Operations Research Letters Operations Research Letters 33 (2005) 467–474 www.elsevier.com/locate/orl

An improvedupper boundfor the TSP in cubic 3-edge-connected graphs DavidGamarnik b, Moshe Lewensteina,∗,1, Maxim Sviridenkob

aDepartment of Computer Science, Bar Ilan University, Room 323, Mailbox 50, Ramat Gan 52900, Israel bIBM T. J. Watson Research Center, Yorktown Heights, P.O. Box 218, NY 10598, USA

Received20 January 2004; accepted3 September 2004

Abstract We consider the travelling salesman problem (TSP) problem on (the metric completion of) 3-edge-connected cubic graphs. These graphs are interesting because of the connection between their optimal solutions andthe subtour elimination LP relaxation. Our main result is an better than the 3/2-approximation algorithm for TSP in general. © 2004 Elsevier B.V. All rights reserved.

Keywords: Approximation algorithms; Travelling salesman problem; Graphs; Regular graphs

1. Introduction In general the TSP problem is NP-complete andnot approximable to any constant. Even the metric variant In this paper we consider the well-known travelling of TSP, where the edge weights of Gc form a metric salesman problem (TSP). Given a complete undirected over the set, is known to be MAXSNP-hard [8]. graph Gc = (V, Ec) with nonnegative edge weights, However, Christofides [5] introduced an elegant ap- the goal is to finda shortest (accordingto the weights) proximation algorithm for metric TSP that produces closedtour visiting each vertex exactly once. A closed a solution with an approximation factor 3/2. Almost tour visiting each vertex exactly once is also known three decades later, there is still no approximation al- as a Hamiltonian cycle. gorithm with a better factor than Christofides’ 3/2 fac- tor. A detailedsurvey of the results andthe history of TSP appears in [7,6]. ∗ Corresponding author. Tel.: +972 3 531 7668; fax: +972 3 736 Let G be an arbitrary graph with edge weights that 0498. do not violate the triangle inequality. A metric com- E-mail addresses: [email protected] (D. Gamarnik), pletion on G is the completion of G into a complete [email protected] (M. Lewenstein), [email protected] graph Gc such that each added edge (u, v) is weighted (M. Sviridenko). G 1 The research in this paper was done when the author was a in a manner so that c forms a metric. A shortest path postdoc at IBM T.J. Watson Research Center. metric completion is a metric completion for which

0167-6377/$ - see front matter © 2004 Elsevier B.V. All rights reserved. doi:10.1016/j.orl.2004.09.005 468 D. Gamarnik et al. / Operations Research Letters 33 (2005) 467–474 each edge added is assigned the weight of the shortest This leads to the following conjecture with regard path between u and v. Since we only consider short- to . est path metric completions in this paper, we will re- fer to them as metric completions. In general, in this Conjecture. For metric TSP, the integrality gap  for paper we study the case when G is a cubic 3-edge SER is 4/3. connectedgraph and Gc is its metric completion. All edge weights of the original graph are 1. There have been numerous attempts to prove this One of the classical integer programming formula- conjecture true; however, so far these attempts have tions of the TSP problem is as follows: been unsuccessful. In fact, not only has the conjec-  ture not been proven so far, there has not even been min w(u,v)x(u,v) (1) an improvement over the 3/2 factor. This has raised (u,v)∈E /  c the counterquestion: perhaps 3 2 is the best factor achievable for the metric TSP problem. Or a some- x(u,v) = 2 for all v ∈ V, (2) what milder variant of the question is, perhaps for (u,v)∈Ec   = IP∗ /LP ∗ ,  = 3/2 is the best achievable. x  ∅ = S ⊂ V, (u,v) 2 for all (3) An optimal solution to the subtour elimination u∈S,v∈V \S LP is closely relatedto k-regular k-edge-connected x(u,v) ∈{0, 1} for all u, v ∈ V. (4) multigraphs in the following sense. Given an optimal solution to the LP, we define the number D to be the We denote optimal solutions of the above integer pro- ∗ ∗ smallest common multiplier of the variables xe for gram with IP andits value with IP∗ . As men- ∗ all edges e, i.e. D satisfies that for each edge e, Dxe tionedabove, finding IP∗ is NP-complete. Changing is integer. On the vertex set V we define a multigraph requirement (4) to require that x(u,v) 0 yields a linear ∗ with Dxe edges between vertices u and v where programming relaxation of the integer program also e = (u, v). This multigraph is 2D-regular and2 D- known as the subtour elimination relaxation (SER for edge-connected by the constraints of the LP. Consider short). the original graph G, which was later completedto a Despite the drawback that there are an exponential metric as a TSP instance. Say the weights of all edges number of constraints (of type (3)), SER can neverthe- were 1, i.e. it was unweighted. If the optimal value of less be solvedin polynomial time since each iteration the LP was n, i.e. G was “fractionally” Hamiltonian, of the separation oracle is one mincut computation. We ∗ then it wouldbe enough to finda tour of value n denote with LP the solution to SER andwith  ∗ LP with  < 3/2 to improve on Cristofides algorithm for its value. A measure for the quality of approximation that special (yet very general) case. of an SER solution is the worst-case ratio (integrality We consider probably the simplest class of graphs gap) satisfying the above properties, the cubic (that is, AP X 3-regular) 3-edge-connected simple graphs with  = max , LP ∗ weights 1. The metric completions of these graphs have weights that can be substantially large. It is where AP X is a value of an SER solution andthe easily seen that LP ∗ = n on this class of graphs. maximum is over all instances of metric TSP. Indeed, consider the following solution: x(u,v) = 2/3 The best-known upper boundto date [10,11] shows if (u, v) ∈ E and x(u,v) = 0 otherwise. The feasibility that 3/2. However, no example showing  to be of this solution follows from the fact that graph G is 3/2 is known. The well-known, andcurrently best, cubic and 3-edge-connected. Since the value of this lower boundon  appears in Fig. 1. The full example solution is n and n is a trivial lower boundon LP ∗ , is a metric completion on this graph. To see the 4/3 this solution is optimal. gap, set x(u,v) to be 1/2 for the weight 2 edges in the Thus, to improve the boundon the value of the op- graph and x(u,v) to be 1 for the weight 1 edges. For all timal Hamiltonian cycle, it is enough to prove that other edges, set x(u,v) = 0. In Fig. 1 the optimal tour there is a Hamiltonian cycle of weight at most n in is shown in the right-handfigure. Gc for  < 3/2. In this paper, we design a polynomial D. Gamarnik et al. / Operations Research Letters 33 (2005) 467–474 469

111 ... 111 ...

2 2 2 11 11 2 2 2 ...... 111 111

Fig. 1. An example for which  is 4/3.

time algorithm which finds a Hamiltonian cycle in Gc Also say we have a spanning tree T = (V, E) on of weight at most (3/2 − 5/389)n. Approximation al- a connectedgraph G. We call the multigraph (V, E ), gorithms with performance guarantees better than 3/2 where E contains two copies of each edge of E,a (or even PTASes) for other special classes of metric double spanning tree. We say that we double T when spaces can be foundin [1–3,8]. we take a double spanning tree according to T. There are several directions for possible improve- A cycle cover is a collection of node disjoint cycles ments of our results. It is conceivable that the im- covering all nodes in a graph. The following result provements over 3/2 approximations can be obtained from 1891 is due to Petersen [9]. Its proof can be for general r-regular graphs. Another possibility is to foundin Behzad,ChartrandandLesniak-Foster [4]. obtain improvedresults for sparse graphs, say graphs with maximum 3. At this point, we cannot even Lemma 1 (Petersen [9]). Let G=(V, E) be a 2-edge- lift the 3-connectivity restriction. connected cubic graph. The edge set E can be parti- tioned into a M and a cycle cover C.

2. Preliminaries Note that both problems of finding a perfect match- ing andfindinga cycle cover are polynomially solv- Consider a cubic 3-edge-connected graph G and able. Using this result we show the following lemma. the corresponding complete weighted graph Gc.Any G=(V, E) Hamiltonian cycle in Gc corresponds to some closed Lemma 2. Let be a 3-edge-connected cu- path in G of the same weight that visits all vertices bic graph. The edge set E can be partitioned into a at least once and, conversely, any such path corre- perfect matching M and a triangle-free cycle cover C sponds to a Hamiltonian cycle in Gc of the same (that is a cycle cover which does not contain cycles of weight. Therefore, our original problem of finding length 3). a shortest Hamiltonian cycle in Gc can be reformu- latedas the problem of findingan Eulerian multi- Proof. We prove the lemma by induction on the size graph with a minimum number of edges on the vertex of the graph G.If|V |=4 the lemma is trivial. Assume set V using edges from G only (possibly more than we provedit for all graphs G with |V |n−2 (note that once). |V | must be even). Consider a graph G with |V |=n.If Let G = (V, E) be a graph and E ⊂ E be a col- G does not contain triangles, then by Lemma 1 graph lection of edges. Let V1,...,Vk be a partition of the G can be split into a cycle cover C (which must be vertices according to the connected components of triangle-free) anda matching M. Otherwise, let E be ∗ (V, E ). The contraction of G according to E is a an arbitrary triangle andlet G be the contraction of G ∗ ∗ ∗ ∗ E G∗ multigraph G = (V ,E ), where V ={V1,...,Vk} according to . is still cubic and3-edge-connected ∗ and E contains an edge (Vi,Vj ) for each original andhas n−2 vertices, andtherefore can be split into a edge (u, v)∈ / E where u ∈ Vi and v ∈ Vj . Note that matching anda triangle-free cycle cover C. Note that ∗ there is a one–one correspondence between the edges in general when contracting to G , parallel edges, and of E from G andthe edges E∗ from G∗. Also note hence loops, may be created. However, since G is 3- that there may be parallel edges in G∗. edge-connected this does not happen. The contracted 470 D. Gamarnik et al. / Operations Research Letters 33 (2005) 467–474 triangle E belongs to some cycle of length at least Let us choose a partition of the edge set E =M ∪C l 4inC. If we uncontract E , we can add two edges according to Lemma 2. We will be interested in the from E to the cycle andget a cycle of length l + 2 cycle cover C. We first consider the situation when C in the original graph G. The remaining edge of the is high-five andthen the, more difficult case, when C triangle E is added to the matching.  is not high-five. Finally we choose the  as a suitable tradeoff between the high-five and non high-five cycle covers as a result we will prove the following theorem. 3. The TSP approximation algorithm Theorem 3. Given an arbitrary cubic 3-edge- 3.1. Algorithm outline connected graph, the metric completion of this graph contains a Hamiltonian cycle of weight at most The algorithm we present utilizes the partition of (3/2 − 5/389)n. the edge set E into M ∪ C impliedby Lemma 2. (We will be interestedin C only andnot in M.) C, the 3.3. High-five cycle covers cycle cover, is triangle-free. Hence, each cycle in C is of length at least four. Obviously, if there is exactly When the cycle cover C is high-five, it is straight- one cycle in the cycle cover, this is a Hamiltonian forwardto show that an improvement over the 3 /2 cycle andwe are done.Likewise, if all the cycles are Christofides’ approximation can be obtained. This im- relatively large then transforming it into a Hamiltonian provement is dependent on  andis as follows. cycle can be done at a small “cost”. The algorithm considers two cases. In the first a large fraction of the Lemma 4. If the cycle cover C is high-five, then Gc cycles are of size 5 or greater (we call this the high-five has a Hamiltonian cycle of weight at most (3/2 − property—to be defined later). If the cycle cover C is /10)n. high-five then we can benefit from the fact that there are many large (greater than length four) cycles and Proof. Let G be the contraction of G according to C. utilize this to finda Hamiltonian cycle at a “reduced Let T be a spanning tree on G . Double T to get T .We cost”. output G = (V, C ∪ T ). Clearly G is Eulerian since The more difficult situation to handle is when the it is connectedandall vertices have even degree and cycle cover C is not high-five. Here C contains a lot therefore it corresponds to a Hamiltonian cycle in Gc. of cycles of length four. In this case we manipulate C We now estimate the number of edges in G . Let in a collection of rounds in order to transform it into V4 ⊆ V be the set of vertices coveredby the cycles of a Eulerian graph from which a Eulerian tour is taken. length 4 in C. Since C satisfies the high-five property, This is the heart of the algorithm andcan be donein |V \V4|n andtherefore the total weight of the tour a manner that improves over the approximation factor is estimatedabove by /     over 3 2. |V | |V \V | 3  n + 2 4 + 4  − n.  4 5 2 10 3.2. The high-five property

Throughout the algorithm presentation we will work 3.4. Cycle covers that are not high-five with a constant , to be set at the endof the analysis. Note that, as mentionedabove, the larger the cycles in We now consider the case where the cycle cover C

C the closer they are to a “real” TSP. The following is not high-five, i.e. when |V \V4| < n. Let C be the definition captures the notion of having many larger edges of C from cycles of length exactly 4. Let G be cycles. the contraction of G according to C . First of all we The high-five property: We say that a cycle cover claim that G does not contain self-loops since other- has the high-five property if the cycles of length five wise it wouldcontain the subgraph from Fig. 2 and andmore in C contain at least n vertices, where n is therefore would not be 3-edge-connected. So G con- the number of vertices in V. tains |V4|/4 vertices of degree four and |V \V4| < n D. Gamarnik et al. / Operations Research Letters 33 (2005) 467–474 471

...... Fig. 3. A chain of 2-cycles.

Fig. 2. Impossible 4-cycle in a 3-edge-connected graph. are 2-cycle-saturatedandhence not incidentto any other edges, implying that the cycle of 2-cycles is a vertices of degree three. The total number of edges in component of G . However, G is a connectedgraph

G is (|V4|+3|V \V4|)/2. (Fig. 3). Denote with C all 2-cycles on chains. For each 3.4.1. Eliminating isolated 2-cycles chain the 2 vertices, at either end, are not 2-cycle satu- In the contractedgraph G there are nodes that cor- ratedandhence are incidentto one or two other edges, respondto a single nodein G andhence still have de- the tail edges. We remove the C andtail edgesfrom gree 3, andnodesthat correspondto cycles of length G . In the case where there is exactly one cycle of 2-

4inC the cycle cover on G. It is easy to verify that cycles, we let C be a Hamiltonian cycle containing these nodes have degree 4. Moreover we may have one edge from each 2-cycle. In this case we remove parallel edges in G . the from G . After removing these

A 2-cycle is a graph consisting of two parallel edges. edges, G does not contain 2-cycles anymore.

We call a 2-cycle isolatedif it doesnot have a com- We now (greedily) find any cycle in G , delete its mon vertex with another 2-cycle. For each isolated2- edges and all edges incident to this cycle from the cycle in G , delete one of the two parallel edges in graph andaddthecycle to C . We repeat this process the 2-cycle (notice that we cannot have three parallel until the graph G becomes acyclic. edges in G since otherwise G wouldnot be 3-edge- Let p be the number of steps (cycles andchains) connected). Let G be the new graph. Since isolated in the above process and Li,i= 1,...,p,bethe 2-cycles do not touch each other the deleted edges number of vertices in the chain or cycle defined in step form a matching in G andtherefore the total number i. At the endof the above algorithm, G contains at G |V |/ +|V \V |− p L of edges in is at least most 4 4 4 k=1 k vertices that are not isolated. Since G is acyclic without parallel edges, i.e. |V |+ |V \V | |V |/ +|V \V | p 4 3 4 4 4 4 |V |/ +|V \V |− L − − a forest, there are at most 4 4 4 k=1 k 2 2 L |V | 1 edges. For each cycle or chain of length k deleted 3 4 G L = +|V \V4|. from , there are at most 3 k edges deleted from 8 G andtherefore 3.4.2. Removing 2-cycle chains and cycles p 3|V4| It is possible that in G some 2-cycles share a node. 3 Lk  +|V \V | 8 4 However, no more than two 2-cycles may share a k=1   node, since the degree of vertices in G is at most 4. p In fact, a node incident on two 2-cycles has no other − |V4|/4 +|V \V4|− Lk − 1 edges incident upon it. Such a node is said to be 2- k=1 p cycle-saturated. Since each 2-cycle saturatednodeis |V4|  + Lk. incident to exactly two 2-cycles, the 2-cycle-saturated 8 nodes form a collection of (disjoint) chains (of 2- k=1 cycles) andcycles (of 2-cycles). We also call the 2- Hence, cycle-saturatednodes, internal nodes as they must be internal to the chain or cycle. However, note that the p |V4| only cycle of 2-cycles possible is one that contains Lk  . (5) 16 all vertices. This is true since all nodes on this cycle k=1 472 D. Gamarnik et al. / Operations Research Letters 33 (2005) 467–474

Now, consider the original G , i.e. the graph we had before removing the edges according to C . Let G be the contraction of G according to C . Remove the self-loops from G andlet T be a spanning tree in G . Let us estimate the total number of vertices in G , which is one more than the size of the spanning tree T. This number is also equal to the number of vertices Fig. 4. Solidedgesare the 4-cycle of C andthe dashededgesare in G minus the total number of vertices in the cycles from the cycle in G . Note that the 4-cycle is a supervertex in G . andchains from C andplus the number of cycles and chains in C , i.e. p |T |+1|V4|/4 +|V \V4|− Lk + p k=1 p 2 k= Lk |V |/4 +|V \V |− 1 , (6) 4 4 3 where |T | denotes the number of edges in tree T and Fig. 5. Solidedgesare the 4-cycle of C andthe dashededgesare the secondinequality follows from the fact that Lk 3 from the cycle in G . for all k = 1,...,p.

3.4.3. Final Eulerian graph (andpossibly removing edgesfrom isolated2-cycles). We now buildthe final Eulerian graph in G. Let Since C contains cycles andchains of 2-cycles in C G be the edges in the graph G corresponding to the G , a vertex v in G can be adjacent to either 0, 1 edges from C (note that even though we have been or 2 edges from CG. The parity of the degree of v transforming G, the edges in every stage are from the is odd if it is adjacent to exactly one edge from C . C G original graph G, so although was defined on a This can happen only if v belongs to a 4-cycle in C. transformedgraph the edgesare still originally from Moreover by construction of C the cycles andchains C ∪ C G). We will transform the graph G so that the of 2-cycles in C are vertex disjoint (relative even to degrees of all vertices will become even to satisfy the G ), so for each 4-cycle in C, which corresponds to a Eulerian requirements. We also guarantee that the new vertex in G , it can be either on a cycle of C ,ona n + p L graph contains at most k=1 k edges and that 2-cycle chain or on neither. But never on both. Hence the connectivity properties are not destroyed, i.e. every we have two cases. connectedcomponent of the transformedgraph is a Case 1: This vertex belongs to some cycle of length connectedcomponent of the original graph andvice four and it is incident to exactly one edge from some versa. After we transform C∪C to maintain the “even G cycle in CG (see left figures in Figs. 4 and 5). degree” requirement we add a doubled spanning tree T For any cycle of length four in C there are either C ∪C to connect all connectedcomponents of G while none or two such vertices. In the latter case, if these maintaining the “even degree” requirement. Hence, the vertices are adjacent in the cycle of length four we graph will be Eulerian. just delete the edge connecting them which decreases their degrees by one, see Fig. 4. This procedure does

3.4.4. Transforming C ∪ CG into a graph with even not change connectivity since we throw away just one degree vertices edge per cycle. If a cycle v1,v2,v3,v4 of length four We begin by analyzing the reasons for an odd de- has two vertices v1,v3 of odd degrees which are not gree vertex in C ∪ CG. Since C is a cycle cover on adjacent we throw away edge (v1,v2) and add edge the original G, each vertex has exactly two edges in (v2,v3), see Fig. 5. Again after this transformation the

C ∪CG from the C cycles. G is obtainedby contract- degrees of all vertices in the cycle are even and the ing cycles of length 4 in C into (super)vertices in G connectivity of the component remains intact. More- D. Gamarnik et al. / Operations Research Letters 33 (2005) 467–474 473 over, the transformation does not increase the total 3.4.5. Upper bounding the size of the Eulerian graph number of edges in a component. In the previous section we explainedhow to trans-

Case 2: The vertex of odd degree belongs to a cycle form graph C ∪ CG into an Eulerian graph. Let us call of length four andthe vertex correspondingto this this graph Gt . We now upper boundthe total number cycle in G belongs to a chain of 2-cycles in C . of edges in Gt .Ifp1 is the number of chains in C

and p2 is the number of cycles in C , then C ∪ CG p p Lemma 5. Every node in this chain corresponds to a contain at most n + 2 1 (Lk − 1) + Lk k=1 k=p1+1 cycle of length four in C. edges where p = p1 + p2. During the transforma- tion we never increase the current number of edges Proof. Indeed, every internal (2-cycle-saturated) node in our graph, but we needa stronger statement since of this chain is a cycle of length four in C since it we wouldlike to show that the number of edgesin G G n + p L has degree four in . Every endnodeof a chain also t is at most k=1 k; we do it by showing that corresponds to a 4-cycle. In fact, otherwise it would we delete enough edges when we transform chains of have degree three in G andtwo edgesof a chain supervertices corresponding to cycles of length four. connecting this vertex to an internal 4-cycle (which Indeed, for each chain of length Lk we delete at least belongs to C) cannot belong to any cycle in C. Since 2(Lk − 2) edges when we transform internal cycles of the end node itself does belong to some cycle in C,it a chain (2 per each internal node of a chain). There- must be incident to two additional edges, but then its fore, the total number of edges in Gt can be upper degree is 4, a contradiction.  bounded by

p p p We distinguish between internal (2-cycle-saturated) 1  1 nodeson these chains andbetween the endnodesof the n + 2 (Lk − 1) + Lk − 2 (Lk − 2) chain. Consider a 4-cycle corresponding to an internal k=1 k=p1+1 k=1 (2-cycle saturated) node of a chain. Let us denote it p p v ,v ,v ,v 1 2 3 4. All these vertices have degree three in n + 2p1 + Lk n + Lk C ∪ C G, 2 edges from the cycle C that they partici- k=p1+1 k=1 pate upon andthe only other edge vi is connectedto G corresponds to a 2-cycle edge in which is chosen since Lk 3 for all k. C in as a chain of 2-cycles (since this 4-cycle is an Let us estimate the total number of edges in the final G internal node in ). Eulerian graph, which contains the Gt anda doubled (v ,v ), (v ,v ) Therefore, by deleting two edges 1 2 3 4 spanning tree T over its connectedcomponents. This (v ,v ), (v ,v ) or 1 4 2 3 we make all degrees of vertices number can be estimatedabove by v1,v2,v3,v4 even. Moreover, one of these two vari- ants does not destroy connectivity, since the initial p chain consisting of the internal 4-cycles, end4-cycles n + Lk + 2|T | andpairs of edgesconnecting the 4-cycles, contains k=1  two vertex disjoint paths, in G, connecting endcycles. p L k=1 k Therefore, if we throw away two opposite edges in n +|V4|/2 + 2|V \V4|− each internal 4-cycle not belonging to these paths we   3 1 1 do not disconnect vertices of the 4-cycle from each n + − |V4|+2|V \V4| other. For two external nodes of a chain we apply the  2 48   3 1 3 1 argument identical to the one from case 1. There are  − n + + n, (7) exactly two nodes of odd degree on a 4-cycle corre- 2 48 2 48 sponding to an end of a chain; if these two nodes are connected by an edge in a 4-cycle, we delete it. Oth- where the first inequality follows from (6), the second erwise, if v1 and v3 are vertices with odd degree in a one follows from (5) andthe last one follows from the 4-cycle v1,v2,v3,v4 we delete the edge (v1,v2) and fact that |V \V4| < n. The inequality (7) implies the add the edge (v2,v3). following lemma. 474 D. Gamarnik et al. / Operations Research Letters 33 (2005) 467–474

Lemma 6. If the cycle cover C is not high-five, then [2] S. Arora, M. Grigni, D. Karger, P. Klein, A. Woloszyn, A Gc has a Hamiltonian cycle of weight at most (3/2 − polynomial-time approximation scheme for weightedplanar 1/48)n + (3/2 + 1/48)n. graph TSP, Proceedings of the Symposium on Discrete Algorithms, 1998, pp. 33–41. [3] D. Arun Kumar, C. Pandu Rangan, Approximation algorithms 3.5. Trading off high-five and non high-five cycle for the traveling salesman problem with range condition, covers Theoret. Inform. Appl. 34 (3) (2000) 173–181. [4] M. Behzad, G. Chartrand, L. Lesniak-Foster, Graphs and To achieve our final result, we needto tradeoff high- Digraphs, PWS Publishers, Massachusetts, 1979. [5] N. Christofides, Worst-case analysis of a new heuristic for five cycle covers andcycle covers which are not high- the travelling salesman problem, Technical Report, Carnegie five. To achieve this we optimize  over inequalities Mellon University, 1976. in Lemmas 4 and6 andobtain Theorem 3. [6] G. Gutin, A. Punnen (Eds.), The Traveling Salesman Problem The construction is obviously achievable in poly- and its Variations, Kluwer Academic Publishers, Dordrecht, nomial time, since all it involves is finding a perfect 2002. [7] E.L. Lawler, J.K. Lenstra, A.H.G. Rinnooy Kan, D.B. Shmoys matching (the cycle cover is then immediately obtain- (Eds.), The Traveling Salesman Problem, John Wiley, New able), detecting cycles of length 4, chains of 2-cycles York, 1985. over them and then greedily finding cycles. Also the [8] C.H. Papadimitriou, M. Yannakakis, The traveling salesman spanning tree can be foundin polynomial time. problem with distances one and two, Math. Oper. Res. 18 (1993) 1–11. [9] J. Petersen, Die Theorie der Regulären Graphen, Acta Math. 15 (1891) 193–220. Acknowledgements [10] D.B. Shmoys, D.P. Williamson, Analyzing the Held-Karp TSP bound: a monotonicity property with application, Inform. We thank the anonymous referees of this paper for Process. Lett. 35 (6) (1990) 281–285. providing comments that improved the exposition of [11] L. Wolsey, Heuristic analysis, linear programming andbranch this paper. and bound, Math. Programming Stud. 13 (1980) 121–134.

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