Approximation Algorithms for Constructing WavelengthRouting Networks

Refael Hassin and Asaf LevinDepartment of Statistics and Operations Research, Tel-Aviv University, Tel-Aviv 69978, Israel

Consider a requirement graph whose vertices representcustomers and an edge represents the need to route aunit of flow between its end vertices along a single path.All these flows are to be routed simultaneously. A solu-tion network consists of a (multi)graph on the same setof vertices, such that it is possible to route simulta-neously all of the required flows in such a way that noedge is used more than K times. The SYNTHESIS OF WAVE-LENGTH ROUTING NETWORK (SWRN) problem is to compute asolution network of a minimum number of edges. Thisproblem has significant importance in the world of fiber-optic networks where a link can carry a limited amountof different wavelengths and one is interested in findinga minimum-cost network such that all the requirementscan be carried in the network without changing thewavelength of a path at any of its internal vertices. In thispaper, we prove that the SWRN problem is NP-hard forany constant K (K ≥ 2). Then, we assume that GR is aclique with n vertices and we find an “almost” optimalsolution network for all values of K (K � o(n)) and presenta Min{(K � 1)/2, 2 � 2/(K � 1)}-approximation algorithmfor the general case and a 2-approximation algorithm ford-regular graphs. © 2002 Wiley Periodicals, Inc.

Keywords: network synthesis; wavelength routing; WDM

1. INTRODUCTION

We study the problem of designing a communicationnetwork composed of optical links, networks that utilizewavelength division multiplexing (WDM). WDM technol-ogy establishes communication between pairs of networkvertices by selecting paths between such pairs and assigningwavelengths to each path so that no two paths going throughthe same fiber link use the same wavelength. Optical band-width is the number of distinct wavelengths. Since state-of-the-art technology allows for a limited optical bandwidth,the following SYNTHESIS OF WAVELENGTH ROUTING NETWORK

(SWRN) problem is a natural one:Given a graph G � (V, ER) and an optical bandwidth

which is an integer K, design a network (multigraph) witha minimum number of links such that using K wavelengthsit is possible to establish communication between every pairof vertices in ER and allocate a path and a wavelength foreach pair of vertices u and v, (u, v) � ER such that if twopaths share an edge they are allocated different wave-lengths. ER corresponds to the set of requirements.

Although we consider this problem as a natural one, weare not aware of any previous discussion of this problem.

In [2] (and references cited therein), a “dual” problem isconsidered, where the solution network is given and theproblem is to compute a routing such that K, the opticalbandwidth used in any edge, is minimized. However, wenote that, in practice, K is not a decision variable but aconstant defined by a technology constraint.

We note that, as appeared in [4] for the undirected caseand in [3] for the directed case (see also [2]), the problem ofallocating paths and wavelengths to a given solution net-work for a given requirements network is NP-complete(even when the input solutions are restricted to simplegraphs).

For K � 1, the optimal solution is G.In this paper, we first prove that the SWRN problem is

NP-hard for any constant K (K � 2). Then, we assume thatGR is a clique with n vertices and we find an “almost”optimal solution network for all values of K (K � o(n))and allocate paths and wavelengths for every pair of verti-ces. Finally, we present a Min{(K � 1)/ 2, 2 � 2/(K� 1)}-approximation algorithm for the general case and a2-approximation algorithm for d-regular graphs.

Note that w.l.o.g. we can assume that GR is connected.Otherwise, we can solve the problem for every connectedcomponent by itself and union the solutions.

2. NP-HARDNESS

In this section, we prove that SWRN is NP-hard for ageneral-requirement graph GR for any constant K (K � 2).

2.1. K � 2

We first prove that SWRN is NP-hard for K � 2.Note that EDGE-PARTITION INTO TRIANGLES (EPT) is NP-

Received November 2001; accepted April 2002Correspondence to: R. Hassin; e-mail: hassin@post.tau.ac.ilPublished online 00 Month 2002 in Wiley InterScience(www.interscience.wiley.com). DOI 10.1002/net.10036© 2002 Wiley Periodicals, Inc.

NETWORKS, Vol. 40(1), 32–37 2002

complete (see [6]). In EPT, we are given an undirectedgraph G � (V, E), �V� � n, �E� � m, and ask if there apartition of E into sets {E1, E2, . . . , Em/3} such that Ei isa 3-edge cycle for every i.

Theorem 1. SWRN with K � 2 is NP-hard.

Proof. Consider the following decision problem:Given a graph G for SWRN with K � 2, is there a solutionof size (2m)/3?

To prove the theorem, we will show that this problem isNP-complete. It is clearly in NP. We claim that this problemis equivalent to EPT for G.

This is so, as any feasible solution to the SWRN for Gwith cost at most (2m)/3 must use any edge in the solutionin a path of length 1 (the edge by itself) and also in a singlepath of length 2. Therefore, any edge in such a solution mustbe in ER. Moreover, for every path with two edges (u, v),(v, w) in the solution, the three edges (u, v), (v, w), (u, w)must be in ER and any edge in ER appears in one suchtriangle. Therefore, ER can be partitioned into triangles.

If {E1, E2, . . . Em/3} is a partition of ER into triangles,then for every triangle, pick two edges into the SWRNsolution and assign to them wavelength 1. For the missingedges, we can use wavelength 2 in the path defined by theother two edges of the corresponding triangle. �

2.2. K � 3

Now, we prove that SWRN is NP-hard for any constantK such that K � 3.

Note that EDGE COLORING (EC) is NP-complete for con-nected K-regular graphs and, in particular, it is NP-completeto decide whether there exists a K-edge-coloring of a K-regular connected graph for any K � 3 (see [7, 8]).

Consider a partition (E1, E2, . . . , EK) of the edge set ofa K-regular graph G � (V, E). Let V� � V be the verticeswhich are incident to one edge from each of Ei. Let s � �V�� �V�� be the number of other vertices.

Lemma 2. If �V� is an even number, then s � 1.

Proof. Assume that s � 1 and let r be the specialvertex. There is i such that Ei does not contain an edgeadjacent to r and Ei should be a perfect matching overV�{r}, but this is an odd size set and therefore it is impos-sible. Therefore, s � 1. �

A nontrivial cut is a cut such that on both of its sidesthere are at least two vertices.

Define a K-(multi)graph as a connected, K-regular (multi)-graph with an even size vertex set such that every nontrivialcut of it has at least K � 1 edges.

Lemma 3. EC on K-multigraphs is NP-complete.

Proof. The proof uses a reduction from EC on generalconnected K-regular graphs: Let G � (V, E) be a K-regulargraph which is an input to the EC problem. If �V� is odd,then G cannot be K-edge-colored because every matchingcontains at most (�V� � 1)/ 2 edges and G has (�V�K)/ 2edges. Therefore, w.l.o.g. �V� is even. Consider a nontrivialcut that has, at most, K edges. Clearly, G has a noncut edgeon each side of the cut. Let (a, b), (c, d) be edges in thedistinct parts of the cut (these are not cut edges). Replacethis pair of edges by the subgraph shown in Figure 1. Thisreplacement increases the size of the cut by 2. It can berepeated until we reach a multigraph G� which is K-regularand connected and every nontrivial cut of it has at least K� 1 edges.

The subgraph shown in Figure 1 has the following prop-erties: A legal K-edge-coloring of it must use identical setsof colors for each of the sets of c1 edges and identical setsof colors for each of the sets of c2 edges (which may or maynot be equal to the set of the c1 pair). Note that since the c1

edges have identical color sets then the c4 edges also havean identical color. Similarly, since the c2 edges have iden-tical color sets, then the c3 edges also have an identicalcolor.

If there is a legal coloring of G, then this induces a legalcoloring of G� by correctly coloring every new subgraphand there will be no conflicts.

If there exists a legal coloring of G�, then the coloring ofG� induces a coloring of G [the color of (a, b) will be thecommon color of the c3 edges]. Therefore, the reduction iscorrect. �

We will use Konig’s lemma (see Lemma 1.4.16 in [9] orTheorem 7 page 139 in [1]):

Lemma 4. A bipartite graph in which the maximum vertexdegree is at most l is a union of l matchings and a partitioninto l matchings can be found in O(n2.5l) time.

Proof. A bipartite graph in which the maximum degreeis at most l can be augmented (using dummy vertices andedges) to a l-regular bipartite graph. A l-regular bipartitegraph has a perfect matching that can be computed inO(n2.5) time by computing a max-flow in a bipartite graphwith unit edge-capacities (see [5]). Using this property ltimes, we partition the graph into l matchings (a dummyedge is removed from the corresponding matching) inO(n2.5l ) time. �

FIG. 1. A subgraph for EC NP-complete result.

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Lemma 5. EC on K-graphs is NP-complete.

Proof. Consider a complete bipartite graph Ga �(Va, Ua, Ea) such that each side contains K vertices. ByLemma 4, Ga is K-edge-colorable. Let v � Va, u � Ua,and remove the edge (v, u) from Ga. Add to the resultinggraph two new vertices: x and y with the edges (v, x), ( y,u). The resulting graph Gres is K colorable (it is bipartitegraph with maximum degree K and, therefore, by Lemma 4,K-edge-colored). Every K edge coloring of Gres must usethe same color for the pair (v, x), ( y, u). [Any K-edgecoloring of Gres must color at least K edges with everycolor. If (v, x) is colored c, then there are at most K � 1edges with color c in Va � Ua, and, therefore, ( y, u) iscolored c, as, otherwise, there will be an uncolored edge.]

We are now ready to present the needed changes in theproof of Lemma 3 to prove the lemma. Replace any multi-edge, ( x, y) used in the subgraph in Figure 1 with K � 2copies (with new vertices) of the graph Gres such that all thecopies corresponding to the same multiedge have the samevertices x and y. For every copy of Gres, the edge incidentsin x have the same color as that of the edge incidents in y.Therefore, the corresponding K � 2 copies of Gres mustuse K � 2 distinct colors incidents in x and y. Hence, acoloring of the original multigraph is induced by the color-ing of the graph, and for every coloring of the multigraph,it is easy to complete it to a coloring of the resulting graph.

�

Theorem 6. SWRN is NP-hard for any K � 3.

Proof. Consider the following decision problem:Given a K-graph G for SWRN, is there a solution of size atmost n � 1?

To prove the theorem, we will show that this problem isNP-complete. It is clearly in NP. We claim that this problemis equivalent to the EC for G.

This is so as any feasible solution to the SWRN for Gmust be a connected graph, and for it to have cost equal ton � 1, it must be a tree.

The only tree that might be feasible is a star: Any othertree has an edge between two inner vertices and this edgemust carry a load which equals the size of the cut (betweenits sides in the tree). But this is at least K � 1 (as G isK-graph) and, therefore, cannot be done with K wave-lengths.

Assume that there is a feasible solution to the SWRNwhich is a star. The K colors in the solution define Kmatchings in the star (a matching is the set of pairs thatcarry the communication between them in the same wave-length). Clearly, the K edges incident with a leaf vertexbelong to distinct matchings. By Lemma 2 (as �V� is evennumber), there is no conflict also at the center of the star andthis is a legal coloring. On the other hand, consider a legalcoloring of G. A feasible solution to the SWRN instancecan be defined by any star, and the wavelength of pathbetween i and j equals the color of (i, j) in the coloring of G.

Therefore, there is a solution for SWRN with K and G ifand only if there is a K-edge-coloring of G. �

3. CLIQUE REQUIREMENTS

We now consider a requirement graph which is a cliquewith n vertices. We will present a solution to this case thathas at most 2n edges more than the optimal solution. Topresent the algorithm, we first need some lemmas.

Let OPT be the edge set in an optimal solution.The following is a lower bound on �OPT�:

Lemma 7.

�OPT� �n�n � 1�

K � 1.

Proof. An edge in the solution can be used in, at most,one single-edge path (the edge itself) and in, at most, K � 1other 2-edge paths. As the number of paths is [n(n � 1)]/ 2,the following must be satisfied:

n�n � 1�

2� �OPT� � �K � 1�

�OPT�2

�K � 1

2�OPT�.

�

The next two lemmas were proved in [1, p. 249]. Forcompleteness, we give their proofs.

Lemma 8. A complete graph Kn with an even number ofvertices is a union of n � 1 (perfect) matchings and apartition into n � 1 matchings can be found in O(n2) time.

Proof. Denote the vertices of Kn by {0, 1, 2, . . . , n� 1} and consider the coloring function c(a, b) given bythe following: c(a, b) � (a � b) mod(n � 1) if a, b � n� 1, c(a, n � 1) � 2a mod(n � 1) if a � n � 1, andc(n � 1, b) � 2b mod(n � 1) if b � n � 1. Then, theedges colored with color i (i � 0, 1, 2, . . . , n � 2) area matching and every edge in Kn is in one of the matchings.This partition into matchings takes O(n2) time. �

Lemma 9. A complete graph Kn with an odd number ofvertices is a union of n matchings and a partition into nmatchings can be found in O(n2) time.

Proof. Kn�1 is a union of n matchings (because n � 1is even and by Lemma 8) that induce n matchings over a Kn

subgraph of it. This set of matchings can be computed inO(n2) time by Lemma 8. �

Lemma 10. If �V� � K � 1, then a star over V is anoptimal solution.

Proof. Since G is connected (a clique), the solutionmust be also a connected graph. Therefore, a tree is optimal

34 NETWORKS—2002

if it is feasible. It remains to show the feasibility of the stargraph. A matching over V defines a set of path assignmentsthrough the root of the star. If �V� is even, by Lemma 8, wecan use �V� � 1 � K disjoint matchings and apply adifferent wavelength for each matching to form a feasiblesolution. If �V� is odd, then use one wavelength to connectthe vertices to the root. Then, using the previous argument,we can use at most K � 1 wavelengths to connect all thepairs of nonroot vertices. �

Lemma 11. Let A and B be disjoint subsets of V such that�A�, �B� � K � 1. If a vertex c � A � B is connected by anedge to every v � A � B, then using, at most, K � 1wavelengths in these edges, it is possible to connect everypair of vertices one from A and the other from B.

Proof. Consider the complete bipartite graph (A, B, E).Then, every vertex has a degree at most K � 1, and byLemma 4, it is a union of K � 1 disjoint matchings. If anedge (a, b) is in the i-th matching, then we use wavelengthi for the path (a, c, b). By definition, we will not use thesame wavelength twice in any edge. �

Lemma 12. Let L � {1, 2, . . . , l}. There exists a mappingm(i, j) : L � L3 L that can be computed in O(l2) time, suchthat

● m(i, j1) � m(i, j2) @j1 � j2

● m(i, j) � m( j, i).

Proof. Consider a complete graph Kl over L.If l is odd, then by Lemma 9, Kl is a union of l matchings

M1, M2, . . . , Ml that can be computed in O(l2) time. Everymatching has one vertex not used and different matchingsdo not use different vertices. W.l.o.g., assume that Mi doesnot use vertex i. For p � 1, 2, . . . , l, if (i, j) � Mp, thendefine m(i, j) � p and define m(i, i) � i @i. m clearlysatisfies all requirements.

If l is even, then by Lemma 8, Kl is a union of l � 1matchings M1, M2, . . . , Ml�1 that can be computed inO(l2) time. For p � 1, 2, . . . , l � 1, if (i, j) � Mp, thendefine m(i, j) � p and define m(i, i) � l @i. m clearlysatisfies all requirements. �

Denote by G(S) � (S, E(S)) the subgraph of G inducedby S.

Algorithm Net_Synt in Figure 2 defines an almost opti-mal solution for the clique requirements case in O(n) time.The algorithm constructs a solution in stages such that ineach stage it adds an edge set of a complete bipartite graphand continues to construct the solution of only one of itssides.

Algorithm Wavelength_Alloc in Figure 3 defines an al-location of paths and wavelengths in the solution returnedby Algorithm Net_Synt. This allocation can be computed inO(n2K1.5) time. The algorithm defines allocation of pathsand wavelengths in stages similar to the stages in Algorithm

Net_Synt. It first constructs a mapping m (as in Lemma 12)such that the vertices of Ai are connected to the vertices ofAj through cm(i, j) and the wavelengths are allocated in anappropriate manner.

Theorem 13. Let Es be the edge set returned by AlgorithmNet_Synt. Then, Gs � (V, Es) is a feasible solution. Its costis �Es� � �OPT�(1 � [2(K � 1)/(n � 1)]). A feasible allo-cation of paths and wavelength can be computed inO(n2K1.5) time by Algorithm Wavelength_Alloc.

Proof. The feasibility of Gs is constructively proved byAlgorithm Wavelength_Alloc. The existence of the map-ping m follows from Lemma 12. The ability to connectevery pair of vertices in Ai follows from Lemma 10. Theability to connect every pair of vertices from distinct sets Ai,Aj using a common center cp follows from Lemma 11. Thegraph Gs(C) is the induced subgraph of Gs over C and alsothe result of Net_Synt(C, K).

Algorithm Wavelength_Alloc can be computed in timeO(n2K1.5) that is dominated by the time required to parti-tion into matchings (n/K)2 bipartite graphs each of size 2K� 2 and maximum degree K � 1.

All the edges are used in paths of length 1 and the otherpaths contains exactly two edges. It follows that if in Gs

there are q edges that are used with K wavelengths then�OPT� � q. It is sufficient to show that �Es� � q � 2n (byLemma 7).

For a vertex u � V, let level(u) be the stage in AlgorithmWavelength_Alloc that u � C, and if u � C (there is onevertex u like this) in all stages, then let level(u) be the last stage.Every vertex is not in C in, at most, one level of the recursiveconstruction and, therefore, level(u) is well defined. Let C(u)be the set C in iteration level(u). Let V(u) � V �C(u).

Consider the edges that are used with at most K � 1wavelengths. Every vertex, u, has in level level(u) at mosttwo adjacent edges with less than K wavelengths. Assumethat u � Ai; then, one edge is (u, cm(i,i)), and if i � l, thenthe other edge is (u, cm(i,l )). Therefore, every vertex has, atmost, two such adjacent edges and, therefore, �Es� � q� 2n. �

FIG. 2. Algorithm Net_Synt.

NETWORKS—2002 35

4. GENERAL REQUIREMENTS

We now consider a general connected graph G � (V,ER) and a bandwidth K. For this problem, we will describea Min{(K � 1)/ 2, 2 � 2/(K � 1)}-approximation algo-rithm. Note that Min{(K � 1)/ 2, 2 � 2/(K � 1)} �

52

forevery K � 2, 3, . . . .

Let V � {1, 2, . . . , n}.Denote by degv the degree of v in G.The following lemma generalizes Lemma 7 and provides

a lower bound on �OPT�:

Lemma 14.

�OPT� �1

K � 1 �v�V

degv .

Proof. An edge in the solution can be used in, at most,one path of length 1 (the edge itself) and in, at most, K � 1other paths of length at least 2. As the number of paths is 1

2¥v�V degv, the following must be satisfied:

1

2 �v�V

degv � �OPT� � �K � 1��OPT�

2�

K � 1

2�OPT�.

�

Note that for a connected graph G, �OPT� � n � 1 asa feasible solution must be a connected graph.

The following corollary is straightforward:

Corollary 15. ER is a feasible solution with cost, at most,(K � 1)/2�OPT�.

We will use the following theorem of Vizing (see, e.g.,[1, p. 260]). The coloring of the edge set can be done inpolynomial time [the constructive proof in [1] leads toO(m2) time algorithm].

Theorem 16. The edge set of a simple graph with maxi-mum degree h can be colored with, at most, h � 1 colors.

Define a solution EA as the following: Let v be a maxi-mum degree vertex in G, and EA is a star rooted at v suchthat an edge (v, i) has degi/(K � 1) copies.

FIG. 3. Algorithm Wavelength_Alloc.

36 NETWORKS—2002

Theorem 17. EA is a feasible solution to SWRN with cost,at most, (2 � 2/(K � 1))�OPT�. The allocation of paths andwavelengths in G can be computed in O(m2) time.

Proof. For the graph G, apply the following procedurewith every vertex i that has degree greater than K � 1:Replace i by �i � degi/(K � 1) vertices i1, i2, . . . , i�i.Each of these new vertices will be adjacent to at most K� 1 neighbors of i (each of the neighbors of i will beadjacent to exactly one of the new vertices). This can bedone in O(m) time. Since in the resulting graph each vertexwill have a degree, at most, K � 1, this graph can beedge-colored by K colors in O(m2) time (by Theorem 16).Assume that the edge (i�, j) has been colored p; then,using the � copy of the edge (i, v) and the copy of theedge (v, j) (in EA) with wavelength p, the pair (i, j) isserved. Therefore, EA is feasible and the allocation of pathsand wavelengths can be computed in O(m2) time.

Using Lemma 14, we obtain

�EA� � �i�V��v

degi

K � 1� �

i�V��v

� degi

K � 1� 1� � n � 1

�K � 1

K � 1 �i�V��v

degi

K � 1 n � 1 �

K � 1

K � 1 �i�V�

degi

K � 1

� �2 �2

K � 1� �OPT�.

�

Corollary 18. There exists an O(m2) time Min{(K � 1)/2,2 � 2/(K � 1)}-approximation algorithm for the SWRNproblem.

Proof. The better solution between ER and EA guaran-tees the claimed bound. This follows from Corollary 15 andTheorem 17. �

4.1. d-Regular Requirements

We now assume that G is d-regular.Define a solution EA(d) as the following: Let v � V;

EA(d) is a star rooted at v such that an edge (v, i) has (d� 1)/K copies.

Theorem 19. Assume that G is d-regular and K � 3. EA(d)

is a feasible solution with cost, at most, 2�OPT� and theallocation of paths and wavelengths in EA(d) can be com-puted in O(m2) time.

Proof. G is d-regular and, therefore, it can be edge-colored by d � 1 colors in polynomial time (by Theorem16). Assume that the edge (i, j) has been colored pK � q;

then, using the p � 1 copy of the edge (i, v) and the p � 1copy of the edge (v, j) [in EA(d)] with wavelength q, thepair (i, j) is served. Therefore, EA(d) is feasible and allo-cation of paths and wavelengths can be computed in O(m2)time.

If d � 2(K � 1), then by Lemma 14, �OPT� � 2n and,therefore, in this case,

�EA�d�� � �i�2

n d � 1

K� �

i�2

n � d

K� 1� � n � 1

��n � 1�d

K� n � 1 �

K � 1

K�OPT�

� �3

2�

1

K� �OPT� � 2�OPT�.

If d � 2K � 1, then �EA(d)� � 2(n � 1), and as �OPT�� n � 1, then �EA(d)� � 2�OPT� in this case as well.

If d � {2K, 2K � 1}, then �EA(d)� � 3(n � 1) and�OPT� satisfies the following:

�OPT� �nd

K � 1�

2nK

K � 1�

3n

2.

Therefore, �EA(d)� � 2�OPT� in this case as well. �

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