 # On the Approximability of the Traveling Salesman Problem with Line Neighborhoods

We study the variant of the Euclidean Traveling Salesman problem where instead of a set of points, we are given a set of lines as input, and the goal is to find the shortest tour that visits each line. The best known upper and lower bounds for the problem in ℝ^d, with d≥ 3, are NP-hardness and an O(log^3 n)-approximation algorithm which is based on a reduction to the group Steiner tree problem. We show that TSP with lines in ℝ^d is APX-hard for any d≥ 3. More generally, this implies that TSP with k-dimensional flats does not admit a PTAS for any 1≤ k ≤ d-2 unless P=NP, which gives a complete classification of the approximability of these problems, as there are known PTASes for k=0 (i.e., points) and k=d-1 (hyperplanes). We are able to give a stronger inapproximability factor for d=O(log n) by showing that TSP with lines does not admit a (2-ϵ)-approximation in d dimensions under the unique games conjecture. On the positive side, we leverage recent results on restricted variants of the group Steiner tree problem in order to give an O(log^2 n)-approximation algorithm for the problem, albeit with a running time of n^O(loglog n).

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## 1 Introduction

In the Euclidean Traveling Salesman problem, one is given points in -dimensional Euclidean space (denoted by ), and the goal is to find the shortest tour visiting all the points. The problem is -hard for  [Pap77], but it has a celebrated polynomial time approximation scheme (PTAS), i.e., a polynomial-time algorithm that produces a tour of length at most times the optimum for any fixed , due to Arora [Aro98] and (independently) by Mitchell [Mit99]. The running time was later improved by Rao and Smith [RS98].

In the past decades, a considerable amount of work has concentrated on finding approximations for variants and generalizations of the Euclidean Traveling Salesman Problem, for example by changing the underlying space [AGK98, KL06, DHK11, BGK16], or the objects being visited [dBGK05, BFG09, CJ18, DM03, Mit07, Mit10, JM19]. In the latter case, which is also known as the Traveling Salesman Problem with Neighborhoods (TSPN), instead of a set of points, the input consists of neighborhoods, and the goal is to find the shortest tour that visits each neighborhood. More formally, we are given the sets , and we wish to compute the shortest closed curve such that for each we have . (It is an easy observation that the optimum curve consists of at most segments.)

In contrast to regular TSP, TSPN is already APX-hard in the Euclidean plane [dBGK05], that is, it does not have a PTAS unless . Worse still, even the basic case in which each neighborhood is an arbitrary finite set of points in the Euclidean plane (the so called Group TSP) does not admit a polynomial-time -approximation (unless P  NP) [SS06]. Even in the case in which each neighborhood consists of exactly two points [DO08] the problem remains APX-hard.

This inherent hardness of TSPN gives rise to studying variants of the problem in which the neighborhoods are restricted in some ways. In a seminal paper, Arkin and Hassin [AH94] looked into the problem for various cases of bounded neighborhoods, including translates of convex regions and parallel unit segments, and gave constant-factor approximation algorithms for them. The best known approximation algorithm for a more general case of bounded neighborhoods in the plane is due to Mata and Mitchell [MM97] and attains an  approximation factor. However, there exist special cases of such bounded neighborhoods in the plane that do allow for -approximation algorithms. These include neighborhoods which are disjoint, fat, or have comparable sizes [dBGK05, BFG09, CJ18, DM03, Mit07, Mit10].

The complementary case of TSPN in which neighborhoods are unbounded regions (which is also the focus of this paper) is, in general, less well understood. Consider neighborhoods that are affine subspaces (flats) of dimension in . On the positive side, and despite the APX-hardness of the general TSPN problem already in , the version with flats (in this case lines) as neighborhoods can be solved exactly in -time via a reduction to the shortest watchman route problem [Jon02, DELM03]. Furthermore, Dumitrescu [Dum12] provides a -approximation algorithm that runs in linear time. In , the problem of line and plane neighborhoods was first raised by Dumitrescu and Mitchell [DM03]. For the line case, they already point out that the problem is -hard as a direct consequence of the -hardness of Euclidean TSP in the plane [Pap77]. Although this leaves the possibility for a PTAS open, the best known approximation algorithm to date for TSPN with line neighborhoods in was given by Dumitrescu and Tóth [DT16] and achieves an -approximation. For the case of -dimensional flats in (which also includes planes in ), they give a linear-time (for any constant dimension  and any constant ) -approximation. This result was subsequently improved by Antoniadis et al. [AFHS19] to an EPTAS that also runs in linear time for fixed and . Whether this variant is -hard or not remains an interesting open problem. As for the case of line neighborhoods in for , a PTAS for -dimensional flats for also remained out of reach.

In this article, we show that unless , there is no PTAS for lines in . As a direct consequence, we can rule out the existence of a PTAS in all remaining open cases of TSPN with flats: there is no PTAS for -dimensional flat neighborhoods for any , unless .

Let us call the Euclidean TSP problem in with -dimensional flat neighborhoods -TSPN. Although ruling out a PTAS for -TSPN is an important step towards settling the approximability of the problem, the inapproximability factor obtained is very close to . It would be desirable to obtain a stronger inapproximability factor, especially given how far we are from any constant-approximation algorithm for the problem. A natural way to obtain such a stronger inapproximability result is to consider the problem in higher dimensional spaces. For example, regarding the classic Euclidean TSP, it is known that the problem becomes APX-hard for  [Tre00]. This result directly implies that TSPN with line neighborhoods in is APX-hard, but this is barely satisfactory, since it again only gives a small inapproximability factor. However, by using a different reduction from the vertex cover problem, we are able to show that the problem has no polynomial -approximation in for any fixed under the Unique Games Conjecture [Kho02].

On the algorithmic side, very little is known about -TSPN. Focusing on , the best known polynomial time approximation for -TSPN is the aforementioned -approximation algorithm due to Dumitrescu and Tóth [DT16]. Their approach is to discretize the problem by selecting a polynomial number of “relevant” points on each line. It is shown that restricting the solution to visiting lines at these points only increases the tour length by a constant factor. The resulting problem instance can now be seen as an instance of group-TSP, where the relevant points of each line comprise a group. By feeding this instance into the -approximation algorithm for general group Steiner tree [GKR00, FRT04] (it is easy to go from the tree solution to a tour by doubling each edge), they obtain the same asymptotic approximation factor for TSPN with line neighborhoods. This is somewhat unsatisfactory, since it ignores that the group Steiner tree instances constructed by the reduction (i) are Euclidean and (ii) all the points of a group are collinear. In other words, although the constructed group Steiner tree instances are highly restricted, there is no known technique to derive algorithmic benefits from this restriction.

However, the reduction from TSPN with line neighborhoods to the group Steiner tree problem implies that, if we allow quasi-polynomial running time, then TSPN with line neighborhoods admits an approximation ratio of in -time due to the result of Chekuri and Pál [CP05]. We would like to point out that this approximation ratio is tight for the class of quasi-polynomial time algorithms due to the recent work of Grandoni, Laekhanukit and Li [GLL19], which holds under the Projection Game Conjecture and . Their hardness result is built on the seminal work of Halperin and Krauthgamer [HK03], who prove that group Steiner tree admits no -approximation for any fixed , unless has quasi-polynomial-time Las Vegas algorithms ().

For the class of polynomial-time approximation algorithms, the group Steiner tree problem admits an approximation ratio of on some special cases, e.g., trees [GKR00] and bounded treewidth graphs [CDLV17, CDE18]. It is still open whether the group Steiner tree problem in general graphs admits a polynomial-time -approximation algorithm; the best running time to obtain an -approximation is [CP05].

This connection between TSPN and group Steiner tree can also be used in the reverse direction: Given an instance of the group Steiner tree problem, one may embed the input metric into a Euclidean space with distortion  [Bou85] and then cast it as TSPN with “set neighborhoods”.

While we cannot improve the approximation factor in polynomial time, we can do so in quasi-polynomial time: we give an -approximation in time. We obtain this result by using Arora’s PTAS for TSP [Aro98], together with the framework of Chalermsook et al. [CDLV17, CDE18], to transform the TSPN problem into a variant of the group Steiner tree problem when the input graph is a tree, and then employing an -approximation algorithm for this problem.

### 1.1 Our Contribution

Our first contribution is to show that unlike the problem with hyperplane neighborhoods, the problem with line neighborhoods is APX-hard.

###### Theorem 1.1.

The TSPN problem for lines in is APX-hard. More specifically, it has no polynomial time -approximation unless .

The reduction is from the vertex cover problem on tripartite graphs. The idea is to represent the graph edges with lines, where two lines intersect if and only if they correspond to incident edges. The main challenge is to keep the pairwise distance between non-intersecting lines large enough. We solve this by carefully placing the intersection points on non-adjacent edges of a cube. For technical reasons, we do not work directly with this placement, but rather on a “flattened” version of this point set. Additionally, we want to restrict the optimal tour so that it visits each line near one of its intersection points with other lines. This is achieved by forcing the optimal tour to follow a certain closed curve using special point gadgets (each consists of polynomially many lines), and to visit the lines representing the edges only at (or close to) intersection points. Visiting an intersection point corresponds to including the corresponding vertex in the vertex cover of the graph. As a direct consequence of Theorem 1.1, we obtain the following.

###### Corollary 1.2.

The Euclidean TSP problem with -dimensional flat neighborhoods in is APX-hard for all .

To prove Corollary 1.2, suppose we are given a set of lines in . We can first change each line into the flat , resulting in -dimensional flats in . Since , we have that is a subspace of , so this is a valid construction for -TSPN. Moreover, any tour in visiting the lines is also a valid tour of the -flats, and a valid tour of the -flats can be projected into a valid tour of in of less or equal length.

Our second contribution is to show a larger inapproximability factor in higher dimensions under the Unique Games Conjecture:

###### Theorem 1.3.

For any , there exists a constant such that there is no -approximation algorithm for TSPN with line neighborhoods in , unless the Unique Games Conjecture is false.

Moreover, for any , there is a constant such that it is -hard to give a -approximation for TSPN with line neighborhoods in .

This reduction is from the general vertex cover problem. Again we are representing the edges of the graph with lines and the vertices correspond to intersection points. This time however the intersection points are almost equidistant: they are obtained via the Johnson-Lindenstrauss lemma applied on an -simplex. This allows the tour to visit the intersection points in any order. To obtain a direct correspondence with vertex cover, we need to ensure that lines are visited near intersection points. To this end, we blow up the underlying graph by replacing each edge by a complete bipartite graph. Just as with -TSPN, we get the following as a direct corollary of Theorem 1.3.

###### Corollary 1.4.

For any there is a number such that the Euclidean TSP problem with -dimensional flat neighborhoods in has no polynomial -approximation for any , unless the Unique Games Conjecture is false.

On the positive side, our third contribution is to develop an -approximation algorithm with slightly superpolynomial running time.

###### Theorem 1.5.

There is an -approximation algorithm for TSPN with line neighborhoods in that runs in time for any fixed dimension .

The algorithm is based on adapting the dynamic program by Arora [Aro98], and reformulating TSPN into the problem of finding a solution in the dynamic programming space that visits all the line neighborhoods. We then build upon the techniques of Chalermsook et al. [CDLV17, CDE18], and show that this task can be reduced to a variant of the group Steiner tree problem that admits an -approximation in slightly superpolynomial running time. The -factor in the exponent of the running time is a consequence of the running time of Arora’s algorithm, and it is possible that we can improve it to polynomial time if an appropriate EPTAS for TSP with running time is discovered.

## 2 Inapproximability in 3 dimensions.

The goal of this section is to prove Theorem 1.1.

##### Overview of the reduction.

The overall setup of our construction is inspired by a reduction in Elbassioni et al. [EFS09] for the planar problem with segment neighborhoods. Our reduction is from vertex cover on -partite graphs (i.e., on graphs where the vertices can be partitioned into three independent sets and ). It is -hard to decide whether a given instance has a vertex cover of size or if all vertex covers have size at least  [CCR99].

In our construction, each vertex of the graph will be assigned to some point

on some edge of a unit cube; the three partition classes are assigned to points on pairwise non-adjacent and non-parallel (i.e., skew) edges of the cube. For each edge

, we add the line ; see Figure 1. Figure 1: Left: Overview of a basic construction with a cube. Right: The optimal tour must visit all points of Q, and it makes detours to some points pv.

Consider now a closed curve of length 10 which is disjoint form the cube, but follows some edges of the cube at a distance for some constant . Let be a set of points along the curve such that any two consecutive points on the curve have a distance from each other.

We define a special point gadget—which consists of a large collection of lines—at each point . This ensures that any TSP tour that has length at most will touch an infinitesimally small ball around each vertex of . Consequently, any not too long TSP tour will have to “trace” . The points in which are placed near the cube edges are arranged so that one can visit each point with a short detour from of length . Given a vertex cover of size in one can create a TSP tour of length at most , namely by folowing and making the short detour at if and only if is in the vertex cover. Conversely, by a careful arrangement of the lines and point gadgets, we can ensure that a tour of length implies the existence of a vertex cover of size at most .

For technical reasons, we need to transform the constructed cube to a very flat parallelepiped; it is convenient to define the point set and the point gadgets only after this flattening transformation takes place. We are now ready to define our construction.

### 2.1 The construction

Let be a tripartite graph on vertices with partition classes . We add dummy vertices (without any incident edges) to G so that each class has vertices; the vertices of are denoted by . Notice that the addition of dummy vertices does not change the set of vertex covers of . Let denote the unit cube , and let be the unit segments , and respectively. We assign each vertex to a point on the middle third of . The assignment is denoted by , and defined as:

 p(vai)=⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩(n+i3n,0,1)Tif a=1(1,n+i3n,0)Tif a=2(0,1,n+i3n)Tif a=3.

We denote by the set of points created this way. For each edge , let be the line through and , and let be the set of lines created this way: . The following technical lemma plays a key role in the contruction.

###### Lemma 2.1.

lem:cube_skew_dist If correspond to non-incident edges, then they are disjoint and their distance is at least .

###### Proof.

Assume without loss of generality that and intersect at the points respectively, and moreover assume that also intersects at . The line intersects either or at some point , see Figure 2

. For a vector

, let and be its coordinates. The line has the vector equation , and similarly . Their distance is therefore

Let and let . Since is parallel to and the -axis, this is

 dist(ℓ,ℓ′)=|p′1−p1||(v×w)1||v×w|=|(p′1−p1)(v2w3−v3w2)|√(v2w3−v3w2)2+(v3w1−v1w3)2+(v1w2−v2w1)2.

Note that , the values and are both in the interval , and . We now consider the cases where intersects and separately. If intersects , then , and and are inside the interval . We can use these facts to bound each term:

 dist(ℓ,ℓ′)≥13n(−2/3+1)√(−1/9+1)2+(2/3+4/9)2+(2/3+4/9)2=12√66n>120n.

If intersects , then and are both in the interval , and . As we now have , it follows that . Let . Then the above formula becomes:

 dist(ℓ,ℓ′)=|δ1δ2|√δ21+δ22+(v1w2−v2w1)2.

The term can be bounded the following way:

 |v1w2−v2w1|=|(v1−w1)(v2+w2)−(v1+w1)(v2−w2)|2=|δ1(v2+w2)−δ2(v1+w1)|2≤|δ1|+|δ2|.

We can substitute this to get a lower bound on the distance:

 dist(ℓ,ℓ′) ≥|δ1δ2|√δ21+δ22+(|δ1|+|δ2|)2>|δ1δ2|√2(|δ1|+|δ2|)2=1√2|δ1δ2||δ1|+|δ2| =12√221|δ1|+1|δ2|≥12√2min(|δ1|,|δ2|)≥16√2n>110n,

where we have used that both and is at least

, and the fact that the harmonic mean of two numbers is at least as large as the smaller number. ∎

##### Flattening.

Due to technical reasons that will become clear in the proof later on, we need to transform the above construction so that the angle of each line with the plane is at most some small constant. Practically, we transform the point set and the line set

with the linear transformation

, where and is the all-ones matrix.

Essentially, the transformation pushes everything closer to the plane in a perpendicular fashion: for a given point and its perpendicular projection on , the point is the point on the segment for which . Note that if is any segment of length , then its length after the transformation is at least and at most . When the transformation is applied to an edge of the cube , then the resulting segment has length . Consequently, and has distance .

Let and be the resulting point set and line set. Using Lemma 2.1 and the above arguments we get the following corollary.

###### Corollary 2.2.

The minimum distance between points of is , and the minimum distance between lines of corresponding to non-incident edges of is at least .

##### Defining the point gadgets, and wrapping up the construction.

For a point set , let denote its image under the flattening transformation . Let and be the planes of the faces of incident to . The following claim shows that and are two planes through whose angle is small. Figure 3: Top left: A plane perpendicular to ¯ea. All lines of ¯L are in the gray area. Bottom left: Defining Qa within the plane Ha, so that all points have distance at least δ∗ from ¯ea. Right: Defining Q so that it has all the required properties. The cylinder Y is perpendicular to the plane x+y+z=0, to which all points of the construction are close to. The cylinder is inside the ”triangle” seen in the picture.
###### Claim 2.3.

cl:small_plane_angle For we have .

###### Proof.

It is sufficient to prove the claim for because of symmetry. First, we compute the normal of , which is the plane through the points , , and . Therefore, it goes through , and the origin; its normal is therefore . Similarly, is the plane through , and . The calculation yields that the normal is . The angle of the planes is therefore

 cos−1(⟨n1,n2⟩)=cos−1(0.33/0.34)<14.\qed

Let be the angle bisector plane of and which does not intersect the image of , see Figure 3(i). Within , we place a set of points , which we define next.

Let , and let be the height of the isoceles triangle with base and two sides of length , that is . Consider a half-plane in whose boundary is parallel to and is at distance from it. Within this half-plane, let be a set of at most points with the following properties: (i) for each there are two points such that , and form an isoceles triangle of side lengths and (ii) there is a unique shortest TSP path of , whose edges are of length exactly ; see Figure 3(ii).

Let be a point set with the following properties:

• [noitemsep]

• For any pair of distinct points , .

• Each segment of the minimum TSP tour of has length , and .

• The minimum distance of points of from is attained only in

• is disjoint form the cylinder of axis and radius .

Such a set is easy to find, for example by following the lines and connecting them far from the origin. See Figure 3(iii) for an illustration.

We will need the following bound on the distance of points of from the lines in . Note that this claim would not hold without the flattening.

###### Claim 2.4.

cl:qdist For any and we have .

###### Proof.

As the distance between points of and lines in is minimized only at points of , we may assume without loss of generality that . If connects and , then its distance from is much more than , so assume that connects a point of to . (The case when goes from to is similar.) Notice that all such lines are separated from by the union of the planes and , so . By the definition of , we have that

 d(q,¯F11)=d(q,¯F12)=dist(q,¯e1)cos(∢(¯Fa1,¯Fa2)/2).

Since by creftypecap 2.3, and , therefore

 d(q,ℓ)>cos(1/8)⋅9.98δ>9.9δ\qed

lem:point_gadget Given a positive integer and a point , there is a set of lines through such that any TSPN tour of which is disjoint from the ball has length at least . Figure 4: Defining the lines of a point gadget for a point q using a grid Γ.
###### Proof.

Let be a plane grid of side length . The grid fits in a square of side length . Let be the origin, and place the grid in the plane , within the axis-parallel square with diagonal vertices and . See Figure 4. Notice that the grid is contained in the ball .

Let be the set of lines through that contain a grid point. This is a set of lines. Consider now a shortest TSPN tour of that is disjoint from . The sets have pairwise distance at least , so the tour must have length at least . ∎

Our construction is the union of the line set together with a point gadget placed at each point ; let denote the resulting line set.

### 2.2 The Reduction

###### Lemma 2.6.

lem:noptas_main If has a vertex cover of size , then there is a tour in of length . If has a tour of length , then has a vertex cover of size .

###### Proof.

To prove the first claim, let be a a vertex cover of of size . We can create a tour by first adding all edges of , and for each vertex , we add a detour: if are the nearest points of to , then we remove the segment of length and add the segments and of length each to the tour. For each vertex , this results in a length increase of , so the resulting tour has length as required. We can verify that touches every line of . It goes through each , thus it goes through all lines in point gadgets. For each line the corresponding graph edge is covered by the vertex cover, so either or . Therefore is touched either at or .

To prove the second claim, let be a tour of length . Since and , we have that the length of is less than . Since touches each line, it is also a valid tour for any subset of lines. In particular, for each it is a tour of length less than of the point gadget of . Consequently, intersects each ball . Note that by the properties of , these balls are disjoint and have pairwise distance more than if is large enough. Let denote the ball .

Without loss of generality, we can assume that is a -dimensional (skew) polygon whose vertices are on the lines of . Consider the vertices of in order, and remove all vertices of the sequence that are only incident to lines of point gadgets, but lie outside the balls . Furthermore, remove entries that fall inside until we get a sequence where there is a unique vertex from each . Let be the sequence of vertices we get this way.333The sequence should be understood as a cyclic sequence, where indices are defined modulo . In particular . As a result, for each there exists a point , and for each , there is some unique entry . Fix an orientation of , and let denote the subpath of from to . The balls partition into subsequences, so can be regarded as the concatenation of sequences where for each we have , , and for each it holds that .

.

###### Proof.

Let be points of such that and . By the definition of , we have . Consequently, . ∎

Consider now a sequence , and let be a sequence of points such that the point is on the line .

###### Lemma 2.8.

There do not exist lines such that connects with , line connects with , and connects with .

###### Proof.

First, we show that is disjoint from the cylinder with axis and radius . The cylinder has distance more than from the points of , so a tour touching the cylinder has length at least . By creftypecap 2.7, we get

 cost(T)=|Q|∑j=1cost(T(gj0,gj+10))≥(|Q|−1)(δ−2/n3)+2σ/3=10+2σ/3−δ−O(1/n2)>10.5,

which is a contradiction as .

Suppose for the sake of contradiction that are lines touched by between touching and . In particular, the portion of the tour between and contains a path that is disjoint from , and goes from to , but touches on the way. Let us project the lines and the tour into the plane perpendicularly, and denote the projection with . We have , and since is perpendicular to , the path is disjoint from the disk , which is a disk of radius in centered at the origin. Notice that form the three non-adjacent sides of a regular hexagon in centered at the origin, see Figure 5(i). Figure 5: (i) Projection into the plane H:x+y+z=0. (ii) Twelve cones, eight of which covers the line π(ℓ1).

Let be the twelve cones centered at the origin whose boundary contain a midpoint and an endpoint of a hexagon side in cyclic order. We can define the indices so that , see Figure 5(ii). For any choice of we have that intersects both and , therefore it is covered by the eight regions . In general is covered by . We claim that intersects at least two non-adjacent regions among . Indeed, if for some , then is disjoint from at least one of , so it cannot touch . Consequently, does not touch , which contradicts the definition of . Therefore intersects at least two non-adjacent regions.

The distance of two non-adjacent regions and is at least , which implies that . Similarly to the calculation seen above for the cylinder, we get , which is a contradiction. ∎

By Lemma 2.8, we may assume without loss of generality that . Note that whenever and correspond to non-incident edges, (equivalently, when ) then we have that by Corollary 2.2, therefore . If there are unique points in the sequence , then . We add each into a set . We execute the same procedure on each sequence . We claim that the resulting set is a vertex cover of size at most .

The set is a vertex cover as each line is visited by , therefore or appears in the sequence for some subinterval of and therefore or gets added to .

It remains to prove the bound on the size of . We give a lower bound on . If the sequence has a single entry (that is, ), then by creftypecap 2.7. Otherwise, by creftypecap 2.4 we have that and both has cost at least , and if there are unique vertices contributed from to , then by the arguments above . Therefore . Putting the case and together, we get that . Consequently, . Since , we have that , so for large enough we have . ∎

We are now ready to prove the main result of this section.

###### Proof of Theorem 1.1.

Suppose that there is a polynomial time algorithm that approximates TSPN with lines in within a factor of . Let be a given -partite graph. If has a vertex cover of size , then the above construction would have a tour of length . On the other hand, if all vertex covers of have size at least , then all tours of the construction have length at least . As , we could use the hypothetical approximation algorithm to distinguish between these two cases in polynomial time, which would imply . ∎

## 3 No (2−ϵ)-approximation Algorithm

We devote this section to proving Theorem 1.3. In particular, we will show that when the objects are lines, TSPN is at least as hard to approximate as the Vertex Cover problem which is known to be hard to approximate to within a factor of , for any constant , under the Unique Games Conjecture (and inapproximable within a factor of unless  [KMS18]).

###### Theorem 3.1 ([Kr08]).

Unless the Unique Games Conjecture is false, for any constant , there is no polynomial-time algorithm that, given a graph and an integer , distinguishes between the cases (i) has a vertex cover of size at most or (ii) has no vertex cover of size less than .

The main idea behind the reduction is to represent a graph in Euclidean space such that:

• [noitemsep]

• Each vertex corresponds to a point ,

• Each edge corresponds to a line going through the points and ,

• An optimal tour visits each line sufficiently close to the points