On the Diameter of the Generalized Graphs UG(B)(n, m), n(2) < m <= n(3)
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(2) 2. UGB (n, m), 2n 2 ≤ m ≤ n 3 Let dG (x, y) denote the distance between two vertices x and y in a graph (or directed graph) G, and let d(G) denote the diameter of the graph G. We use < u, . . . , v > to denote a path from u to v in G. Imase and Itoh [6] proved that the diameter of the generalized de Bruijn digraph GB (n, m) is bounded above by logn m, where x denotes the smallest integer not less than x. Since for any two distinct vertices u and v in UGB (n, m), the distance from u to v in the corresponding GB (n, m) provides an upper bound for the distance between u and v, we have d(UGB (n, m)) ≤ d(GB (n, m)). Therefore, the following bound is immediate. Lemma 2.1. The diameter of the generalized undirected de Bruijn graph UGB (n, m) is at most logn m. On the other hand, in UGB (n, m), the degree of every vertex is at most 2n. Therefore, from a vertex u of degree 2n − 1, one can reach at most (2n − 1) + (2n − 1)2 + · · · + (2n − 1)d vertices via a path of length d. With this observation, we get the following lower bound for the diameter. Lemma 2.2. log2n−1 m ≤ d(UGB (n, m)) for n + 1 ≤ m. Corollary 2.3. n2 ≤ m ≤ n3 .. The diameter of UGB (n, m) is 2 or 3 for. Proof. By Lemma 2.1 and log2n−1 n2 = 2 for n > 2, we ■ have the result. Now, we are ready to show our main results. Theorem 2.4. For positive integers n ≥ 2 and n3 , the diameter of UGB (n, m) is 3.. 2n2. ≤m≤. Proof. Let [0, m − 1] be the vertex set of G = UGB (n, m). We claim that either dG (0, m − n) = 3 or dG (0, m − n − 1) = 3. For convenience, let j1 = m − n and j2 = m − n − 1. By inspection, we have j1 ∈ N(0) and j2 ∈ N(0). Therefore, it suffices to prove that either N(0) ∩ N(j1 ) = ∅ or N(0) ∩ N(j2 ) = ∅, which implies that d(G) ≥ 3. Then, by Corollary 2.3, the result follows. By definition, N(0) = R(0) ∪ L(0) and N(j) = R(j) ∪ L(j) where j = j1 or j2 as the case may be. Therefore, it is equivalent to show that [R(0) ∪ L(0)] ∩ [R(j) ∪ L(j)] = ∅. We split the proof into four cases with the first three cases dealing with j = j1 or j2 . Case 1. R(0) ∩ L(j) = ∅. Since i∈R(0) R(i) = 2 − 1], neither j nor j are in R(i) = [n, n 1 2 i∈[1,n−1] i∈R(0) R(i). This implies that R(0) ∩ L(j) = ∅. Case 2. R(0) ∩ R(j) = ∅. By the definition of R(j), R(j) = {jn + α (mod m) : α ∈ [0, n − 1]}. Hence, it is clear that R(0) ∩ R(j) = ∅.. Case 3. L(0) ∩ L(j) = ∅. Assume that L(0) ∩ L(j) = ∅. Then there exists a k such that 0 ∈ R(k) and j ∈ R(k). This implies that there exist α and β where 0 ≤ α, β ≤ n − 1 satisfying kn + α ≡ 0 (mod m), (2.1) kn + β ≡ j (mod m). Therefore, β−α ≡ j (mod m) and −(n−1) ≤ β−α ≤ n−1. Since β − α = j if β − α ≥ 0 and (−β + α) + m − n < m or (−β + α) + m − n − 1 < m, we conclude that no solution (α, β) exists for (2.1). Hence the case is proved. Case 4. L(0) ∩ R(j) = ∅, j = j1 or j2 . First, we define δ(j1 ) = 0 and δ(j2 ) = 1. We claim that either 0 ∈ i∈R(j1 ) R(i) or 0 ∈ i∈R(j2 ) R(i). Assume that the above assertion is not true. Then, there exist 0 ≤ α, β, γ , ≤ n − 1 such that ((m − n − δ(j1 ))n + α)n + β ≡ 0 (mod m), ((m − n − δ(j2 ))n + γ )n + ≡ 0 (mod m). Thus,. . −n3 + αn + β ≡ 0 −n3 − n2 + γ n + ≡ 0. (mod m), (mod m).. This implies that n2 + (α − γ )n + (β − ) ≡ 0 (mod m). Since both α − γ and β − are integers between −(n − 1) and (n − 1), we have 2n2 > n2 + (α − γ )n + (β − ) > 0. Therefore, we are not able to find (α, β, γ , ) to satisfy n2 + (α − γ )n + (β − ) ≡ 0 (mod m). Hence, we conclude that either 0 ∈ i∈R(j1 ) R(i) or 0 ∈ i∈R(j2 ) R(i) and thus either L(0) ∩ R(j1 ) = ∅ or L(0) ∩ R(j2 ) = ∅. Now, combining the above four cases and j ∈ N(0), we have either dG (0, j1 ) = 3 or dG (0, j2 ) = 3. ■ 3. UGB (n, m), n 2 < m < 2n 2 Similar to Theorem 2.4, if we can find two vertices i, j ∈ [0, m − 1] such that dG (i, j) ≥ 3, then we can show that d(G) ≥ 3. First, we find the diameter of UGB (n, n2 + 1). Proposition 3.1.. d(UGB (n, n2 + 1)) = 3 for n ≥ 4.. Proof. Let m = n2 + 1 and n ≥ 4. Consider i = n − 2 and j = n2 − n + 2 in G = UGB (n, m). We claim dG (i, j) ≥ 3. Since (n2 − n + 2)n + α ≡ n + 1 + α (mod m) > n − 2 = i and (n − 2)n + α ≤ n2 − n − 1 < n2 − n + 2 = j, i ∈ R(j) and j ∈ R(i) follow. Hence, it suffices to show that [R(i) ∪ L(i)] ∩ [R(j) ∪ L(j)] = ∅ which can be broken down into four cases. • R(i) ∩ R(j) = ∅ Since (n − 2)n + α ≡ (n2 − n + 2)n + β (mod m), α − β ≡ 3n + 2 (mod m). Clearly, there are no solutions for α and β when n ≥ 4. • R(i) ∩ L(j) = ∅ Since (ni + α)n + β ≡ j (mod m), we have αn + β ≡ i + j = n2 (mod m). By the fact |αn + β| ≤ n2 − 1, there are no solutions for α and β.. NETWORKS—2008—DOI 10.1002/net. 181.
(3) • L(i) ∩ R(j) = ∅ Since (nj + α)n + β ≡ i (mod m), we have αn + β ≡ n2 (mod m) and we are not able to find solutions for α and β. • L(i) ∩ L(j) = ∅ Suppose not. Then there must exist k ∈ [0, n2 ] satisfying kn + α ≡ i (mod m) and kn + β ≡ j (mod m). Therefore, |α − β| = |i − j| = |n2 − 2n + 4| > n − 1. Again, this is not possible.. (n, n2. We note here that d(UGB + 1)) = 2 for n = 2, 3. To show the diameter of UGB (n, m) is equal to 2 for some n2 < m < 2n2 , we have to make sure that for each pair of vertices i and j, N(i) ∩ N(j) = ∅ or i ∈ N(j). Surprisingly, if m = n2 + 2, then the diameter of UGB (n, m) is equal to 2. Proposition 3.2.. d(UGB (n, n2 + 2)) = 2 for n ≥ 3.. Proof. Let m = n2 + 2. For any two distinct vertices x and y in UGB (n, m), we claim that dG (x, y) ≤ 2. It suffices to show that N(x) ∩ N(y) = ∅. Since N(x) = R(x) ∪ L(x) and N(y) = R(y) ∪ L(y), we have to prove that one of the following four conditions holds: (1) R(x) ∩ L(y) = ∅, (2) R(y) ∩ L(x) = ∅, (3) R(x) ∩ R(y) = ∅ or (4) L(x) ∩ L(y). = ∅. Observe that R(x) ∩ L(y) = ∅ if and only if (nx + α)n + β ≡ y (mod m) for some 0 ≤ α, β ≤ n − 1. Therefore, y + 2x ≡ αn + β ∈ [0, n2 − 1] (mod m). In fact, {αn + β : 0 ≤ α, β ≤ n − 1} = [0, n2 − 1]. This implies that if y + 2x ∈ [0, n2 − 1] (mod m), then d(x, y) ≤ 2. On the other hand, by considering R(y) ∩ L(x) = ∅, we have that if x + 2y ∈ [0, n2 − 1] (mod m), then d(x, y) ≤ 2. So, assume x + 2y and 2x + y are equal to either n2 or 2 n + 1 (mod m). Since 0 ≤ x = y ≤ n2 + 1, there are only six possible cases to consider. But, if 2x +y = n2 and 2y+x = 2n2 +2, then 3n2 +2 ≡ 0 (mod 3) which is not possible. By the same reason, 2x + y = n2 + 1 and 2y + x = 2n2 + 3 are not possible. Furthermore, if 2x +y = n2 and 2y +x = 2n2 +3, then y −x = n2 +3, which is not possible, either. Thus, we have exactly three cases to check. • 2x + y = n2 and 2y + x = n2 + 1 In this case, since 2n2 +1 ≡ 0 (mod 3), we may let n = 3p+1. Then x = 3p2 + 2p and y = 3p2 + 2p + 1. Hence, we have a path < 3p2 + 2p, p, 3p2 + 2p + 1 > from x to y, which concludes the proof. • 2x + y = n2 + 1 and 2y + x = 2n2 + 2 We have x = 0 and y = n2 + 1. Therefore, the path < 0, n, n2 + 1 > connects x and y for n ≥ 3, giving the result. • 2x + y = 2n2 + 2 and 2y + x = 2n2 + 3 Since 4n2 + 5 ≡ 0 (mod 3), it suffices to consider the cases n ≡ 1, 2 (mod 3). First, if n = 3p+1, then let x = 6p2 +4p+2. 182. NETWORKS—2008—DOI 10.1002/net. and y = 6p2 + 4p + 1. It is easy to see that < 6p2 + 4p + 1, 2p, 6p2 + 4p + 2 > is a path from x to y. If n = 3p + 2, the proof follows by letting x = 6p2 + 8p + 4 and y = 6p2 + 8p + 3. ■. Acknowledgments The authors are grateful to the referees and editors for their very valuable comments and suggestions which yielded an improved version of this article.. REFERENCES [1]. [2]. [3]. [4] [5]. [6]. [7]. [8]. [9] [10]. J.D.L. Caro, L.R. Nochefranca, and P.W. Sy, On the diameter of the generalized de Bruijn graphs UGB (n, n2 +1), 2000 International Symposium on Parallel Architectures, Algorithms, and Networks (ISPAN ’00), Washington, DC, USA, December 07–07, 2000, pp. 57–63. J.D.L. Caro, L.R. Nochefranca, P.W. Sy, and F.P. Muga, II, The wide-diameter of the generalized de Bruijn graphs UGB (n, n(n + 1)), 1996 International Symposium on Parallel Architectures, Algorithms, and Networks (ISPAN ’96), Washington, DC, USA, June 12–14, 1996, pp. 334–336. J.D.L. Caro and T.W. Zeratsion, On the diameter of a class of the generalized de Bruijn graphs, 2002 International Symposium on Parallel Architectures, Algorithms, and Networks (ISPAN ’02), Washington, DC, USA, May 22–24, 2002, pp. 197–202. D.Z. Du and F.K. Hwang, Generalized de Bruijn digraphs, Networks 18 (1988), 27–38. H.E. Escuadro and F.P. Muga, II, Wide-diameter of generalized undirected de Bruijn graph UGB (n, n2 ), 1997 International Symposium on Parallel Architectures, Algorithms, and Networks (ISPAN ’97), Washington, DC, USA, December 18–20, 1997, pp. 417–420. M. Imase and M. Itoh, Design to minimize diameter on building-block network, IEEE Trans Comput C 30 (1981), 439–442. L.R. Nochefranca and P.W. Sy, The diameter of the generalized de Bruijn graph UGB (n, n(n2 + 1)), 1997 International Symposium on Parallel Architectures, Algorithms, and Networks (ISPAN ’97), Washington, DC, USA, December 18–20, 1997, pp. 421–423. S.M. Reddy, D.K. Pradhan, and J.G. Kuhl, Directed graphs with minimal diameter and maximal connectivity, School of Engineering, Oakland University Technical Report, Oakland, USA, July 1980. D.B. West, Introduction to graph theory, Prentice-Hall, Upper Saddle River, NJ, 2001. J.M. Xu, Wide diameters of cartesian product graphs and digraphs, J Combin Optim 8 (2004), 171–181..
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