On Two-label Interval Routing

HKU CSIS Tech Report TR-96-04On Two-label Interval RoutingSavio S.H.Tse and Francis uDepartment of Computer ScienceThe University of Hong Kongsshtse,fcmlau@cs.hku.hkJuly1996AbstractWe prove a lower bound of for the longest path in arbitrary graphs thatuse two-label interval routing(2-IRS)as the routing method,where is thediameter of the graph.1IntroductionInterval routing is a kind of compact routing which tries to keep as little amountof routing information in every node as possible[2][5].In one-label interval rout-ing scheme(1-IRS),every outgoing edge of a node is labeled with at most one labelwhich is actually an interval of node numbers.Routing is performed by comparingthe destination node number with these intervals at every node along the path.Forsome of the more popular types of graphs,optimal1-IRSs exist—i.e.,all the routingpaths are shortest paths[5].For arbitrary graphs,Ruˇz iˇc ka proved that it is not pos-sible to have an optimal1-IRS[1].He gave a lower bound of for the longestpath in a family of graphs.This bound was recently improved to by the au-thors[4].The natural question then arises:Using more labels per edge,is it possibleto have an optimal IRS for arbitrary graphs?In this paper,we consider the case ofusing two labels per outgoing edge.We derive a lower bound of for the longestpath,which again is greater than.The answer to the question is negative for twolabels.This new bound also represents an improvement of the lower bound resultfor2-IRS presented in[3].In the following,we assume that interval labels are cyclic.In addition to inter-val labels,there could be null labels and complement labels[1].However,it can beeasily seen that null labels and complements labels will not affect the result(see forexample the arguments in[4]).12PreliminariesThe network in question is a connected graph,,where is the set of edges,and the set of the nodes.Every edge in is bidirectional.There are nodes in.To implement interval routing,each node is labeled with a unique integer, called node number,from the set.In the following,we identify a node by its node number.Every edge in each direction is labeled with two interval labels,each of which of the form where.For that are directly connected,and denote the two interval labels for the edge that goes from to.We use as the short-hand notation for.A node is said to be contained in if(1)for,or(2)or, otherwise.During routing,the routing mechanism at a node compares the destina-tion node number of a message that is being routed with the interval labels of the node’s outgoing edges;the edge of which one of its labels contains the destination node number is selected.A valid IRS is one that can route a message from any node to any other node.We use the notation,to denote the cyclic ordering of node numbers, for.Naturally,.As in[4],the expression means that and are contained in some interval and that they are ordered after and before,but the order of and is not shown.Property1(Completeness)The set of interval labels for edges directed from a node u is complete.That is,every node in V u must be contained in one of u’s intervals.Property2(No ambiguity)The interval labels for edges directed from a node are disjoint. That is,for,is contained in exactly one of these intervals.Property3(No bouncing)For each,there exists no node,such that is contained in both and.Note that these properties are necessary but not sufficient for a valid IRS.For2-IRS,Property2seems to be unnecessary for the two interval labels that are attached to the same outgoing edge.It seems,however,not practical to have two such labels that are overlapping;two overlapping labels can always be com-bined into one label,thus saving some space.In fact,a2-IRS allows a stronger kind of disjointness for the dual labels—that is,there could exist a gap between the two labels’intervals such that they cannot be concatenated to form a single valid label. This property is vital in the proof of the lower bound.23The Lower BoundThe following two facts will be used in the main proof.Fact1Given objects,the number of ways to partition them into two equal-size groups is.Fact2Among all the different ways of partitioning objects that are arranged in a circle into two equal-size groups(each of size),there is one by which all the objects in the same group will not be neighbors in the circle.We are going to construct the graph based on which we will derive our lower bound.Figure1:The graph3Define of size and of diameter for, where and are as follows.ForForwhere represents the-th way of partitioning the set into two groups—in one group,and in the other group.An exam-ple of the graph is shown in Figure1,where.This particular graph has six “flaps”(whose roots are the nodes),and within eachflap,two columns and ten layers.The subscripts are used to denote theflap,the column,and the layer respectively.Note that the’s are connected to threeflaps and the’s are connected to the remaining three.In every layer,we partition the sixflaps into two groups,each group having threeflaps connected to either the’s or the’s.By Fact1,there are different ways of partitioning.Therefore,we need ten layers in the graph ,of which every layer represents a different way of partitioning theflaps.We can now prove the lower bound.Theorem1There is a graph such that for any valid-label interval labeling scheme, the longest routing path will be no shorter than.4Proof:Assume the contrary that there exists a labeling scheme for such that the longest path is shorter than.By the definition of,for all, the-th,the-th and the-thflaps are connected to’s and the-th,the-th and the-thflaps are connected to’s.For all,it must be the case thatand;otherwise,if, then by Property1,,and the length of the path from to through the edge would be at least.Hence,for all,In each of the above,since three elements are contained in two interval labels,by the pigeon-hole principle,one of the intervals contains at least two elements.In fact,by Property2,we canfind two such elements in one interval such that none of the other four elements falls between them.According to the definition of,all the different ways of partitioning the six ’s—which are in a cycle ordered by“”—are represented by the ten layers.By Fact2,there exists one way of partitioning,say the one represented by the-th layer, such thatandwhere,.Clearly,it is impossible to satisfy the condition that at least two elements in each of the groups must be in the same in-terval.4ConclusionNote that the proof is valid even if the edges and have only one la-bel(or the two labels are not disjoint).For such a one-label case,fourflaps,and hence three layers,would be enough for achieving the lower bound.This shows the power of having two(strong)disjoint labels per edge,which forces the graph to have at least sixflaps and ten layers.In this paper,we have derived a lower bound for the two-label interval routing scheme in arbitrary networks,which bining the lower bound results in5[3][4],we can present the lower bound for IRS in arbitrary graphs as a function,, of the number of labels,,as follows.ififififNote that all these bounds are greater than.References[1]P.Ruˇz iˇc ka,A note on the efficiency of an interval routing algorithm.The Com-puter Journal34(1991)475–476.[2]N.Santoro and R.Khatib,Labeling and implicit routing in networks.The Com-puter Journal28(1985)5–8.[3]S.S.H.Tse and u,Lower bounds for multi-label interval routing.Pro-ceedings of2nd Colloquium on Structural Information&Communication Complexity (SIROCCO’95)(1995).123–134.[4]S.S.H.Tse and u,A lower bound for interval routing in general net-works,to appear.[5]J.van Leeuwen and R.B.Tan,Interval routing.The Computer Journal30(1987)298–307.6。