1 Introduction
1.1 Motivation
In wireless network routing, a common approach is to select a set of nodes as the virtual backbone. The virtual backbone is responsible for relaying packets. Specifically, when a node generates a packet destined to , the packet is routed through path , where every internal node belongs to the virtual backbone. To realize this idea, we can model the wireless network as a graph , where is the set of nodes in the wireless network, and if and only if and can communicate with each other directly. Thus, a connected dominating set of is a virtual backbone for the wireless network.^{1}^{1}1A set is a dominating set of if every vertex in is adjacent to . Furthermore, if induces a connected subgraph of , then is called a connected dominating set of . One of the concerns in constructing the virtual backbone is the routing cost. Specifically, the routing cost of sending a packet from the source to the destination is the number of internal nodes (relays) in the routing path from to . For example, the routing cost is if the routing path is . The routing cost should not be too high even if packets are only allowed to be routed through the virtual backbone. Next, we give the formal definition of the problem.
1.2 Problem Definition
Let be the subgraph of induced by . Let be the number of internal vertices on the shortest path between and in . For example, if and are adjacent, then . If and are not adjacent and have a common neighbor, then . Furthermore, given a vertex subset of , is defined as , i.e., the number of internal vertices on the shortest path between and through . We use to denote the number of vertices in graph . When the graph we are referring to is clear from the context, we simply write , , and instead of , , and , respectively.
Definition 1.
Given a connected graph and a positive integer , the Connected Dominating set problem with Routing cost constraint (CDR) asks for the smallest connected dominating set of , such that, for every two vertices and , if and are not adjacent in , then .
1.3 Preliminary
1.3.1 An Equivalent Problem
In the CDR problem, we need to consider all the pairs of nonadjacent nodes. Ding et al. discovered that to solve the CDR problem, it suffices to consider only vertex pairs such that , i.e., and are not adjacent but have a common neighbor [5]. We call the corresponding problem the 1DR problem.
Definition 2.
Given a connected graph and a positive integer , the 1DR problem asks for the smallest dominating set of , such that, for every two vertices and , if , then .
We say that and form a target couple, denoted by , if . We say that a set covers a target couple if . Hence, the 1DR problem asks for the smallest dominating set that covers all the target couples. Notice that any feasible solution of the 1DR problem must induce a connected subgraph of . The equivalence between the CDR problem and the 1DR problem is stated in the following theorem.
Theorem 1 (Ding et al. [5]).
is a feasible solution of the CDR problem with input graph if and only if is a feasible solution of the 1DR problem with input graph .
Corollary 1.
Any approximation algorithm of the 1DR problem is an approximation algorithm of the CDR problem.
In this paper, we thus focus on the 1DR problem.
1.3.2 Feasibility of the 1DR Problem for
Next, we give the basic idea of finding a feasible solution of the 1DR problem for used in previous researches, e.g., in [11]. One of our algorithms still uses this idea. First, find a dominating set . Thus, for any target couple , there exist and in , such that and dominate and , respectively.^{2}^{2}2 dominates if or and are adjacent. Let . For any two vertices and in , if , then we add the internal vertices of the shortest path between and on to . Observe that . Hence, and is a feasible solution of the 1DR problem for .
Lemma 1.
Let be a dominating set of . Let be a vertex subset of such that, for any two vertices and in , if , then . Then, is a feasible solution of the 1DR problem with input and .
1.4 Previous Result
Previous result on general graphs
When , the 1DR problem can be transformed to the set cover problem, i.e., cover all the vertices (to form a dominating set) and cover all the target couples. Observe that each target couple can be covered by a single vertex. The resulting approximation ratio is [5]. When is sufficiently large, e.g., , any connected dominating set is feasible for the CDR problem. Note that, for any , the size of the minimum connected dominating set is a lower bound of the CDR problem. Since the connected dominating set can be approximated within a factor of [12, 21], the CDR problem can be approximated within a factor of . If falls between these two extremes, e.g., , the only known previous result is the trivial approximation algorithm. On the hardness side, it has been proved that, unless , there is no polynomialtime algorithm that can approximate the CDR problem within a factor of () for [5] and [8, 10], where is the maximum degree of .
Open Question 1 (Du and Wan [8]).
Is there a polynomialtime approximation algorithm for the CDR problem for ?
Previous result on Unit Disk Graph (UDG)
Most of the studies on the CDR problem focused on UDG [5, 8, 10, 11, 19]. UDG exhibits many nice properties that enable constant factor approximation algorithms (or PTAS) in many problems where only approximation algorithms (or worse) are known in general graphs, e.g., the minimum (connected) dominating set problem and the maximum independent set problem [3, 4, 20]. All the previous research on the CDR problem on UDG leveraged constant bounds of the maximum independent set or the minimum dominating set. However, all the previous research only solved the case where (by Lemma 1), and the best result so far is a PTAS by Du et al. [11]. When , the only known previous result is the trivial approximation algorithm.
1.5 Our Result and Basic Ideas
In this paper, we first give an approximation algorithm of the 1DR problem on general graphs for any constant . A critical observation is that the 1DR problem is a special case of the Set Cover with Pairs (SCP) problem [13]. Hassin and Segev proposed an approximation algorithm for the SCP problem, where is the number of targets to be covered. However, since there are target couples to be covered, directly applying the approximation bound yields a trivial upper bound for the 1DR2 problem. We reexamine the analysis in [13] and find that, when applying the algorithm to the 1DR2 problem, the approximation ratio can also be expressed as . Nevertheless, in this paper, we give a slightly simplified algorithm with an easier analysis for the SCP problem. The algorithm and analysis also make it easy to solve the generalized SCP problem. We obtain the following result, which is the first nontrivial result of the CDR problem for on general graphs and for on UDG.
Theorem 2.
For any constant , there is an approximation algorithm for the 1DR problem.
Apparently, the above performance guarantee deteriorates quickly as increases. In our second algorithm, we apply the aforementioned idea of finding a feasible solution when , i.e., Lemma 1. We have the following result.
Theorem 3.
When , there is an approximation algorithm for the 1DR problem.
Finally, we answer Open Question 1 negatively. We improve upon the hardness result for the 1DR2 problem (albeit under a stronger hardness assumption) [8, 10]. In this paper, we give a reduction from the MINREP problem [15].
Theorem 4.
Unless , for any constant , the 1DR2 problem admits no polynomialtime approximation algorithm, even if the graph is trianglefree^{3}^{3}3If the graph is trianglefree, then any two vertices with a common neighbor form a target couple. or the constraint that the feasible solution must be a dominating set is ignored^{4}^{4}4One may drop the constraint that the solution must be a dominating set, and focuses on minimizing the number of vertices to cover all the target couples. This theorem also applies to such a problem..
1.6 Relation with the Basic Spanner Problem
When we ignore the constraint that any feasible solution must be a connected dominating set, the CDR problem is similar to the basic spanner problem. For completeness, we give the formal definition of the basic spanner problem. Given a graph , a spanner of is a subgraph of such that for all and in , where is the number of edges in the shortest path between and in . The basic spanner problem asks for the spanner that has the fewest edges. The CDR problem differs with the basic spanner problem in the following three aspects: First, in the CDR problem, we find a set of vertices , and all the edges in the subgraph induced by can be used for routing; while in the basic spanner problem, only edges in can be used. Second, in the CDR problem, the objective is to minimize the number of chosen vertices; while in the basic spanner problem, the objective is to minimize the number of chosen edges. Finally, in the basic spanner problem, the distance is measured by the number of edges; while in the CDR problem, the distance is measured by the number of internal nodes. Despite the above differences, these two problems share similar approximability and hardness results. Althöfer et al. proved that every graph has a spanner of at most edges, and such a spanner can be constructed in polynomial time [1, 7]. Since the number of edges in any spanner is at least , this yields an approximation algorithm for the basic spanner problem. For , there is an approximation algorithm due to Kortsarz and Peleg [17], and this is the best possible [15]. For , Berman et al. proposed an approximation algorithm [2]. For , Dinitz and Zhang proposed an approximation algorithm [7]. On the hardness side, it has been proved that for any constant and for , unless , there is no polynomialtime algorithm that approximates the basic spanner problem to a factor better than [6].
2 Two Algorithms for the 1DR Problem
2.1 The First Algorithm
We first give the formal definition of the Set Cover with Pairs (SCP) problem.
Definition 3.
Let be a set of targets. Let be a set of elements. For every pair of elements , denotes the set of targets covered by . The Set Cover with Pairs (SCP) problem asks for the smallest subset of such that .
Let be the number of elements in the optimal solution. We only need to consider the case where and .
2.1.1 Approximating the SCP Problem
Our algorithm is a simple greedy algorithm: in each round, we choose at most two elements and that maximize the number of covered targets. Specifically, is an empty set initially. In each round, we select a set such that and increases the number of covered targets the most, i.e., , where
We then add to and repeat the above process until all the targets are covered. The algorithm terminates once all targets are covered.^{5}^{5}5In [13], in each round, a set is added to , where .
Theorem 5.
The above algorithm is an approximation algorithm for the SCP problem.
Proof.
Let be the number of uncovered targets after round . In the first round, some pair of elements in the optimal solution can cover at least targets. Since we choose a pair of elements greedily in each round, . In the second round, there exists a pair of elements in the optimal solution that can cover at least targets among the uncovered targets. Again, we choose the pair of elements that increases the number of covered targets the most. Hence, . In general, . After rounds, the number of uncovered targets is at most . Hence, after rounds, all targets are covered. Let be the number of elements chosen by the algorithm. Since we choose at most two elements in each round, . Finally, since , . ∎
Note that, in Theorem 5, we can replace with any upper bound of the size of solutions obtained by any polynomialtime algorithm for the SCP problem. This is achieved by executing both and our algorithm. Choosing the best between the two outputs yields the desired approximation ratio. An example is replacing with .
2.1.2 Approximating the 1DR2 Problem
To transform the 1DR2 problem to the SCP problem, we treat each target couple as a target. Moreover, we treat each vertex as a target so that the output is a dominating set. The set of elements in the SCP problem is the vertex set of . consists of all the vertices that are dominated by in and all the target couples that are covered by in . In this SCP instance, and . It is easy to verify the following result.
Theorem 6.
There is an approximation algorithm for the 1DR2 problem.
2.1.3 The Set Cover with Tuples (SCT) Problem
In the 1DR2 problem, every target couple can be covered by no more than two vertices. In the 1DR problem, every target couple can be covered by no more than vertices. Hence, we consider the following generalization of the SCP problem.
Definition 4.
Let be a set of targets. Let be a set of elements. Let be a positive integer constant greater than one. For every tuple , denotes the set of targets covered by . The Set Cover with Tuples (SCT) problem asks for the smallest subset of such that .
We only need to consider the case where and ( is a constant).
2.1.4 Approximating the SCT Problem and the 1DR Problem
The algorithm for the SCT problem is a straightforward generalization of the algorithm for the SCP problem. The difference is that, in each round, we choose a set of at most elements that increases the number of covered targets the most. The transformation from the 1DR problem to the SCT problem is also similar to the previous transformation. The value of in the constructed SCT instance is equal to that in the 1DR instance. Again, and in the constructed SCT instance. Theorem 2 is a direct result of the following theorem.
Theorem 7.
There is an approximation algorithm for the SCT problem.
We have the following claim, whose proof can be found in the appendix.
Claim 1.
When , .
Proof of Theorem 7: Let be the number of uncovered targets after round . By a similar argument in the proof of Theorem 5, we get that . After rounds, the number of uncovered targets is at most one. Hence, after rounds, all targets are covered. Let be the number of elements chosen by the algorithm. Since we choose at most elements in each round, . Since , .
Let . When , the approximation ratio is . When , . The proof then follows from Claim 1 and . ∎
2.2 The Second Algorithm
The second algorithm is designed for the 1DR problem when . It has a better approximation ratio than that of the previous algorithm when . The algorithm is suggested in Lemma 1: We first find a dominating set by any approximation algorithm. Let . For any two vertices and in , if , we then add at most three vertices to so that .
Proof of Theorem 3: Let be the size of the minimum dominating set in . Let be the size of the optimum of the 1DR problem. Since any feasible solution of the 1DR problem must be a dominating set, . . Since , we have . ∎
3 Inapproximability Result
3.1 The MINREP Problem
We prove Theorem 4 by a reduction from the MINREP problem [15]. The input of the MINREP problem consists of a bipartite graph , a partition of , , and a partition of , , such that and . Every (respectively, ) has size (respectively, ). and are called super nodes, and two super nodes and are adjacent if some vertex in and some vertex in are adjacent in . If and are adjacent, then and form a super edge. In the MINREP problem, our task is to choose representatives for super nodes so that if and form a super edge, then some representative for and some representative for are adjacent in . Note that a super node may have multiple representatives. Specifically, the goal of the MINREP problem is to find the smallest subset such that if and form a super edge, then must contain two vertices and such that , and . In this case, we say that covers the super edge . The inapproximability result of the MINREP problem is stated as the following theorem.
Theorem 8 (Kortsarz et al. [16]).
For any constant , unless , there is no polynomialtime algorithm that can distinguish between instances of the MINREP problem with a solution of size and instances where every solution is of size at least , where is the number of vertices in the input graph of the MINREP problem.
3.2 The Reduction
Given inputs , , and of the MINREP problem, we construct a corresponding graph of the 1DR2 problem. When , , and are clear from the context, we simply write instead of . Initially, . Hence, contains , , and . For each super node (respectively, ), we create two corresponding vertices and (respectively, and ) in . If is in super node (respectively, is in super node ), then we add two edges and (respectively, and ) in . If and form a super edge, then we add two vertices and to , and we add four edges , , , to . (respectively, ) is called the relay of and (respectively, and ).
Before we complete the construction of , we briefly explain the idea behind the construction so far. If two super nodes and form a super edge, then and () have a common neighbor in , i.e., the relay . Because and are not adjacent, and form a target couple. To transform a solution of the 1DR2 problem to a solution of the MINREP problem, we need to transform to another feasible solution for the 1DR2 problem so that none of the relays is chosen, and only vertices in are used to connect and . This is the reason that we have two corresponding vertices for each super node (and thus two relays for each super edge). Under this setting, to connect to and to , choosing two vertices in is no worse than choosing the relays.
Let be the set of vertices in corresponding to the super nodes in . Similarly, let . Let be the set of all relays. To complete the construction, we add four vertices (hubs) , , , and to . In , all the vertices in , , , and are adjacent to , , , and , respectively. Moreover, every relay is adjacent to and . These four hubs induce a 4cycle in . Finally, for each hub , we create two dummy nodes and , and add two edges and to . This completes the construction of . Fig. 1 shows an example of the reduction. Let and be the set of hubs and the set of dummy nodes, respectively. Hence, the vertex set of is . Let be the set of neighbors of in . We then have
Observe that . We have the following lemma.
Lemma 2.
.
It is easy to check that, for any two adjacent vertices and in , and have no common neighbor. Hence, we have the following lemma.
Lemma 3.
is trianglefree.
We say that a target couple is in if and . It is easy to verify the following two lemmas.
Lemma 4.
Only can cover the target couples in .
Lemma 5.
is a dominating set of .
The proof of the following lemma can be found in the appendix.
Lemma 6.
covers all the target couples except those in .
Let and be vertices in and , respectively. Observe that, if is a path in , then and . We then have the following lemma.
Lemma 7.
covers target couples and if and only if at least one of the following conditions is satisfied.

There exist and such that and are paths in and .

.
3.3 The Analysis
Let be an instance of the MINREP problem with inputs , , and . Let be the instance of the 1DR2 problem with input . To prove the inapproximability result, we use the following two lemmas.
Lemma 8.
If has a solution of size , then has a solution of size .
Lemma 9.
If every solution of has size at least , then every solution of has size at least .
Proof of Theorem 4: By Theorem 8, for any constant , unless , there is no polynomialtime algorithm that can distinguish between instances of the MINREP problem with a solution of size and instances where every solution is of size at least . By the above two lemmas, it is hard to distinguish between instances of the 1DR2 problem with a solution of size and instances in which every solution is of size at least . Therefore, for any constant , unless , there is no polynomialtime algorithm that can approximate the 1DR2 problem by a factor better than . Lemma 2 implies that, for any constant , unless , there is no approximation algorithm for the 1DR2 problem. By considering sufficiently large instances and a small enough , we have the hardness result claimed in Theorem 4. On the other hand, let 1DR be the problem obtained by removing the constraint that any feasible solution must be a dominating set from the 1DR2 problem. Thus, in the 1DR problem, we only focus on covering target couples. By Lemmas 4 and 5, a solution is feasible for the 1DR problem with input if and only if is a feasible solution of . Thus, the inapproximability result also applies to the 1DR problem. Finally, the proof follows from Lemma 3. ∎
Lemma 8 is a direct result of the following claim.
Claim 2.
If is a feasible solution of , then is a feasible solution of .
Proof.
Since is a dominating set, by Lemma 6, it suffices to prove that every target couple in is covered by . Note that cannot be a target couple if . This is because and do not have a common neighbor if . If , then the common neighbor must be . By the construction of , this implies that and form a super edge. Since is a feasible solution of , there exists and such that and are adjacent in and . Again, by the construction of , is a path in . Hence, covers . ∎
To prove Lemma 9, we use the following claim.
Claim 3.
has an optimal solution , such that is a feasible solution of .
Proof of Lemma 9: Let be the optimal solution of . By the assumption, we have . It suffices to prove that is an optimal solution for , which implies that every feasible solution of has size at least . The feasibility of follows from Claim 2. For the sake of contradiction, assume that the optimal solution of has size smaller than . Claim 3 and Lemma 4 then imply that is not an optimal solution of , which is a contradiction. ∎
Proof of Claim 3: Let be any optimal solution of . By Lemmas 4, 6, and 7, . If , by Lemma 7, each target couple is covered by some and some . By the construction of , such and also cover the super edge in . Because each super edge in has a corresponding target couple in , is a feasible solution of .
If , then some . We can further assume that both and are in ; otherwise, by Lemma 7, we can remove from , the resulting solution is smaller and is still feasible. We then replace and with some and some satisfying the first condition in Lemma 7. By Lemma 7, the resulting solution is still feasible, and the size remains the same. Repeat the above replacing process until the resulting solution does not contain any relay. The proof then follows from the argument of the case where . ∎
4 Transforming the 1DR Problem to Other Related Problems
Submodular Cost Set Cover Problem: The 1DR problem can also be considered as a special case of the submodular cost set cover problem [9, 14, 23]. In the set cover problem, we are given a set of targets and a set of objects . Each object in can cover a subset of (specified in the input). The goal is to choose the smallest subset of that covers . In the submodular cost set cover problem, there is a nonnegative submodular function that maps each subset of to a cost, and the goal is to find the set cover with the minimum cost. To transform the 1DR problem with input to the submodular cost set cover problem, let be the union of and the set of all target couples, and let be the set of all subsets of with size at most . Hence, each object in is a subset of . An object can cover a vertex if is adjacent to some vertex in or . An object can cover a target couple if . The cost of a subset of is simply the size of the union of objects in , i.e., the number of distinct vertices specified in .
Iwata and Nagano proposed a approximation algorithm and an approximation algorithm, where is the maximum frequency, [14]. Koufogiannakis and Young also proposed an approximation algorithm when the cost function is nondecreasing [18]. It is easy to see that these algorithms give trivial bounds for the 1DR problem. When the cost function is integervalued, nondecreasing, and satisfies , Wan et al. proposed a approximation algorithm, where , is the largest number of targets that can be covered by an object in , and is the th Harmonic number [23]. Du et al. applied this algorithm to the 1DR problem on UDG for and obtained a constant factor approximation algorithm [9]. It is unclear whether or not can be upper bounded by for some when applied to the 1DR problem on general graphs.
Minimum Rainbow Subgraph Problem on Multigraphs: Given a set of colors and a multigraph , where each edge is colored with one of the colors, the Minimum Rainbow Subgraph (MRS) problem asks for the smallest vertex subset of , such that each of the colors appears in some edge induced by . The 1DR2 problem can be transformed to the MRS problem as follows. Let be the input graph of the 1DR2 problem. Let be the union of and the set of all target couples. The set of colors for the MRS problem is . The input multigraph of the MRS problem has the same vertex set as . To form a dominating set, for each , is incident to loops in , where is the degree of in . Each of these loops receives a different color in . For each target couple in , if is a common neighbor of and in , we add a loop with color to . Finally, for each target couple in , if is a path in , we add an edge with color to . The MRS problem can be transformed to the SCP problem. When the input graph is simple, Tirodkar and Vishwanathan proposed an approximation algorithm [22].
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Appendix A Proof of Claim 1
When ,
(1)  
(2)  
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