New Bounds for Multi-Label Interval Routing Λ

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1 New Bounds for Multi-Label Interval Routing Λ Savio S.H. Tse and Francis C.M. Lau y Department of Computer Science and Information Systems The University of Hong Kong fsshtse,fcmlaug@csis.hku.hk November 2000 Abstract Interval routing is a well-known space-efficient routing method for computer networks. For longest routing path analysis, researchers have focused on lower bounds for many years. For any graph, there exists an upper bound of 2D for using one or more labels, and an upper bound of d 3 2 De for using O(p n log n) or more labels, where D is the diameter of the graph, and n the number of nodes. We present in this paper an upper bound of d 5 3 De for using O(p n) or more labels, and an upper bound of d 17 9 De for using O( 3p n) or more labels. The results can be 2F (i) 1 p generalized, leading to an upper bound of d F De for using O( 2+i (i) n) or more labels, where F (i +1) = 3F (i), F (0) = 3, and i is any non-negative integer constant. We also present an upper bound of (1 + ff)d for using O( D n ) or more labels, where ff is any constant in (0; 1). The second part of the paper offers some lower bounds for planar graphs. For any M-label interval routing scheme (M-IRS), where M covers the range of M = O( p n), we derive a lower bound of 2M+1 2M D 1 on the longest path for M = O( 3p n), and a lower bound of 2(1+ffi)M+1 2(1+ffi)M D, where ffi 2 (0; 1], for M = O( p n). The latter result implies a lower bound of Ω( p n) on the number of labels needed to acheive optimality. Keywords: Compact routing, computational complexity, distributed systems, graph theory, interval routing, network protocols, planar graphs. 1 Introduction Interval routing as a research topic has been under study for many years. For a detailed survey on results up to 1999, one can refer to [1]. Interval routing is attractive because of its simplicity: Λ A preliminary version of the planar graph results has appeared under the title Lower Bounds for Multi-Label Interval Routing in Proc. 2nd International Colloquium on Structural Information & Communication Complexity (SIROCCO 95), , y Correspondence: F.C.M. Lau, Department of Computer Science and Information Systems, The University of Hong Kong, Hong Kong / fcmlau@csis.hku.hk / Fax: (+852)

2 every node is assigned a unique integer ID from a cyclicly-ordered set, and every outgoing link is assigned an interval label which is a range of integers from the same set. Message routing is carried out by comparing the destination ID with interval labels as the message moves from node to node in the network. This is the one-label interval routing scheme, or1-irs. A valid IRS is one that can route a message from any node to any other node along a deterministic path. There are advantages to attaching more labels to an edge. An M-label IRS, or M-IRS, is an IRS where we can attach up to M labels to any edge. One way of measuring the quality of interval routing scheme is to look at the longest routing path. In this paper, we say that the an IRS is optimal if the resulting longest path is equal to the diameter, D, in length. For arbitrary graphs, there exists a 1-IRS such that the longest path is bounded by 2D [8], and an p O( n log n)-irs such that the longest path is bounded by d 3 De [4]. 2 Comparing with the lower bounds of 2D 3 and b 3 2 Dc 1 in [13, 2], the 2D and d 3 De upper 2 bounds are very close to the optimal for 1-IRS and p O( n log n)-irs, respectively. Between M =2 and M p = O( n log n), there has been a lack of upper bound results for years. A trivial upper bound for this range is 2D based on the fact that the path lengths cannot be longer with using 2F (i) 1 p more labels. In this paper, we improve the upper bound to d De for O( 2+i F (i) n)-irs, where F (i + 1) = 3F (i), F (0) = 3, and i is any non-negative integer constant. A summary of the existing upper and lower bounds, including the ones derived in this paper, is given in Figure 1. The upper bounds for i 3 are marked by a in the figure. 1 length of longest path 2D [8] D [*] ( ) [*] 3 2 D-0.5 [13] D upper bounds [*] 17 9 D [*] 5 3 D [*] [*] lower bounds 3 2 D [*]--- this paper [4] trivial upper bound trival bound D θ ( n) θ n) θ n) k ( 4 ( 3 ( θ n) θ( n log n) n D log n/d θ ( ) θ( n ) D θ(n) number of labels Figure 1: Spectrum of upper and lower bounds (not to scale). We also present an O( n )-IRS of which the longest path is bounded by (1 + ff)d, for any ffd ff 2 (0; 1). This result is applicable to those graphs with large diameters. If a small constant is 1 The figure describes the case of D = O( p n log3 n ). 2

3 chosen for ff, this result is close to the lower bound result in [14]: there exists a graph such that q for any M-IRS, if M» n 18D O( n ), the longest path is no shorter than D + ( p D D M ), where D =Ω( 3p n). As shown in Figure 1, for the cases of one label, p n ( n log n) to ( D log D n ) labels, and then more than ( n ) labels, the upper bounds and the lower bounds are very close to each other. But D between two to p n ( n log n) labels, and ( D log D n ) to ( n ) labels, there are an appreciable gap D between the upper bounds and the lower bounds, such as a gap of 1 D from the best known lower 6 bound of b 3 2 Dc 1 for the case of (p n) to p ( n log n) labels [2]. One could hope for a narrower gap in the future. Since many graph algorithms perform better in planar graphs than in non-planar graphs, we would like to know how interval routing would perform in planar graphs. Several lower bounds have been proposed for non-planar graphs (e.g., [10, 11]), but for planar graphs, there exists only one lower-bound result 3D 1 which is due to Ružička 2 [7].2 His proof is based on a simple planar graph, which he later referred to as a globe graph [4] (see Figure 2 for an example). In [12], the authors improved the bound to 3D 1. In the second part of this paper, we present two lower 2 2 bounds for planar graphs: 2M D 2M 1 for M = O( 3p n), and 2. 2(1+ffi)M +1 2(1+ffi)M D for M = O(p n), for any constant ffi 2 (0; 1]. The second bound directly implies a lower bound of p Ω( n) on the number of labels needed to achieve optimality i.e., where the longest path is at least equal to D in length. It also implies a lower bound of p Ω( n) on the number of labels needed to achieve shortest-path routing, coinciding with a result due to Gavoille and Pérennès in [3]. A spectrum of lower bounds for planar graphs is given as the bottom solid line in Figure 1. It is smoother than the spectrum of lower bounds for non-planar graphs (the dashed line), although we do have yet have an idea about the optimality of these planar graph lower bounds. As for upper bounds, planar and non-planar graphs share the same spectrum. We cannot conclude that interval routing performs better in planar graphs than in non-planar graphs until there exist some upper bounds for planar graphs. 2 Definitions and Properties A graph we consider here for interval routing is an undirected connected graph, G =(V;E), where V is the set of nodes and E the set of directed edges. An undirected edge is treated as two directed 2 The proof has some minor defects which we corrected in [12]. 3

4 edges i.e. (u; v) 6= (v; u). There are n nodes in V and each node has a unique label from the set V = f0; 1;:::;n 1g. The node labels are cyclicly ordered, denoted as 0 ffi 1 ffi ffin 1 ffi 0. Definition 1 Let k 2 [2;n]. For any distinct v 1 ;v 2 ;:::;v k 2 f0; 1;:::;n 1g, v 1 ffi v 2 ffi ffi v k if 9p 1 ;p 2 ;:::;p k 1 such that P i=k 1 i=1 p i <nand v i+1 v i + p i (mod n) for i =1to k 1. We further define the expression u ffi fv; wg ffi x to be two simultaneous relations based on the cyclic order: u ffi v ffi x and u ffi w ffi x. Definition 2 An interval ha; bi is the set fa; a +1;:::;b (mod n)g. The elements a; b are called the marginal elements of the interval. In particular, ha; ai = hai = fag, and ; is the empty interval. We refer to such a set as an interval set. A set A ρ V is not an interval if and only if A is a proper subset of every interval set containing it. Definition 3 Given that B is an interval. A set A is a sub-interval of an interval B if it is an interval and is a subset of B. A is a proper sub-interval of B if A is a sub-interval of B and the marginal elements of A are non-marginal elements of B. Definition 4 Two intervals A and B are non-overlapping if A B = ;. Definition 5 Two intervals A and B are disjoint if A [ B are not an interval. Any two disjoint intervals are non-overlapping. Definition 6 Let I be the set containing every possible interval subset of V. E.g., V = f0; 1; 2; 3; 4g; I = f;; h0i; h1i; h2i; h3i; h4i; h0; 1i; h1; 2i; h2; 3i; h3; 4i; h4; 0i; h0; 2i; h1; 3i; h2; 4i; h3; 0i; h4; 1i; h0; 3i; h1; 4i; h2; 0i; h3; 1i; h4; 2i; h0; 4ig: Definition 7 For M 2, let I M be the set containing every union of M elements of I. Referring to the previous example, I 2 = I[fh0i[h2i; h0i [h3i; h1i [h3i; h1i [h4i; h2i [h4i; h0; 1i[h3i; h1; 2i [h4i; h2; 3i [h0i; h3; 4i [h1i; h4; 0i [h2ig: Some unions of intervals are intervals; for example, h0i[h1i = h0; 1i, h0; 1i[h4i = h4; 1i. Definition 8 For any V, a node labeling function L is a one-one mapping L : V! V. Definition 9 Let M 2. AnM-label edge labeling function L Λ is a mapping L Λ : E!I M. 4

5 For each (u; v) 2 E, L Λ (u; v) is a union of M intervals. Each of these M intervals is an interval label of u on (u; v). Since the union of two non-disjoint intervals is an interval, in other words, L Λ (u; v) is a union of at most M disjoint intervals. Definition 10 An M-interval routing scheme, or M-IRS, on a graph G = (V;E) is an ordered pair (L; L Λ ) where L is a node labeling function and L Λ is an M-label edge labeling function such that the following are satisfied. ffl 8u; v 2 V, u 6= v, 9 a simple path u; x 1 ;x 2 ;:::;x k ;vin G such that L(v) 2 L Λ (u; x 1 ) L Λ (x 1 ;x 2 ) ::: L Λ (x k ;v), and ffl 8u 2 V,if(u; v 1 ); (u; v 2 ) 2 E, and v 1 6= v 2, then L Λ (u; v 1 ) L Λ (u; v 2 )=;. Directly from Definition 10, we have the following two properties. Property 1 (Complete) The set of interval labels for edges directed from a node u is complete. That is, 8u 2 V, V fl(u)g ρ[ (u;v)2e L Λ (u; v). Property 2 (Deterministic) The interval labels for edges directed from a node u are disjoint. That is, for u 6= v, L(v) is contained in exactly one of these interval labels. It should be noted that these two properties are necessary but not sufficient for a valid IRS for general graphs. 3 Upper Bounds on Multi-Label Interval Routing 3.1 Basic Lemmas Definition 11 Given a graph G =(V;E), x 2 V, ff and p i some positive constants, i a positive integer, and p 0 =1. Define ffl i = ffd p i p i 1 p 1 p 0, ffl V i x = fv 2 V jd(x; v)»b icg, and ffl d(x; A) = minfd(x; a)ja 2 Ag, for A ρ V. In particular, V 0 x = fv 2 V jd(x; v)»bffdcg. Lemma 1 For any graph G =(V;E), if there exists a subset of V, C, such that jcj = O(M ) and 8v 2 V, d(v; C)» D 0, then there exists an O(M )-IRS such that the longest path is bounded by D + D 0. 5

6 Proof: Given a graph G = (V;E). We partition G into at most O(M ) connected subgraphs such that each subgraph has a spanning tree of depth D 0. Assume that there are m = O(M ) subgraphs denoted as G i =(V i ;E i ), with corresponding roots R i, i =1;:::;m. Consider the nodes labels. The nodes labels of V 1 are from 0 to jv 1 j 1. For i =2; 3;:::;m, the nodes labels of V i are from P i 1 j=1 jv jj to P i j=1 jv jj 1. The labels of each V i, i 2 [1;m], form an P i 1 interval, denoted as I i. Consider G i. The root R i is labeled with a number jv j=1 jj, or0 if i =1. Then, we label the remaining nodes in a pre-order fashion based on its spanning tree. 3 For routings inside G i, we use the edges of the spanning tree only. We assign the edges in the spanning tree with at most two interval labels. For routing downwards, one interval label per downward edge is enough, because with the pre-order numbering, the nodes labels for a downward edge form an interval. Hence, each downward edge will have one interval label. The upward edge will then have two interval labels. The reason is that for a node u, the labels of u and its descendants form an interval due to the pre-order numbering; that means that this set s complement forms two intervals in I i. For routings from x to y, x 2 V i, y 2 V j, i; j» m, i 6= j, we first find a shortest path from x to R j. Let x; a 1 ;a 2 ;:::;a k ;R j be the shortest path. If a i ; 8i 2 [1;k], are not in the the spanning tree of G j, we simply assign the directed edges (x; a 1 ); (a k ;R j ); (a i ;a i+1 ), 8i 2 [1;k 1], an interval I j. If the path x; a 1 ;a 2 ;:::;a k is not disjoint with the spanning tree, we choose the minimum r such that a r is in the spanning tree of G j. We now label the directed edges (a; a 1 ); (a 1 ;a 2 );:::;(a r 1 ;a r ) with an interval I j. We count the maximum number of interval labels used by the edges. A directed edge (u; v) which is not in any spanning tree and u 2 V i, i 2 [1;m], has at most m 1 interval labels which are I 1 ;I 2 ;:::;I i 1 ;I i+1 ;:::;I m. 4 (I 1 [ I 2 [ :::[ I m = f0; 1;:::;n 1g.) For a directed edge which is in one of the spanning trees, the edge has two more intervals i.e., m +1interval labels. We then consider the routing paths lengths. For routings inside each G i, i 2 [1;m], the routing paths are at most two times the depth of the spanning tree, which are no longer than 2D 0, which is less than D + D 0. For a routing from x to y, x 2 V i, y 2 V j, i; j 2 [1;m], i 6= j, we have two cases: ffl The routing path will pass through R j. The path from x to R j takes at most D steps, and the path from R j to y takes at most D 0 steps, and so totally the routing path takes less than D + D 0 steps. ffl The routing path will not pass through R j. It will reach the first node in V j, say u. Ify 6= u, the routing path will be x;:::;u;:::;z;:::;y, where z is the root of the smallest subtree containing u and y. The path from x to z takes 3 A similar technique was used in [6, 8]. 4 We can use at most m interval labels because any two adjacent interval labels can be combined into one. 2 6

7 less than D steps, and the path from z to y takes less than D 0 steps, and so the whole routing path takes less than D + D 0 steps. For simplicity, we skip the routine to show the validity of this IRS with respect to the definitions and properties. 2 Lemma 2 Given that 9v 2 V such that jvv i j»m, where i is a non-negative integer. Then, 9C ρ V such that jcj»m and for any w 2 V, d(w; C)» 2b ic or d(w; C)»dD ie. Proof: We find a BFS tree rooted at v. At the b ic-th level of the tree, there must be less than M nodes; otherwise, jvv i j >M. We assign C to be the set containing the nodes at the b ic-th level. For w (2 V ) above the b ic-th level in the tree, d(w; C)» 2b ic; for w below the b ic-th level in the tree, d(w; C)» D b ic = dd ie. 2 Lemma 3 Given that 8v 2 V, jvv 0 j > M. Then, 9C ρ V such that jcj < n, and for any w 2 V, M d(w; C)» 2b 0c. Proof: There are at most n M elements of V forming a subset C such that for any distinct x; y 2 C, V 0 x V 0 y = ;. Then, 8w 2 V=C, 9c 2 C such that V 0 w V 0 c 6= ;. Since 9t 2 V 0 w V 0 c, d(w; c)» d(w; t) +d(t; c)» 2b 0c. 2 Lemma 4 Given that 9v 2 V, M 0 < jvv i j»m, and 8a 2 Vv i, jva i+1 j >M 0, where i is a non-negative integer. Then, 9C ρ V such that jcj < M, and for any w 2 V, d(w; M 0 C)» 2b i+1 c or d(w; C)» dd i +3 i+1 e. Proof: Since jv i v j»f i (n), there exist at most M M 0 elements of V i v forming a subset C such that for any x; y 2 C, Vx i+1 Vy i+1 = ;, and Vx i+1 ;Vy i+1 ρ Vv i. We consider a BFS tree rooted at v. Consider a w 2 V=C such that d(w; v)» b ic b i+1 c. In this case, w is a node in Vv i and V i+1 w ρ V i v. The reason that w 62 C is that 9c 2 C such that V i+1 w V i+1 c 6= ;. Therefore, 9t 2 Vw i+1 Vc i+1 such that d(w; c)» d(w; t) +d(t; c)» 2b i+1 c. For a node w 2 V=C such that d(w; v) > b ic b i+1 c, since the tree must have more than b ic b i+1 c levels, we can find a (b ic b i+1 c)-th level element t such that d(t; w)» dd i + i+1 e. If t 2 C, the lemma is proved. If t 62 C, 9c 2 C such that V i+1 t Vc i+1 6= ;; and 9t 0 2 V i+1 t V i+1 c such that d(t; c)» d(t; t 0 )+d(t 0 ;c)» 2b i+1 c. Hence, d(w; c)» d(w; t)+d(t; c)» dd i +3 i+1 e An O( p n)-irs Theorem 1 For any graph G, there exists an O( p n)-irs such that the longest path is bounded by d 5 3 De. 7

8 Proof: We have two cases. First, consider that 9v 2 V such that jv 0 v j» p n. By Lemma 2, 9C ρ V such that jcj» p n and for any w 2 V, d(w; C)» 2b 0c or d(w; C)»dD 0e. Second, consider that 8v 2 V, jv 0 v j > p n. By Lemma 3, 9C ρ V such that jcj» p p n n = n and for any w 2 V, d(w; C)» 2b 0c. Take ff = 1, we have C ρ V such that jcj 3 = O(p n) and d(v; C)»d 2 De, 8v 2 V. By Lemma 1, 3 the result follows An O( 3p n)-irs Theorem 2 For any graph G, there exists an O( 3p n)-irs such that the longest path is bounded by d 17 9 De. Proof: We have three cases. First, consider that 9v 2 V such that jv 0 v j» 3p n. By Lemma 2, 9C ρ V such that jcj» 3p n and for any w 2 V, d(w; C)» 2b 0c or d(w; C)»dD 0e. Second, consider that 8v 2 V, jv 0 v j > 3p n 2. By Lemma 3, 9C ρ V such that jcj» 3pn n and for any w 2 V, 2 d(w; C)» 2b 0c. Third, consider that 9v 2 V, 3 p n<jv 0 v j» 3p n 2. We have two sub-cases. 1. 9a 2 Vv 0 such that jva 1 j» 3p n. By Lemma 2, 9C ρ V such that jcj» 3p n and for any w 2 V, d(w; C)» 2b 1c or d(w; C)» dd 1e. 2. 8a 2 Vv 0, jva 1 j > 3p n. By Lemma 4, 9C ρ V such that jcj» 3p p n 2 3 n = 3p n and for any w 2 V, d(w; C)» 2b 1c or d(w; C)»dD e. Take ff = 4 and p 9 1 =4, we have C ρ V such that jcj = O( 3p n) and 8v 2 V, d(v; C)»d 8 De. By 9 Lemma 1, the result follows. 2 p 3.4 An O( 2+i n)-irs p Lemma 5 For any graph G =(V;E), there exists a subset C of V such that jcj = O( 2+i n) and 8v 2 V, d(v; C) is no greater than one of the following terms: 2b 0c; dd ie; and dd j 1 + je; 8j 2 [1;i]; where i is any integer constant greater than one. p Proof: Let f j (n) = 2+i n i+1 j ;j 2 [0;i]. If9v 2 V such that jvv 0 j»f i (n), by Lemma 2, the lemma is proved. Similarly, if 8v 2 V, jvv 0 j > f 0 (n) (by Lemma 3). If neither is the case, there exists a v 2 V such that f i (n) < jv 0 v j» f 0 (n). Since 8j 2 [1;i], f j (n) < f j 1 (n), there exists a j 2 [1;i] such that f j (n) < jv 0 v j»f j 1 (n). 8

9 We consider some Va 1 s, a 2 Vv 0. If 9a 2 Vv 0 such that jva 1 j» f i (n), by Lemma 2, we have the proof. Since v 2 Vv 0 and jvv 1 j < jvv 0 j» f j 1 (n), we need not consider the case 8a 2 Vv 0, jva 1 j >f j 1 (n). For 8a 2 Vv 0, jva 1 j >f j (n), by Lemma 4, the lemma is proved. If j = i, one of the above cases must be true. If j <iand none of the above cases is true, there exists an a 2 V v 0 such that f i (n) < jva 1 j»f j (n). Then, there exists a k 2 [j +1;i] such that f k (n) < jva 1 j»f k 1 (n). We now consider some V 2 b s, b 2 V a 1. Similarly, if k = i, either 9b 2 Va 1 such that jv 2 b j»fi (n), or 8b 2 Va 1, jvb 2j > f k (n). By applying Lemmas 2 and 4 respectively, these two cases are taken care of. If k < i, there may exist a b 2 Va 1 such that f i (n) < jvb 2j» f k (n). Then, there exists an l 2 [k +1;i] such that f l (n) < jv 2 b j»f l 1 (n). We will then consider some V 3 c s, c 2 V 2 b. Inductively, the lemma is proved. 2 Theorem 3 For any graph G, there exists an O( 2+i p n)-irs such that the longest path is bounded by d 2F (i) 1 De, where F (i + 1) = 3F (i), F (0) = 3 and i is any non-negative integer constant. F (i) Proof: For the case of i = 0; 1, by Theorems 1 and 2, the theorem is proved. Consider i > 1. By p Lemma 5, there exists C ρ V such that jcj = O( 2+i n) and 8v 2 V, d(v; C) is no greater than the maximum of 2b 0c; dd ie; and dd j 1 + je; 8j 2 [1;i]: To find the minimum upper bound, we use some standard technique to make the above terms ffd equal to each other. Recall that i = p i p i 1 p 1 p 0. We take p i =4and p i 1 = 13. Let the denominators of p j be q j, j 2 [1;i 1]. We take p j = 3q j +1 4 q j, 8j 2 [1;i 1], and q j 1 =3q j +1, 8j 2 [2;i 1]. And, we take ff = 3q 1+1. Therefore, d(v; 2(3q 1 +1)+1 C)» d 2(3q 1+1) De. Setting F 2(3q 1 +1)+1 (i) = 2(3q 1 +1)+1, d(v; C)»d F (i) 1 De. F (i) We now show that if i is replaced by i +1, d(v; C)» d F (i+1) 1 De, where F (i +1) = 3F (i). F (i+1) We set p i+1 = 4, and re-set p i = 13, p 4 j = 3q0 j +1 for j 2 [1;i], where q 0 j 1 = 3q0 j +1for j 2 [2;i]. Setting F (i +1)=2(3q 0 1 q 0 j F (i+1) 1 +1)+1, d(v; C) is then bounded by d De. Obviously, q 0 F (i+1) 2 = q 1, and F (i +1)=2(3q )+1=2(3(3q0 +1)+1)+1=2(3(3q )+1)+1=3F (i). By Lemma 1, the result follows An O( n )-IRS for any ff 2 (0; 1) ffd Lemma 6 For any graph G, there exists a partition such that each connected component has a spanning tree of depth from K to 2K inclusively, where K» D. Proof: Given a graph G = (V;E), we partition V into V 1 ;V 2 ;:::;V m and V 0 such that (1) for all i 2 [1;m], the induced subgraph of V i is a connected graph and has a spanning tree of depth K, for some m»b n c; and (2) the induced subgraph of V 0 is a disconnected graph having m 0 connected K components, with none of them having a spanning tree of depth K. Let G i be the induced 9

10 subgraph of V i, for all i 2 [1;m], and let G 0 j = (V j 0;E0 j ) be the j-th connected component of the induced subgraph of V 0, for all j 2 [1;m 0 ]. For any j 2 [1;m 0 ], 9i 2 [1;m] and 9(u; v) 2 E such that u 2 Vj 0, v 2 V i. Intuitively, a G 0 j is a neighbour of one or more G i s, but not a neighbour of G 0 k, k 6= j. For each j 2 [1;m0 ], we attach G 0 j to one of its neighbours, G i, i 2 [1;m]. For those G i s not having been attached any G 0 j, i 2 [1;m];j 2 [1;m0 ], a spanning tree of depth K exists. For those G i s that have been attached G 0 i 1 ;G 0 i 2 ;:::;G 0, i 2 [1;m];i ip s 2 [1;m 0 ];s 2 [1;p];p 2 [1;m 0 ], we consider the subgraph G i induced by V i [ Vi 0 1 [ Vi 0 2 [ ::: [ V 0. In G ip i, there exists a spanning tree of depth K, and with root r. Let the edge connecting V i and V 0 be (u is is;v ), i s s 2 [1;p] and u i s 2 V i ;v i s 2 V 0 0. For all x 2 V, there exists a path from r to x, passing through the is is spanning tree of G i, the edge (u i s;v ), and a path from v i s is to x. The length of this path is at most K +1+(K 1) = 2K. Hence, a spanning tree of depth between K and 2K for G i exists. 2 Theorem 4 For any graph G, there exists an (b n c +1)-IRS such that the longest path is bounded by dffde (1 + 2ff)D, for any ff 2 (0; 1 2 ). Proof: By Lemmas 1 and Some Remarks We have presented two results on upper bounds: p 1. An O( 2+i n)-irs whose longest path is bounded by d 2F (i) 1 De, where F (i +1) = 3F (i), F (i) F (0) = 3, and i is any constant non-negative integer. Note that 8m 2 Z +, 2m 1 and m are relatively prime, meaning that the upper bounds as given here are in their simplest forms. 2. An O( n )-IRS whose longest path is bounded by (1 + ff)d, for any constant ff 2 (0; 1). D According to Definition 11, our first result is meaningful if ffd p i p i 1 p 1. This means that if i is a constant, we can apply our result to arbitrary graphs of any diameter which can be as small as O(1). Our second result is mainly for graphs of large diameter, preferably p Ω( n). For graphs of smaller diameter, this scheme uses more labels although the longest path is shorter. For example, log n if D = Ω(2 log log n ), this scheme gives an O(n 1 1 log log n )-IRS whose longest path is slightly longer than D. The other scheme above would give an O(log n)-irs whose longest path is slightly shorter than 2D. We can easily generate O(n 3 )-time labeling algorithms for each scheme. First, we can use O(n 3 ) time to build an n n all-to-all distance matrix. Then, using this matrix, we can use O(n 2 ) time 10

11 to build an n D matrix, where each cell (i; j) stores the number of nodes in V having distance j from node i. With the second matrix, we can easily find the set C in each case. Then, we can build the disjoint spanning trees rooted at elements in C and label the nodes, which will take O(n 2 ) time. Labeling the edges requires constructing the shortest paths from all the nodes to each element in C. This will take O(jCj(jEj + n log n)) = O(n 2+ffi ) time, where ffi is some constant in (0; 1). Hence, we have an O(n 3 )-time algorithm for each IRS. We can see that the dominating part is to build the all-to-all distance matrix. 4 Lower Bounds for Planar Graphs 4.1 The Graph We use the globe graph G S;C;K, as shown in Figure 2, to prove our lower bounds. We define K K v_0,2 v_0,c-2 t_l x_0,1,1 x_1,1,1 X_l v_0,1 v_1,1 v_1,2.... v_1,c-2 v_0,c-1 x_0,c,k-1 v_1,c-1 x_1,c,k-1 t_r X_r I_1 I_C-1 I_2 I_C-2 Figure 2: The skeleton of G S;C;K. G S;C;K =(V S;C;K ;E S;C;K ) which is of diameter D = CK, and size n = SCK + CK S +1, where 11

12 C is even, and V S;C;K and E S;C;K are as follows. V S;C;K = fv s;c j0» s» S; 1» c» C 1g [ fx s;c;k j0» s» S; 1» c» C; 1» k» K 1g [ ft l ;t r g E S;C;K = f(x s;c;k ;x s;c;k+1 )j0» s» S; 1» c» C; 1» k» K 2g [ f(v s;c ;x s;c+1;1 )j0» s» S; 1» c» C 1g [ f(x s;c;k 1 ;v s;c )j0» s» S; 1» c» C 1g [ f(t l ;x s;1;1 )j0» s» Sg [ f(x s;c;k 1 ;t r )j0» s» Sg There are S +1rows. The 0-th row is the base row which plays a different role in the proof. In each row, there are C 1 v s;c s. All v s;c s are grouped into C 1 columns, as shown in Figure 2. With the columns formed by x s;1;1 s and x s;c;k 1 s, we have C +1columns participating in the proof. For the convenience of discussion, let X l be the set fx s;1;1 j1» s» Sg, X r be fx s;c;k 1 j1» s» Sg, and I c be fv s;c j» s» Sg, 8c 2 [1;C 1]. In this section, we consider the interval structure of the elements in X l [ X r [ I 1 [ I 2 [ :::[ I C 1. Also for convenience, we let L + c L Λ(v 0;c ;x 0;1;c+1 ) and L c L Λ (v 0;c ;x 0;K 1;c ), for all c 2 [1;C 1]. For the sake of simplicity, but without loss of generality, we assume that V S;C;K f0; 1;:::;n 1g and the node labeling function L is an identity function i.e., 8v 2 V S;C;K ;L(v) =v. 4.2 Basic Lemmas Our approach is to prove by contradiction. If there is an M-IRS such that the longest path is shorter than C+1 D 1, the following lemmas (7 to 9) must hold. C Lemma 7 9p l 2 [1;M] such that there are p l disjoint intervals which contain all elements in X l but not any elements in X r [ I 2 [ I 4 [ :::[ I C 2. Proof: Consider a base node v 0;1. By the assumption on the path length, we have X l [ I 2 [ I 4 [ :::[ I C 2 ρ L, and X 1 r ρ L + 1. By the definition of M-IRS, we have at most M disjoint intervals containing X r but not any elements in X l [ I 2 [ I 4 [ :::[ I C 2. Therefore, the existence of a p l in the lemma statement is guaranteed. 2 Lemma 8 9p r 2 [1;M] such that there are p r disjoint intervals which contain all elements in X r but not any elements in X l [ I 2 [ I 4 [ :::[ I C 2. Proof: Similar to the proof of Lemma 7. 2 Lemma 9 For each c 2 f2; 4;:::;C 2g, 9p c 2 [1; 2M ] such that there are p c disjoint intervals which contain all elements in I c but not any elements in I c 0, c 0 2f2; 4;:::;C 2g, c 0 6= c, and not any elements in X l [ X r. 12

13 Proof: By the assumption of path length, and considering v 0;1, we have X r ρ L + 1 and X l [ I 2 [ I 4 [ :::[I C 2 ρ L 1. Similarly, considering v 0;3, we have X r [I C 2 ρ L + 3 L 3. Inductively, we have Table 4.2. and X l [I 2 [I 4 [:::[I C 4 ρ c base node L + c contains L c contains 1 v 0;1 X r X l [ I 2 [ :::I C 6 [ I C 4 [ I C 2 3 v 0;3 X r [ I C 2 X l [ I 2 [ :::I C 6 [ I C 4 5 v 0;5 X r [ I C 2 [ I C 4 X l [ I 2 [ :::I C 6 C fl 1 v 0;C fl 1 A B [ I fl C fl +1 v 0;C fl+1 A [ I fl B C 3 v 0;C 3 X r [ I C 2 [ I C 4 [ :::I 4 X l [ I 2 C 1 v 0;C 1 X r [ I C 2 [ I C 4 [ :::I 4 [ I 2 X l Table 1: Table of intervals. In the table, A is a short form of X r [ I C 2 [ I C 4 [ :::[ I fl+2 and B is a short form of X l [ I 2 [ I 4 [ :::[ I fl 2. Consider v 0;C fl 1. Since A ρ L + and B [ I C fl 1 fl ρ L C fl 1, 9p 2 [1;M] such that there are p disjoint intervals A 1 ; A 2 ;:::;A p containing all elements in A but not any elements in B [ I fl, and there are p disjoint intervals BΛ 1 ; BΛ 2 ;:::;BΛ p containing all elements in B [I fl but not any elements in A. Then, we have A 1 ffibλ 1 ffia 2 ffibλ 2 ffi ffia p ffibλ p : The A s and BΛ s alternate (Figure 3); otherwise, we can group two A s or two BΛ s together and choose a smaller p. Similarly, by considering v 0;C fl+1, 9q 2 [1;M] such that there are q disjoint intervals B 1 ; B 2 ;:::;B q containing all elements in B but not any elements in A [ I fl. For the convenience of discussion, we can restrict the marginal elements of B s to be in B. Then, the q B s may intersect with p BΛ s only; they cannot have any intersections with any one of A s. Therefore, these p + q intervals p A s and q B s are non-overlapping (Figure 3). All elements of I fl can not be in the p A s, nor in the q B s. They can only be in the gap between two A s, or between two B s, or between one A and one B. There are p + q such gaps. In 13

14 \cal B*_1 \cal A_1 \cal A_3 \cal B*_2 \cal A_2 \cal B*_3 \cal B_j \cal A_i Figure 3: Cyclic structures of three A s, three BΛ s and four B s. other words, they belong to the set f0; 1; 2;:::;n 1g=([ p i=1 A i [[ q j=1 B j) which are in at most p + q disjoint intervals. Hence, all elements in I fl are in at most p + q» 2M disjoint intervals which do not contain any elements of A [ B. 2 Lemmas 7 to 9 show the interval structure of the elements in X l [ X r [ I 2 [ I 4 [ :::[ I C 2.By the similar argument, we have the following Lemma 10 which states the interval structure of I 1 [ I 3 [ :::[ I C 1. If there is an M-IRS such that the longest path is shorter than C+1 D, the following lemma, C Lemma 10, holds. Note that the additive term 1 is not necessary here. Lemma 10 (1) 9p 1 2 [1;M] such that there are at least p 1 disjoint intervals which contain I 1 but not any elements in I 3 [ I 5 [ :::[ I C 1. (2) 9p C 1 2 [1;M] such that there are at least p C 1 disjoint intervals which contain I C 1 but not any elements in I 1 [ I 3 [ ::: [ I C 3. (3) For each c 2 f3; 5;:::;C 3g, 9p c 2 [1; 2M ] such that there are at least p c disjoint intervals which contain I c but not any elements in I c 0, c 0 2f1; 3;:::;C 1g, c 0 6= c. Proof: Similar to the proof of Lemmas 7 to

15 4.3 The First Bound: 2M +1 2M D 1 for M = O( 3p n) Theorem 5 There exists a planar graph such that for any valid M-IRS, the longest path will be no shorter 2M +1 than D 1. 2M Proof: We use the graph G S;C;K and set C = 2M. Assume that there exists a valid M-IRS such 2M +1 that every path is shorter than D 1. Then, Lemmas 7 to 10 hold. 2M Let A be the set X l [ X r [[ M 1 I c=1 2c. By Lemmas 7, 8, and 9, we have: (1) p l (» M ) disjoint intervals which contain all elements in X l but not A=X l, (2) p r (» M ) disjoint intervals which contain all elements in X r but not A=X r, and (3) for each c 2f2; 4;:::;2M 2g, P p c (» 2M ) disjoint M 1 intervals which contain all elements in I c but not A=I c, where these p l + p r + p c=1 2c (» 2M 2 ) intervals, called A s intervals hereafter, are non-overlapping. For the convenience of discussion, the marginal elements of A s intervals are assumed to be in A; mathematically, if any one of these intervals has marginal element(s) not in A, we can replace it by its largest sub-interval such that its marginal elements are in A. Consider the set B = [ M I c=1 2c 1. By Lemma 10, for each c 2f1; 3;:::;2M 1g, there are p c disjoint intervals which contain all elements in I c but not B=I c, where p 1 ;p 2M 1» M, p c» 2M;8c 2 P M f3; 5;:::;2M 3g and these p c=1 2c 1 (» 2M 2 2M ) intervals, called B s intervals hereafter, are non-overlapping. By a similar argument as that for A s intervals, B s intervals marginal elements belong to B. We are now going to show that the two sets of intervals will lead to a contradiction. Since there are at most two marginal elements in an interval, there are at most 8M 2 4M rows, each of which having at least one marginal element in any one of A s intervals or in any one of B s intervals. Assume there is a sufficiently large number of rows. We take a row, say the i-th row, which has marginal elements neither in A s intervals nor in B s intervals. Consider L Λ (t l ;x i;1;1 ). L Λ (t l ;x i;1;1 ) contains x i;1;1 ;v i;2 ;v i;4 ;:::;v i;2m 2 ; otherwise, a routing path from t l will be longer than 2M +1 2M D 1. Since these M elements x i;1;1 ;v i;2 ;v i;4 ;:::;v i;2m 2 are not marginal elements of A s intervals, an interval containing any two of them (v i;2 ;v i;4, say) will contain the marginal elements (v i 0 ;2;v i 00 ;4, say) of the A s intervals to which the two elements belong (Figure 4). According to the assumpmarginal elements Figure 4: Two marginal elements are grouped. tion on the path length, L Λ (t l ;x i;1;1 ) cannot contain any elements in A=X r except that from the 15

16 i-th row; hence it cannot contain any marginal elements of those A s intervals containing A=X r because these marginal elements are not from the i-th row. In order to contain x i;1;1 ;v i;2 ;v i;4 ;:::;v i;2m 2, L Λ (t l ;x i;1;1 ) must be a union of M disjoint intervals which contain x i;1;1 ;v i;2 ;v i;4 ;:::;v i;2m 2,respectively. Since v i;1 ;v i;3 ;:::;v i;2m 1 2 L Λ (t l ;x i;1;1 ), by similar argument, the M disjoint intervals of L Λ (t l ;x i;1;1 ) must contain v i;1 ;v i;3 ;:::;v i;2m 1, respectively. Hence, 9q 2 f1; 3;:::;2M 1g such that v i;q and x i;1;1 belong to the same interval label of L Λ (t l ;x i;1;1 ), say L 1 (t l ;x i;1;1 ). Let the A s interval which contains x i;1;1 be Xl o. Since L 1(t l ;x i;1;1 ) contains x i;1;1 but not the marginal elements of X o l, L 1(t l ;x i;1;1 ) is a proper sub-interval of X o l. Hence, v i;q is a non-marginal element of X o l (although it is not an element of X l ). Consider L Λ (t r ;x i;c;k 1 ). v i;q ;v i;2 ;v i;4 ;:::;v i;2m 2 ;x i;c;k 1 2 L Λ (t r ;x i;c;k 1 ); otherwise, the assumption on the path length will be violated. Hence, L Λ (t r ;x i;c;k 1 ) contains M +1nonmarginal elements of different A s intervals. By the Pigeon Hole Principle, one of the interval labels of L Λ (t r ;x i;c;k 1 ), say L 1 (t r ;x i;c;k 1 ), will contain two elements from v i;q ;v i;2 ;v i;4 ;:::; v i;2m 2 ;x i;c;k 1, and one of them must be from v i;2 ;v i;4 ;:::;v i;2m 2 ;x i;c;k 1 (2 A=X l ). L 1 (t r ;x i;c;k 1 ) will therefore contain a marginal element of the A s interval containing A=X l. Hence, the assumption on the path length is violated. 2 Corollary 1 There exists a planar graph such that if we use 3 q n 32 or fewer labels, the longest path will be of length equal to at least 2M +1 2M D 1. Proof: To reach a contradiction in the proof of Theorem 5, we set C = 2M, S =8M 2 4M +1and K =2. Recall that n = SCK + CK S +1, and so we have M > 3 q n 32, by some standard technique The Second Bound: 2(1+ffi)M +1 2(1+ffi)M D for M = O(p n) By extending the length of the chain in G S;C;K, we can arrive at a different lower bound on the longest path and a different requirement on the number of labels. We again prove the lower bound by contradiction. Unlike the previous proof, here we make use of Lemma 10, and the following lemma, Lemma 11. Lemma 11 Given (1 + ffi)m objects arranged in a single file and a gap between two adjacent objects, where ffim is an integer. Dividing them into M sub-files (some of them may be empty) would result in at least ffim gaps being in the sub-files. Proof: (Outline) A sub-file containing K objects will contain K 1 gaps. 2 An example is shown in Figure 5. 16

17 First Example Second Example Object Gap in a subfile Gap not in a subfile Figure 5: Two examples of Lemma 11 with M =5and ffi =4=5. Theorem 6 There exists a planar graph such that for any valid M-IRS, the longest path will be no shorter 2(1+ffi)M +1 than D for any constant ffi 2 (0; 1]. 2(1+ffi)M Proof: We use the graph G S;C;K and set C = 2(1 + ffi)m. Assume the contrary that there is an M-IRS such that the longest routing path is shorter than 2(1+ffi)M +1 2(1+ffi)M D. Let B be the set [ (1+ffi)M c=1 I 2c 1. By Lemma 10, for each c 2 f1; 3;:::;2(1 + ffi)m 1g, there are p c disjoint intervals which contain I c but not B=I c, where p 1 ;p 2(1+ffi)M 1» M;p c» 2M, 8c 2 P (1+ffi)M f3; 5;:::;2(1 + ffi)m 3g, and these c=1 p 2c 1 (» 2(1 + ffi)m 2 2M ) intervals, called B s intervals hereafter, are non-overlapping. Consider L Λ (t l ;x i;1;1 ). fv i;j jj =1; 3;:::;2(1 + ffi)m 1g ρl Λ (t l ;x i;1;1 ); otherwise the assumption on the path length will be violated. The elements fv i;j jj = 1; 3;:::;2(1 + ffi)m 1g all fall into different (1 + ffi)m B s intervals, but L Λ (t l ;x i;1;1 ) is a union of at most M disjoint intervals. By Lemma 11, at least ffim gaps between B s intervals are covered by L Λ (t l ;x i;1;1 ) (Figure 6). By gap Figure 6: Two gaps between B s intervals are covered. Property 2, these ffim covered gaps cannot be covered again by L Λ (t l ;x i 0 ;1;1), for i 6= i 0. Hence, each row will cover at least ffim gaps, but there are 2(1 + ffi)m 2 2M B s intervals and hence 2(1 + ffi)m 2 2M gaps in between. If we set s to be 2(1+ffi)M 2 ffi since we cannot provide ffim gaps for each row to cover , we have a contradiction ffim Corollary 2 There exists a planar graph such that if we useq ffin or fewer labels, the longest path will 4(1+ffi) 2 2(1+ffi)M +1 have a length equal to at least D, for any constant ffi 2 (0; 1]. 2(1+ffi)M Proof: To reach a contradiction in the proof of Lemma 6, we set C = 2(1+ffi)M, S = 2(1+ffi)M ffi ffim and K =1. Recall that n = SCK + CK S +1, and so we have M > q ffin 4(1+ffi)

18 References [1] C. Gavoille, A Survey on Interval Routing, Theoretic Computer Science, 245(2): , [2] C. Gavoille, On Dilation of Interval Routing, The Computer Journal, 43(1), 1 7, [3] C. Gavoille and S. Pérennès, Lower Bounds on Interval Routing on 3-Regular Networks, Proc. 3rd International Colloquium on Structural Information & Communication Complexity (SIROCCO 96), , [4] R. Kráľovič, P. Ružička, D. Štefankovič, The Complexity of Shortest Path and Dilation Bounded Interval Routing, Theoretic Computer Science, 234(1-2), , [5] Kranakis, E., Krizanc, D. and Ravi, S.S. On Multi-Label Linear Interval Routing Schemes, The Computer Journal, 39: , [6] J. van Leeuwen and R.B. Tan, Interval Routing, The Computer Journal, 30: , [7] P. Ružička, A Note on the Efficiency of an Interval Routing Algorithm, The Computer Journal, 34: , [8] N. Santoro and R. Khatib, Labelling and Implicit Routing in Networks, The Computer Journal, 28:5 8, [9] S.S.H. Tse and F.C.M. Lau, Lower Bounds for Multi-label Interval Routing, Proc. 2nd International Colloquium on Structural Information & Communication Complexity (SIROCCO 95), , [10] S.S.H. Tse and F.C.M. Lau, A Lower Bound for Interval Routing in General Networks, Networks, 29:49 53, [11] S.S.H. Tse and F.C.M. Lau, Two Lower Bounds for Multi-Label Interval Routing, Proc. Computing: The Australasian Theory Symposium (CATS 97), 36 43, [12] S.S.H. Tse and F.C.M. Lau, More on the Efficiency of Interval Routing, The Computer Journal, 41(4): , [13] S.S.H. Tse and F.C.M. Lau, An Optimal Lower Bound for Interval Routing in General Networks, Proc. 4th International Colloquium on Structural Information & Communication Complexity (SIROCCO 97), , [14] S.S.H. Tse and F.C.M. Lau, On the Space Requirement of Interval Routing, IEEE transactions on Computers, 48(7): ,