (4s 4)x 1 (mod 8s 4). This is a contradiction. If Eq. (11) holds, then (4s 4)x 4(5s 4s 2 ) 5 (mod 8s 4). This is also a contradiction. For Eq. (9), let us assume that γ(g b (6s,2s 1)) = 3. We should consider three cases; (1) any two vertices are not consecutive in the dominating set, (2) exactly two vertices are consecutive in the dominating set, and (3) vertices in the dominating set are consecutive. Case (1): let x be a vertex in the dominating set, then other vertices in the dominating set are x + 2s and x + 4s. Thus x satisfies (2s 1)x x + 2ts + 1 (mod 6s)

6 84 Y. Kikuchi, Y. Shibata / Information Processing Letters 86 (2003) for some t(t= 0, 1, 2). From this equation, we obtain 2(s 1)x 2ts + 1 (mod 6s), and this is a contradiction. Case (2): let x and x + 1 be consecutive vertices in the dominating set, then another vertex in the dominating set is x + 2s + 1 or x + 4s. If the dominating set is {x,x+ 1,x+ 2s + 1},then { (2s 1)(x + 2s + 1) x + 2 (mod 6s), (2s 1)(x + 1) + 2s 2 x 1 (mod 6s). By these simultaneous equations, we obtain 4s(s 1) 1 (mod 6s). Then this is a contradiction. If the dominating set is {x,x + 1,x+ 4s} then (2s 1)(x + 4s) + 2s 2 x 1 (mod 6s). By this equation, we obtain 2(s 1)x 8s 8s (mod 6s) and this is a contradiction. Case (3): let us assume that x, x + 1andx + 2 constitute a dominating set. Then (2s 1)x x + 3 (mod 6s). Hence we obtain 2(s 1)x 3 (mod 6s), and this is also a contradiction. 4. Domination numbers of generalized Kautz digraphs We can see the following lemma similar to Lemma 3.1. Lemma 4.1. γ(g K (n, d)) n/(). Besides the difficulty of determining domination numbers of generalized de Bruijn digraphs, it is easy to determine the domination number of the generalized Kautz digraph G K (n, d). Theorem 4.2. γ(g K (n, d)) = n/(). Proof. D = {0, 1, 2,..., n/() 1} V(G K (n, d)) is a dominating set of G K (n, d). 5. Conclusion We investigated domination numbers of generalized de Bruijn digraphs and generalized Kautz digraphs. The consecutive dominating set is important in our investigation. Finding a dominating set of a generalized de Bruijn digraph is more difficult than that of generalized Kautz digraph. We conjecture that there is a consecutive minimum dominating set of G B (n, d). Acknowledgement The authors would like to thank anonymous referees for their helpful suggestions. References [1] R.B. Allan, R. Laskar, On domination and independent domination number of a graph, Discrete Math. 23 (1978) [2] A.E. Barkauskas, L.H. Host, Finding efficient dominating sets in oriented graphs, Congr. Numer. 98 (1993) [3] R. Bar-Yehuda, U. Vishkin, Complexity of finding k-path-free dominating sets in graphs, Inform. Process. Lett. 14 (1982) [4] C. Berge, A.R. Rao, A combinatorial problem in logic, Discrete Math. 17 (1977) [5] J.-C. Bermond, C. Peyrat, De Bruijn and Kautz networks: a competitor for the hypercube?, in: F. Andrè, J.P. Verjus (Eds.), Hypercube and Distributed Computers, Elsevier/North- Holland, Amsterdam, [6] P. Duchet, H. Meyniel, Kernels in directed graphs: a poison game, Discrete Math. 115 (1993) [7] D. Fisher, J.R. Lundgren, S.K. Merz, K.B. Reid, Domination graphs of tournaments and digraphs, Congr. Numer. 108 (1995) [8] A.S. Fraenkel, Y. Yesha, Complexity of problems in games, graphs, and algebraic equations, Discrete Appl. Math. 1 (1979) [9] J. Ghoshal, R. Laskar, D. Pillone, Topics on domination in directed graphs, in: T.W. Haynes, S.T. Hedetniemi, P.J. Slater (Eds.), Domination in Graphs, Marcel Dekker, New York, 1998, pp [10] T.W. Haynes, S.T. Hedetniemi, P.J. Slater, Fundamentals of Domination in Graphs, Marcel Dekker, New York, [11] M. Imase, M. Itoh, Design to minimize diameter on buildingblock network, IEEE Trans. Comput. C-30 (1981) [12] M. Imase, M. Itoh, A design for directed graphs with minimum diameter, IEEE Trans. Comput. C-32 (1983) [13] Y. Kikuchi, Y. Shibata, On the domination numbers of generalized de Bruijn digraphs and generalized Kautz digraphs, in: Proc. 7th Annual International Conference, COCOON 2001,

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