# r-Gather Clustering and r-Gathering on Spider: FPT Algorithms and Hardness

We consider min-max r-gather clustering problem and min-max r-gathering problem. In the min-max r-gather clustering problem, we are given a set of users and divide them into clusters with size at least r; the goal is to minimize the maximum diameter of clusters. In the min-max r-gathering problem, we are additionally given a set of facilities and assign each cluster to a facility; the goal is to minimize the maximum distance between the users and the assigned facility. In this study, we consider the case that the users and facilities are located on a "spider" and propose the first fixed-parameter tractable (FPT) algorithms for both problems, which are parametrized by only the number of legs. Furthermore, we prove that these problems are NP-hard when the number of legs is arbitrarily large.

## Authors

• 4 publications
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## 1 Introduction

### 1.1 Background and Motivation

We consider min-max -gather clustering problem and min-max -gathering problem. In the min-max -gather clustering problem, we are given a metric space and users located on . The goal of the problem is to find a partition of the users such that each cluster contains at least users, and the maximum diameter of the clusters is minimized111Several variations of -gather clustering problem are known. At first appearance, it is the special case of -gathering problem, where the set of locations of the users are the subset of the set of locations of the facilities [1]. In recent researches on the restricted metrics, the same definition as this paper is often used. If the metric is a subclass of the tree, that corresponds to the situation, where we can open facilities at any points on the metric. As the clustering problem, this is a more natural way to define the cost, since we do not need the set of facilities as an input. Thus we use this definition.  [3, 7, 2]. In the min-max -gathering problem, we are additionally given facilities on . The goal of the problem is to open a subset of facilities and assign each user to an opened facility so that all opened facilities have at least users and the maximum distance from users to the assigned facilities is minimized [5].

Both problems have plenty of practical applications. One typical application is a privacy protection [9]: Consider a company that publishes clustered data about their customers. If there is a too small cluster, one can easily identify the individuals in a such cluster. To guarantee the anonymity, we require the clusters to have at least individuals, which is obtained via -gather clustering problem. Another example is the formation of sport-teams. Suppose that we want to divide people into several teams for a football game. Then, each team must contain at least eleven people. Such grouping can be obtained via -gather clustering problem.

Several tractability and intractability results are known for both problems. If is a general metric space, there is a -approximation algorithm for -gathering problem, and no algorithm can achieve a better approximation ratio unless P=NP [5]. If the set of locations of the users is a subset of that of the facilities222This version problem is originally called -gather clustering problem [1]., there is a -approximation algorithm [1], and no algorithm can achieve a better approximation ratio unless P=NP [5]. If is a line, there are polynomial time exact algorithms by dynamic programming for -gathering problem [3, 6, 7, 4], where the fastest one runs in linear time [4]. This technique can also be applied to the -gather clustering problem. If is a spider, which is a metric space constructed by joining half-line-shaped metrics together at endpoints, these are polynomial-time algorithms if is a constant [2] (i.e., these are XP algorithms).

Thus far, the best tractability result is the XP algorithms on a spider, and the best intractability is NP-hardness on a general metric space. The purpose of this study is to close this gap.

### 1.2 Our Contribution

In this study, we close the gap between the tractability and intractability of the problems on a spider. We first propose fixed-parameter tractable algorithms for both -gather clustering problem and -gathering problem parameterized by . There is an algorithm to solve the -gather clustering problem on a spider in time, where is the degree of the center of the spider. Similarly, there is an algorithm to solve the -gathering problem on a spider in time. The proof is given in Section 3.

We then show that the problem is NP-hard; this is our main theoretical contribution. The min-max -gather clustering problem and min-max gathering problem are NP-hard even if the input is a spider. The proof is given in Section 4. Note that, on a spider, -gather clustering problem is the special case of -gathering problem because we can reduce the former into latter by putting facilities for all midpoints of two users. Thus we only prove the NP-hardness only for -gather clustering problem.

## 2 Preliminaries

A spider is a set of half-lines that share the endpoint (see Figure (a)). Each half-line is called a leg and is called the center. We denote by the point on leg whose distance from the center is . Note that is the center for all . The metric on is defined by if and if .

Let be a set of users on . A cluster is a subset of users and the diameter of a cluster is the distance between two farthest users in the cluster.

The min-max -gather clustering on spider is a problem to find a partition of users into disjoint clusters so that each cluster contains at least users (Figure (b)). The goal is to minimize the maximum diameter of the clusters (Figure (c)). When the meaning is clear, we simply call it -gather clustering problem. The min-max -gathering on spider additionally specifies a set of facilities on . The goal of this problem is to open a subset of facilities and assign each user to an opened facility so that each cluster that corresponds to a facility contains at least users (Figure (d)). The goal is to minimize the maximum distance from users to the assigned facilities (Figure (e)). When the meaning is clear, we simply call it -gathering.

## 3 FPT Algorithm for r-Gather Clustering and r-Gathering on Spider

We give an FPT algorithm to solve min-max -gather clustering problem and -gathering problem on spider parametrized by the number of legs . First, we exploit the structure of optimal solutions. After that, we give a brute-force algorithm. Finally, we accelerate it by dynamic programming.

We denote the coordinate of user by . Without loss of generality, we assume that . We refer this order to explain a set of users: for example, “the first/last users on leg ” means the users with smallest/largest index among all users on leg . Users on the center are located on every leg, but we choose an arbitrary leg and consider that all the users are located on this leg. Thus every user is considered to be located on exactly one leg.

We introduce a basic lemma about the structure of a solution. A cluster is single-leg if it contains users from a single leg; otherwise, it is multi-leg. Ahmed et al. [2] showed that there is an optimal solution that has a specific single-leg/multi-leg structure as follows.

[[2, Lemma 2]] For both -gather clustering problem and -gathering problem, there is an optimal solution such that, for all leg , first some users on are contained in multi-leg clusters and rest of them are contained in single-leg clusters.

For a while, we concentrate on the structure of multi-leg clusters. Let be a multi-leg cluster. Let be the last user in and be the last user with in . A ball part of is the set of users whose indices are at most and a segment part of is the set of the remaining users. is special if contains all the users on and the ball part is for some integer . The list of multi-leg clusters are suffix-special if for all , is a special when we only consider the users in .

The following lemma is the key to our algorithm. Since it is a reformulation of Lemma 3 and Lemma 8 in Ahmed et al. [2] using Lemma 2 in Nakano [7], we omit the proof.

[Reformulation of [2, Lemmas 3 and 8] by [7, Lemma 2]] Suppose that there exists an optimal solution without any single-leg cluster. Then there is an optimal solution such that all the clusters contain at most users, and there exists a special cluster.

By definition, the segment part of a cluster is non-empty and contains the users from a single leg. By removing a special cluster and applying the lemma repeatedly, we can state that there is an optimal solution consists of a suffix-special family of multi-leg clusters.

We first present a brute-force algorithm that enumerates all the suffix-special families of multi-leg clusters in Algorithm 1. The correctness is clear from the definition. Once all such families of clusters are enumerated, the remaining thing is to deal with single-leg clusters. It can be done by solving the line-case problem on the remaining users, independently for each leg, which can be solved by pre-calculated dynamic programming.

Then, we accelerate Algorithm 1 by dynamic programming. Similar to Algorithm 1, we make a special cluster of remaining users one by one, by looking through the users and decide whether to use them into the current cluster. In Algorithm 1, all information we should remember to construct the current cluster is , , and . We claim that, for the dynamic programming, we only need to remember (1) the size of in order not to make a too-small cluster and (2) the index of the last user in the ball part of in order to calculate the diameter/cost of the cluster. Assume that we know the last user in the ball part of and last user in the segment part of . Then, the diameter/cost of is spanned by and ; Then, we can compute the diameter/cost of the cluster. Here we denote the diameter/cost of the multi-leg cluster by for both of the problems.

We should also deal with single-leg clusters. For single-leg clusters, we can apply the results for line case, which are well studied. For all leg and for all integers from to the number of users on leg , we first compute the optimal objective value, only considering the last users on leg . For each user , we denote the optimal objective value for the set of users on leg whose indices are greater than and no less than by and , respectively. All these values can be computed in linear time for both -gather clustering problem and -gathering problem, by using the technique in the latest method [4].

The algorithm is shown in Algorithm 2. The correctness is clear from the construction. Thus, we analyze the time complexity. A naive implementation of the algorithm requires evaluations of and a preprocessing for and . Each evaluation of requires time for -gather clustering problem and time for -gathering problem. The preprocessing requires time for -gather clustering problem and time for -gathering problem [4]. Thus the time complexities are for -gather clustering problem and for -gathering problem.

We can further improve the complexity. The loop for is reduced to see only first users from each leg since other users cannot be contained in the ball part of multi-leg clusters. Thus we can replace to in the complexity so as we obtain the complexities for -gather clustering problem and for -gathering problem. In -gather clustering problem, this is a linear-time algorithm when are sufficiently smaller than .

In -gathering problem, we can further improve the complexity by improving the algorithm to calculate (see Appendix A). When are sufficiently smaller than , this is also a linear-time algorithm. Therefore, Theorem 1.2 is proved.

## 4 NP-hardness of min-max r-Gather Clustering on Spider

We prove the min-max -gather clustering problem is NP-hard even on a spider. We first propose arrears problem (Problem 1) as an intermediate problem. Then, we reduce the arrears problem to the min-max -gather clustering problem on a spider. Finally, we prove the strong NP-hardness of the arrears problem.

###### Problem 1 (Arrears Problem).

We are given sets of pairs of integers, i.e., for all . We are also given pairs of integers . The task is to decide whether there is integers such that

 ∑ai,zi≤bjpi,zi≤qj

holds for for all .

The name of the “arrears problem” comes from the following interpretation. Imagine a person who is in arrears in his payment duties . Each payment duty has multiple options such that he can choose a payment amount of dollar with the payment date for some . Each pair corresponds to his budget constraint such that he can pay at most dollar until -th day.

The arrears problem itself may be an interesting problem, but here we use this problem just for a milestone to prove the hardness of min-max -gather clustering problem on a spider. The proof follows the following two propositions.

###### Proposition 1 (Reduction from the arrears problem).

If the arrears problem is strongly NP-hard, the min-max -gather clustering problem on a spider is NP-hard.

###### Proposition 2 (Hardness of the arrears problem).

The arrears problem is strongly NP-hard.

Without loss of generality, we assume that and . Also, we assume that and for all .

### 4.1 Reduction from Arrears Problem

We first prove Proposition 1. In this subsection, let be the number of payment duties and be the number of budget constraints.

Let be an instance of the arrears problem. We define and . We construct an instance of the decision version of the -gather clustering problem on a spider that requires to decide whether there is a way to divide vertices into clusters all of which has size at least and diameter at most .

In construction, we distinguish two types of legs — long and short. Each long leg corresponds to a payment duty and each short leg corresponds to a budget constraint.

For each payment duty , we define a long leg . We first put users on . Then, we put users on for all . Finally, we put users on .

A short leg has only one user. The distance from the center to the user is referred to as the length of the short leg. For each , we define short legs of length , where we set . We also define short legs of length .

This construction can be done in pseudo-polynomial time. Now we prove that has a feasible solution if and only if is a yes-instance of the arrears problem. We start by looking some basic structures of clusters in a feasible solution of .

In a feasible solution to , there is no cluster that contains users from two different long legs.

###### Proof.

By definition, the distance between the center and a user on a long leg is larger than . So, the distance between users from two different long legs exceeds ; hence, they cannot be in the same cluster. ∎

An end cluster of long leg is a cluster that contains the farthest user of . The above lemma implies that, in a feasible solution, end clusters of different long legs are different. Intuitively, the “border” of end cluster of long leg corresponds to the choice from the options of payment duty .

For each long leg , the following three statements hold. (a) An end cluster of only contains the users from . (b) There is exactly one end cluster of , and no other cluster consists of only users from leg . (c) Some users on are not contained in tend cluster.

###### Proof.

(a) The endpoint of is distant by more than from center. (b) There are less than users on so they cannot form more then one clusters alone. (c) Users on the point are distant from the endpoint of by more than , thus they cannot be in the same cluster. ∎

Lemma 4.1 and the third statement of Lemma 4.1 implies that users who are not contained in end clusters should form a cluster together with users from short legs. Now, we prove Proposition 1.

###### Proof of Proposition 1.

Suppose that we have a feasible solution to the instance of the min-max -gathering problem on a spider that is constructed as the above. For each long leg , let be the last user that is not contained in end clusters, and be the cluster that contains . Then, the location of is represented as by using an integer . We choose payment date for payment duty . We prove that these choices of payment dates are a feasible solution to the arrears problem.

As described above, consists of users from leg and short legs. Since on leg there are only users on the path from the center to the location of , should contain at least users on short legs with length at most . For -th budget constraint, by the rule of construction, there are users on short legs whose length is at most . Suppose . We use at least users on short legs whose lengths are at most in the cluster . Thus, the sum of among all with is at most the number of users on short legs whose length is at most , that is, . That means the budget constraint holds.

Conversely, suppose that we are given a feasible solution to the instance of the arrears problem. First, for each payment duty we make a cluster with all users located between and , inclusive. This cluster contains at least users since there are users on point and has diameter at most . We renumber the payment duties (thus so do long legs) in the non-decreasing order of and proceed them through the order of indices: For a payment duty , we make a cluster by all remaining users on leg and all users from remaining shortest short legs. By the construction, these clusters have exactly users. We show that the diameter of is at most . The diameter is spanned by a long leg and the longest short leg. The distance to the long leg in is . The longest short leg in is the -th shortest short leg. We take the smallest such that . Then, since given solution is a feasible solution to , holds. Since there are users on short legs with length less than , the length of longest short leg in is at most . This gives the diameter of is at most . Finally, we make a cluster with all remaining users. Since there are short legs of length and all these users are located within the distance from the center, we can just put them into a cluster. Then we obtain a feasible solution to , which completes the proof. ∎

### 4.2 Strong NP-Hardness of Arrears Problem

Now we give a proof outline of Proposition 2; the full proof is given in Appendix B. We reduce the 1-IN-3SAT problem, which is known to be NP-complete [8].

###### Problem 2 (1-IN-3 SAT problem [8]).

We are given a set of clauses, all of them contains exactly three literals. Decide whether there exists a truth assignment such that all clause has exactly one true literal.

###### Proof Outline of Proposition 2.

Let and be the number of boolean variables and clauses, respectively. For each variable , we prepare items for the positive literal and items for negative literal . Let be the set of all items. Each item corresponds to a payment duty having two options. Then, a solution to the arrears problem is specified by a set of items such that is chosen. We denote by the complement of . We want to construct a solution to the 1-IN-3SAT problem from a solution to the arrears problem by if for some ; otherwise . To make this construction valid, we define payment dates and payment amounts suitably as follows.

The payment days consist of two periods: the first period is and the second period is . For each item , belongs to the first period and belongs to the second period. Let and . Then, the payment amount is given in the form where is a sufficiently large integer, and is a non-negative integer, where holds for all . We define for all .

Let . We make two budget constraints and . Then, these constraints hold in equality: Let be the total payment until . Then the total payment until is . These inequalities imply . (see Lemma B on Appendix B).

We use the first period to ensure that the truth assignment produced by is well-defined, i.e., if for some then for all . First, for each , we add a budget constraint . By comparing the coefficients of and , we have

 ∑y∈¯X∩⋃ij=1(Tj∪¯Tj)(B4+ayB3)≤iNB4+iNB3.

We can prove that for all these inequalities hold in equality, i.e.,

 ∑y∈y∈¯X∩(Ti∪¯Ti)(B4+ayB3)=NB4+NB (1)

holds for all as follows. By using the relation between the coefficients of , we have (see Proposition 3 on Appendix B). Since the budget constraint is fulfilled in equality, and the coefficients of and in are both , this inequality holds in equality, which implies (1). Then, we define values appropriately so that the only or satisfy equation (1) (see Proposition 3 on Appendix B). This ensures the well-definedness of the truth assignment.

The second period represents the clauses. Let be the set of items with . We put budget constraint for each , where are non-negative integers determined later. Then, by a similar argument to the first period, we can prove that

 |¯X|+2|X∩(Zn+1∪⋯∪Zi)|=nN+2i∑j=n+1Kj

for each (see Proposition 4 in Appendix B). This implies that for each . The budget constraint on day corresponds to the -th clause. For , we set . Then, we have , i.e., exactly one literal in -th clause is . The budget constraints on day and are used for the adjustment. Since forms a partition of items, we have . Moreover, since the constant term of is for all and , we have . By solving these equations, we obtain and . Since all value appears in is at most and we can take in a polynomial of . Thus, The hardness proof is completed. ∎

## References

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• [2] Shareef Ahmed, Shin-ichi Nakano, and Md Saidur Rahman. r-gatherings on a star. In Proceedings of International Workshop on Algorithms and Computation, pages 31–42. Springer, 2019.
• [3] Toshihiro Akagi and Shin-ichi Nakano. On r-gatherings on the line. In Proceedings of International Workshop on Frontiers in Algorithmics, pages 25–32. Springer, 2015.
• [4] Sarker Anik, Sung Wing-kin, and Rahman Mohammad Sohel. A linear time algorithm for the r-gathering problem on the line (extended abstract). In Proceedings of International Workshop on Algorithms and Computation, pages 56–66. Springer, 2019.
• [5] Amitai Armon. On min–max r-gatherings. Theoretical Computer Science, 412(7):573–582, 2011.
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• [7] Shin-ichi Nakano. A simple algorithm for r-gatherings on the line. In Proceedings of International Workshop on Algorithms and Computation, pages 1–7. Springer, 2018.
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Proceedings of the tenth annual ACM symposium on Theory of computing

, pages 216–226. ACM, 1978.
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## Appendix A Calculation of Cost in r-gathering

In this section, we show how to calculate the in the -gathering problem efficiently. The number of candidates of pair is at most , so we calculate the for all candidates in advance and store them. Now, we describe how to calculate these values. We assume that the facilities are given in increasing order of the distances from the center.

There are two cases of the location of the facility which will be assigned to the cluster – whether the facility is located on the leg or not. If it is not located on the leg , we should simply choose the facility which is nearest to the center. This case can be processed in constant time for each pair of .

For the facility located on , we should choose the facility which is nearest to the midpoint of the coordinates of and . We enumerate the midpoints for all pairs and sort them by distance from the center, for each leg. We can apply the two-pointer technique to find the optimal facility by seeing the midpoints in sorted order. This case can be processed in time, where came from the sorting operation.

We can refer each pre-calculated value in time, so the total time complexity of Algorithm 2 is reduced to .

## Appendix B Proof of NP-Hardness of Arrears Problem

In this section, we construct an instance of arrears problem from given instance of 1-IN-3SAT . In our construction, every payment duty has exactly two payment dates. Let us fix the variable . For all and , we prepare two items and . We also prepare auxiliary items and for each . In this way, We prepare items in total for each .

We name some important sets of items in following way:

• .

We prepare a payment duty for all item .

Before defining the value of these values, we take an integer . All integers which appear as is represented in the form by non-negative integers . Similarly, all integers which appear as is represented in the form by non-negative integers

. We represent these values as if like a vector

and . In our construction is always equal to , so can be represented in the form .

We take to be sufficiently large value (but still in a polynomial of ) so that for all , sum of over all item is still less than . That means that we can compare the sum of payment amount and budget constraint just by comparing sum of (or ) and by lexicographical order of five-dimensional vector. The concrete value of is .

Let us start to set the value of payment duties. Let us fix a variable . For , we set and

 au,2=⎧⎪⎨⎪⎩n+2+j(u=ui,j,kandcj,k=xi)n+2+j(u=¯ui,j,kandcj,k=¯xi)n+1(otherwise).

is always equal to . Remaining task is define the value . We set by following formula.

For each , let be and be . We set and

 aw,2=⎧⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪⎩n+2(w=wi,land1≤l≤3m(m+1)−Ki)n+1(w=wi,land3m(m+1)−Ki+1≤l≤3m(m+1)+1)n+2(w=¯wi,land1≤l≤3m(m+1)−¯Ki)n+1(w=¯wi,land3m(m+1)−¯Ki+1≤l≤3m(m+1)+1).

is also always equal to . Remaining task is define the value . We set by following formula.

Note that, since by definition,

 ∑t∈Tipt,1=∑t∈¯Tipt,1=(3m(m+2)+1)(B2+i)(B+1)B+3m(m+1)

holds for all . We call this value and set

 R=n∑i=1Ri=(3m(m+2)+1)(B+1)(nB2+n(n+1)2)B+3m(m+1)n.

We now remark that is sufficiently large. We can calculate each values just by expanding the definition formula. It can be calculated that, for all the sum of over all is at most . It is sufficiently small to avoid carry. Then, we set budget constraints. We set a budget constraints for each . Value of is

 ⎧⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩(3m(m+2)+1)(B+1)iB3+(B3−1)(1≤i≤n−1)R(i=n)((3m(m+2)+1)n+6mn+2n+m(m+1))B4+(B4−1)(i=n+1)(3(3m(m+2)+1)n−2(n+m+2−i))B4+(B4−1)(n+2≤i≤n+m+1)3R(i=n+m+2).

We represent .

Now we complete our construction. All appearing values are at most , which is bounded in a polynomial of . We prove that is a yes-instance of 1-IN-3SAT if and only if is a yes-instance of arrears problem.

For feasible solution of , Let be the set of items , such that payment date is chosen for payment duty . We define by complement of . Intuitively, for or , means is true and means is false. Following proposition guarantees that this type of truth assignment is well-defined.

###### Proposition 3.

In a feasible solution of , for all , one of the following condition holds.

• .

• .

Before proving this proposition, we prove the following basic property.

In a feasible solution of ,

 ∑y∈¯Xpy,1=∑y∈Xpy,1=R.
###### Proof.

We only needs budget constraints for and to prove this lemma. Note that,

 ∑y∈¯Xpy,1+∑y∈Xpy,1=n∑i=1⎛⎝∑y∈Tipy,1+∑y∈¯Tipy,1⎞⎠=2R

holds. From budget constraint for ,

 ∑y∈¯Xpy,1≤R

holds. From budget constraint for ,

 ∑y∈¯Xpy,1=4R−⎛⎝∑y∈¯Xpy,1+2∑y∈Xpy,1⎞⎠≥4R−3R=R

holds. It means equality holds for both of the inequality, and thus the lemma holds. ∎

###### Proof.

Let us start by rephrasing some budget constraints. First, we only concern about the coefficients of and . From budget constraint for ,

 ∑