Absorbant of generalized de Bruijn digraphs

Information Processing Letters 105 (2007) 6–11 www.elsevier.com/locate/ipl

Absorbant of generalized de Bruijn digraphs Erfang Shan a,b,∗ , T.C.E. Cheng b , Liying Kang a a Department of Mathematics, Shanghai University, Shanghai 200444, PR China b Department of Logistics, The Hong Kong Polytechnic University, Hung Hom, Kowloon, Hong Kong

Received 3 June 2007; received in revised form 5 July 2007 Available online 14 July 2007 Communicated by L. Boasson

Abstract The generalized de Bruijn digraph GB (n, d) has good properties as an interconnection network topology. The resource location problem in an interconnection network is one of the facility location problems. Finding absorbants of a digraph corresponds to solving a kind of resource location problem. In this paper, we establish bounds on the absorbant number for GB (n, d), and we give some sufficient conditions for the absorbant number of GB (n, d) to achieve the bounds. When d divides n, the extremal digraphs achieving the upper bound are characterized by determining their absorbants. © 2007 Elsevier B.V. All rights reserved. MSC: 05C20; 05C69 Keywords: Absorbant; Generalized de Bruijn digraph; Interconnection networks

1. Introduction and notation In this paper, we deal with simple digraphs which admit self-loops but no multiple arcs. Let D = (V , A) be a digraph with the vertex set V and the arc set A. There is an arc from x to y if (x, y) ∈ A. The vertex x is called a predecessor of y and y is called a successor of x. For a vertex v ∈ V , the out-neighborhood of v is O(v) = {w | (v, w) ∈ A} and its in-neighborhood is the set I (v) = {u | (u, v) ∈ A}. The closed out-neighborhood of v is the set O[v] = O(v) ∪ {v} and its closed inneighborhood is the set I [v] = I (v) ∪ {v}.  For S ⊆ V , its out-neighborhood is the set O(S) = s∈S O(s) and * Corresponding author at: Department of Mathematics, Shanghai University, Shanghai 200444, PR China. E-mail address: [email protected] (E. Shan).

0020-0190/$ – see front matter © 2007 Elsevier B.V. All rights reserved. doi:10.1016/j.ipl.2007.07.007

 its in-neighborhood is the set I (S) = s∈S I (s). O[S] and I [S] are defined similarly. An absorbant of a digraph D is a set S of vertices of D such that for all v ∈ V − S, O(v) ∩ S = ∅, i.e., I [S] = V . A set S ⊆ V is independent if for any x, y ∈ S, (x, y) ∈ / A. In particularly, if S is both independent and absorbant, then it is called a kernel of D. The absorbant number of D, denoted by γa (D), is defined as the minimum cardinality of an absorbant of D. An absorbant of D of cardinality γa (D) we call a γa set. A set S ⊆ V is a dominating set of D if for all v ∈ V − S, I (v) ∩ S = ∅, i.e., O[S] = V . Similarly, the domination number of D, denoted by γ (D), is defined as the minimum cardinality of a dominating set of D. For standard graph theory terminology not given here we refer to [12].

E. Shan et al. / Information Processing Letters 105 (2007) 6–11

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graphs. The purpose of this paper is to investigate the absorbants of generalized de Bruijn digraphs. We establish bounds on γa (GB (n, d)) for GB (n, d), and we give some sufficient conditions for the absorbant number of GB (n, d) to be the bounds. In particular, if d divides n, the extremal digraphs achieving the upper bound are characterized by a method for determining an absorbant of GB (n, d). Fig. 1. The graph GB (6, 3).

Domination and other related topics in digraphs have not received much attention. However, there is some work on domination in tournaments [5,16,17,19], and kernels in digraphs are now well studied in graph theory [1,6–11,18]. A survey on domination in digraphs has been written by Ghoshal et al. [13]. The resource location problem in an interconnection network is one of the facility location problems. Constructing the absorbants and dominating sets corresponds to solving two kinds of resource location problems [13,15]. For example, each vertex in an absorbant or a dominating set provides a service (file-server, and so on) for a networks. In this case, every vertex has a direct access to file-servers. Since each file-server may cost a lot, the number of an absorbant or a dominating set has to be minimized. The generalized de Bruijn digraph GB (n, d) is defined in [4,14] by congruence equations as follows: ⎧   ⎨ V  GB (n, d) = {0, 1, 2, . . . , n − 1}, A GB (n, d) = (x, y) | y ≡ dx + i (mod n), ⎩ 0i
B (n, d) where G

B (n, d) is the digraph obtained to G from GB (n, d) by reversing the direction of all arcs in GB (n, d), an absorbant differs from a dominating set. Indeed, it is easy to check that γa (GB (11, 3)) = 4 for GB (11, 3) by the numeration. But the set {1, 2, 3} is a dominating set of cardinality 3 of GB (11, 3). Hence γa (GB (11, 3)) > γ (GB (11, 3)). Du et al. [3] studied the Hamiltonian property of generalized de Bruijn digraphs. Recently, Kikuchi and Shibata [15] investigated the dominating sets in the di-

2. Absorbant of GB (n, d) For positive integers m, n, m | n means that m divides n. In what follows we may assume d  2 and n  d, since GB (n, 1) is empty graph for the case d = 1. We first present a lower bound and an upper bound on absorbant number of GB (n, d). Theorem 1. n/(d + 1)  γa (GB (n, d))  n/d. Proof. Let S be a γa -set of GB (n, d). Then |S|+d|S|  n by the definition of GB (n, d), so the lower bound follows. To obtain the upper bound of γa (GB (n, d)), we construct an absorbant S as follows.    S = 0, d, 2d, . . . , n/d − 1 d . By the choice of S, then for each vertex x ∈ V − S, O(x) ∩ S = ∅, since the elements of O(x) are d consecutive integers. So γa (GB (n, d))  |S| = n/d. 2 For 2  d  4, by giving a method to determine absorbants of GB (n, d), we present a sufficient condition for the absorbant number of GB (n, d) to be the lower bound n/(d + 1). Theorem 2. If d = 2, 4, and d + 1 | n, or d = 3 and 8 | n, then γa (GB (n, d)) = n/(d + 1). Proof. From Theorem 1, it suffices to construct an absorbant of order n/(d + 1) of GB (n, d). Clearly, V = n/(d+1)−1 {(d + 1)i, (d + 1)i + 1, . . . , (d + 1)i + d}. i=0 Depending on the value of d, we distinguish the following cases. n/3−1 Case 1. d = 2. Then V = i=0 {3i, 3i + 1, 3i + 2}. n/3−1 Let S = i=0 {3i + 1}. Then S is an absorbant of GB (n, 2). Indeed, for every vertex x ∈ V − S, if x = 3i + 2 for some 0  i  n/3 − 1, then 6i + 4 (mod n) ∈ O(x)∩S, while if x = 3i for some 0  i  n/3−1, then 6i + 1 (mod n) ∈ O(x) ∩ S. Hence S is an absorbant of order n/3 of GB (n, 2). n/4−1 Case 2. d = 3. Then V = i=0 {4i, 4i + 1, 4i + 2, 4i + 3}. Let

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S1 =

E. Shan et al. / Information Processing Letters 105 (2007) 6–11 n/8−1 

{8j + 1} = {1, 9, . . . , 8i + 1, . . . , n − 7},

j =0

S2 =

n/8−1 

{8i + 6} = {6, 14, . . . , 8j + 6, . . . , n − 2}.

j =0

Set S = S1 ∪ S2 . We claim that S is an absorbant of GB (n, 3). Let x be any vertex of V − S. We prove that x has an out-neighbor in S. If x = 4i for 0  i  n/4 − 1, then 3(4i) + 1 (mod n) ∈ O(x) ∩ S1 if i is even, or 3(4i) + 2 (mod n) ∈ O(x) ∩ S2 if i is odd. If x = 4i + 3 for 0  i  n/4 − 1, then 3(4i + 3) (mod n) ∈ O(x) ∩ S1 if i is even, or 3(4i + 3) + 1 (mod n) ∈ O(x) ∩ S2 if i is odd. If x = 4i + 1 for 0  i  n/4 − 1, then, by the definition of S1 , i is odd, thus there exists a positive integer k such that x = 4(2k + 1) + 1, so 3(4(2k + 1) + 1) + 2 = 8(3k +2)+1 (mod n) ∈ O(x)∩S1 . Finally, if x = 4i +2 for 0  i  n/4 − 1, then, by the definition of S2 , i is even, hence 3(4i + 2) = 12i + 6 (mod n) ∈ O(x) ∩ S2 . Therefore, S is an absorbant of order n/4 of GB (n, 3). n/5−1 Case 3. d = 4. Then V = i=0 {5i, 5i + 1, 5i + 2, n/5−1 5i +3, 5i +4}. Let S = i=0 {5i +2}. Then S is an absorbant of GB (n, 4). Indeed, for every vertex x ∈ V −S, we can show that O(x) ∩ S = ∅. For some 0  i  n/5 − 1, if x = 5i, then 4(5i) + 2 (mod n) ∈ O(x) ∩ S; if x = 5i + 1, then 4(5i + 1) + 3 = 5(4i + 1) + 2 (mod n) ∈ O(x) ∩ S; if x = 5i + 3, then 4(5i + 3) = 5(4i + 2) + 2 (mod n) ∈ O(x) ∩ S; if x = 5i + 4, then 4(5i + 4) + 1 = 5(4i + 3) + 2 (mod n) ∈ O(x) ∩ S. Therefore, S is an absorbant of order n/5 of GB (n, 4). 2 From Theorem 2, one can see that the minimum cardinality of kernels in GB (n, d) is equal to its absorbant number if d = 2 or 4, and d + 1 | n, or d = 3 and 8 | n. Moreover, Theorem 2 is not true if d  5, or d = 3 and 8 does not divide n. For example, it can be easily checked that γa (GB (12, 5)) = 12/5 = 3 and γa (GB (12, 3)) = 12/3 = 4. It is likely difficult to determine the minimum absorbant for an arbitrary generalized de Bruijn digraph, so we restrict mostly our attention to the case d divides n. Lemma 3. If d | n and n  d(2d −1), then γa (GB (n, d))  n/d − 1. Proof. From the definition of GB (n, d), it is clear that

 n O 1 + j = d, d + 1, . . . , d + (d − 1) , d j = 0, 1, 2, . . . , d − 1.

And n  d(2d − 1) implies that 1 + n/d > 2d − 1 and n/d > d. So we can choose a set S of vertices of GB (n, d) satisfying simultaneously the following conditions.  d−1  n 1 + j ⊆ S. (1) S1 = d j =0

(2)

d−1 

{d + j } ∩ S = ∅.

j =0

(3)

d−1      {id + j } ∩ S  = 1  j =0

for i = 0 and 2  i  n/d − 1. We show that S is an absorbant of GB (n, d). From the definition of GB (n, d), we have  n/d−1  d−1  n i+ j , V= d i=0 j =0

where each value i + (n/d)j is taken modulo n. Clearly, all vertices in d−1 j =0 {i + (n/d)j } have the same outneighborhood. Hence, for any x = i + (n/d)j ∈ V − S1 , it follows that d−1  {id + j }. O(x) = O(i) = j =0

According to the choice of S, we have O(x) ∩ S = ∅. So S is an absorbant of GB (n, d). Thus γa (GB (n, d))  |S| = n/d − 1. 2 The following result shows a sufficient and necessary condition for absorbant number of GB (n, d) to attain the upper bound n/d where d divides n. Theorem 4. If d | n, then γa (GB (n, d)) = n/d if and only if n  d 2 , or n = d(d + l) for 1  l  d − 2 and l + 1 | (d − 1). Proof. Let γa (GB (n, d)) = n/d. Since d divides n, it follows from Lemma 3 that n  d(2d − 2). We may assume that n > d 2 . Then there exists an integer l, 1  l  d − 2, such that n = d(d + l) as d | n. We claim that l + 1 | (d − 1). Otherwise, d − 1 = (l + 1)k + r, 1  r  l. Let i0 = (d − r − 1)/(l + 1) + 1. It is easy to see that 1 < i0  d − 1 < n/d, and we have n n i0 + (d − 1) = i0 + n − < n (1) d d and i0 d + (d − 1) < n.

(2)

E. Shan et al. / Information Processing Letters 105 (2007) 6–11

Then, for 0  j  i0 − 1, we have i0 +

Then, for each i ∈

n j  i0 + (i0 − 1)(d + l) d = i0 d + i0 (l + 1) − d − l

n j  i0 + i0 (d + l) d = i0 d + (d − r + l)  i0 d + d.

(4)

By (1), (2), (3) and (4), we obtain  d−1  d−1    n i0 + j = ∅. {i0 d + j } ∩ d j =0

Note the fact that n/d = d + l > d. As the case in Lemma 3, we can choose a vertex set S of GB (n, d) as follows.  d−1  n i0 + j ⊆ S. (1) S1 = d j =0

(2)

{i0 d + j } ∩ S = ∅.

 d−1      {id + j } ∩ S = 1    j =0

n − 1, i = i0 . d It can been easily verified that S is an absorbant of GB (n, d). Thus γa (GB (n, d))  |S| = n/d − 1, a contradiction. This completes the proof of the “only if” part. Conversely, suppose that n  d 2 , or n = d(d + l) for 1  l  d − 2 and l + 1 | (d − 1). In the former case, if γa (GB (n, d)) = n/d, then there exists an absorbant S of GB (n, d) such that |S|  n/d − 1 by Theorem 1. Thus n  |S|(d + 1)  (n/d − 1)(d + 1). This implies that d 2 + d  n, which contradicts our assumption. In the latter case, it suffices to prove that each absorbant of GB (n, d) has order at least n/d by Theorem 1. Obviously, j (d + l)/(l + 1) = j ((d − 1)/(l + 1) + 1) for 0  j  l, and thus it is an integer. Moreover, note that    l (d−1)/(l+1)    d +l n k+ j = 0, 1, . . . , − 1 . l+1 d for 0  i 

j =0

k=0

d+l l+1 j },

we have

(5)

(6)

and i + (i − j )

n = id + i(l + 1) − j (d + l) d d − l + j (d + l)  id + (l + 1) l+1 − j (d + l)  id + (d − 1).

(7)

Then, by (5), (6) and (7), we obtain for each 0  j  l that n  ∈ {id + k}, d d−1

i + (i − j )

k=0

j =0

(3)

{k +

Furthermore, we have n i + (i − j ) = id + i(l + 1) − j (d + l) d j (d + l) (l + 1) − j (d + l)  id + l+1 = id,

j =0

d−1 

k=0

d − 1 + j (d + l) −j l+1 (d − 1)(1 + j ) = l+1  d − 1.

(3)

and for i0  j  d − 1, we have i0 +

(d−1)/(l+1)

i −j 

= i0 d − r < i0 d

9

for each i ∈

(d−1)/(l+1)   k=0

 d +l k+ j . l+1

So, for each i, 0  i  n/d − 1, we have  d−1    d−1   i + (n/d)k = ∅. {id + k} ∩ k=0

(8)

k=0

Let S be an  arbitrary absorbant of GB (n, d). We claim that S ∩ ( d−1 k=1 {id + k}) = ∅ for each 0  i  n/d − 1. Suppose to the contrary that there exists an integer  {i i0 such that S ∩ ( d−1 k=1 0 d + k}) = ∅. Then by (8),  d − 1, such that there exists an  integer l0 , 0  l0  d−1 {i d + k}) ∩ {i + (n/d)k}. i0 + (n/d)l0 ∈ ( d−1 k=1 0 d−1 k=1 0 Note that O(i0 + (n/d)l0 ) = k=1 {i0 d + k} and i0 + / S. Hence O[i0 + (n/d)l0 ] ∩ S = ∅. Thus there (n/d)l0 ∈ is no vertex in S that is a successor of i0 + (n/d)l0 , which contradicts our assumption. Therefore, |S|  n/d. Combining this with Theorem 1, the sufficiency follows. 2 As an consequence of the above results, we present a sufficient condition for the absorbant number to be n/d − 1.

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E. Shan et al. / Information Processing Letters 105 (2007) 6–11

Theorem 5. Let d | n. If d(2d − 1)  n < 2d(d + 1), or n = d(d + l) for 1  l  d + 1 and l + 1 does not divide d − 1, then γa (GB (n, d)) = n/d − 1. Proof. Let d | n and d(2d − 1)  n < 2d(d + 1). Then γa (GB (n, d))  n/d − 1 by Lemma 3. Suppose γa (GB (n, d))  n/d − 2. Then (n/d − 2)(d + 1)  n, or equivalently, n  2d(d + 1), contrary to the assumption. Let n = d(d + l), 1  l  d + 1, and l + 1 does not divide (d − 1). By Lemma 3 and Theorem 4, then γa (GB (n, d))  n/d − 1. Suppose γa (GB (n, d)) < n/d − 1. Then (n/d − 2)(d + 1)  d(d + l). This implies that l  d + 2, which contradicts our assumption. 2 When d does not divide n, we now consider the special case n = kd + 1 for some integer k  1. Theorem 6. If n = kd + 1 and n/d  d, then γa (GB (n, d))  n/d − 1. Proof. From the definition of GB (n, d), the d elements in O(i) are consecutive integers for each i ∈ V . In particular, we have  O(d − 1) = d(d − 1) + j | j = 0, 1, 2, . . . , d − 1 . Let S = {d − 1, (d − 1) + d, . . . , (d − 1) + (d − 2)d, d(d − 1) + d, d(d − 1) + 2d, . . . , d(d − 1) + ( n/d − d)d}. Clearly, d(d − 1) + ( n/d − d)d = n − 1, and S ∩ O(d − 1) = ∅. We show that S is an absorbant of GB (n, d). For each i ∈ V − S, we claim that O(i) = O(d − 1). Suppose to the contrary that O(i) = O(d − 1). Then, there exists an integer j ∈ {0, 1, . . . , d − 1} such that id = (d − 1)d + j (mod n). If j = 0, then (i − (d − 1))d = 0 (mod n). From n = 1 (mod d), it follows that i = d − 1 (mod n). On the other hand,

n/d  d implies that n > (d − 1)d, so d − 1 < i < n. This means that there exits a positive integer m such that i − (d − 1) = mn. So, mn + d − 1 < n (m > 0). Thus, d − 1 < 0, a contradiction since d  2. If j = 0, it then follows that id + (d − 1) = (d − 1)d + (j − 1) (mod n), since the elements in O(i) (resp., O(d − 1)) are consecutive integers. Hence, (d − 1)d + j + (d − 1) = (d − 1)d + (j − 1) (mod n) from the choice of j . So d = 0, this contradiction implies our claim. From the choice of S, it immediately follows that O(i) ∩ S = ∅ for each i ∈ V − S. Thus γa (GB (n, d))  |S| = n/d − 1, as required. 2 Remark 1. If n = 1 (mod d), S constructed from Theorem 6 is not necessarily an absorbant of GB (n, d). For example, S = {2, 5, 9} is not an absorbant of GB (11, 3), since O(7) = {10, 0, 1}. In fact, if n = 1 (mod d),

then it is possible to γa (GB (n, d)) = n/d. Indeed, γa (GB (11, 3)) = 11/3 and γa (GB (12, 5)) = 12/5. Using an argument analogous to that given in Theorem 5, we obtain the following result by Theorem 6. Corollary 7. If n = kd + 1, d  n/d  d + 1, then γa (GB (n, d)) = n/d − 1. 3. Conclusion and open problems We investigated the absorbant number of a generalized de Bruijn digraph. We presented some sufficient conditions for the absorbant number of GB (n, d) to be the bounds. If d divides n, then the extremal digraphs achieving the upper bound were characterized. By somewhat improving the methods in this paper, one can deal with the generalized Kautz digraphs. We finally close this paper with a list of open problems. 1. Is it true that if GB (n, d) is a generalized de Bruijn digraph with d  2 and n  d, γ (GB (n, d))  γa (GB (n, d))? If it is not so, does there exist a generalized de Bruijn digraph GB (n, d) satisfying γa (GB (n, d)) < γ (GB (n, d))? 2. Find sufficient conditions for the absorbant number of GB (n, d) to be the lower bound n/(d + 1). 3. Find a sufficient and necessary condition for the absorbant number of GB (n, d) to be its domination number. 4. Is it true that γa (GB (8k − 4, 4k − 3)) = 3 for k  2? 5. Is it true that γa (GB (6k, 2k − 1)) = 4 for k  2? Acknowledgements This research was partially supported by The Hong Kong Polytechnic University under grant number GYX69, the National Nature Science Foundation of China under grant 10571117, the ShuGuang Plan of Shanghai Education Development Foundation under grant 06SG42 and the Development Foundation of Shanghai Education Committee under grant 05AZ04. The authors would like to thank the referee for his/her valuable suggestions to improve this paper. References [1] R. Aharoni, R. Holzman, Fractional kernels in digraphs, J. Combin. Theory Ser. B 73 (1998) 1–6. [2] J.-C. Bermond, C. Peyrat, de Bruijn and Kautz networks: a competitor for the hypercube?, in: F. André, J.P. Verjus (Eds.), Hypercube and Distributed Computers, Elsevier Science Publishers B.V. (North-Holland), 1989, pp. 279–293.

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