Network Optimization on Partitioned Pairs of Points

Given n pairs of points, S = {{p_1, q_1}, {p_2, q_2}, ..., {p_n, q_n}}, in some metric space, we study the problem of two-coloring the points within each pair, red and blue, to optimize the cost of a pair of node-disjoint networks, one over the red points and one over the blue points. In this paper we consider our network structures to be spanning trees, traveling salesman tours or matchings. We consider several different weight functions computed over the network structures induced, as well as several different objective functions. We show that some of these problems are NP-hard, and provide constant factor approximation algorithms in all cases.

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1 Introduction

We study a class of network optimization problems on pairs of sites in a metric space. Our goal is to determine how to split each pair, into a “red” site and a “blue” site, in order to optimize both a network on the red sites and a network on the blue sites. In more detail, given pairs of points, , in the Euclidean plane or in a general metric space, we define a feasible coloring of the points in to be a coloring, , such that if and only if . Among all feasible colorings of , we seek one which optimizes the cost function over a pair of network structures, spanning trees, traveling salesman tours (TSP tours) or matchings, one on the red set and one on the blue set. Let be a certain structure computed on point set and let be the longest edge of a bottleneck structure, , computed on point set . For each of the aforementioned structures we consider the objective of (over all feasible colorings ) minimizing , minimizing and minimizing . Here, denotes the cost (e.g., sum of edge lengths) of the structure.

The problems we study are natural variants of well-studied network optimization problems. Our motivation comes also from a model of secure connectivity in networks involving facilities with replicated data. Consider a set of facilities each having two (or more) replications of their data; the facilities are associated with pairs of points (or -tuples of points in the case of higher levels of replication). Our goal may be to compute two networks (a “red” network and a “blue” network) to interconnect the facilities, each network visiting exactly one data site from each facility; for communication connectivity, we would require each network to be a tree, while for servicing facilities with a mobile agent, we would require each network to be a Hamiltonian path/cycle. By keeping the red and blue networks distinct, a malicious attack within one network is isolated from the other.


Our results.

We show that several of these problems are NP-hard and give O(1)-approximation algorithms for each of them. Table 1 summarizes our O(1)-approximation results.

min min-max min-max
Spanning tree
9
3 for
Matching 2 3 3
TSP tour 18
Table 1: Table of results: is the Steiner ratio and the best approximation factor of the TSP in the underlying metric space. Unless specified otherwise, all other results in this table apply to general metric spaces.

Related work.

Several optimization problems have been studied of the following sort: Given sets of tuples of points (in a Euclidean space or a general metric space), select exactly one point or at least one point from each tuple in order to optimize a specified objective function on the selected set. Gabow et al. [10] explored the problem in which one is given a directed acyclic graph with a source node and a terminal node and a set of pairs of nodes, where the objective was to determine if there exists a path from to that uses at most one node from each pair. Myung et al. [15] introduced the Generalized Minimum Spanning Tree Problem: Given an undirected graph with the nodes partitioned into subsets, compute a minimum spanning tree that uses exactly one point from each subset. They show that this problem is NP-hard and that no constant-factor approximation algorithm exists for this problem unless . Related work addresses the generalized traveling salesperson problem [5, 16, 17, 18], in which a tour must visit one point from each of the given subsets. Arkin et al. [3] studied the problem in which one is given a set and a set of subsets of , and one wants to select at least one element from each subset in order to minimize the diameter of the chosen set. They also considered maximizing the minimum distance between any two elements of the chosen set. In another recent paper, Consuegra et al. [7] consider several problems of this kind. Abellanas et al. [1], Das et al. [8], and Khantemouri et al. [13] considered the following problem. Given colored points in the Euclidean plane, find the smallest region of a certain type (e.g., strip, axis-parallel square, etc.) that encloses at least one point from each color. Barba et al. [4] studied the problem in which one is given a set of colored points (of

different colors) in the Euclidean plane and a vector

, and the goal is to find a region (axis-aligned rectangle, square, disk) that encloses exactly points of color for each . Efficient algorithms are given for deciding whether or not such a region exists for a given .

While optimization problems of the “one of a set” flavor have been studied extensively, the problems we study here are fundamentally different: we care not just about a single structure (e.g., network) that makes the best “one of a set” choices on, say, pairs of points; we must consider also the cost of a second network on the “leftover” points (one from each pair) not chosen. As far as we know, the problem of partitioning points from pairs into two sets in order to optimize objective functions on both sets has not been extensively studied. One recent work of Arkin et al. [2] does address optimizing objectives on both sets: Given a set of pairs of points in the Euclidean plane, color the points red and blue so that if one point of a pair is colored red (resp. blue), the other must be colored blue (resp. red). The objective is to optimize the radii of the minimum enclosing disk of the red points and the minimum enclosing disk of the blue points. They studied the objectives of minimizing the sum of the two radii and minimizing the maximum radius.

2 Spanning Trees

Let be a minimum spanning tree over the point set , and be the cost of the tree, i.e. sum of edge lengths. Let be the longest edge in a bottleneck spanning tree on point set and be the cost of that edge. Given pairs of points in a metric space, find a feasible coloring which minimizes the cost of a pair of spanning trees, one built over each color class.

2.1 Minimum Sum

In this section we consider minimizing .

Theorem 1.

The Min-Sum 2-MST problem is NP-hard in general metric spaces. [The proof is in the appendix.]

An -approximation algorithm for Min-Sum 2-MST problem.
Compute , a minimum spanning tree on all points. Imagine removing the heaviest edge, , from . This leaves us with two trees; and . Perform a preorder traversal on , coloring nodes red as long as there is no conflict. If there is a conflict ( is reached in the traversal and was already colored to red) then color the node blue. Repeat this for . We then return the coloring as our approximate coloring.

  • Case 1: All nodes in are of the same color and all nodes in are of the same color.

    This partition is optimal. To see this, note that the weight of is a lower bound on the cost of the optimal solution as it is the cheapest way to create two trees, the union of which span all of the input nodes. Since each tree is single colored, we know that each tree must have points, exactly one from each pair, and thus is also feasible to our problem.

  • Case 2: One tree is multicolored and the other is not.

    Let be the optimal solution. Suppose without loss of generality that contains only red nodes and contains both blue and red nodes. Then, there must be a pair with both nodes in . Imagine also constructing an MST on each color class of an optimal coloring. By definition, in the MSTs built over each color class, at least one point in must be connected to a point in . This implies that the weight of the optimal solution is at least as large as , as is the cheapest edge which spans the cut . Therefore, .

    Consider . By the Steiner property, we have that an MST over a subset has weight at most where is the Steiner ratio of the metric space. Recall that . In this case, since , we have that .

    Next, consider building . Since no blue node exists in , there does not exist an edge that crosses the cut in , and thus we have that . Therefore, .

  • Case 3: Both trees are multicolored.

    In this case, there are two pairs one with both nodes contained in and one with both nodes contained in . Imagine, again, constructing an MST on each color class in this optimal coloring. In this case, there must be at least two edges crossing the cut , one edge belonging to each tree. Note that each of these edges has weight at least as is the cheapest edge spanning the cut , implying that . Thus, as and .

    Using our approximate coloring, one can compute and , each with weight at most . Therefore , where is again the Steiner ratio of the metric space.

Using the above case analysis, we have the following theorem.

Theorem 2.

There exists a -approximation for the Min-Sum 2-MST problem.

Remark:The Steiner ratio is the supremum of the ratio of length of an minimum spanning tree and a minimum Steiner tree over a point set. In a general metric space and in the Euclidean plane  [12].

2.2 Min-max

In this section the objective is to }.

Theorem 3.

The Min-Max 2-MST problem is strongly NP-hard in general metric spaces.

Proof.

The reduction is from a problem which we will call connected partition [9]. In connected partition one is given a graph , where , and asked if it is possible to remove a set of edges from which breaks it into two connected components each of size .

Given an instance of connected partition, , we will create an instance of min-max 2-MST as follows. For each vertex create an input pair . For each edge set the distance between the corresponding points, and to be one. Set the distances to be zero for all , and the distances to be two for all . In order to complete the construction, set all remaining distances to be the shortest path length among the distances defined above.

Claim: can be partitioned into two connected components of size if and only if there is a solution to the corresponding instance of min-max 2-MST with value .

To show the first direction, suppose that the graph can be split into two connected components, , of size exactly . Without loss of generality suppose and . Then, it is easy to verify based on the pairwise distances in the metric space described above that the coloring , , achieves a cost of .

To show the opposite direction, suppose that there is a solution to the instance of min-max 2-MST of cost . Notice that the minimum distance from point to any other point is at least one; therefore, there can be at most points from the set colored either red or blue in the solution which achieves this cost. Thus there are at least points from the set colored either red or blue in this solution in order for it to be a feasible coloring. This implies that there will be at least one edge crossing the cut in both the red and blue MST which realize the cost of this solution, and this edge has cost two. Then, of the remaining budget of units in order to complete the trees which realize the cost of this solution, it must be the case that we can utilize edges of length one which interconnect exactly nodes from the set in each color class.

The edges of length one in our metric space correspond directly to original edges of the graph in connected partition thus showing that there exists two spanning trees each of which spans exactly nodes of and thus can be partitioned into two connected components of size exactly . ∎

Theorem 4.

There exists a 4-approximation for the Min-Max 2-MST problem.

Proof.

We use the same algorithm as we did for the Min-Sum 2-MST problem. The approximation factor is dominated by case 2 in the Min-Sum 2-MST analysis. For the Min-Max objective function, we have that and that . Thus, . ∎

2.3 Bottleneck

In this section the objective is to .

Lemma 1.

Given pairs of points on a line in where consecutive points on the line are unit separated, there exists a feasible coloring of the points, such that .

Proof.

The proof will be constructive, using Algorithm 1. We partition the points into disjoint buckets, where a bucket consists of two consecutive points on the line.

Color the leftmost point, , red
Let be the point that is in ’s bucket
Let be a set of red points and be a set of blue points
;
while There exists an uncolored point do
       while  is uncolored do
             if  is red then
                   Color ’s pair, , blue
                  
                  
                  
            else
                   Let be the point in ’s bucket
                   Color red
                  
                  
                  
             end if
            
       end while
      Find the leftmost uncolored point and color it red. Let be the point in ’s bucket
       ;
      
end while
return
Algorithm 1 Coloring points on a line.

Observe that at the end of Algorithm 1, each bucket has exactly one red point and one blue point. Thus, the maximum distance between any two points of the same color is 3. ∎

Theorem 5.

There exists a 3-approximation algorithm for the Bottleneck 2-MST problem on a line.

Proof.

Note that if the leftmost points do not contain two points from the same pair, then it is optimal to let be the leftmost points and be the rightmost points. Suppose now that the leftmost points contain two points from the same pair. We run Algorithm 1 on the input. Imagine building two bottleneck spanning trees over the approximate coloring as well as over an optimal coloring. Let be the longest edge (between two points of the same color) in our solution and be the longest edge in the optimal solution.

Consider any two consecutive input points and on the line. We first show that by arguing that the optimal solution must have an edge that covers the interval . Suppose to the contrary that no such edge exists. This means that is connected to points only to its left and is connected to points only to its right. This contradicts the assumption that the leftmost points contain two points from the same pair.

Let the longest edge in our solution be defined by two points, and . Consider the number of input points in interval . Input points in this interval other than and will have a different color than and . It is easy to see that if consists of two input points, that , and if consists of three input points, that . We know by lemma 1 that can consist of no more than four input points. In this last case, must be at least the length of the longest edge of the three edges in . Thus, we see that . ∎

Theorem 6.

There exists a 9-approximation algorithm for the Bottleneck 2-MST problem in a metric space.

Proof.

First, we compute and consider the heaviest edge, . The removal of this edge separates the nodes into two connected components, and . If for and , then we let and and return and . Let be the heaviest edge in the bottleneck spanning trees built on an optimal coloring. Note that lexicographically minimizes the weight of the th heaviest edge, , among all spanning trees over , and thus the weight of the heaviest edge in is a lower bound on . Thus, in this case, our solution is clearly feasible and is also optimal as and are subsets of .

Now suppose . This means that . In this case, we compute a bottleneck TSP tour on the entire point set. It is known that that a bottleneck TSP tour with bottleneck edge can be computed from so that .

Next we run Algorithm 1 on the TSP tour and return two paths, each having the property that the largest edge has weight no larger than . ∎

Remark: Consider the problem of computing a feasible partition which minimizes the bottleneck edge across two bottleneck TSP tours. Let the heaviest edge in the bottleneck TSP tours built on the optimal partition be . The above algorithm gives a 9-approximation to this problem as well because the algorithm returns two Hamilton paths and we know that (using the notation in the above proof) . Thus, .

The following is a generalization of Lemma 1. Let be a set of -tuples of points on a line. Each set , , must be colored with colors. That is, no two points in set can be of the same color.

Consider two consecutive points of the same color, and . We show that there exists a polynomial time algorithm that colors the points in so that the number of input points in interval is .

Lemma 2.

There exists a polynomial time algorithm to color so that for any two consecutive input points of the same color, and , the interval contains at most input points.

Proof.

The algorithm consists of steps, where in the th step, we color of the yet uncolored points with color . We describe the first step.

Divide the points into disjoint buckets, each of size , where the first bucket consists of the leftmost points, the second bucket consists of the points in places , etc. Let be the bipartite graph, with node set , in which there is an edge between and if and only if at least one of ’s points lies in bucket . According to Hall’s theorem [11], there exists a perfect matching in . Let be such a matching and for each edge in , color one of the points in with color 1. Now, remove from each tuple the point that was colored 1, and remove from each bucket the point that was colored 1. In the second step we color a single point in each bucket with the color 2, by again computing a perfect matching between the buckets (now of size ) and the -tuples. It is now easy to see that for any two consecutive points of the same color, and , at most points exist in interval . ∎

3 Matchings

Let be the minimum weight matching on point set and be the cost of the matching. Let be the longest edge in a bottleneck matching on point set and be the cost of that edge edge. Given pairs of points in a metric space, find a feasible coloring which minimizes the cost of a pair of matchings, one built over each color class.

3.1 Minimum Sum

In this section the objective is to minimize .

Theorem 7.

There exists a 2-approximation for the Min-Sum 2-Matching problem in general metric spaces.

Proof.

First, note that the weight of the minimum weight perfect matching on , , which forbids edges for all is a lower bound on . Next, we define the minimum weight one of a pair matching, , to be a minimum weight perfect matching which uses exactly one point from each input pair (that is, a matching using the most advantageous point from each input pair to minimize the total weight of the matching.) Observe that , is a lower bound on the weight of the smaller of the matchings of OPT and therefore has weight at most .

Our algorithm is to compute , and color the points involved in this matching red, and the remainder blue. We return the coloring as our approximate solution.

We have that . To bound , consider the multigraph . All have degree 1 (from ), and all have degree 2 (from and ). For each , either is matched to by , or is matched to by . In the former case we can consider and matched in and charge the weight of this edge to . In the latter case, note that each is part of a unique cycle, or a unique path. If is part of a cycle then no vertex in that cycle belongs to due to the degree constraint. Thus, if is matched to , is part of a unique path whose other terminal vertex belongs to , due to the degree constraint. We can consider , and matched and charge the weight of this edge to the unique path connecting and in . Thus, can be charged to and has weight at most .

Therefore, our partition guarantees . Figure 1 shows the approximation factor using our algorithm is tight. ∎

Figure 1:

3.2 Min-max

In this section the objective is to .

Theorem 8.

The Min-Max 2-Matching problem is weakly NP-hard in the Euclidean plane.

Proof.

The reduction is from Partition: given a set of integers, decide if there exists a partition , with . Let . Given any instance of Partition, we create a geometric instance of the Min-Max 2-Matching problem, as shown in Figure 2.

We place point pairs along two -separated horizontal lines, such that are vertically adjacent, with horizontal separation of between consecutive pairs. Then, for each in the instance of Partition we place a point at distance from , and its corresponding pair at distance from both and .

Notice that any solution which minimizes the weight of the larger matching created only uses edges between points of the same “cluster” . Any edge between two clusters , costs at least and if matching edges are chosen within clusters the entire matching can be constructed with cost at most for chosen small enough.

Within each cluster an assignment will have to be made, that is, without loss of generality, or . Therefore, any algorithm that minimizes the maximum weight of either matching also minimizes across all partitions . Thus, for chosen small enough the instance of partition is solvable if and only if the weight of the larger matching created is at most .

Figure 2: Set up of the Min-Max 2-Matching instance given an instance of Partition: .

Remark: The above reduction can also be used to show that the Min-Max 2-MST problem is weakly NP-hard in the Euclidean plane. Given an instance of partition we create the exact same instance for Min-Max 2-MST as described above, and note that there exists a solution for partition if and only if there is a solution to Min-Max 2-MST with value .

Theorem 9.

The approximation algorithm for the Min-Sum 2-Matching problem serves as a 3-approximation for the Min-Max 2-Matching problem in general metric spaces. [The proof is in the appendix.]

3.3 Bottleneck

In this section the objective is to .

Theorem 10.

There exists a 3-approximation to the Bottleneck 2-Matching problem in general metric spaces. [The proof is in the appendix]

4 TSP Tours

Let be a TSP tour on point set and be the cost of the tour. Let be the longest edge in a bottleneck TSP tour on point set and be the cost of that longest edge. Given pairs of points in a metric space, find a feasible coloring which minimizes the cost of a pair of TSP tours, one built over each color class.

It is interesting to note the complexity difference emerging here. In prior sections, the structures to be computed on each color class of a feasible coloring were computable exactly in polynomial time. Thus, the decision versions of these problems, which ask if there exists a feasible coloring such that some cost function over the pair of structures is at most , are easily seen to be in NP. However, when the cost function is over a set of TSP tours or bottleneck TSP tours, this is no longer the case. That is, suppose that a non-deterministic Touring machine could in polynomial time, for a point set and , return a coloring for which it claimed the cost of the TSP tours generated over both color classes is at most . Unless , the verifier cannot in polynomial time confirm that this is a valid solution, and therefore the problem is not in NP. Thus, the problems considered in this section are all NP-hard.

4.1 Minimum Sum

In this section the objective is to minimize .

Let denote a -factor approximate TSP tour on set .
  1. Compute .

  2. Let be the largest even number not exceeding . Enumerate all ways of decomposing into connected components: for each decomposition, color the nodes from consecutive components red and blue alternately (i.e. color all nodes in component one red, all nodes in component two blue, etc.). If this coloring is infeasible, then skip to the next decomposition; otherwise compute and

  3. Compute a random feasible coloring, , and compute and

  4. Among all pairs of tours produced in steps 2 and 3, choose the pair of minimum sum.

Algorithm 2 Algorithm . and .

We will show for and for the proper choice of , that Algorithm 2 gives a -approximation for the Min-Sum 2-TSP problem. Fix a constant . Let be the optimal (feasible) coloring . Let be the minimum point-wise distance between sets and . We call an instance of the problem -separable if there exists a feasible coloring

Let be the coloring returned by our algorithm. We will show that if is not -separable, then (see Lemma 3) and that if is -separable, then (see Lemma 4). Supposing both of these are true, then the approximation factor of our algorithm is . One can easily verify that is the minimizer which gives the desired factor. The following lemma states that if is not -separable, then any feasible coloring yields a “good” approximation.

Lemma 3.

If is not -separable, then .

Proof.

If is not -separable, then for any feasible coloring we have . In particular, for the coloring induced by the optimal solution, , Then,

Hence, when , . Let be the random feasible coloring computed by . Then, as we are returning the best coloring between and all colorings of , we have . ∎

The following lemma states that if is -separable, then any witness coloring to the -separability of gives a “good” approximation.

Lemma 4.

If is -separable, then .

Proof.

Suppose we successfully guessed a coloring that is a “witness” to the -separability of (we will show how to guess later).

  • Case 1: Then .

  • Case 2: Then which means each tour in OPT must contain at least 2 edges crossing the cut (), hence the optimal solution must contain at least 4 edges crossing the cut (). So Equivalently,

The next two lemmas show how to guess a witness coloring in polynomial time. First, we show that if is -separable with a witness coloring , then cannot cross the red/blue cut defined by this coloring “too many” times.

Lemma 5.

Let be an -factor approximation for Also, suppose is -separable with witness Then crosses the cut , at most times.

Proof.

One can construct a TSP tour for by adding two bridges to and , thus we have Also, suppose crosses the cut (, ) times. Then, Combining the above two inequalities, we obtain . ∎

The next lemma completes our proof.

Lemma 6.

Suppose is -separable. Let be any coloring which serves as a “witness”. Then, in step 2 of , we will encounter at some stage.

Proof.

Given a nonnegative integer and a TSP tour , define {: is a feasible coloring and crosses at most times}. By Lemma 5, we know Since step 2 of is actually enumerating all colorings in this completes the proof. ∎

Note that step 2 considers decompositions and for each coloring that is feasible, we compute two approximate TSP tours. Suppose the running time to compute a -factor TSP tour on points is . Then the worst case running time of Algorithm 2 is . Thus, we have the following Theorem.

Theorem 11.

For any , the algorithm is a -approximation for the Min-Sum 2-TSP problem with running time .

Remark: If is in the Euclidean plane then for some [14] yielding a -approximation and if is in a general metric space then [6] yielding a 4.5-approximation. In both cases is polynomial.

4.2 Min-Max

In this section the objective is to .

Theorem 12.

There exists a -approximation to the Min-Max 2-TSP problem, where is the approximation factor for TSP in a certain metric space. [The proof is in the appendix.]

4.3 Bottleneck

In this section the objective is to .

Theorem 13.

There exists an 18-approximation algorithm for the Bottleneck 2-TSP problem. [The proof is in the appendix.]

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Appendix A Additional Proofs

Theorem 1. The Min-Sum 2-MST problem is NP-Hard in general metric spaces.

Proof.

The reduction is from Max 2SAT where one is given variables and clauses . Each clause contains at most two literals joined by a logical or. The objective is to maximize the number of clauses that evaluate to true.

For each variable we create a variable gadget that consists of two pairs of points: and (see Figure 3). Setting to true is equivalent to using edges and . Setting to false is equivalent to using edges and . Variable gadgets will be arranged on a line with distance between consecutive variable gadgets for (see Figure 4).

For every pair of variable gadgets corresponding to variables we place a cluster of points near point . Each of these points is paired to a point in a cluster of points near point . Any two points in the same cluster, or , are separated by distance two from each other and by distance one from point respectively. Note that this enforces points and to be in different trees for all . Otherwise, if and were placed in the same tree, then connecting the points in clusters to the trees would cost at least more than it would to have and in different trees.

Now we argue that the optimal solution uses edges and , , as “backbones” of the two MSTs. To see this, observe that if any other edge was used to connect two consecutive variable gadgets, then we would need to use at least one edge of length . Since and will be in different trees for all and since points and will be connected to points and (), we have a set of “lower” components that must be connected and a set of “upper” components that need to be connected. No upper component can be connected to a lower component. Any edge of length at least connecting any of these components can thus be replaced by an edge of length .

The remaining variable gadget points, {} (), must be connected to the backbones. That is, for variable , points and will be picked up either by using edges and (green in Figure 5) or edges and (red in Figure 5). As mentioned, the green edges correspond to setting to true and the red edges correspond to setting to false.

Figure 3: Variable gadget.
Figure 4: Metric distances between variable gadgets.
Figure 5: Truth assignment.

A clause gadget consists of a configuration of 3 point pairs surrounding variable gadgets corresponding to the variables in that clause (see Figure 6). The placement of the 3 point pairs depends on whether the literals appear positively or negatively.

Consider clause which consists of variables and . We create a pair of points , each of which will be placed next to variable . If appears negatively in , then we place at distance 1 away from an endpoint of a green edge of and place at distance 1 away from the other endpoint of the green edge (see Figure 6c). If appears positively in , then we do the same thing at the endpoints of a red edge (see Figure 6a). Then, we create a pair of points and follow the same procedure for variable . Finally, for each clause gadget we create a pair of points and place them at a distance of 1 from certain variable gadget points chosen based on how many literals appear positively in clause ; zero, one, or two. The placement of for all three cases can be found in Figure 6.

As a technical note, to complete the construction, all clause gadget points placed around the same variable gadget vertex are separated from each other by distance 2, and these points are at distance only 1 from the nearest variable gadget vertex. This will ensure that the optimal solution will not link any two clause gadget points to each other. Also, note that we use the shortest path distance to define the weights of the rest of the edges in the graph.

Connecting all points except those associated with clause gadgets into two MSTs has a base cost of . Now note that a clause evaluates to true if and only if it costs 7 units to attach the clause gadget points to the backbones. A clause evaluates to false if and only if it costs 9 units to attach the clause gadget points to the backbones (see Figure 6). Thus, it is now apparent that there exists a truth assignment in 2SAT with clauses satisfied if and only if there exists a solution to the Min-Sum 2-MST problem with cost .

Figure 6: Placement of clause gadget points and extra cost incurred to incorporate clause gadget points into two MSTs once a truth assignment over the variables is fixed.

Theorem 9. The approximation algorithm for the Min-Sum 2-Matching problem serves as a 3-approximation for the Min-Max 2-Matching problem in general metric spaces.

Proof.

In this case we are concerned only with the larger of the two matchings returned by our approximation, , which, as described above, has weight bounded above by . However, in this case, under the new cost function , and . Therefore, The example illustrated in Figure 1 shows the approximation factor achieved by this algorithm is tight. ∎

Theorem 10. There exists a 3-approximation to the Bottleneck 2-Matching Problem in general metric spaces.[See section B of the appendix.]

Theorem 12. There exists a -approximation to the Min-Max 2-TSP problem, where is the approximation factor for TSP in a certain metric space.

Proof.

We use the same algorithm, and return the same coloring, , as in Section 4.1. Let be the coloring returned, and be the cost of the larger TSP on both sets of . Let be the optimal solution. Note that as for any . ∎

Theorem 13. There exists an 18-approximation algorithm for the Bottleneck 2-TSP problem.

Proof.

We remarked in section 2.3 that there exists a 9-approximation to the problem of finding a partition that minimizes the weight of the bottleneck edge on two Hamilton paths built on the partition. A Hamilton path can be converted into a Hamilton cycle by at most doubling the weight of the bottleneck edge in the Hamilton path. This yields an 18-approximation to the Bottleneck 2-TSP problem. ∎

Appendix B Bottleneck 2-Matching

We begin with a lemma concerning the structure of a feasible solution. Let be any feasible coloring to the Bottleneck 2-Matching instance. Let be a minimum bottleneck matching on point set . Construct a graph where and is the union of any pair of optimal bottleneck matchings on , and and the edges .

Lemma 7.

is a 2-factor such that each input pair is contained in exactly one cycle, and each cycle contains an even number of input pairs.

Proof.

The edge set of is the union of two disjoint perfect matchings over . Therefore, each node has degree exactly 2 and is by definition a 2-factor. Also, by definition of a 2-factor, each input point is part of a unique cycle, and in this case, as each node has an edge of the form incident to it, therefore, the point must be contained in the same cycle as . Thus, each input pair is contained in the same unique cycle in .

By definition, two nodes defining an edge of must be of the same color, and two nodes defining an edge of the form must be of different color. This together with the fact that the edges of this cycle alternate between the form and edges in implies that if we were to contract edges of the form we would still get a cycle which strictly alternates color. This is only possible if there are the same number of red and blue nodes in the contracted cycle. Thus an even number of input pairs. ∎

Using the above structure lemma we will argue that we can compute a graph with the same properties and extract a feasible partition with constant factor approximation guarantees. Let be the minimum weight (exactly) one of a pair bottleneck matching over ; note that edges of this matching go between points of . Let be the minimum weight bottleneck matching over (excluding the edges ). Let (resp. ) be the heaviest edge used in (resp. ) and let be the heaviest edge in a minimum weight bottleneck matching computed over each of the two sets in . Note that and .

Begin by constructing a graph , which is a 2-factor as its edge set is the union of two disjoint perfect matchings. Note, it will be the case that each input pair exists in the same unique cycle. If each cycle contains an even number of input pairs then this graph has the same structure as that described in Lemma 7 and thus we can extract a feasible partition from . We will describe how to obtain this partition later. As

, this graph induces an optimal partition. On the other hand, if there exists a cycle with an odd number of input pairs (there must be an even number of such cycles) then we “merge” cycles of

together into larger cycles until a point is reached in which each new “super-cycle” contains an even number of pairs. From this graph we can extract a constant factor approximation to an optimal coloring.

Lemma 8.

If contains at least one cycle with an odd number of input pairs, then it is possible to merge cycles of into super-cycles, each of which contains an even number of input pairs, such that the heaviest edge excluding in any super-cycle has weight at most .

Proof.

(sketch) Superimpose a subset of the edges in over the nodes of in the following way. Consider only edges in which have endpoints in different cycles. Treat each cycle in as a node and run Kruskal’s algorithm until all of the aforementioned edges of are exhausted. This yields a forrest on the cycles of . It is easy to see that every cycle of containing an odd number of input pairs has an edge of connecting it to some other cycle. This implies that it is possible to merge all cycles which are connected by an edge of