On shortest-path all-optical networks without wavelength conversion requirements

Transcription

1 Research Collection Working Paper On shortest-path all-optical networks withot waelength conersion reqirements Athor(s): Erlebach, Thomas; Stefanakos, Stamatis Pblication Date: 2002 Permanent Link: Rights / License: In Copyright - Non-Commercial Use Permitted This page was generated atomatically pon download from the ETH Zrich Research Collection. For more information please conslt the Terms of se. ETH Library

2 Eidgenössische Technische Hochschle Zürich Swiss Federal Institte of Technology Zrich Thomas Erlebach, Stamatis Stefanakos On Shortest-Path All-Optical Networks withot Waelength Conersion Reqirements TIK-Report Nr. 153, October 2002 Institt für Technische Informatik nd Kommnikationsnetze Compter Engineering and Networks Laboratory

3 Thomas Erlebach, Stamatis Stefanakos On Shortest-Path All-Optical Networks withot Waelength Conersion Reqirements October 2002 Version 2 (Reised: ) TIK-Report Nr. 153 Compter Engineering and Networks Laboratory, Swiss Federal Institte of Technology (ETH) Zrich Institt für Technische Informatik nd Kommnikationsnetze, Eidgenössische Technische Hochschle Zürich Gloriastrasse 35, ETH Zentrm, CH-8092 Zürich, Switzerland

4 On Shortest-Path All-Optical Networks withot Waelength Conersion Reqirements Thomas Erlebach Stamatis Stefanakos Compter Engineering and Networks Laboratory (TIK) ETH Zürich, CH-8092 Zürich, Switzerland {erlebach Abstract In all-optical networks with waelength diision mltiplexing, eery connection is roted along a certain path and assigned a waelength sch that no two connections se the same waelength on the same link. For a gien set P of paths (a roting), let χ(p) denote the minimm nmber of waelengths in a alid waelength assignment and let L(P ) denote the maximm link load. We always hae L(P) χ(p). Motiated by practical concerns, we consider rotings containing only shortest paths. We gie a complete characterization of ndirected networks for which any set P of shortest paths admits a waelength assignment with L(P) waelengths. These are exactly the networks that do not benefit from the se of (expensie) waelength conerters if shortest-path roting is sed. We also gie an efficient algorithm for compting a waelength assignment with L(P) waelengths in these networks. 1 Introdction In all-optical networks that employ waelength diision mltiplexing, a connection is established by first choosing a path from the sender to the receier and then assigning a waelength for that connection to all the links of the path. The waelength assignment has to be done so that no two connections that share a link are transmitted throgh the same waelength. Since the nmber of aailable waelengths is limited, one is interested in minimizing the nmber of tilized waelengths for a gien set of connections. One techniqe that helps ct down the nmber of necessary waelengths for operating a network is that of waelength conersion. A waelength conerter is placed in some node of the network and has the ability of altering the transmitting waelength of any incoming signal. An interesting algorithmic problem that arises then is that of placing as few conerters as possible in sitable positions of the network in order to optimize its Research partially spported by the Swiss National Science Fondation nder Contract No (Project AAPCN) and the EU Thematic Network APPOL II (IST ), with fnding proided by the Swiss Federal Office for Edcation and Science (BBW). 1

5 capacity. In this paper we focs on networks where connections are always established throgh shortest paths and we characterize the networks that do not profit from the se of conerters. The network can be natrally modeled by a graph G = (V, E), a connection in the network can be seen as a path on G, and waelengths can be regarded as colors. If the network does not employ waelength conersion then a waelength assignment for a set P of paths is an assignment of a color to each path in P. A alid coloring is one in which no two paths that se the same edge get assigned the same color. For a gien set P of paths (a roting) we denote by χ(p) the minimm nmber of colors needed for a alid coloring of P. A triial lower bond for χ(p) is the congestion or load L(P) = max e E L e (P) of the network, where L e (P) is the nmber of paths in P that se e. If the network has waelength conerters a coloring is an assignment of a color to eery edge of each path. In this case, a alid coloring has to satisfy the additional constraint that the color assignments to two consectie edges of a path can only differ if there is a conerter between the two edges. Obiosly, een with the se of conerters, we will always need at least L(P) colors for a alid coloring of P. Howeer, the placement of conerters in a network can redce the nmber of waelengths needed for a fixed set of connections. It can be the case, for example, that while there is no single waelength, from the already sed ones, aailable along a path there are different aailable waelengths along parts of that path. If there are conerters in sitable positions along that path then these different waelengths can be exploited in order to sere the connection which otherwise wold hae had to be assigned a new waelength. Ideally, by placing conerters in some nodes of the network we can garantee that any roting P can be accommodated with L(P) waelengths. This is easily seen to be the case if all nodes of the network are eqipped with waelength conerters. Howeer, the cost of these deices is prohibitie for sch improident se. Therefore, the network designer is typically interested in eqipping only a small sbset of the network terminals with conerters, while still achieing the same capacity sage as if there where conerters eerywhere. Motiated by this, Wilfong and Winkler [11] introdced the following problem: Minimm Sfficient Set. Gien a graph G = (V, E), find a sfficient set S for G, i.e., a set S V sch that any set P of paths on G can be colored with L(P) colors if we place waelength conerters on the ertices of S. The goal is to minimize the size of S. Wilfong and Winkler [11] proed that Minimm Sfficient Set is N P-hard een for planar bidirected graphs (a bidirected graph is a directed graph where (, ) E (, ) E). Moreoer they showed that the only bidirected graphs that admit the empty sfficient set are spiders, i.e., trees with at most one ertex of degree greater than two, and that rings admit a sfficient set of size 1. Finally, they described an efficient way of determining whether a set S is sfficient for a bidirected graph G: one modifies G by exploding each node s S into degree-of-s-many copies, each of which is made adjacent to one of the old neighbors of s. S is sfficient for G if and only if eery component of the graph obtained after this modification is a spider. Extending this work, Kleinberg and Kmar [7] gae -approximation algorithm for directed graphs and a polynomial time approximation scheme for directed planar graphs sing techniqes based on the ndirected feedback ertex set problem. They also showed that any improement on the approximation ratio for Minimm Sfficient Set on bidirected graphs wold lead to a corresponding improement for ertex coer. The approach of Kleinberg and Kmar can be extended to gie a linear time algorithm for Minimm Sfficient Set in directed graphs of 2

6 bonded treewidth [4]. 1.1 Or Contribtion As described aboe, the preios work concentrated on the stdy of Minimm Sfficient Set in bidirected or directed graphs. These graphs sere as models for networks with nidirectional links, i.e., networks that spport only one-way commnication. In this paper, we trn to ndirected graphs. Undirected graphs model networks where the physical links are bidirectional or networks with nidirectional links with the additional property that wheneer a connection is established in one way, the reerse connection mst also be established throgh the same path and mst be transmitted oer the same waelength. It is easy to see that the only ndirected graphs that admit the empty sfficient set are chains. Frthermore, sing the techniqe deeloped by Wilfong and Winkler, one can show that Minimm Sfficient Set is polynomial in ndirected graphs: we simply hae to place a conerter in eery node of degree greater than or eqal to 3 or in any single node if the graph is a cycle (for more details see [4]). Neertheless, sch placement of conerters is not satisfying. For example, in the case where the network is a cliqe we will need to place a conerter in eery node; howeer, it is nlikely that we wold need any conerter at all in practice since in a cliqe most connections wold be carried oer a single link. In order to captre this real-world scenario we restrict orseles to shortest-path rotings, i.e., we are interested in placing as few conerters as possible so that any set P of shortest paths can be colored with L(P) colors. More formally, we introdce the following problem: Minimm SP-Sfficient Set. Gien a graph G = (V, E), find an SP-sfficient set S for G, i.e., a set S V sch that any set P of shortest paths on G can be colored with L(P) colors if we place waelength conerters on the ertices of S. The goal is to minimize the size of S. We note that this problem is of significant practical importance since shortest-path roting is a common strategy in optical networks, see for example [12]. Moreoer, it imposes a weaker condition than Minimm Sfficient Set (any sfficient set is an SP-sfficient set) and therefore can help decrease the cost of conerters in the design of optical networks. In this paper we gie a complete characterization of the networks that admit the empty SPsfficient set. We show that the block graph of sch networks is a chain and all internal blocks are cliqes. The oter blocks are a special case of co-bipartite graphs. If the graph consists of only one block we show that its diameter is less than or eqal to 3. For the case of diameter 3 we proe that the graph belongs to a special class of co-bipartite graphs. For the case where the diameter is less than 3 we show that the graph admits the empty SP-sfficient set if and only if its edges can be 2-colored in a certain way. In all cases or proofs proide efficient algorithms for recognizing graphs that admit the empty SP-sfficient set and for optimally coloring shortest-path rotings on these graphs. An interesting aspect of or work is that we do not need to consider complicated rotings in order to obtain the characterization: all graphs that reqire a conerter admit a witness roting P with L(P) = 2 and χ(p) = 3 whose conflict graph is an odd cycle. 3

7 1.2 Other Related Work Perhaps the main algorithmic problem that arises in optical networks is that of roting and waelength assignment, i.e., gien a set of connections in a network find a roting and a alid coloring of that roting so that the nmber of colors is minimized. Most research has focsed on the case where the network does not hae any conersion capabilities. In that case the problem is known to be N P-hard een for simple topologies like rings and trees [2, 11], and in the case of rings, een if the roting is part of the inpt [5]. Conseqently, a lot of effort has been pt in designing approximation algorithms for specific network topologies, see e.g., [10, 9, 3]. There has been another line of research on waelength conerters of bonded degree. A conerter of bonded degree does not hae fll conersion capabilities, i.e., it can transform color i to only a few other colors. We refer the reader to [1, 6] and the references therein. 1.3 Otline The rest of the paper is strctred as follows. In the following section we proide the necessary notation and gie some insight to the techniqes sed in or proofs. In Section 3 we characterize graphs that do not need conerters and contain ct-ertices while in Section 4 we trn to biconnected graphs. Finally, in Section 5 we smmarize or reslts and discss abot ftre work. 2 Preliminaries Throghot this paper, a graph G = (V, E) is finite, simple, and ndirected nless explicitly stated otherwise. For a graph G we will denote by V (G), E(G) its ertex-set and edge-set respectiely. In some cases, in order to simplify notation, we will write G instead of V (G). For U V, G[U] denotes the graph on U whose edges are the edges of G with both endpoints in U, i.e., G[U] is the sbgraph of G indced by U. The distance d(, ) in G of two ertices, is the nmber of edges in a shortest path in G (if no sch path exists then d(, ) := ). The eccentricity of a ertex in G, ecc() = max V d(, ), is the maximm distance of to all other ertices of G. The diameter of G, diam(g) = max V ecc(), is the maximm eccentricity oer all ertices of G. The degree, deg(), of a ertex in G is the nmber of edges incident to. We denote by (G) = max V deg() the maximm degree of G. We denote an indced cycle with k edges and k ertices by C k and an indced path with k ertices and k 1 edges by P k. A maximal connected sbgraph of G withot a ct-ertex is a block of G. The block graph of G is the bipartite graph on A B and edges ab for a A and B B if a B, where A is the set of ct-ertices of G and B is the set of blocks of G. The conflict graph of a gien set of paths is the graph with one ertex for each path and an edge between two ertices if the corresponding paths share an edge. Obiosly, a graph that does not need conerters can not contain a configration that allows s to constrct a set P of shortest paths with L(P) < χ(p). The configrations of this type that we will mainly se in this paper are shown in Fig. 1. In the text, in order to exhibit sch a configration we will refer to the ertices that indce the configration; for example, we say that we hae an antenna arond zwyx (Fig. 1(b)). For the case of the claw we say that we hae a claw at x. All these configrations allow s to constrct a set of shortest paths with load 2 whose conflict graph is an odd cycle and hence reqire 3 colors for a alid coloring. For example, in the case of the tent we can take 1 2, 2 5, 5 6, 1 2 3, 2 3 4, 3 4 5, and Notice that for some of 4

8 PSfrag replacements x z w y w w y x z x (a) claw (b) antenna (c) net w x y (d) tent z (e) satellite Figre 1: Some configrations that allow the constrction of a set P of shortest paths with L(P) < χ(p). the configrations shown in Fig. 1, in order to find sch a set of paths, we need to be able to find a shortest path of length 3 on them. This is the case for the net, the antenna and the satellite. For the latter for example, we can take the following shortest paths of length 2: w, wx, wxy, yxz, xz and z. In order to hae an odd nmber of paths we need d(, x) = 3 or d(, y) = 3 so that we can exchange two paths of length 2 with one of length 3. Other configrations which allow s to constrct a set P of shortest paths with L(P) < χ(p) are indced cycles of length greater than or eqal to 5. On an indced odd cycle we can jst take all paths of length two. On an indced een cycle the set of all paths of length two has een cardinality and ths we need one path of length three. In some of the proofs that follow, in order to exhibit the strctre of the graphs that admit the empty SP-sfficient set, we will do case analyses regarding the presence or absence of some edges of the graph. In these case analyses we will refer to bad configrations, i.e., configrations that allow s to constrct a set of shortest paths P with L(P) < χ(p), sing the terminology introdced aboe, een thogh some ftre edges that hae not yet been considered might not allow the constrction of sch a set P. For example, in Fig. 1(a), and assming that we hae eliminated the presence of edges, w already by the case analysis, we say that we hae a claw at x een thogh edge w might be there. In this case this actally forces w E since otherwise this wold gie a bad configration. Alternatiely we also say that the claw can only be broken by w. 3 Graphs with Ct-Vertices First, assme that G = (V, E) is connected bt not biconnected and admits the empty SP-sfficient set. Lemm The block graph of G is a chain. 5

9 x PSfrag replacements x P 1 P 2 P 3 y z P 1 P 2 P y 3 z Figre 2: Proof of Lemm: not all three paths can se the same edge. Proof. We will show that eery block of G contains at most two ct-ertices and eery ct-ertex is contained in exactly two blocks. For the first part assme for a contradiction that there exists a block C of G containing three ct-ertices x, y, z. Let x, y, z be neighbors of x, y and z respectiely otside C. Consider three shortest x y, x z and y z paths. The only edges otside C that are toched by these paths are xx, yy, zz and all three paths pairwise intersect on these edges. Frthermore, not all three paths se the same edge. To see this assme to the contrary that there is an edge e = sed by all three paths (consider Fig. 2). Since is on a shortest x y path we hae that d(, x ) d(, x ). Assme that d(, x ) < d(, x ). Then, d(, y ) < d(, y ) and also d(, z ) < d(, z ) since is also on the x z path. Bt this is a contradiction since is on the y z shortest path. Hence, we can find three shortest paths with load 2 that all pairwise intersect and ths reqire three colors for a alid coloring, a contradiction. For the second part, notice that if there exists a ct-ertex contained in three or more blocks, we hae a claw and hence we can find a set of shortest paths P with L(P) < χ(p). Lemm Let C be a block of G containing a ct-ertex x. Then the following hold: (i) For all C, d(, x) 2. (ii) For all, C sch that, are not ct-ertices, d(, ) 2. (iii) If C contains a second ct-ertex y then d(x, y) = 1. Proof. (i) Let C and assme d(, x) > 2. Let x a k, x be two disjoint x paths, each of length at least 3 sch that k +l is minimm, and let x be a neighbor of x otside C. Notice that if there are edges a i b j, a i b j with i < i and j < j then i = i + 1 and j = j + 1 since otherwise k + l is not minimm. Also, if there are two sch edges then we also hae a i b j, a i+1 b j+1 E since otherwise we wold hae a claw. We distingish cases depending on the edges between, and,. In all cases we hae that E becase of a claw at x. If both, E then we hae that E becase of claws at and. Also, b 3 / E becase if this was not the case, k + l wold not be minimm. Hence, we hae a satellite arond x x b 3 with a path of length 3 on it, say x x, a contradiction (the same holds if b 3 = ). If there are no edges between, 6

10 and, and frthermore / E we hae a net arond x x with a path of length 3 on it, say x x, a contradiction. If E we hae an antenna arond x x with a path of length 3 on it, say x x, a contradiction. Consider the case where there is only one edge between, and,, say. We hae that E becase of a claw at. We hae a satellite arond x x b 3 with a path of length 3 on it, namely the path x x. If b 3 = the satellite can not be broken and hence we reach a contradiction. Otherwise, we hae that b 3 E. We hae a tent arond x x b 3 b 4. If b 4 = the tent can not be broken and we reach a contradiction; otherwise we hae that b 4 E bt this creates a claw at that can not be broken, a contradiction. (ii) Let,, be two arbitrary ertices in C that are not ct-ertices and assme to the contrary that d(, ) > 2. Let x be a neighbor of x otside C. By (i) we hae that d(x, ) 2 and d(x, ) 2. We distingish cases according to the distance of, from x. We can not hae d(x, ) = d(x, ) = 1 since then d(, ) 2, a contradiction to or assmption. Assme d(x, ) = 2 and let xa be an x path of length 2. If d(x, ) = 1 we mst hae a E becase of a claw at x and hence d(, ) 2, a contradiction. For the case where d(x, ) = 2 we can assme that there exists an x path of length 2, disjoint from xa, since otherwise we hae that d(, ) 2. Let xb be sch a path. We hae that ab E becase of a claw at x. This creates a net arond x xba with a path of length 3 on it, say x xb, and hence we shold hae a E or b E or E. In all cases we hae that d(, ) 2, a contradiction. (iii) Assme to the contrary that d(x, y) 2. By (i) we hae that d(x, y) 2 and hence there exist two disjoint x y paths sch that one of them is of length 2. Let xy and x a k y be two sch paths with k minimm and let x, y be neighbors of x, y respectiely otside C. We reach a contradiction by exhibiting a set of paths P with L(P) = 2 and χ(p) = 3. Consider the paths x x, x, a 3,, a k 1 a k y, a k yy, x xy, yy. All of them are shortest paths, their load is 2 and their intersection graph is a cycle. If k is odd, the cycle is of odd length and hence we need 3 colors for a alid coloring of P. Otherwise replace the path x xy with two paths of length 2. Lemma 3 Let C be a block of G containing two ct-ertices x, y. For all C, d(, x) = 1 or d(, y) = 1. Proof. Let be an arbitrary ertex in C. Assme for a contradiction that neither x nor y are adjacent to. From Lemm we hae that d(, x) 2, d(, y) 2 and d(x, y) = 1. Therefore, d(, x) = d(, y) = 2 and there exists a ertex w adjacent to both x and. Since y is adjacent to x we hae that wy E since otherwise there wold be a claw at x. Let x, y be neighbors of x, y respectiely otside C. We hae a net arond x xwyy with the path x xyy of length 3 on it and no possibility to break it, a contradiction. Corollary 4 Eery block C of G containing two ct-ertices is a cliqe. Proof. Let C be a block of G containing two ct-ertices x, y. Consider an arbitrary ertex C. From the preios lemma we hae that is adjacent to x or to y. Assme w.l.o.g. x E. We mst hae y E since otherwise we get a claw at x. Hence, any ertex in C that is not a ct-ertex is adjacent to both x and y. Therefore, any two ertices in C are adjacent to each other since otherwise we wold hae claws at x and y. 7

11 Lemma 5 Let C be a block of G containing exactly one ct-ertex x and let N 1, N 2 be the sbgraphs of G indced by the neighborhoods of x in C in distance 1, 2 respectiely. The following hold: (i) Both N 1, N 2 are cliqes. (ii) There is no C 4 in C. (iii) If there exists a ertex w N 2 adjacent to two ertices, N 1, then all ertices in N 2 are adjacent to or to. Proof. For the first part, notice that if N 1 is not a cliqe we get a claw at x. Consider now two ertices, N 2 and assme / E. If, are adjacent to the same ertex of N 1 we hae a claw at that ertex. If they are adjacent to different ertices we hae a net (since N 1 is a cliqe) with a path of length 3 on it. We proceed to show the second part. Assme there is a C 4 in C. Since N 1, N 2 are cliqes the cycle shold contain two ertices, say a, b from N 1 and two ertices, say c, d from N 2. Let abcd be the cycle and let x be a neighbor of x otside C. We hae an antenna arond x xabcd with a path of length 3 on it, a contradiction. For the last part of the lemma assme that there exists a ertex w N 2 adjacent to two ertices, N 1 and that there exists a ertex b N 2 adjacent neither to nor to. As before, let x be a neighbor of x otside C. We hae a satellite arond bwxx with a path of length 3 on it, say x xw, a contradiction. Now we are able to characterize those graphs that are not biconnected and admit the empty SP-sfficient set: Theorem 1 Let G be an ndirected graph that is connected bt not biconnected. The empty set is SP-sfficient for G if and only if the following hold: (i) The block graph of G is a chain. (ii) Eery block of G that contains two ct-ertices is a cliqe. (iii) Eery block C of G that contains only one ct-ertex x does not contain a C 4, N1 C and N2 C are cliqes and if there exists a ertex w N2 C adjacent to two ertices, N1 C, then all ertices in N2 C are adjacent to or to, where N1 C, N 2 C are the sbgraphs of G indced by the neighborhoods of x in C in distances 1, 2 respectiely. Moreoer, no ertex in C is in distance 3 from x. Proof. The only if part is clear from the preios lemmas. For the if part we show how to color a set P of shortest paths on a gien graph G that satisfies the conditions of the statement with L(P) colors. A high leel description of or approach follows. We will first modify P by shortening some paths and discarding some others. We will then constrct a directed graph G and a set P of directed paths on G in 1-1 correspondence with P, with L(P ) = L(P) and with the same conflict graph. We will obtain a coloring for P with L(P) colors by coloring P with L(P ) colors. The coloring of P will be done by compting many different local colorings that will be merged to gie a single global coloring for P. A coloring for the initial set of paths will be obtained by coloring greedily the paths that were discarded in the first phase. Let C 1,, C k be the blocks of G ordered so that C 1, C k are the two blocks containing only one ct-ertex and C i C i+1 = 1 for all 1 i < k. Let x 1,, x k 1 be the ct-ertices of G ordered 8

12 so that C i C i+1 = {x i }. Let P 1 P be the set of all paths in P of length one. We remoe these paths from P. Now consider all paths that se an edge in one of N C 1 2, N C k 2. Since we are dealing with shortest paths no sch path can se an edge incident to x 1 or x k 1. Therefore, all these paths are of length at most two. We claim that eery edge in N C 1 2, N C k 2 will only be sed by identical paths. To see this assme that there is an edge in, say N C 1 2 (the case for C k is similar), that is sed by two paths that are non-identical. Since N C 1 2 is a cliqe and these paths are shortest they mst go from or to two different ertices of N C 1 1, say w, w. There are two cases: either these paths are w and w or w and w. Since these paths are shortest, in the first case C 1 mst contain a C 4, while in the second case N C 1 2 is adjacent to both w, w N C 1 1 and there exists a ertex in N C 1 2, namely, that is not adjacent to any of w, w. Both cases contradict the third condition of the statement and hence we obtain that eery edge in N C 1 2, N C k 2 will only be sed by identical paths of length 2. Hence, we can shorten all sch paths so that they do not se any edge in N C 1 2, N C k 2 withot modifying the conflict graph of P. If, after this modification, we obtain any paths of length one we remoe them from P and add them to P 1. Now we proceed to show how we constrct G and the set P of directed paths on G with the properties described aboe. An example is shown in Fig. 3. For each block C i of G containing two ct-ertices x i 1, x i we constrct its corresponding gadget as follows. We take two copies in, ot of eery ertex of C i that is not a ct-ertex and connect x i 1 with a directed edge to each in ertex and each ot ertex with a directed edge to x i. Finally, we connect x i 1 with a directed edge to x i. The gadget for the blocks that contain only one ct-ertex, say for C 1, is constrcted by starting from C 1, deleting all edges within each of N C 1 1, N C 1 2 and orienting the rest of the edges from N C 1 2 to N C 1 1 and from N C 1 1 to x 1. Finally we add edges from each ertex of N C 1 1 to all other ertices of N C 1 1. The gadget for C k is constrcted similarly as for C 1 bt with reerse orientations for the edges. To complete the constrction of G we connect all gadgets by identifying those ertices that correspond to the same ct-ertices in G. Since we hae remoed all paths of length one and hae modified P so that no path ses an edge in N C 1 2 and N C k 2, eery path in P has a niqe correspondent in G and hence the constrction of the set of paths P on G is straightforward. Notice that eery edge in N C 1 1 and N C k 1 corresponds to two oppositely directed edges in G. Howeer, since C 1 and C k contain no C 4, all paths that se an edge e in N C 1 1 or N C k 1 correspond to paths in G that all se the edge, corresponding to e, in the same direction and hence the conflict graph does not change. We contine with the coloring of P. Let y 1,, y l be the ertices in G that correspond to ertices in N C 1 1 and let z 1,, z m be the ertices in G that correspond to ertices in N C k 1. Define V := {x 1,, x k 1, y 1, y l, z 1,, z m }. For eery ertex V, let Q ot be the set of edges directed ot of and Q in be the set of edges directed into. For 1 i < k let P x i be the set of paths that toch x i (i.e., start at x i, end at x i or go oer x i ), for 1 i l let P y i be the set of paths that go oer y i or start at y i, and for 1 i m let P z i be the set of paths that go oer z i or end at z i. Notice that V P = P. We show how to color P for all V with L(P ) colors. Or method is similar to the one gien in [3]. We bild a bipartite mltigraph H with one ertex for each edge incident to. The one side of the partition consists of the ertices corresponding to edges in Q ot and the other consists of the ertices corresponding to edges in Q in. Each path in P that goes oer ses one edge in Q in, one in Qot and no other edge incident to. For each sch path we add one edge in H connecting its ot ertex to its in ertex. For eery path that starts or ends at we add a loop to the ertex of H that corresponds to the edge incident to sed by this path. By König s classical reslt [8], H has a proper edge-coloring with (H ) colors which 9

13 PSfrag replacements x 1 x 2 x k 2 x k 1 N C1 N C1 2 1 N Ck 1 N Ck 2 Figre 3: Example of the constrction sed in the proof of Theorem 1. we can find in polynomial time. Since two paths in P intersect in G if and only if they intersect in an edge incident to this reslts in a alid coloring of P. Frthermore, this coloring ses L(P ) colors since (H ) = L(P ) (we assme that a loop contribtes 1 to the degree of the ertex). Let S be the coloring obtained for P. We show how we can merge all local colorings S into a global coloring S for P withot increasing the maximm nmber of colors sed. We start by merging S x1 with each of S for {y 1,, y l }. The merging for some S yi is done as follows. The only paths that are in both S x1 and S yi, and ths might case conflicts when merging, are the paths that se edge (y i, x 1 ). Notice that if a path p colored in S yi intersects a path q that is colored in S x1 bt not in S yi, then p ses (y i, x 1 ) and both p, q se the same otgoing edge incident to x 1, and are both colored with different colors in S x1. Therefore, to combine the two colorings we maintain S x1 as is and modify S yi : we permte S yi (i.e., we rename the colors) so that the paths that se edge (y i, x 1 ) get the color they hae in S x1. Since S yi was a alid coloring, it remains alid after the permtation. Now the two colorings are compatible and can be merged in the obios way. After the merging, S x1 is the new coloring. After l mergings we hae extended S x1 to inclde the colorings S y1,, S yl withot increasing the maximm nmber of colors sed, while maintaining its alidity. We repeat the same procedre to merge coloring S xk 1 with the colorings S z1,, S zm. Now, we can merge S x1,, S xk 1 to one global coloring S for P. To do this we initially set S = S x1 and contine to the processing of S x2. After processing S xi we contine to S xi+1 and merge the preios global coloring S (which we hae obtained by merging colorings S x1 S xi ) with S xi+1. The merging of S with S xi+1 is done as follows. Consider the paths that se edge (x i, x i+1 ) on G. These paths are the only paths that are in both S and S xi+1 and therefore might case conflicts in the merging. To combine the two colorings we maintain S as is and modify S xi+1 : we permte S xi+1 so that the paths that se edge (x i, x i+1 ) get the color they hae in S. Since S xi+1 was a alid coloring before the modification, it remains alid and now we can merge the two colorings in the obios way. This way we extend coloring S withot increasing the nmber of colors sed in S and S xi+1 while maintaining its alidity. After k 2 mergings we obtain a global coloring S for P that ses max 1 i k 1 L(P i ) = L(P ) colors. We obtain a coloring for the initial set P with L(P) colors by extending S greedily to the paths in P 1 which we hae preiosly discarded. 10

14 4 Biconnected Graphs Now, consider the case where G is biconnected. We assme that G admits the empty SP-sfficient set. Lemma 6 For all, V : d(, ) 3. Proof. Assme to the contrary that d(, ) > 3 and let a k and be two disjoint paths, each of length at least 4, sch that k + l is minimm. Notice that if there are edges a i b j, a i b j with i < i and j < j then i = i + 1 and j = j + 1 since otherwise k + l is not minimm. Also, if there are two sch edges then we also hae a i b j, a i+1 b j+1 E since if this were not the case we wold hae a claw. We obtain a contradiction by case analysis depending on the edges between, and,, as in the proof of Lemm(i). There are for main cases: Case 1:, E. We hae that, E. Notice that b 3, a 3 / E since otherwise k+l wold not be minimm. Therefore, we hae that d(, b 3 ) = 3 since by or choice of the two paths the only shortct between and b 3 can be throgh. We distingish cases depending on the edges between, b 3 and, a 3, as shown in Fig. 4(a). Case 1.1: b 3, a 3 / E. We hae a net arond b 3 a 3 or an antenna arond the same ertices if a 3 b 3 E. Moreoer, we can find a path of length 3 on the net or the antenna, namely the path b 3 (recall that d(, b 3 ) = 3), a contradiction. Case 1.2: b 3 E, a 3 / E. We hae a satellite arond b 3 b 4 and a path of length 3 on it, namely the path b 3. If b 4 = (i.e., if l = 3), then the satellite can not be broken and we obtain a contradiction. Otherwise we shold hae b 4 E bt this creates a tent arond b 3 b 4 b 5 and if b 5 = we hae a contradiction. Otherwise b 5 E and we obtain a contradiction de to a claw at. Case 1.3: b 3 / E, a 3 E. Similar to case 1.2. Case 1.4: b 3, a 3 E. We hae a 3 b 3 E and we get a satellite arond b 3 b 4 with a path of length 3 on it, namely the path b 3. This gies s a contradiction since b 4 / E becase otherwise k + l wold not be minimm and hence the satellite can not be broken. Case 2: E, / E. We hae E becase of a claw at. Also, notice that we can not hae a 3 E since in that case k + l wold not be minimm. By or choice of the two paths the only shortct from to a 3 cold be throgh. It follows that d(, a 3 ) = 3. We distingish 4 cases depending on the edges between, b 3 and, a 3, as shown in Fig. 4(b). Case 2.1: b 3, a 3 E. We hae a satellite arond a 3 a 4. Moreoer there exists a path of length 3 on it, namely a 3, since d(, a 3 ) = 3 as was shown aboe. Hence, if a 4 = we hae a contradiction. Otherwise we mst hae a 4 E bt this can not be the case since then k+l is not minimm, a contradiction. Case 2.2: b 3 E, a 3 / E. We hae a satellite arond b 3 b 4. If b 4 = the satellite can not be broken and we can find 11

15 2.1 a 3 a k 1.1 a 3 a k 4.1 a 3 a k 2.2 b 3 a 3 a k PSfrag replacements 1.2 b 3 a 3 a k 4.2 b 3 a 3 a k 2.3 b 3 a 3 a k 1.4 b 3 a 3 a k 4.4 b 3 a 3 a k 2.4 b 3 a 3 a k b 3 b 3 b 3 (a) (b) (c) PSfrag replacements Figre 4: Case analysis in the proof of Lemma a 3 a 4 a k a 3 a a k b 3 b 4 b 3 b 4 (a) (b) (c) a 3 a 4 a k a 3 a 4 a k b 3 b 4 b 3 b 4 Figre 5: Case analysis (case 2.2) in the proof of Lemma 6. 12

16 the path b 3 of length 3 on it, since d(, ) 4 and hence b 3 / E. So we can assme l 4. We distingish more cases, according to the edges that are present between, b 4 and a 3, b 3. We hae 4 cases shown in Fig. 5. Case 2.2.1: b 4, a 3 b 3 E. We hae a 3 b 4 E and there is a satellite arond b 3 b 4 b 5. Moreoer there is a path of length 3 on it, namely the path b 3 b 4 since b 4 / E and a i / E for all i 2. Hence, if b 5 = the satellite can not be broken and we obtain a contradiction. Otherwise, we shold hae b 5 E bt this gies a tent arond b 3 b 4 b 5 b 6 and a contradiction in the case where b 6 =. If b 6 then we shold hae b 6 E bt this gies a claw at, a contradiction. Case 2.2.2: b 4 E, a 3 b 3 / E. Similar to case Case 2.2.3: b 4 / E, a 3 b 3 E. We hae that b 3 E de to a satellite arond b 3 b 4. This creates another satellite arond b 3 a 3 a 4 with a path of length 3 on it, namely the path a 3. If a 4 = this gies s a contradiction; otherwise we shold hae b 3 a 4 E. This creates a tent arond b 3 a 3 a 4 a 5. If a 5 = we hae reached a contradiction; otherwise we shold hae a 5 b 3 E bt this creates a claw at b 3, a contradiction. Case 2.2.4: b 4, a 3 b 3 / E. We hae a net arond a 3 b 3 b 4 or an antenna arond the same ertices if a 3 b 4 E. Moreoer there is a path of length 3 on the net or the antenna, namely the path b 3 b 4 since b 4 / E and a i / E for all i 2. Hence, a contradiction. Case 2.3: b 3 / E, a 3 E. We hae a claw at that can not be broken, a contradiction. Case 2.4: b 3, a 3 / E. We hae a claw at that can not be broken, a contradiction. Case 3: / E, E. Similar to case 2. Case 4:, / E. Again, we distingish for cases according to the edges between, b 3 and, a 3. Three of these cases are shown in Fig. 4(c). Case 4.1: b 3, a 3 E. We hae that, a 3 b 3 E. Becase of the C 5 arond we shold hae E. We hae an antenna arond a 3 a 4 with a path of length 3 on it, namely the path a 3 a 4 since a 5, a 4, a 3, a 4 / E becase otherwise k + l wold not be minimm. Moreoer the antenna can not be broken and we reach a contradiction. Case 4.2: b 3 E, a 3 / E. We distingish two cases according to whether E. Case 4.2.1: E. We hae a C 5 arond b 3. Ths, E or b 3 E. Notice that if E we get a claw at that can not be broken. Hence, b 3 E and we get claws at b 3 that force b 4, b 4 E (and hence we mst hae l 5; otherwise d(, ) < 4). This creates an antenna arond b 3 b 4 b 5 with a path of length 3 on it, namely the path b 5 b 4, since b 5, a j / E for all j 2 becase otherwise k + l wold not be minimm. Frthermore, the antenna can not be broken and ths we reach a contradiction. 13

17 Figre 6: Examples of graphs of diameter 3 that admit the empty SP-sfficient set. Case 4.2.2: / E. We hae a C 6 arond b 3 and if b 3 / E we can find a path of length 3 on it, namely the path b 3. Note that / E de to a claw at. Hence, b 3 E. Then we hae a C 5 arond b 3 that can not be broken, a contradiction. Case 4.3: b 3 / E, a 3 E. Similar to case 4.2. Case 4.4: b 3, a 3 / E. We hae that b 3, a 3 / E since otherwise we wold hae a claw at, respectiely. Hence, the paths of length 3 b 3 and a 3 are shortest. Therefore, if / E we hae a C 5 or a longer indced cycle with a path of length 3 on it. It follows that E. We claim that the paths b 3 b 4 and a 3 a 4 are also shortest. We will show it for b 3 b 4 (it follows with similar argments for a 3 a 4 ). Recall that a 3, b 3 / E. We hae that b 4 / E becase otherwise we hae a C 5 arond b 3 b 4. Also, if a 4 E we hae a C 5 arond a 3 a 4. Ths, if there exists a shortct from to b 4 it shold be throgh a j with j 5. Howeer, in this case we wold hae a C 6 or greater with a path of length 3 on it, namely a 3 a 4 a 5 since there cold not be another shortct from to a 5 throgh some b i becase then k + l wold not be minimm. Hence, the paths b 3 b 4, a 3 a 4 are shortest. Ths, we shold hae E or a 3 b 3 E since otherwise we wold hae a C 5 or a longer indced cycle with a path of length 3 on it. If E we hae claws at and. Hence, a 3 b 3 E and this gies a C 6 arond a 3 b 3. We reach a contradiction since we can find a path of length 3 on the cycle between and b 3. We considered all possible cases, reaching a contradiction for each of them; the lemma follows. Notice that the bond in the statement is tight: there exist graphs that admit the empty SPsfficient set and hae diameter 3. Two sch graphs are shown in Fig. 6. On the other hand, if the diameter is at most two (or if we restrict to shortest-path rotings of length at most two), then we can efficiently check whether a graph admits the empty SP-sfficient set as the following theorem illstrates. Theorem 2 Let G = (V, E) be an ndirected graph of diameter at most 2. The empty set is SPsfficient for G if and only if E can be 2-colored sch that eery shortest path of length 2 in G ses edges of different colors. Proof. For sfficiency we show how to color a set P of shortest paths on G with L(P) colors. We assme that P contains only paths of length two since any paths of length one can be colored greedily. We constrct a mltigraph H on E and for each path w in P we add an edge in H between and w. If the condition holds then H is bipartite and by König s theorem [8] it can be edge-colored with (H) colors in polynomial time. Since the degree of each ertex of H is eqal to the load of the corresponding edge of G this gies a coloring for P with L(P) colors. 14

18 PSfrag replacements 1 x 3 x x y y y c d 2 a b x x c c x c y y y Figre 7: Case analysis in the proof of Lemma 7. To see that the condition is necessary assme that it does not hold and let P be the set of all shortest paths of length two on G. We constrct a mltigraph H on E as before. Since E can not be 2-colored sch that eery path in P ses edges of different colors, H is not bipartite and hence contains an odd cycle. The paths corresponding to the edges of the odd cycle form a set of paths P with load L(P ) = 2 and χ(p ) = 3. With this reslt at or disposal, we can assme from now on that diam(g) = 3. For V, we denote with N i () the sbgraph of G indced by the neighborhood of in distance i. With the following 5 lemmas we establish that, for a ertex of eccentricity 3, N 1 (), G[V (N 2 ()) V (N 3 ())] are cliqes. Lemma 7 If has eccentricity 3, then N 3 () is a cliqe. Proof. Let be a ertex of eccentricity 3, x, y two ertices in distance 3 from, and assme to the contrary that xy / E. Let p, q be two paths of length 3 connecting to x and y respectiely. We distingish cases according to the nmber of common edges and ertices between p and q. The different cases are shown in Fig. 7. Case 1: p and q share 2 edges. We mst hae xy E becase of a claw. Case 2: p and q share one edge. Let abx, acy be the two paths. We hae a claw at a and hence bc E. This creates a net and so we shold hae cx E or by E. Assme w.l.o.g. cx E. We get a claw at c which can not be broken, a contradiction. Case 3: p and q share no edges bt hae a common ertex. Again, we shold hae xy E becase of a claw. Case 4: p and q share no edges and no ertices. Let, be the internal ertices of p and, the internal ertices of q. If we hae any of the edges,, y, x, then we can apply one of the cases 1-3. So we can assme now that none of these edges is present. Also, we can not hae E either, since that wold create a claw. We distingish two cases depending on d(x, y). 15

19 Case 4.1: d(x, y) = 2. We hae to consider only the case where there is a path of length two between x, y, disjoint from p and q, since otherwise we can apply case 1. Let c be the internal ertex of sch a path. If c E or c E we can apply case 1 and ths we can assme that c, c / E. Hence, the only possible edges are, c, c. Note also that d(, c) = 3 becase otherwise we can apply case 1. It is easy to see that in all possible cases there is an indced odd cycle or an indced een cycle with a path of length 3 on it. The only case where there is an indced een cycle with no obios path of length 3 on it, is when, c E (the case where, c E is identical). Howeer, for this case we hae that d(, y) = 3 since otherwise we can apply case 2. Case 4.2: d(x, y) = 3. We consider two different cases depending on whether there is an x y path, disjoint from p and q. Case 4.2.1: there is a path xcdy disjoint from p and q. If d E or c E then we can apply case 2 and ths we can assme that d, c / E. If c E or d E then we get claws at c, d respectiely. If d E or c E then we get claws at, respectiely. Ths, the only possibilities are, c, d. In all cases we hae an indced odd cycle or an indced een cycle with a path of length 3 on it. The only exception is the case where all three edges are there, i.e.,, c, d E. For this case we show that d(, ) = 3. Assme to the contrary that there exists an path of length 2. Since we can not hae any other edges between existing ertices, this path ses a new ertex, say w. We hae a C 5 arond w and therefore we shold hae w E or w E. Assme w.l.o.g. that w E. If wy E we can apply case 2 and ths we can assme that wy / E. De to a claw at we get w E. De to the C 5 arond w dc we shold hae wc E or wd E. In both cases we get a claw at w with no possibility to break it, a contradiction. Case 4.2.2: there is no x y path of length 3 disjoint from p and q. There is a path of length 3 that has one common edge with p or q, say xc y. We hae that cy / E since d(x, y) = 3. Also c / E since otherwise we can apply case 2. Therefore we hae a claw at that can not be broken, a contradiction. We considered all possible cases, reaching a contradiction for each of them; the lemma follows. Lemma 8 If has eccentricity 3, then N 2 () is a cliqe. Proof. Let be a ertex of eccentricity 3, a ertex at distance 3 from, and x, y two ertices in distance 2 from. Assme to the contrary that xy / E. There exist two disjoint paths of length 2 from to x and y; otherwise we hae a claw that forces xy E. Let ax, by be two sch paths. We distingish cases depending on the distance of x, y from. From Lemma 6 we hae that their distance from is at most 3. Notice that if d(x, ) = d(y, ) = 3 then, since by Lemma 7 N 3 () is a cliqe, we hae xy E, a contradiction. Some of the cases are shown in Fig. 8. Notice that in all cases ay, bx / E de to claws. Assme, w.l.o.g., d(x, ) d(y, ). Case 1: d(x, ) = d(y, ) = 1. Depending on whether ab E we hae a C 5 or a C 6 with a path of length 3 on it, a contradiction. Case 2: d(x, ) = 1, d(y, ) = 2. Let yc be a path of length two connecting y and. Notice that ac / E becase of a claw at c. 16

20 (b) 1. d(x, ) = d(y, ) = 1 2. d(x, ) = 1, d(y, ) = 2 4. d(x, ) = d(y, ) = 2 (a) x x a x a a e c d(x, ) = 1, d(y, ) = 3 d(x, ) = 2, d(y, ) = 3 c b y b b y y d Figre 8: Case analysis in the proof of Lemma 8. The only possible edges are ab, bc, xc. We distingish cases according to which of these edges are present. Case 2.1: ab, bc, xc E. We can find a set of paths P with L(P) = 2 and χ(p) = 3, namely the paths bc, cy, ycx, cxa, xab, aby, yb. Case 2.2: two of ab, bc, xc are present. We hae a C 5. Case 2.3: one of ab, bc, xc is present. We consider the three cases. Case 2.3.1: ab E. We hae a C 6 and we need to find a path of length 3 on it, in order to obtain a contradiction. We show that d(b, ) = 3. Assme for a contradiction that there exists a b path of length 2. This path mst se a new ertex, say w. Since d(, ) = 3 we hae that w / E. De to a claw at b we get wy E. This creates an antenna arond bwyc with a path of length 3 on it, namely the path bw. In order to break it we shold hae wc E since w / E. We need wa E or wx E de to the C 5 arond abwx. In both cases we get a claw at w that can not be broken. Case 2.3.2: bc E. We hae a C 6 with a path of length 3 on it, namely the path bc, a contradiction. Case 2.3.3: xc E. We hae a C 6 and we need to find a path of length 3 on it, in order to obtain a contradiction. We show that d(, c) = 3. Assme, as before, that there exists a c path of length 2. This mst se a new ertex, say w. Since d(, ) = 3 we hae w / E. Becase of a claw at c we get wy E. This creates an antenna arond cywb with a path of length 3 on it, namely the path cw and hence we shold hae wb E. Becase of the C 5 arond wcxa we shold hae wa E or wx E. In both cases we get a claw at w that can not be broken, a contradiction. Case 2.4: none of ab, bc, xc is present. We hae a C 7. Case 3: d(x, ) = 1, d(y, ) = 3. We hae that y, N 3 () and therefore by Lemma 7 we get y E, a contradiction. Case 4: d(x, ) = d(y, ) = 2. If two paths of length two connecting x to and y to share a common edge then we hae a claw. Ths, we can assme that there exist two disjoint paths xc and yd. We hae that d, c / E becase d(, ) = 3. If xd, yc E we get claws at d and c respectiely. If ad, bc E we hae again 17

21 claws at d, c. Ths, the only possibilities are ac, bd, ab, cd. We distingish cases according to which of these edges exist. Case 4.1: ac, bd, ab, cd E. We can find a set of paths P with L(P) = 2 and χ(p) = 3 similarly as in case 2.1. Case 4.2: three of ac, bd, ab, cd are present. We hae a C 5. Case 4.3: two of ac, bd, ab, cd are present. We distingish frther cases. Case 4.3.1: ac, bd E. We get a C 6 with a path of length 3 on it, namely the path ac. Case 4.3.2: ab, ac E. We take the following paths: ac, cd, dy, dyb, yba, bax and xa. Case 4.3.3: ab, bd E or ac, cd E or bd, cd E. Similar to case Case 4.3.4: ab, cd E. We hae a C 6 arond abydcx and need to find a shortest path of length 3 on it in order to obtain a contradiction. If d(x, y) = 3 we are done. Otherwise there exists an x y path of length two in G. This path mst se a new ertex, say w. De to the C 5 arond axwyb we shold hae aw E or bw E. Assme w.l.o.g. bw E. De to the C 5 arond xcdyw we shold hae wc E or wd E. If wd E we get a claw at w. Hence, wc E. Also, w, w, aw / E becase of claws at w. We hae two antennas arond cwxab and bwydc and hence if d(, c) = 3 or d(, a) = 3 we are done. We will show that it can not be d(, a) = d(, c) = 2. Assme for a contradiction that there exist a, c paths of length 2. The a, c paths can not se any of the existing ertices. Let z 1 c, z 2 a be the two paths. We hae a claw at a and x, z 2 / E, hence xz 2 E. This creates an antenna arond axz 2 c with a path of length 3 on it, namely the path az 2 and therefore we shold hae z 2 c E. With similar argments we obtain that xz 1, z 1 a E. We can now constrct a set of paths P with L(P) = 2 and χ(p) = 3. If z 1 z 2 / E take z 1 x, z 1 xz 2, xz 2, z 2 a, ax, axc, xc, cz 1, and cz 1. If z 1 z 2 E take z 1 z 2, z 2 a, z 2 a, ax, axc, xc and cz 1. Case 4.4: one of ac, bd, ab, cd is present. We hae a C 7. Case 4.5: ac, bd, ab, cd / E. In this case we hae a C 8 and we need to find a shortest path of length 3 on it in order to obtain a contradiction. If d(, c) = 3 we are done. Otherwise there exists a c path of length two in G. This path mst se a new ertex, say w. We hae a claw at c and w / E since d(, ) = 3. Therefore, wx E. We hae an antenna arond cxwa with a path of length 3 on it, namely the path cw and so we shold hae wa E. We hae a C 7 arond wcdyb bt no possibility to break it becase of claws at w, a contradiction. Ths, d(, c) = 3 and we hae a contradiction. Case 5: d(x, ) = 2, d(y, ) = 3. We hae that y, N 3 () and therefore by Lemma 7 we get y E, a contradiction. We considered all possible cases, reaching a contradiction for each of them; the lemma follows. Lemma 9 Let, V with d(, ) = 3. There exist two disjoint paths, each of length 3. Proof. Since G is biconnected there exists a pair of disjoint paths. We will first show that there exists sch a pair of paths sch that at least one of them is of length 3. Assme to the 18