Antibandwidth and Cyclic Antibandwidth of Hamming Graphs

Antibandwidth and Cyclic Antibandwidth of Hamming Graphs

Electronic Notes in Discrete Mathematics 34 (2009) 295–300 www.elsevier.com/locate/endm Antibandwidth and Cyclic Antibandwidth of Hamming Graphs Stef...

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Electronic Notes in Discrete Mathematics 34 (2009) 295–300 www.elsevier.com/locate/endm

Antibandwidth and Cyclic Antibandwidth of Hamming Graphs Stefan Dobrev 1 Institute of Mathematics, Slovak Academy of Sciences D´ ubravsk´ a 9, 841 04 Bratislava, Slovakia [email protected]

Rastislav Kr´aloviˇc, Dana Pardubsk´a Department of Computer Science, Comenius University Mlynsk´ a dolina, 842 48 Bratislava, Slovakia {kralovic,pardubsk}@dcs.fmph.uniba.sk

L’ubom´ır T¨or¨ok 2 Institute of Mathematics and Computer Science, Slovak Academy of Sciences and Matej Bel University ˇ Dumbierska 1, 974 11, Bansk´ a Bystrica, Slovakia [email protected]

Imrich Vrt’o Institute of Mathematics, Slovak Academy of Sciences D´ ubravsk´ a 9, 841 04 Bratislava, Slovakia [email protected]

Abstract The antibandwidth problem is to label vertices of graph G(V, E) bijectively by integers 0, 1, ..., |V | − 1 in such a way that the minimal difference of labels of adjacent vertices is maximised. In this paper we study the antibandwidth of Hamming graphs. We provide labeling algorithms and tight upper bounds for general Hamming graphs Πdk=1 Knk . We have exact values for special choices of ni ’s and equality between antibandwidth and cyclic antibandwidth values. Keywords: antibandwidth, Hamming graph

1571-0653/$ – see front matter © 2009 Elsevier B.V. All rights reserved. doi:10.1016/j.endm.2009.07.048

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Introduction

The antibandwidth problem is to label vertices of graph G(V, E) by integers 0, 1, ..., |V | − 1 in such a way that the minimal difference of labels of adjacent vertices is maximised. This problem was originally introduced in [6] in connection with multiprocessors scheduling problems. Another motivation comes from the area of frequency assignment problems [4]. This problem is dual to the well-known bandwidth minimization problem [3] and also belongs to the large family of graph labelling problems. The antibandwidth problem is N P -complete. So far there exists polynomial algorithms for 3 classes of graphs: the complements of interval, arborescent comparability and treshold graphs [2,5]. Known results on antibandwidth include exact or at least asymptotically exact values for paths, cycles, complete trees[1], meshes [9,10], tori and hypercubes [8]. The cyclic antibandwidth is a natural and typical extension of the original problem when the guest graph is a cycle. The value of the cyclic antibandwidth is determined for meshes, toroidal meshes and hypercubes in [8]. Most known results are for bipartite graphs. In this paper we provide antibandwidth values for Hamming graphs. This class of graphs is interesting because of its connection to the area fo the error-correcting codes and association schemes. Let f be a one-to-one labelling f : V → {0, 1, 2, 3, ...|V | − 1} Define the antibandwidth of G according to f as ab(G, f ) = min |f (u) − f (v)|. uv∈E

The antibandwidth of G is defined as ab(G) = max ab(G, f ). f

Define the cyclic antibandwidth of a connected graph G according to f as cab(G, f ) = min {|f (u) − f (v)|, |V | − |f (u) − f (v)|}. uv∈E

The cyclic antibandwidth of G is defined as cab(G) = max cab(G, f ). f

Πdk=1 Knk

The d-dimensional Hamming graph is defined as the Cartesian product of d complete graphs Knk , for k = 1, 2, ..., d. The vertices of Πdk=1 Knk are d-tuples (i1 , i2 , ..., id ), where ik ∈ {0, 1, 2, ..., nk − 1}. Two vertices (i1 , i2 , ..., id ) and (j1 , j2 , ..., jd ) are adjacent iff the two tuples differ in precisely one coordinate. In case nk = n, for all k, we denote the graph as Knd . Define the value of Nk as follows. Set N0 = 1 and for k = 1, 2, ..., d, denote Nk = n1 n2 ...nk .

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Hamming Graphs

Theorem 2.1 For d ≥ 2 and 2 ≤ n1 ≤ n2 ≤ ... ≤ nd , ab(Πdk=1 Knk ) = n1 n2 ...nd−1 , if nd−1 = nd , d ≥ 2 ab(Πdk=1 Knk ) = n1 n2 ...nd−1 − 1, if nd−1 = nd and nd−2 = nd−1 , d ≥ 3 1

Research of all authors ecxept for the 4th one was supported partially by the APVV grant No. 0433/06, research of the 1st, 4th and 5th author was supported by Vega grant No. 2/0111/09. 2 Supported by Vega grant No. 2/0722/08 and APVV grant No. 51-009605.

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and n1 n2 ...nd−1 − min(n1 n2 ...nd−2 , nq+1 ...nd−1 ) ≤ ab(Πdk=1 Knk ) ≤ n1 n2 ...nd−1 − 1, where nd−2 = nd−1 = nd , d ≥ 3 and q is minimal index such that q ≤ d − 2 and nq = nd . Proof. Upper bound. We use an alternative definition of the antibandwidth problem: the vertices of Πdk=1 Knk are placed on a line bijectively into integer points 0, 1, 2, ..., Nd − 1, such that the minimum distance of adjacent vertices is maximized. Every vertex belongs to a clique Knd . Consider the vertex placed at the position ab(Πdk=1 Knk ) − 1 and the corresponding clique Knd . Clearly, all vertices of this clique must lie in the interval [ab(Πdk=1 Knk )−1, Nd −1]. Observe that Nd − 1 − (ab(Πdk=1 Knk ) − 1) ab(Πdk=1 Knk ) ≤ , nd − 1 which gives ab(Πdk=1 Knk ) ≤ Nd−1 . In case nd−1 = nd we can slightly improve the argument. Every vertex belongs to two cliques: Knd and Knd−1 , whose intersection is precisely that vertex. Consider the vertex placed at the position ab(Πdk=1 Knk ) − 1 and the corresponding cliques Knd and Knd−1 . Clearly, all vertices of these cliques must lie in the interval [ab(Πdk=1 Knk ) − 1, Nd − 1]. Because nd−1 = nd , one of the cliques, say Knd , must lie in a shorter interval, otherwise the two cliques would have 2 vertices in common. Hence Nd − 2 − (ab(Πdk=1 Knk ) − 1) 1 ab(Πdk=1 Knk ) ≤ ≤ Nd−1 − , nd − 1 nd which implies ab(Πdk=1 Knk ) ≤ Nd−1 − 1. Lower bound. Case I. Let nd−1 = nd . Define a labelling of the vertices of the Hamming graph as f (i1 , i2 , ..., id ) = ((Nd−1 + Nd−1 /N1 )i1 + (Nd−1 + Nd−1 /N2 )i2 + (Nd−1 + Nd−1 /N3 )i3 + ... + (Nd−1 + Nd−1 /Nd−1 )id−1 + Nd−1 id ) mod Nd . We show that f is a bijection. Assume that for two vertices (i1 , i2 , ..., id ) = (j1 , j2 , ..., jd ) f (i1 , i2 , ..., id ) = f (j1 , j2 , ..., jd ). It follows that (Nd−1 + Nd−1 /N1 )(i1 − j1 ) + ... + (Nd−1 + 1)(id−1 − jd−1 ) + Nd−1 (id − jd ) = 0 mod Nd . Observe that id−1 − jd−1 is divisible by nd−1 . As |id−1 − jd−1 | < nd−1 , we have id−1 = jd−1 . Divide the congruence by nd−1 . Then id−2 − jd−2 is divisible by nd−2 . As |id−2 − jd−2 | < n2 , we have id−2 = jd−2 . Divide the congruence by n2 and continue in a similar way up to i1 = j1 . Finally, this implies that id = jd , which is a contradiction. Distinguish 2 cases. a)Consider an edge of the k-th dimension, k ≤ d − 1, i.e., an edge between (i1 , i2 , ..., ik , ..., id ) and (i1 , i2 , ..., jk , ..., id ). Wlog assume that f (i1 , i2 , ..., ik , ..., id ) ≥ f (i1 , i2 , ..., jk , ..., id ). Then compute f (i1 , i2 , ..., ik , ..., id ) − f (i1 , i2 , ..., jk , ..., id ) = (Nd−1 + Nd−1 /Nk )(ik − jk ) mod Nd . If ik > jk then (Nd−1 + Nd−1 /Nk )(ik − jk ) mod Nd ≥ Nd−1 + Nd−1 /Nk > Nd−1 . If ik < jk then

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(Nd−1 + Nd−1 /Nk )(ik − jk ) mod Nd = Nd − (Nd−1 + Nd−1 /Nk )(jk − ik ) ≥ Nd − (Nd−1 + Nd−1 /Nk )(nk − 1) = Nd−1 (nd − nk + 1 − 1/Nk−1 + 1/Nk ) > Nd−1 . b) Consider an edge of the d-th dimension. Wlog assume that f (i1 , i2 , ..., id ) ≥ f (i1 , i2 , ..., jd ). Compute f (i1 , i2 , ..., id ) − f (i1 , i2 , ..., jd ) = Nd−1 (id − jd ) mod Nd . If id > jd then Nd−1 (id − jd ) mod Nd ≥ Nd−1 . If id < jd then Nd−1 (id − jd ) mod Nd = Nd − Nd−1 (jd − id ) ≥ Nd − Nd−1 (nd − 1) = Nd−1 . Case II. Let nd−1 = nd . Let q be the minimal index s.t. q ≤ d − 1 and nq = nd . Define a labelling of the vertices of the Hamming graph as f (i1 , i2 , ..., id ) = ((Nd−1 + Nd−1 /N1 )i1 + ... + (Nd−1 + Nd−1 /Nq−1 )iq−1 + (Nd−1 − Nd−1 /Nq )iq + ... + (Nd−1 − Nd−1 /Nd−1 )id−1 + Nd−1 id ) mod Nd . Using the same method as above, we can easily show that f is a bijection. Distinguish 3 subcases. a) Consider an edge of the k-th dimension, k ≤ q − 1. The analysis is the same as in Case I.a). b) Consider an edge of the k-th dimension, q ≤ k ≤ d − 1. Wlog assume that f (i1 , i2 , ..., ik , ..., id ) ≥ f (i1 , i2 , ..., jk , ..., id ). Compute f (i1 , i2 , ..., ik , ..., id ) − f (i1 , i2 , ..., jk , ..., id ) = (Nd−1 − Nd−1 /Nk )(ik − jk ) mod Nd . If ik > jk then (Nd−1 − Nd−1 /Nk )(ik − jk ) mod Nd ≥ Nd−1 − Nd−1 /Nk ≥ Nd−1 − Nd−1 /Nq . An important observation is that if q = d − 1 we get an optimal lower bound of Nd−1 − 1. If ik < jk then (Nd−1 − Nd−1 /Nk )(ik − jk )) mod Nd = Nd − (Nd−1 − Nd−1 /Nk )(jk − ik ) ≥ Nd − (Nd−1 − Nd−1 /Nk )(nk − 1) = Nd−1 (1 + 1/Nk−1 − 1/Nk ) > Nd−1 . c) Consider an edge of the k = d-th dimension. The analysis is the same as in Case I.b). Case III. Let nd−1 = nd . We define a new labelling of the vertices of the Hamming graph which gives the alternative lower bound in Theorem 2.1. f (i1 , i2 , ..., id ) = ((Nd−1 − N0 )i1 + (Nd−1 − N1 )i2 + ... + (Nd−1 − Nd−2 )id−1 + Nd−1 id ) mod Nd . (1) We show that f is a bijection. Assume that for two vertices (i1 , i2 , ..., id ) = (j1 , j2 , ..., jd ) f (i1 , i2 , ..., id ) = f (j1 , j2 , ..., jd ). It follows that (Nd−1 − N0 )(i1 − j1 ) + ... + (Nd−1 − Nd−2 )(id−1 − jd−1 ) + Nd−1 (id − jd ) = 0 mod Nd . Observe that i1 − j1 is divisible by n1 . As |i1 − j1 | < n1 , we have i1 = j1 . Divide the congruence by n1 . Then i2 − j2 is divisible by n2 . As |i2 − j2 | < n2 , we have i2 = j2 . Divide the congruence

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by n2 and continue in a similar way up to id = jd , which is a contradiction. Distinguish 2 subcases. a) Consider an edge of the k-th dimension, k ≤ d − 1. Wlog assume that f (i1 , i2 , ..., ik , ..., id ) ≥ f (i1 , i2 , ..., jk , ..., id ). Then compute f (i1 , i2 , ..., ik , ..., id ) − f (i1 , i2 , ..., jk , ..., id ) = (Nd−1 − Nk−1 )(ik − jk ) mod Nd . If ik > jk then (Nd−1 − Nk−1 )(ik − jk ) mod Nd ≥ Nd−1 − Nk−1 ≥ Nd−1 − Nd−2 . If ik < jk then (Nd−1 − Nk−1 )(ik − jk ) mod Nd = Nd − (Nd−1 − Nk−1 )(jk − ik ) ≥ Nd − (Nd−1 − Nk−1 )(nk − 1) > Nd−1 . b) Consider an edge of the d-th dimension. The analysis is the same as in Case I.b). For the cyclic antibandwidth of Hamming graphs we have (proof omitted):

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Theorem 2.2 For d ≥ 2 and 2 ≤ n1 ≤ n2 ≤ ... ≤ nd , cab(Πdk=1 Knk ) = ab(Πdk=1 Knk ) Concluding Remarks. To fill the gap in Theorem 2.1 remains a hard open problem. Note that the d-dimensional hypercube graph Qd is a special√subcase. Recall that ab(Qd ) = 2d−1 (1+o(1)) d−1 was proved − 2d / 2πd(1 + o(1)) in [8] and finally ab(Qd ) =  m[7].  Then ab(Qd ) = 2 d−2 in d−1 2 − m=0  m  in [11]. An interesting open problem is the antibandwidth of Kn1 ×Kn2 ×Kn3 . 2 Theorem 2.1 says that the only remaining unclear case is n1 = n2 = n3 = n. We conjecture that ab(Kn × Kn × Kn ) = n2 − n. For d > 3, we can prove that ab(Knd ) ≤ nd−1 − Θ(d2 ) suggesting that the upper bound in Theorem 2.1 is strictly less than nd−1 − 1.

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