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The Parameterized Complexity of the Unique Coverage Problem 3 Hannes Moser 1,2 Institut f¨ ur Informatik, Friedrich-Schiller-Universit¨ at Jena, Ernst-Abbe-Platz 2, D-07743 Jena, Germany
Venkatesh Raman and Somnath Sikdar 2 The Institute of Mathematical Sciences, C.I.T Campus, Taramani, Chennai 600113, India
Abstract We consider the parameterized complexity of the Unique Coverage problem: given a family of sets and a parameter k, we ask whether there exists a subfamily that covers at least k elements exactly once. This NP-complete problem has applications in wireless networks and radio broadcasting and is also a natural generalization of the well-known Max Cut problem. We show that this problem is fixed-parameter tractable with respect to the parameter k. We also show a 4k kernel for this problem. However a more general weighted version, with costs associated with each set and profits with each element, turns out to be much harder. The question here is whether there exists a subfamily with total cost at most a prespecified budget B such that the total profit of uniquely covered elements is at least k. In the most general setting, assuming real costs and profits, the problem is not fixed-parameter tractable unless P = NP. Assuming integer costs and profits we show the problem to be W[1]-hard with respect to B as parameter. However, under some reasonable restriction, the problem becomes fixed-parameter tractable with respect to both B and k as parameters. Key words: Parameterized Complexity, Covering Problems, Kernelization
Email addresses: [email protected] (Hannes Moser), {vraman|somnath}@imsc.res.in (Venkatesh Raman and Somnath Sikdar). 1 Supported by the Deutsche Forschungsgemeinschaft, project ITKO (iterative compression for solving hard network problems), NI 369/5. 2 Supported by a DAAD-DST exchange program, D/05/57666. 3 An extended abstract of this work appeared in the proceedings of the 18th Inter-
Preprint submitted to Elsevier
25 April 2008
1
Introduction
In this paper, we consider the parameterized complexity of the Unique Coverage problem. This problem was introduced by Demaine et al. [4] as a natural maximization version of Set Cover and has applications in several areas including wireless networks and radio broadcasting. Unique Coverage is defined as follows. Given a ground set U = {1, 2, . . . , n}, a family of subsets S = {S1 , . . . , St } of U and a nonnegative integer k, we ask whether there exists a subcollection S 0 ⊆ S such that at least k elements are covered uniquely by the members in S 0 . An element is covered uniquely if it appears in exactly one set of S 0 . The optimization version requires to maximize the number of uniquely covered elements. The weighted version of Unique Coverage is called Budgeted Unique Coverage and is defined as follows. Given a ground set U = {1, 2, . . . , n}, a profit pi for each element i ∈ U, a family of subsets S of U, a cost ci for each set Si ∈ S, a budget B and a nonnegative integer k, we ask whether there exists a subset S 0 ⊆ S such that the total cost of S 0 is at most B and such that the profit of the uniquely covered elements is at least k. The optimization version asks for a subset S 0 of total cost at most B such that the profit of uniquely covered elements is maximized. The original motivation for this problem is a real-world application arising in wireless networks [4]. Assume that we are given a map of the densities of mobile clients along with a set of possible base stations, each with a specified building cost and a specified coverage region. The goal is to choose a set of base stations, subject to a budget on the total building cost, in order to maximize the density of served clients. The difficult aspect of this problem is the interference between base stations. A mobile client’s reception is better when it is within the range of a few base stations. An ideal situation is when every mobile client is within the range of exactly one base station. This is the situation modelled by the Budgeted Unique Coverage problem. The Unique Coverage problem is closely related to a single “round” of the Radio Broadcast problem [1]. For more about this relation, see Demaine et al.’s work [4]. One can also view the Unique Coverage problem as a generalization of the Max Cut problem [4]. The input to the Max Cut problem consists of a graph G = (V, E) and the goal is to find a cut (T, T 0 ), where ∅ = 6 T ⊂V 0 and T = V \ T , that maximizes the number of edges with one endpoint in T and the other endpoint in T 0 . Let U denote the set of edges of G and for each vertex v ∈ V define Sv = {e ∈ E : e is incident to v}. Finally let S = ∪v∈V {Sv }. Then G has a cut (T, T 0 ) with at least k edges across it if and only if S 0 = ∪v∈T {Sv } uniquely covers at least k elements of the ground set. national Symposium on Algorithms and Computation (ISAAC’07), Springer, LNCS 4835, pages 621–631.
2
Known Results. (Budgeted) Unique Coverage was introduced by Demaine et al. [4]. They have considered the approximability of this problem. On the positive side, they give an Ω(1/ log n)-approximation for Budgeted Unique Coverage. Moreover, if the ratio between the maximum cost of a set and the minimum profit of an element is bounded by B, then there exists an Ω(1/ log B)approximation. They show that Unique Coverage is hard to approximate to within a factor of O(1/ logc n) for some constant c depending on , assum ing NP * BPTIME(2n ) for some > 0. They strengthen this inapproximability to 1/( log n) for some > 0 based on a hardness hypothesis for Balanced Bipartite Independent Set. Erlebach and van Leeuwen [6] study the approximability of geometric versions of the Unique Coverage problem. Among the many versions that they consider is Unique Coverage on Unit Disks, a variant in which each set is a unit disk in R2 , for which they give a factor-18 approximation algorithm. They also consider a variant called Unique Coverage on Disks of Bounded Ply and design an asymptotic fully polynomial-time approximation scheme (FPTASω ) for it. Our Results. In this paper, we give first-time results on the parameterized complexity of Unique Coverage and Budgeted Unique Coverage. As in many problems in parameterized complexity, the Unique Coverage problem can be parameterized in a number of ways. We first consider an extensive list of plausible parameterizations of the problem in Section 3 and discuss their parameterized complexity. Our results show that barring the standard parameterized version (where the parameter is the number of uniquely covered elements) the remaining parameterized problems are unlikely to be fixed-parameter tractable. In Section 4 we consider the standard parameterized version of Unique Coverage. We show that a special case of this version where any two sets in the input family intersect in at most c elements is fixed-parameter tractable by demonstrating a polynomial kernel of size k c+1 . This leads to a problem kernel of size k k in the general case, proving that Unique Coverage is fixed-parameter tractable. Then using results from extremal combinatorics on strong systems of distinct representatives we obtain a 4k kernel. In Section 5 we consider the Budgeted Unique Coverage problem. For this problem too there are several variants. If the profits and costs are allowed to be arbitrary positive real numbers, then Budgeted Unique Coverage, with parameters B and k, is not fixed-parameter tractable unless P = NP. If we restrict the costs and profits to be positive integers and parameterize by B, then the problem is W [1]-hard. In the case when the number of sets intersecting any given set of the input family is bounded by a function of k, 3
Unique Coverage (Parameter: k) Result
Sect.
Each element occurs in at most b sets Intersection size bounded by c
(k − 1)b kernel
4.1
k c+1 kernel
4.2
General case
4k kernel
4.3
Each set of size at most b
2b+k kernel
4.3
Arbitrary costs/profits (parameters B and k)
Not FPT (unless P = NP)
5.1
Integer weights (parameter B)
W [1]-hard
5.1
Integer weights (intersection number f (k); parameters B and k)
O∗ ((B 2 · f (k) · 2f (k) )min{B,k} ) 5.2 time algorithm
Integer weights (parameters B and k)
Open
Budgeted Unique Coverage
Fig. 1. Main results in this paper.
the problem is fixed-parameter tractable with parameters B and k. The main results of this paper along with the relevant sections in which they appear are depicted in Figure 1. The version in which the profits and costs are positive integers and the parameters are both B and k is open. We note that the parameterized versions of all variants of Unique Coverage considered in [6] are fixed-parameter tractable. Unique Coverage on Unit Disks is fixed-parameter tractable with respect to the number of uniquely covered elements as parameter since it is a restriction of the problem that we show to be fixed-parameter tractable in this paper. It is well-known that an optimization problem with an FPTAS is fixed-parameter tractable [5] and it can be easily shown that the same result holds for problems with an asymptotic FPTAS. Therefore variants such as Unique Coverage on Disks of Bounded Ply are fixed-parameter tractable (with respect to the cost and budget as parameters). 4
2
Preliminaries
We briefly introduce the necessary concepts concerning parameterized complexity. A parameterized problem is a subset of Σ∗ × N, where Σ is a finite alphabet and N is the set of natural numbers. An instance of a parameterized problem is therefore a pair (I, k), where k is the parameter. In the framework of parameterized complexity, the running time of an algorithm is viewed as a function of two quantities: the size of the problem instance and the parameter. A parameterized problem is said to be fixed parameter tractable (FPT) if there exists an algorithm for the problem with running time f (k) · |I|O(1) , where f is a computable function only depending on k. We sometimes refer to such an algorithm as an FPT algorithm. A common method to prove that a problem is fixed-parameter tractable is to provide data reduction rules that lead to a problem kernel. A data reduction rule is a polynomial-time algorithm which takes a problem instance (I, k) and either • outputs yes or no according as (I, k) is a yes or a no-instance, or • replaces (I, k) by an equivalent instance (I 0 , k 0 ) such that |I 0 | ≤ |I| and k 0 ≤ k, where two problem instances (I, k) and (I 0 , k 0 ) are equivalent if they are both yes-instances or both no-instances. An instance to which none of a given set of data reduction rules applies is called reduced with respect to this set of rules. A parameterized problem is said to have a problem kernel if the resulting reduced instance has size f (k) for a function f depending only on k. If a parameterized problem has a kernel, then it is clearly fixed-parameter tractable: simply use brute-force on the kernel to decide whether the given instance is a yes-instance or not. The converse is also true: if a parameterized problem is fixed-parameter tractable, then it has a kernel. A parameterized problem π1 is fixed-parameter reducible to a parameterized problem π2 if there exist functions f, g : N → N, Φ : Σ∗ × N → Σ∗ and a polynomial p(·) such that for any instance (I, k) of π1 , (Φ(I, k), g(k)) is an instance of π2 computable in time f (k) · p(|I|) and (I, k) ∈ π1 if and only if (Φ(I, k), g(k)) ∈ π2 . The basic complexity class for fixed-parameter intractability is W [1] as there is strong evidence that W [1]-hard problems are not fixed-parameter tractable [5]. To show that a problem is W [1]-hard, one needs to exhibit a fixed-parameter reduction from a known W [1]-hard problem to the problem at hand. For more on parameterized complexity see [5,7,10]. We write O∗ (f (k)) to denote a running time of O(f (k) · poly(n, k)), where n is the input size and k is the parameter. That is, we use the O∗ (·) notation to suppress polynomial factors in the running time. 5
3
The Unique Coverage Problem: Which Parameterization?
The Unique Coverage problem can be parameterized in a number of ways. We begin with a general discussion on different parameterizations of the problem and show that except for the standard parameterized version (where the parameter is the solution size) and its variant, the remaining parameterizations are unlikely to be fixed-parameter tractable. An interesting analog is the Longest Common Subsequence problem which can be parameterized in a number of ways, some of which are fixed-parameter tractable and the others hard [2,11]. First consider the following parameterized problem: given (U, S), find a subfamily S 0 ⊆ S that covers all of U, each element being covered at most k times and at least once, assuming k as parameter. This is a practical parameterization for mobile-computing applications. Unfortunately, this problem is not fixed-parameter tractable unless P = NP as the case k = 1 reduces to the NP-complete Exact Cover problem [8]. An alternate parameterization with a similar motivation of covering each element a small number of times is as follows: given (U, S), find a subfamily S 0 ⊆ S that covers all of U, each element being covered at most |S| − k times. This problem can be rephrased as follows: Is there a subfamily S 00 ⊆ S of size at most k such that S 0 = S −S 00 covers all elements of U at most |S| − k times? Call this problem All But k Coverage. Lemma 1 Suppose an instance (U, S, k) of All But k Coverage is such that there exists an element e ∈ U that is contained in every set of S. Then the given instance is a yes-instance if and only if there exists a subfamily S 00 ⊆ S of size exactly k such that S − S 00 covers all elements of U at most |S| − k times. PROOF. The proof consists in observing that if an element occurs in every set of S then any solution must necessarily exclude at least k sets of S. Since |S| − k sets can cover any element at most |S| − k times, excluding k sets is sufficient. 2 We show that this problem is W[1]-hard by a fixed-parameter reduction from the W[1]-complete Red/Blue Nonblocker problem [5]: A bipartite graph G = (R ] B = V, E) with its vertex set partitioned into a red set R and a blue set B, and a positive integer k. Parameter: The positive integer k. Question: Does there exist a set T of k red vertices such that the vertices in R − T dominate all vertices in B? In order to show that All But k Coverage is W[1]-hard, we consider a special case of the Red/Blue Nonblocker problem where we are guaran-
Input:
6
teed that there exists a blue vertex that is adjacent to all red vertices. This special case is W[1]-hard too. Given an instance (G, k) of Red/Blue Nonblocker create a special blue vertex v ∗ and add edges from v ∗ to all red vertices. Then the original instance has a size-k solution if and only if the transformed instance has a size-k solution. Given an instance (G = (R ] B, E), k) of this special case of Red/Blue Nonblocker, create an instance (U, S, k) of All But k Coverage by identifying the universe U with the blue vertices and the sets of the input family S with the red vertices. A set contains an element if and only if the corresponding red-blue vertex-pair is connected by an edge. Lemma 2 There exists a set R0 ⊆ R of size exactly k such that R − R0 dominates all of B if and only if there exist a subfamily S 00 of size k such that S − S 00 covers all of U, each element being covered at most |S| − k times. PROOF. Given any set R0 of size exactly k such that R − R0 dominates all of B, define S 00 to be the family consisting of the sets corresponding to the vertices of R0 . Clearly S − S 00 covers all of U. As |S − S 00 | = |S| − k, each element of U is covered at most |S| − k times. Conversely, suppose S 00 ⊆ S is of size k such that S − S 00 covers all of U, each element being covered at most |S| − k times. Define R0 to be the set of red vertices corresponding to S 00 . Then |R0 | = k and R − R0 dominates all of B. This completes the proof of the lemma. 2 We therefore have the following: Theorem 3 All But k Coverage is W[1]-hard with respect to k as parameter. We also note that the following parameterized versions of Unique Coverage are not fixed-parameter tractable. Given (U, S) and nonnegative integers k and t as parameters, (1) Does there exist a subfamily S 0 ⊆ S of size at most k that covers all of U with each element being covered at most t times? This version is not fixed-parameter tractable as the case t = 1 is W[1]-hard by a reduction from Perfect Code [3] as shown below. (2) Does there exist a subfamily S 0 ⊆ S of size at most k that covers all of U with each element being covered at most |S| − t times? This is W[2]-hard as the case t = 0 reduces to the Set Cover problem which is W[2]-complete. Call the version in Item 1 above with t = 1 the Disjoint Set Cover problem which we now show to be W[1]-hard by a reduction from Perfect Code [3]. Perfect Code Input: A graph G = (V, E) and an integer k. Parameter: The integer k. 7
Question:
Does G have a k-element perfect code? A perfect code is a vertex subset V 0 ⊆ V such that for all u ∈ V we have |N [u] ∩ V 0 | = 1.
Given an instance (G = (V, E), k) of Perfect Code construct an instance (U, S, k) of Disjoint Set Cover as follows: U = V and S = {N [v] | v ∈ V }. Lemma 4 The graph G has a k-element perfect code if and only if there exists a subfamily S 0 ⊆ S of pairwise disjoint sets and of size at most k that covers all of U. PROOF. Let {v1 , . . . , vk } ⊆ V be a k-element perfect code of G. Then S clearly ki=1 N [vi ] covers all of U and the sets N [v1 ], . . . , N [vk ] are pairwise disjoint. For if x ∈ N [vi ] ∩ N [vj ], i 6= j, then |N [x] ∩ {v1 , . . . , vk }| ≥ 2, which contradicts the definition of a perfect code. Conversely if N [v1 ], . . . , N [vk ] is a collection of pairwise disjoint sets that covers all of U then clearly v1 , . . . , vk is a k-element perfect code for G. 2 Theorem 5 Disjoint Set Cover is W[1]-hard with respect to the number of sets in the solution as parameter. We next consider a generalization of the standard parameterized version: • Gen Unique Coverage: Given (U, S) and nonnegative integers k and t, does there exist a subfamily of S that covers k elements at least once and at most t times? Setting t = 1 gives us the standard parameterized version of Unique Coverage. We note that Gen Unique Coverage is fixed-parameter tractable as the kernel result for the standard parameterized version works for this problem also. We elaborate this further in Section 4.3. From now on we work with the standard parameterized version of Unique Coverage: Given (U, S) and a nonnegative integer k, is there a subfamily of S that uniquely covers at least k elements?
4
The Unique Coverage Problem: The Standard Parameterization
Let (U = {1, 2, . . . , n}, S = {S1 , S2 , . . . , Sm }, k) be an instance of Unique Coverage. Apply the following data reduction rules on (U, S, k) until no longer applicable. R1 If there exists Si ∈ S such that |Si | ≥ k, then the given instance is a yes-instance. R2 If there exists S1 , S2 ∈ S such that S1 = S2 , then delete S1 . These reduction rules are obviously correct. In the first case Si gives the solution. In the second case, it is clear that no solution need have both S1 and S2 . In the following we always assume that the given instance of Unique Coverage is reduced with respect to the above rules. 8
4.1
Bounded Number of Occurrences
We begin with the simple case where each element e ∈ U is contained in at most b sets of S. A special case of this situation is Max Cut (b = 2). Lemma 6 If each element e ∈ U occurs in at most b sets of S then the Unique Coverage problem admits a kernel of size b(k − 1). PROOF. Find a maximal collection S 0 of pairwise disjoint sets in S. If |∪S∈S 0 S| ≥ k, we are done. Therefore assume that | ∪S∈S 0 S| ≤ k − 1. Since every set in S − S 0 intersects some set in S 0 and since every element occurs in at most b sets in S, we have |S − S 0 | ≤ (k − 1)(b − 1). But |S 0 | ≤ k − 1 and so |S| ≤ b(k − 1). 2 4.2
Bounded Intersection Size
Consider the situation where for all Si , Sj ∈ S we have |Si ∩ Sj | ≤ c, for some constant c. In this case we say that the problem instance has bounded intersection size c and show that the problem admits a polynomial kernel of size O(k c+1 ). First consider the case when |Si ∩ Sj | ≤ 1. Lemma 7 Suppose that for all Si , Sj ∈ S, i 6= j, we have |Si ∩ Sj | ≤ 1. If an element e ∈ U is covered by at least k + 1 sets, then one can obtain a solution covering k elements uniquely in polynomial time. PROOF. Suppose an element e ∈ U is covered by the sets S1 , . . . , Sk+1 . Then by reduction rule R2, at most one of these sets can have size 1. The remaining k sets uniquely cover at least one element each. 2 One can now easily obtain a kernel of size k 2 for the case when the intersection size is at most 1. Lemma 8 Suppose that for all Si , Sj ∈ S, |Si ∩ Sj | ≤ 1. If |S| ≥ k 2 , then there exists T ⊆ S that covers at least k elements uniquely. PROOF. If an element appears in at least k + 1 sets then we are done by Lemma 7. Otherwise every element appears in at most k sets and by Lemma 6 we have a kernel of size k(k − 1) < k 2 . 2 Next, we generalize these observations to the case when |Si ∩ Sj | ≤ c, for some constant c. Theorem 9 Suppose that for all Si , Sj ∈ S we have |Si ∩ Sj | ≤ c, for some positive constant c. If |S| ≥ k c+1 then there exists T ⊆ S that covers k elements uniquely. PROOF. By induction on c. For c = 1, this follows from Lemma 8. Assume the theorem to hold for c > 1. Greedily obtain a maximal collection S 0 = {S1 , . . . , Sp } of pairwise disjoint sets. If | ∪Si ∈S 0 Si | ≥ k then we are done. 9
Therefore assume | ∪S∈S 0 S| ≤ k − 1 (this also implies p ≤ k − 1). Since |S| ≥ k c+1 , and since every set in S intersects with at least one set in S 0 , there exists e ∈ ∪S∈S 0 S such that at least k c + 1 sets in S − {S1 , . . . , Sp } contain e. For otherwise, |S| ≤ (k − 1)k c + p < k c+1 , a contradiction. Let T1 , . . . , Tkc +1 be some k c + 1 such sets. Delete e from each of these sets. We obtain at least k c nonempty distinct sets T10 , . . . , Tk0 c (there is at most one set consisting only of the element e which is deleted in this process). Note that any two of these sets intersect in at most c − 1 elements. By induction hypothesis, there exists a collection T 0 ⊆ {T10 , . . . , Tk0 c } that uniquely covers at least k elements, and thus there exists a collection T ⊆ {T1 , . . . , Tkc } that uniquely covers at least k elements (just take the solution for T 0 and add e to every set in it). This proves the theorem. 2 Corollary 10 Unique Coverage admits a kernel of size k c+1 for bounded intersection size c. By reduction rule R1 we have c ≤ k − 1 and therefore for the general case we have a kernel of size k k . Corollary 11 The Unique Coverage problem is fixed-parameter tractable and admits a problem kernel of size k k . An algorithm that checks all possible subsets of a family of size k k to see whether any of them uniquely covers at least k elements is an FPT algorithm k with time complexity O∗ (2(k ) ). But note that we can assume without loss of generality that every set in the solution covers at least one element uniquely. Thus it suffices to check whether subfamilies of size at most k uniquely cover 2 2 at least k elements. This can be done in time O∗ (k k ) = O∗ (2k log k ). However, this kernelization result is tailored especially for the bounded intersection size case. It turns out that a much better kernel can be obtained for the general case. 4.3
A Better Kernel for the General Case
We now show that Unique Coverage has a kernel of size 4k using a result on strong systems of distinct representatives. Given a family of sets S = {S1 , . . . , Sm }, a system of distinct representatives for S is an m-tuple (x1 , . . . , xm ) where the elements xi are distinct and xi ∈ Si for all i = 1, 2, . . . , m. Such a system is strong if we additionally have xi ∈ / Sj for all i 6= j.
Theorem 12 ([9]) In any family of more than r+s sets of cardinality at s most r, at least s + 2 of its members have a strong system of distinct representatives. Given an instance (U = {1, . . . , n}, S = {S1 , . . . , Sm }, k) of Unique Coverage, put r = k − 1 and s = k in the statement of the above theorem and we have a kernel of size 2k−1 ≤ 2k ≤ 22k . k−1 k 10
Corollary 13 Unique Coverage admits a problem kernel of size 4k . 2
Corollary 14 There is an O∗ (4k ) time algorithm for the Unique Coverage problem. PROOF. A subfamily that covers k elements uniquely has size at most k. Therefore it is sufficient to consider all possible size-k subfamilies of the 4k 2 kernel. This takes time O∗ (4k ). 2 At this point, we note that Gen Unique Coverage (see Section 3) is fixedparameter tractable: an instance (U, S, k, t) of Gen Unique Coverage is trivially a yes-instance if |S| > 4k . Otherwise |S| ≤ 4k and we have a kernel. Corollary 15 Gen Unique Coverage is fixed-parameter tractable. For the case where each set of the input family has size at most b, for some constant b, there is a better kernel. By Theorem 12, if there exists at least b+k k sets in the input family, then there exists at least k sets with a strong system of distinct representatives. Corollary 16 If each set S ∈ S has size at most b then the Unique Coverage problem has a kernel of size O(2b+k ). Another interesting case is when each set of the input family S has a nonempty intersection with at most f (k) sets in S, where f is a function of k alone. We consider this case in Section 5.2, Corollary 25, and show that this version can be solved in time O∗ ((k 2 · f (k) · 2f (k) )k ).
5
The Budgeted Unique Coverage Problem
In this section we consider the Budgeted Unique Coverage problem where each set in the input family has a cost and each element in the universe has a profit; the goal is to decide whether there exists a subcollection of total cost at most B that uniquely covers elements of total profit at least k. By parameterizing on k or B or both we obtain different parameterized versions of this decision question. 5.1
Budgeted Unique Coverage: Intractable Parameterized Versions
We first consider the Budgeted Max Cut problem which is a specialization of the Budgeted Unique Coverage problem. An instance of this problem is an undirected graph G = (V, E) with a cost function c : V → R+ on the vertex set and a profit function p : E → R+ on the edge set; positive real numbers B and k. The question is whether there exists a cut (T, T 0 ) such that the total cost of the vertices in T is at most B and the total profit of the edges crossing the cut is at least k. 11
We first show that the Budgeted Max Cut problem with arbitrary positive real costs and profits is probably not FPT. Lemma 17 The Budgeted Max Cut problem with arbitrary positive costs and profits with parameters B and k is not FPT, unless P = NP. PROOF. Suppose there exists an algorithm for the Budgeted Max Cut problem (with arbitrary positive costs and profits) with run-time O(f (k, B) · p(n)), where p is a polynomial in n. We will use this to solve the decision version of Max Cut in polynomial time. Let (G = (V, E), k) be an instance of the Max Cut problem, where |V | = n. Assign each vertex of the input graph cost 1/n and each edge profit 1/k. Let the budget B = 1/2 and the profit k 0 = 1. Clearly, G has a maximum cut of size at least k if and only if there exists S ⊆ V of total cost at most B such that the total profit of the edges crossing the cut (S, V − S) is at least k 0 . And this can be answered in time O(f (1, 1/2) · p(|V |)), implying P = NP. 2 Theorem 18 The Budgeted Unique Coverage problem with arbitrary positive costs and profits is not FPT, unless P = NP. Henceforth by the ‘budgeted’ version we mean the case when the costs and profits are positive integers. We next show that the Budgeted Max Cut problem parameterized by the budget B alone is W [1]-hard. Lemma 19 The Budgeted Max Cut problem parameterized by the budget is W [1]-hard. PROOF. To show W [1]-hardness, we exhibit a fixed-parameter reduction from the Independent Set problem to the Budgeted Max Cut problem with unit costs and profits. Let (G = (V, E), B) be an instance of Independent Set with |V | = n. For every vertex u ∈ V add |V | − 1 − deg(u) new vertices and connect them to u. Call the resulting graph G0 . Define c(v) = 1 for all v ∈ V (G0 ) and p(e) = 1 for all e ∈ E(G0 ). Note that every vertex u ∈ G has degree |V | − 1 in G0 . We let (G0 = (V 0 , E 0 ), B, k = B(n − 1)) be the instance of Budgeted Max Cut. Claim 20 G has an independent set of size B if and only if G0 has a cut (S, V 0 − S) such that |S| = B and at least k = B(n − 1) edges lie across it. If G has an independent set S of size B, then clearly S is independent in G0 . The cut (S, V 0 − S) does indeed have B(n − 1) edges crossing it, as every vertex of S has degree n − 1. Next suppose that G0 has a cut (S, V 0 − S) such that |S| = B and B(n − 1) edges cross the cut. Note that every vertex in S must be a vertex from G. Otherwise the cut cannot have B(n − 1) edges crossing it. Suppose two vertices u and v in S are adjacent. Then both u and v contribute less than n − 1 edges to the cut. Since each vertex in S contributes at most n − 1 edges to the cut, the number of edges crossing the cut must be less than B(n − 1), a contradiction. Hence S is independent in G0 and hence G has an independent set of size B. 2 12
Since the Budgeted Unique Coverage problem is a generalization of Budgeted Max Cut we have Theorem 21 The Budgeted Unique Coverage problem parameterized by the budget B is W [1]-hard. 5.2
Budgeted Unique Coverage: A Special Case
In this subsection, we give an FPT algorithm for Budgeted Unique Coverage, when B and k are parameters, assuming that for every set S in the input family the number of sets with a non-empty intersection with S is at most some function of k. This is a natural situation in real-world applications; for example, in wireless networks. For the Budgeted Max Cut problem, for instance, every set is intersected by at most k − 1 sets. This follows because every vertex has degree at most k − 1 and any two vertices share at most one edge. Let (U = {1, . . . , n}, S = {S1 , . . . , Sm }, c, p, B, k) be an instance of the Budgeted Unique Coverage problem where c : S → N and p : U → N. P For T ⊆ S, define c(T ) = S∈T c(S) and p(T ) to be the total profit of the elements uniquely covered by T . If Si ∈ S, define N [Si ] to be the set of all members of S that have a nonempty intersection with Si . We can without loss of generality assume that c(Si ) ≤ B and |Si | ≤ k − 1 for all 1 ≤ i ≤ m. In what follows, we assume that for all Si ∈ S, we have |N [Si ]| ≤ f (k) for some function f . The FPT algorithm that we describe here builds the solution in stages. Note that if we decide to include a set S in the solution, there is no way of deciding how many elements S covers uniquely unless we make choices for each set in N [S]. To get around this, the algorithm, at any stage, decides whether or not to include a subfamily A ⊆ N [S] for some set S. If it includes A in the solution, then it automatically excludes N [S] \ A from it. The current solution is a pair (T , T 0 ), where T , T 0 ⊆ S and T ∩ T 0 = ∅. The sets included by the algorithm in the solution till the current stage are in T ; those excluded from the solution are in T 0 . Call a pair (T , T 0 ) a feasible solution for an instance of Budgeted Unique Coverage if T 0 = S − T , c(T ) ≤ B and p(T ) ≥ k. A pair (T , T 0 ) is a partial solution if T , T 0 ⊆ S and T ∩ T 0 = ∅. A partial solution (T , T 0 ) can be extended to a feasible solution if there exist X , X 0 ⊆ S − (T ∪ T 0 ) such that X ∩ X 0 = ∅ and (T ∪ X , T 0 ∪ X 0 ) is a feasible solution. A partial solution (T , T 0 ) is isolated if for each set Si ∈ T , N [Si ] ⊆ T ∪ T 0 . Note that if a partial solution (T , T 0 ) is isolated then for all S ∈ S − (T ∪ T 0 ) we S have S ∩ T ∈T T = ∅. If a partial solution does not satisfy this property we call it non-isolated. Given an isolated partial solution (T , T 0 ), let U1 , . . . , Ut be a partition of S − (T ∪ T 0 ) according to costs. That is, all members in any set Ui have the same cost ci and for all i 6= j, ci 6= cj . Note that t ≤ B. For 13
each Ui , let Uimax denote a member of Ui with maximum total profit. Lemma 22 Let (T , T 0 ) be an isolated partial solution and let U1 , U2 , . . . , Ut be a partition of S − (T ∪ T 0 ) according to costs. Suppose (T , T 0 ) can be extended to a feasible solution by adding a member of Ui to T . Then there exists an extension of (T , T 0 ) into a feasible solution (T ∪ X , T 0 ∪ X 0 ) such that X ∩ N [Uimax ] 6= ∅. PROOF. Suppose (T , T 0 ) can be extended to a feasible solution (T ∪X , T 0 ∪ X 0 ) by adding a member U ∈ Ui to T and that X ∩ N [Uimax ] = ∅. This means that N [Uimax ] ⊆ X 0 . Remove U from X and replace it by Uimax . Note that every element of Uimax is uniquely covered and that the total profit of these newly uniquely covered elements is at least as that of those covered by U . Since c(U ) = c(Uimax ), the new solution continues to be feasible. 2 One can use Lemma 22 to develop an FPT algorithm with time complexity O∗ ((B 2 · f (k) · 2f (k) )min{B,k} ). This algorithm works in stages; each stage consists of two phases: the building phase and the isolation phase. We will argue that at the end of each stage, the cost of the solution being constructed and the profit of the elements uniquely covered are strictly greater than what they were at the end of the previous stage. The number of stages required is therefore at most min{B, k}. We will then show that the algorithm makes at most B 2 · f (k) · 2f (k) branches at each stage which would then prove that the overall running time of the algorithm is O∗ ((B 2 · f (k) · 2f (k) )min{B,k} ). At the beginning of each stage, the algorithm starts with an isolated partial solution. At the beginning of the first stage, this isolated partial solution is (T = ∅, T 0 = ∅). The algorithm then partitions the family S − (T ∪ T 0 ) according to costs into the subfamilies U1 , . . . , Ut . If we assume that there exists a feasible solution, it has to include a set from one of the subfamilies Ui . For each choice of a subfamily, Lemma 22 assures us that it is sufficient to consider a set S ∈ N [Smax ] where Smax is a set in the subfamily that maximizes profit. Since t ≤ B and |N [Smax ]| ≤ f (k) there are at most B · f (k) ways of choosing this set S. The algorithm makes at most B · f (k) branches and chooses which set to include in the solution. If there is no subfamily Ui such that for U ∈ Ui , c(U ) ≤ B − c(T ), the algorithm aborts this branch. This ends the building-phase. Adding a set to an isolated solution may cause it to become non-isolated. In the isolation phase, the algorithm makes the current non-isolated partial solution isolated. It chooses a set T ∈ T such that N [T ] − (T ∪ T 0 ) 6= ∅ and considers all bipartitions (A, A0 ) of N [T ] − (T ∪ T 0 ) satisfying the property that each set in T uniquely covers at least one element in the family T ∪ A. It branches on each such bipartition and sets T ← T ∪ A and T 0 ← T 0 ∪ A0 . If there is no bipartition satisfying this property the algorithm aborts this branch. Since |T | can never exceed B and |N [T ]| ≤ f (k) for any T ∈ T , the number 14
of branches that the algorithm creates in an isolation phase at any stage is at most B · 2f (k) . Since the building phase produces at most B · f (k) branches and the isolation phase works on each branch generated by the building phase, the total number of branches at any stage is at most B 2 · f (k) · 2f (k) . To prove the claimed time complexity of our algorithm, we have to show that at the end of each stage both the cost and profit of the solution being constructed increase. To argue this it is helpful to introduce some terminology. Call a set S ∈ T distinguished if N [S] ⊆ T ∪ T 0 . Thus if (T , T 0 ) is an isolated partial solution then every set in T is distinguished. At the beginning of each building stage, the algorithm adds a set S to T and this increases the cost of the solution. Since the algorithm considers an isolated partial solution at the beginning of each building stage, no set in N [S] has a nonempty intersection with a set in T . In the isolation phase S becomes distinguished. The algorithm considers all bipartitions (A, A0 ) of N [S] that satisfy the property that each set in T (which includes the newly added set S as well) uniquely covers at least one element in the family T ∪A. This ensures that the profit of the uniquely covered elements increases at least in this first step of the isolation phase. Subsequent additions to T do not change the elements uniquely covered by the distinguished sets of T . Thus the profit does not decrease during the isolation phase. Theorem 23 Let (U, S, c, p, B, k) be an instance of the Budgeted Unique Coverage problem such that for every set S ∈ S we have |N [S]| ≤ f (k). Then there is an algorithm with time complexity O∗ ((B 2 · f (k) · 2f (k) )min{B,k} ) for this problem. The Budgeted Max Cut problem is a special case where |N [S]| ≤ k − 1 for all S ∈ S, and the following corollary is immediate. Corollary 24 The Budgeted Max Cut problem with positive integer costs and profits is fixed-parameter tractable when parameterized by B and k. There is an algorithm with time complexity O∗ ((B 2 · k · 2k )min{B,k} ) for this problem. This algorithm can be used for the special case of the (unweighted) Unique Coverage problem where every set in the input family has a nonempty intersection with at most f (k) sets. Note that B ≤ k and for the unweighted case we have the following result. Corollary 25 Let (U, S, k) be an instance of (unweighted) Unique Coverage where every set in S intersects with at most f (k) distinct sets of the family, where f is a function of k alone. Then this instance can be solved in time O∗ ((k 2 · f (k) · 2f (k) )k ). Hence this algorithm is better than that developed in Section 4.3 (Corollary 14) whenever log f (k) + f (k) < 2(k − log k). 15
6
Concluding Remarks
In this paper, we considered the parameterized complexity of the Unique Coverage problem. We note that our FPT algorithms are applicable even under the generalized objective function defined in terms of satisfactions s0 = 0, s1 ≥ s2 ≥ . . . ≥ 0 [4,6], where an element u of the ground set yields a satisfaction modulated profit si · p(u) if it is covered by exactly i sets of the solution. For the (unweighted) Unique Coverage problem, the objective P function is then i si · ni where ni elements are covered exactly i times and P for the Budgeted Unique Coverage problem it is u si · p(u). It is not difficult to see that our kernel result for the (unweighted) Unique Coverage problem shows that the (unweighted) version with this generalized objective function is FPT too. For the budgeted version, if we assume that every set in the input family has a nonempty intersection with at most f (k) sets in the family, then Lemma 22 holds and the algorithm outlined in Section 5.2 works here as well. There are several directions in which to proceed. Firstly, the reduction rules that we give for the unweighted case are simple and the kernel that we obtain is exponential in k. Kernelization is a very important topic in the design of FPT algorithms and the challenge is to devise reduction rules to obtain a polynomial (linear?) kernel or prove that no such kernel exists under some plausible complexity-theoretic assumption. Are there reduction rules that lead to a better problem kernel? In particular, is there a polynomial kernel for the Unique Coverage problem? What about the variant Unique Coverage on Unit Disks introduced by Erlebach et al. [6]? We can show that with respect to a broader setqof reduction rules, which we state below, the kernel size is at least Ω(2k / k/2). The input instance is (U = {1, 2, . . . , n}, S = {S1 , S2 , . . . , Sm }, k). R3 If S covers at most k−1 elements, then the given instance is a no-instance. R4 Suppose the input satisfies the following: (1) S covers exactly k elements, and (2) any two sets in the family have a non-empty intersection. Then the given instance is a no-instance. R5 If there exists distinct sets Si , Sj ∈ S such that |Si ∆Sj | ≥ k, then the given instance is a yes-instance. Here ∆ denotes the symmetric difference. R6 If there exists S ∈ S such that for all S 0 ∈ S \ S, we have S ∩ S 0 = ∅, then delete S and set k := k − |S|. R7 If an element e ∈ U appears in each set of S, then delete e from U and all the sets in S. This element cannot be uniquely covered. The following example illustrates the situation where the kernel-size can be q k Ω(2 / k/2) under reduction rules R1 through R7. Let U = {1, 2, . . . , k}, S = S1 ∪ S2 , where S1 consists of all subsets of U of size exactly dk/2e + 1 and S2 is some collection of subsets of U of size at most k/4. Note that |S1 | = 16
k dk/2e+1
q
, which by Stirling’s approximation is, Ω(2k / k/2). If S2 = ∅ then the given instance is a no-instance by Rule R4. But if we take S2 = {{dk/2e+ 2}, . . . , {k}}, then this is a yes-instance and one can show that the reduction rules do not change theqinput size. Thus there are yes and no instances of the problem with Ω(2k / k/2) sets unaltered by the reduction rules. Another important question is whether there exists a good branching algorithm for Unique Coverage and, in particular, for Unique Coverage on Unit Disks. The algorithm that we gave for Unique Coverage runs in time 2 O∗ (4k ). Finally, is the Budgeted Unique Coverage problem with positive integer costs/profits, with parameters B and k, fixed-parameter tractable? Acknowledgements We thank Saket Saurabh for pointing out the connection between Unique Coverage and strong systems of distinct representatives and for several fruitful discussions.
References [1] R. Bar-Yehuda, O. Goldreich, and A. Itai. On the time-complexity of broadcast in multi-hop radio networks: An exponential gap between determinism and randomization. Journal of Computer and System Sciences, 45:104–126, 1992. [2] H. L. Bodlaender, R. Downey, M. R. Fellows, M. Hallett, and H. Wareham. Parameterized complexity analysis in computational biology. Computer Applications in the Biosciences, 11(1):49–57, 1995. [3] M. Cesati. Perfect code is W[1]-complete. Information Processing Letters, 81(3):163–168, 2002. [4] E. D. Demaine, M. T. Hajiaghayi, U. Feige, and M. R. Salavatipour. Combination can be hard: approximability of the unique coverage problem. In Proceedings of the Seventeenth Annual ACM-SIAM Symposium on Discrete Algorithms (SODA 2006), pages 162–171. SIAM, 2006. [5] R. Downey and M. R. Fellows. Parameterized Complexity. Springer Verlag, New York, 1999. [6] T. Erlebach and E. J. van Leeuwen. Approximating geometric coverage problems. In Proceedings of the Nineteenth Annual ACM-SIAM Symposium on Discrete Algorithms (SODA 2008), pages 1267–1276. SIAM, 2008. [7] J. Flum and M. Grohe. Parameterized Complexity Theory. Springer Verlag, Berlin Heidelberg, 2006. [8] M. R. Garey and D. S. Johnson. Computers and Intractability: A Guide to the Theory of NP-completeness. W. H. Freeman., 1979. [9] S. Jukna. Extremal Combinatorics. Springer-Verlag, Berlin, 2001.
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[10] R. Niedermeier. An Invitation to Fixed-Parameter Algorithms. University Press, 2006.
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[11] K. Pietrzak. On the parameterized complexity of the fixed alphabet shortest common supersequence and longest common subsequence problems. Journal of Computer and System Sciences, 67(4):757–771, 2003.
18
Venkatesh Raman and Somnath Sikdar 2 The Institute of Mathematical Sciences, C.I.T Campus, Taramani, Chennai 600113, India
Abstract We consider the parameterized complexity of the Unique Coverage problem: given a family of sets and a parameter k, we ask whether there exists a subfamily that covers at least k elements exactly once. This NP-complete problem has applications in wireless networks and radio broadcasting and is also a natural generalization of the well-known Max Cut problem. We show that this problem is fixed-parameter tractable with respect to the parameter k. We also show a 4k kernel for this problem. However a more general weighted version, with costs associated with each set and profits with each element, turns out to be much harder. The question here is whether there exists a subfamily with total cost at most a prespecified budget B such that the total profit of uniquely covered elements is at least k. In the most general setting, assuming real costs and profits, the problem is not fixed-parameter tractable unless P = NP. Assuming integer costs and profits we show the problem to be W[1]-hard with respect to B as parameter. However, under some reasonable restriction, the problem becomes fixed-parameter tractable with respect to both B and k as parameters. Key words: Parameterized Complexity, Covering Problems, Kernelization
Email addresses: [email protected] (Hannes Moser), {vraman|somnath}@imsc.res.in (Venkatesh Raman and Somnath Sikdar). 1 Supported by the Deutsche Forschungsgemeinschaft, project ITKO (iterative compression for solving hard network problems), NI 369/5. 2 Supported by a DAAD-DST exchange program, D/05/57666. 3 An extended abstract of this work appeared in the proceedings of the 18th Inter-
Preprint submitted to Elsevier
25 April 2008
1
Introduction
In this paper, we consider the parameterized complexity of the Unique Coverage problem. This problem was introduced by Demaine et al. [4] as a natural maximization version of Set Cover and has applications in several areas including wireless networks and radio broadcasting. Unique Coverage is defined as follows. Given a ground set U = {1, 2, . . . , n}, a family of subsets S = {S1 , . . . , St } of U and a nonnegative integer k, we ask whether there exists a subcollection S 0 ⊆ S such that at least k elements are covered uniquely by the members in S 0 . An element is covered uniquely if it appears in exactly one set of S 0 . The optimization version requires to maximize the number of uniquely covered elements. The weighted version of Unique Coverage is called Budgeted Unique Coverage and is defined as follows. Given a ground set U = {1, 2, . . . , n}, a profit pi for each element i ∈ U, a family of subsets S of U, a cost ci for each set Si ∈ S, a budget B and a nonnegative integer k, we ask whether there exists a subset S 0 ⊆ S such that the total cost of S 0 is at most B and such that the profit of the uniquely covered elements is at least k. The optimization version asks for a subset S 0 of total cost at most B such that the profit of uniquely covered elements is maximized. The original motivation for this problem is a real-world application arising in wireless networks [4]. Assume that we are given a map of the densities of mobile clients along with a set of possible base stations, each with a specified building cost and a specified coverage region. The goal is to choose a set of base stations, subject to a budget on the total building cost, in order to maximize the density of served clients. The difficult aspect of this problem is the interference between base stations. A mobile client’s reception is better when it is within the range of a few base stations. An ideal situation is when every mobile client is within the range of exactly one base station. This is the situation modelled by the Budgeted Unique Coverage problem. The Unique Coverage problem is closely related to a single “round” of the Radio Broadcast problem [1]. For more about this relation, see Demaine et al.’s work [4]. One can also view the Unique Coverage problem as a generalization of the Max Cut problem [4]. The input to the Max Cut problem consists of a graph G = (V, E) and the goal is to find a cut (T, T 0 ), where ∅ = 6 T ⊂V 0 and T = V \ T , that maximizes the number of edges with one endpoint in T and the other endpoint in T 0 . Let U denote the set of edges of G and for each vertex v ∈ V define Sv = {e ∈ E : e is incident to v}. Finally let S = ∪v∈V {Sv }. Then G has a cut (T, T 0 ) with at least k edges across it if and only if S 0 = ∪v∈T {Sv } uniquely covers at least k elements of the ground set. national Symposium on Algorithms and Computation (ISAAC’07), Springer, LNCS 4835, pages 621–631.
2
Known Results. (Budgeted) Unique Coverage was introduced by Demaine et al. [4]. They have considered the approximability of this problem. On the positive side, they give an Ω(1/ log n)-approximation for Budgeted Unique Coverage. Moreover, if the ratio between the maximum cost of a set and the minimum profit of an element is bounded by B, then there exists an Ω(1/ log B)approximation. They show that Unique Coverage is hard to approximate to within a factor of O(1/ logc n) for some constant c depending on , assum ing NP * BPTIME(2n ) for some > 0. They strengthen this inapproximability to 1/( log n) for some > 0 based on a hardness hypothesis for Balanced Bipartite Independent Set. Erlebach and van Leeuwen [6] study the approximability of geometric versions of the Unique Coverage problem. Among the many versions that they consider is Unique Coverage on Unit Disks, a variant in which each set is a unit disk in R2 , for which they give a factor-18 approximation algorithm. They also consider a variant called Unique Coverage on Disks of Bounded Ply and design an asymptotic fully polynomial-time approximation scheme (FPTASω ) for it. Our Results. In this paper, we give first-time results on the parameterized complexity of Unique Coverage and Budgeted Unique Coverage. As in many problems in parameterized complexity, the Unique Coverage problem can be parameterized in a number of ways. We first consider an extensive list of plausible parameterizations of the problem in Section 3 and discuss their parameterized complexity. Our results show that barring the standard parameterized version (where the parameter is the number of uniquely covered elements) the remaining parameterized problems are unlikely to be fixed-parameter tractable. In Section 4 we consider the standard parameterized version of Unique Coverage. We show that a special case of this version where any two sets in the input family intersect in at most c elements is fixed-parameter tractable by demonstrating a polynomial kernel of size k c+1 . This leads to a problem kernel of size k k in the general case, proving that Unique Coverage is fixed-parameter tractable. Then using results from extremal combinatorics on strong systems of distinct representatives we obtain a 4k kernel. In Section 5 we consider the Budgeted Unique Coverage problem. For this problem too there are several variants. If the profits and costs are allowed to be arbitrary positive real numbers, then Budgeted Unique Coverage, with parameters B and k, is not fixed-parameter tractable unless P = NP. If we restrict the costs and profits to be positive integers and parameterize by B, then the problem is W [1]-hard. In the case when the number of sets intersecting any given set of the input family is bounded by a function of k, 3
Unique Coverage (Parameter: k) Result
Sect.
Each element occurs in at most b sets Intersection size bounded by c
(k − 1)b kernel
4.1
k c+1 kernel
4.2
General case
4k kernel
4.3
Each set of size at most b
2b+k kernel
4.3
Arbitrary costs/profits (parameters B and k)
Not FPT (unless P = NP)
5.1
Integer weights (parameter B)
W [1]-hard
5.1
Integer weights (intersection number f (k); parameters B and k)
O∗ ((B 2 · f (k) · 2f (k) )min{B,k} ) 5.2 time algorithm
Integer weights (parameters B and k)
Open
Budgeted Unique Coverage
Fig. 1. Main results in this paper.
the problem is fixed-parameter tractable with parameters B and k. The main results of this paper along with the relevant sections in which they appear are depicted in Figure 1. The version in which the profits and costs are positive integers and the parameters are both B and k is open. We note that the parameterized versions of all variants of Unique Coverage considered in [6] are fixed-parameter tractable. Unique Coverage on Unit Disks is fixed-parameter tractable with respect to the number of uniquely covered elements as parameter since it is a restriction of the problem that we show to be fixed-parameter tractable in this paper. It is well-known that an optimization problem with an FPTAS is fixed-parameter tractable [5] and it can be easily shown that the same result holds for problems with an asymptotic FPTAS. Therefore variants such as Unique Coverage on Disks of Bounded Ply are fixed-parameter tractable (with respect to the cost and budget as parameters). 4
2
Preliminaries
We briefly introduce the necessary concepts concerning parameterized complexity. A parameterized problem is a subset of Σ∗ × N, where Σ is a finite alphabet and N is the set of natural numbers. An instance of a parameterized problem is therefore a pair (I, k), where k is the parameter. In the framework of parameterized complexity, the running time of an algorithm is viewed as a function of two quantities: the size of the problem instance and the parameter. A parameterized problem is said to be fixed parameter tractable (FPT) if there exists an algorithm for the problem with running time f (k) · |I|O(1) , where f is a computable function only depending on k. We sometimes refer to such an algorithm as an FPT algorithm. A common method to prove that a problem is fixed-parameter tractable is to provide data reduction rules that lead to a problem kernel. A data reduction rule is a polynomial-time algorithm which takes a problem instance (I, k) and either • outputs yes or no according as (I, k) is a yes or a no-instance, or • replaces (I, k) by an equivalent instance (I 0 , k 0 ) such that |I 0 | ≤ |I| and k 0 ≤ k, where two problem instances (I, k) and (I 0 , k 0 ) are equivalent if they are both yes-instances or both no-instances. An instance to which none of a given set of data reduction rules applies is called reduced with respect to this set of rules. A parameterized problem is said to have a problem kernel if the resulting reduced instance has size f (k) for a function f depending only on k. If a parameterized problem has a kernel, then it is clearly fixed-parameter tractable: simply use brute-force on the kernel to decide whether the given instance is a yes-instance or not. The converse is also true: if a parameterized problem is fixed-parameter tractable, then it has a kernel. A parameterized problem π1 is fixed-parameter reducible to a parameterized problem π2 if there exist functions f, g : N → N, Φ : Σ∗ × N → Σ∗ and a polynomial p(·) such that for any instance (I, k) of π1 , (Φ(I, k), g(k)) is an instance of π2 computable in time f (k) · p(|I|) and (I, k) ∈ π1 if and only if (Φ(I, k), g(k)) ∈ π2 . The basic complexity class for fixed-parameter intractability is W [1] as there is strong evidence that W [1]-hard problems are not fixed-parameter tractable [5]. To show that a problem is W [1]-hard, one needs to exhibit a fixed-parameter reduction from a known W [1]-hard problem to the problem at hand. For more on parameterized complexity see [5,7,10]. We write O∗ (f (k)) to denote a running time of O(f (k) · poly(n, k)), where n is the input size and k is the parameter. That is, we use the O∗ (·) notation to suppress polynomial factors in the running time. 5
3
The Unique Coverage Problem: Which Parameterization?
The Unique Coverage problem can be parameterized in a number of ways. We begin with a general discussion on different parameterizations of the problem and show that except for the standard parameterized version (where the parameter is the solution size) and its variant, the remaining parameterizations are unlikely to be fixed-parameter tractable. An interesting analog is the Longest Common Subsequence problem which can be parameterized in a number of ways, some of which are fixed-parameter tractable and the others hard [2,11]. First consider the following parameterized problem: given (U, S), find a subfamily S 0 ⊆ S that covers all of U, each element being covered at most k times and at least once, assuming k as parameter. This is a practical parameterization for mobile-computing applications. Unfortunately, this problem is not fixed-parameter tractable unless P = NP as the case k = 1 reduces to the NP-complete Exact Cover problem [8]. An alternate parameterization with a similar motivation of covering each element a small number of times is as follows: given (U, S), find a subfamily S 0 ⊆ S that covers all of U, each element being covered at most |S| − k times. This problem can be rephrased as follows: Is there a subfamily S 00 ⊆ S of size at most k such that S 0 = S −S 00 covers all elements of U at most |S| − k times? Call this problem All But k Coverage. Lemma 1 Suppose an instance (U, S, k) of All But k Coverage is such that there exists an element e ∈ U that is contained in every set of S. Then the given instance is a yes-instance if and only if there exists a subfamily S 00 ⊆ S of size exactly k such that S − S 00 covers all elements of U at most |S| − k times. PROOF. The proof consists in observing that if an element occurs in every set of S then any solution must necessarily exclude at least k sets of S. Since |S| − k sets can cover any element at most |S| − k times, excluding k sets is sufficient. 2 We show that this problem is W[1]-hard by a fixed-parameter reduction from the W[1]-complete Red/Blue Nonblocker problem [5]: A bipartite graph G = (R ] B = V, E) with its vertex set partitioned into a red set R and a blue set B, and a positive integer k. Parameter: The positive integer k. Question: Does there exist a set T of k red vertices such that the vertices in R − T dominate all vertices in B? In order to show that All But k Coverage is W[1]-hard, we consider a special case of the Red/Blue Nonblocker problem where we are guaran-
Input:
6
teed that there exists a blue vertex that is adjacent to all red vertices. This special case is W[1]-hard too. Given an instance (G, k) of Red/Blue Nonblocker create a special blue vertex v ∗ and add edges from v ∗ to all red vertices. Then the original instance has a size-k solution if and only if the transformed instance has a size-k solution. Given an instance (G = (R ] B, E), k) of this special case of Red/Blue Nonblocker, create an instance (U, S, k) of All But k Coverage by identifying the universe U with the blue vertices and the sets of the input family S with the red vertices. A set contains an element if and only if the corresponding red-blue vertex-pair is connected by an edge. Lemma 2 There exists a set R0 ⊆ R of size exactly k such that R − R0 dominates all of B if and only if there exist a subfamily S 00 of size k such that S − S 00 covers all of U, each element being covered at most |S| − k times. PROOF. Given any set R0 of size exactly k such that R − R0 dominates all of B, define S 00 to be the family consisting of the sets corresponding to the vertices of R0 . Clearly S − S 00 covers all of U. As |S − S 00 | = |S| − k, each element of U is covered at most |S| − k times. Conversely, suppose S 00 ⊆ S is of size k such that S − S 00 covers all of U, each element being covered at most |S| − k times. Define R0 to be the set of red vertices corresponding to S 00 . Then |R0 | = k and R − R0 dominates all of B. This completes the proof of the lemma. 2 We therefore have the following: Theorem 3 All But k Coverage is W[1]-hard with respect to k as parameter. We also note that the following parameterized versions of Unique Coverage are not fixed-parameter tractable. Given (U, S) and nonnegative integers k and t as parameters, (1) Does there exist a subfamily S 0 ⊆ S of size at most k that covers all of U with each element being covered at most t times? This version is not fixed-parameter tractable as the case t = 1 is W[1]-hard by a reduction from Perfect Code [3] as shown below. (2) Does there exist a subfamily S 0 ⊆ S of size at most k that covers all of U with each element being covered at most |S| − t times? This is W[2]-hard as the case t = 0 reduces to the Set Cover problem which is W[2]-complete. Call the version in Item 1 above with t = 1 the Disjoint Set Cover problem which we now show to be W[1]-hard by a reduction from Perfect Code [3]. Perfect Code Input: A graph G = (V, E) and an integer k. Parameter: The integer k. 7
Question:
Does G have a k-element perfect code? A perfect code is a vertex subset V 0 ⊆ V such that for all u ∈ V we have |N [u] ∩ V 0 | = 1.
Given an instance (G = (V, E), k) of Perfect Code construct an instance (U, S, k) of Disjoint Set Cover as follows: U = V and S = {N [v] | v ∈ V }. Lemma 4 The graph G has a k-element perfect code if and only if there exists a subfamily S 0 ⊆ S of pairwise disjoint sets and of size at most k that covers all of U. PROOF. Let {v1 , . . . , vk } ⊆ V be a k-element perfect code of G. Then S clearly ki=1 N [vi ] covers all of U and the sets N [v1 ], . . . , N [vk ] are pairwise disjoint. For if x ∈ N [vi ] ∩ N [vj ], i 6= j, then |N [x] ∩ {v1 , . . . , vk }| ≥ 2, which contradicts the definition of a perfect code. Conversely if N [v1 ], . . . , N [vk ] is a collection of pairwise disjoint sets that covers all of U then clearly v1 , . . . , vk is a k-element perfect code for G. 2 Theorem 5 Disjoint Set Cover is W[1]-hard with respect to the number of sets in the solution as parameter. We next consider a generalization of the standard parameterized version: • Gen Unique Coverage: Given (U, S) and nonnegative integers k and t, does there exist a subfamily of S that covers k elements at least once and at most t times? Setting t = 1 gives us the standard parameterized version of Unique Coverage. We note that Gen Unique Coverage is fixed-parameter tractable as the kernel result for the standard parameterized version works for this problem also. We elaborate this further in Section 4.3. From now on we work with the standard parameterized version of Unique Coverage: Given (U, S) and a nonnegative integer k, is there a subfamily of S that uniquely covers at least k elements?
4
The Unique Coverage Problem: The Standard Parameterization
Let (U = {1, 2, . . . , n}, S = {S1 , S2 , . . . , Sm }, k) be an instance of Unique Coverage. Apply the following data reduction rules on (U, S, k) until no longer applicable. R1 If there exists Si ∈ S such that |Si | ≥ k, then the given instance is a yes-instance. R2 If there exists S1 , S2 ∈ S such that S1 = S2 , then delete S1 . These reduction rules are obviously correct. In the first case Si gives the solution. In the second case, it is clear that no solution need have both S1 and S2 . In the following we always assume that the given instance of Unique Coverage is reduced with respect to the above rules. 8
4.1
Bounded Number of Occurrences
We begin with the simple case where each element e ∈ U is contained in at most b sets of S. A special case of this situation is Max Cut (b = 2). Lemma 6 If each element e ∈ U occurs in at most b sets of S then the Unique Coverage problem admits a kernel of size b(k − 1). PROOF. Find a maximal collection S 0 of pairwise disjoint sets in S. If |∪S∈S 0 S| ≥ k, we are done. Therefore assume that | ∪S∈S 0 S| ≤ k − 1. Since every set in S − S 0 intersects some set in S 0 and since every element occurs in at most b sets in S, we have |S − S 0 | ≤ (k − 1)(b − 1). But |S 0 | ≤ k − 1 and so |S| ≤ b(k − 1). 2 4.2
Bounded Intersection Size
Consider the situation where for all Si , Sj ∈ S we have |Si ∩ Sj | ≤ c, for some constant c. In this case we say that the problem instance has bounded intersection size c and show that the problem admits a polynomial kernel of size O(k c+1 ). First consider the case when |Si ∩ Sj | ≤ 1. Lemma 7 Suppose that for all Si , Sj ∈ S, i 6= j, we have |Si ∩ Sj | ≤ 1. If an element e ∈ U is covered by at least k + 1 sets, then one can obtain a solution covering k elements uniquely in polynomial time. PROOF. Suppose an element e ∈ U is covered by the sets S1 , . . . , Sk+1 . Then by reduction rule R2, at most one of these sets can have size 1. The remaining k sets uniquely cover at least one element each. 2 One can now easily obtain a kernel of size k 2 for the case when the intersection size is at most 1. Lemma 8 Suppose that for all Si , Sj ∈ S, |Si ∩ Sj | ≤ 1. If |S| ≥ k 2 , then there exists T ⊆ S that covers at least k elements uniquely. PROOF. If an element appears in at least k + 1 sets then we are done by Lemma 7. Otherwise every element appears in at most k sets and by Lemma 6 we have a kernel of size k(k − 1) < k 2 . 2 Next, we generalize these observations to the case when |Si ∩ Sj | ≤ c, for some constant c. Theorem 9 Suppose that for all Si , Sj ∈ S we have |Si ∩ Sj | ≤ c, for some positive constant c. If |S| ≥ k c+1 then there exists T ⊆ S that covers k elements uniquely. PROOF. By induction on c. For c = 1, this follows from Lemma 8. Assume the theorem to hold for c > 1. Greedily obtain a maximal collection S 0 = {S1 , . . . , Sp } of pairwise disjoint sets. If | ∪Si ∈S 0 Si | ≥ k then we are done. 9
Therefore assume | ∪S∈S 0 S| ≤ k − 1 (this also implies p ≤ k − 1). Since |S| ≥ k c+1 , and since every set in S intersects with at least one set in S 0 , there exists e ∈ ∪S∈S 0 S such that at least k c + 1 sets in S − {S1 , . . . , Sp } contain e. For otherwise, |S| ≤ (k − 1)k c + p < k c+1 , a contradiction. Let T1 , . . . , Tkc +1 be some k c + 1 such sets. Delete e from each of these sets. We obtain at least k c nonempty distinct sets T10 , . . . , Tk0 c (there is at most one set consisting only of the element e which is deleted in this process). Note that any two of these sets intersect in at most c − 1 elements. By induction hypothesis, there exists a collection T 0 ⊆ {T10 , . . . , Tk0 c } that uniquely covers at least k elements, and thus there exists a collection T ⊆ {T1 , . . . , Tkc } that uniquely covers at least k elements (just take the solution for T 0 and add e to every set in it). This proves the theorem. 2 Corollary 10 Unique Coverage admits a kernel of size k c+1 for bounded intersection size c. By reduction rule R1 we have c ≤ k − 1 and therefore for the general case we have a kernel of size k k . Corollary 11 The Unique Coverage problem is fixed-parameter tractable and admits a problem kernel of size k k . An algorithm that checks all possible subsets of a family of size k k to see whether any of them uniquely covers at least k elements is an FPT algorithm k with time complexity O∗ (2(k ) ). But note that we can assume without loss of generality that every set in the solution covers at least one element uniquely. Thus it suffices to check whether subfamilies of size at most k uniquely cover 2 2 at least k elements. This can be done in time O∗ (k k ) = O∗ (2k log k ). However, this kernelization result is tailored especially for the bounded intersection size case. It turns out that a much better kernel can be obtained for the general case. 4.3
A Better Kernel for the General Case
We now show that Unique Coverage has a kernel of size 4k using a result on strong systems of distinct representatives. Given a family of sets S = {S1 , . . . , Sm }, a system of distinct representatives for S is an m-tuple (x1 , . . . , xm ) where the elements xi are distinct and xi ∈ Si for all i = 1, 2, . . . , m. Such a system is strong if we additionally have xi ∈ / Sj for all i 6= j.
Theorem 12 ([9]) In any family of more than r+s sets of cardinality at s most r, at least s + 2 of its members have a strong system of distinct representatives. Given an instance (U = {1, . . . , n}, S = {S1 , . . . , Sm }, k) of Unique Coverage, put r = k − 1 and s = k in the statement of the above theorem and we have a kernel of size 2k−1 ≤ 2k ≤ 22k . k−1 k 10
Corollary 13 Unique Coverage admits a problem kernel of size 4k . 2
Corollary 14 There is an O∗ (4k ) time algorithm for the Unique Coverage problem. PROOF. A subfamily that covers k elements uniquely has size at most k. Therefore it is sufficient to consider all possible size-k subfamilies of the 4k 2 kernel. This takes time O∗ (4k ). 2 At this point, we note that Gen Unique Coverage (see Section 3) is fixedparameter tractable: an instance (U, S, k, t) of Gen Unique Coverage is trivially a yes-instance if |S| > 4k . Otherwise |S| ≤ 4k and we have a kernel. Corollary 15 Gen Unique Coverage is fixed-parameter tractable. For the case where each set of the input family has size at most b, for some constant b, there is a better kernel. By Theorem 12, if there exists at least b+k k sets in the input family, then there exists at least k sets with a strong system of distinct representatives. Corollary 16 If each set S ∈ S has size at most b then the Unique Coverage problem has a kernel of size O(2b+k ). Another interesting case is when each set of the input family S has a nonempty intersection with at most f (k) sets in S, where f is a function of k alone. We consider this case in Section 5.2, Corollary 25, and show that this version can be solved in time O∗ ((k 2 · f (k) · 2f (k) )k ).
5
The Budgeted Unique Coverage Problem
In this section we consider the Budgeted Unique Coverage problem where each set in the input family has a cost and each element in the universe has a profit; the goal is to decide whether there exists a subcollection of total cost at most B that uniquely covers elements of total profit at least k. By parameterizing on k or B or both we obtain different parameterized versions of this decision question. 5.1
Budgeted Unique Coverage: Intractable Parameterized Versions
We first consider the Budgeted Max Cut problem which is a specialization of the Budgeted Unique Coverage problem. An instance of this problem is an undirected graph G = (V, E) with a cost function c : V → R+ on the vertex set and a profit function p : E → R+ on the edge set; positive real numbers B and k. The question is whether there exists a cut (T, T 0 ) such that the total cost of the vertices in T is at most B and the total profit of the edges crossing the cut is at least k. 11
We first show that the Budgeted Max Cut problem with arbitrary positive real costs and profits is probably not FPT. Lemma 17 The Budgeted Max Cut problem with arbitrary positive costs and profits with parameters B and k is not FPT, unless P = NP. PROOF. Suppose there exists an algorithm for the Budgeted Max Cut problem (with arbitrary positive costs and profits) with run-time O(f (k, B) · p(n)), where p is a polynomial in n. We will use this to solve the decision version of Max Cut in polynomial time. Let (G = (V, E), k) be an instance of the Max Cut problem, where |V | = n. Assign each vertex of the input graph cost 1/n and each edge profit 1/k. Let the budget B = 1/2 and the profit k 0 = 1. Clearly, G has a maximum cut of size at least k if and only if there exists S ⊆ V of total cost at most B such that the total profit of the edges crossing the cut (S, V − S) is at least k 0 . And this can be answered in time O(f (1, 1/2) · p(|V |)), implying P = NP. 2 Theorem 18 The Budgeted Unique Coverage problem with arbitrary positive costs and profits is not FPT, unless P = NP. Henceforth by the ‘budgeted’ version we mean the case when the costs and profits are positive integers. We next show that the Budgeted Max Cut problem parameterized by the budget B alone is W [1]-hard. Lemma 19 The Budgeted Max Cut problem parameterized by the budget is W [1]-hard. PROOF. To show W [1]-hardness, we exhibit a fixed-parameter reduction from the Independent Set problem to the Budgeted Max Cut problem with unit costs and profits. Let (G = (V, E), B) be an instance of Independent Set with |V | = n. For every vertex u ∈ V add |V | − 1 − deg(u) new vertices and connect them to u. Call the resulting graph G0 . Define c(v) = 1 for all v ∈ V (G0 ) and p(e) = 1 for all e ∈ E(G0 ). Note that every vertex u ∈ G has degree |V | − 1 in G0 . We let (G0 = (V 0 , E 0 ), B, k = B(n − 1)) be the instance of Budgeted Max Cut. Claim 20 G has an independent set of size B if and only if G0 has a cut (S, V 0 − S) such that |S| = B and at least k = B(n − 1) edges lie across it. If G has an independent set S of size B, then clearly S is independent in G0 . The cut (S, V 0 − S) does indeed have B(n − 1) edges crossing it, as every vertex of S has degree n − 1. Next suppose that G0 has a cut (S, V 0 − S) such that |S| = B and B(n − 1) edges cross the cut. Note that every vertex in S must be a vertex from G. Otherwise the cut cannot have B(n − 1) edges crossing it. Suppose two vertices u and v in S are adjacent. Then both u and v contribute less than n − 1 edges to the cut. Since each vertex in S contributes at most n − 1 edges to the cut, the number of edges crossing the cut must be less than B(n − 1), a contradiction. Hence S is independent in G0 and hence G has an independent set of size B. 2 12
Since the Budgeted Unique Coverage problem is a generalization of Budgeted Max Cut we have Theorem 21 The Budgeted Unique Coverage problem parameterized by the budget B is W [1]-hard. 5.2
Budgeted Unique Coverage: A Special Case
In this subsection, we give an FPT algorithm for Budgeted Unique Coverage, when B and k are parameters, assuming that for every set S in the input family the number of sets with a non-empty intersection with S is at most some function of k. This is a natural situation in real-world applications; for example, in wireless networks. For the Budgeted Max Cut problem, for instance, every set is intersected by at most k − 1 sets. This follows because every vertex has degree at most k − 1 and any two vertices share at most one edge. Let (U = {1, . . . , n}, S = {S1 , . . . , Sm }, c, p, B, k) be an instance of the Budgeted Unique Coverage problem where c : S → N and p : U → N. P For T ⊆ S, define c(T ) = S∈T c(S) and p(T ) to be the total profit of the elements uniquely covered by T . If Si ∈ S, define N [Si ] to be the set of all members of S that have a nonempty intersection with Si . We can without loss of generality assume that c(Si ) ≤ B and |Si | ≤ k − 1 for all 1 ≤ i ≤ m. In what follows, we assume that for all Si ∈ S, we have |N [Si ]| ≤ f (k) for some function f . The FPT algorithm that we describe here builds the solution in stages. Note that if we decide to include a set S in the solution, there is no way of deciding how many elements S covers uniquely unless we make choices for each set in N [S]. To get around this, the algorithm, at any stage, decides whether or not to include a subfamily A ⊆ N [S] for some set S. If it includes A in the solution, then it automatically excludes N [S] \ A from it. The current solution is a pair (T , T 0 ), where T , T 0 ⊆ S and T ∩ T 0 = ∅. The sets included by the algorithm in the solution till the current stage are in T ; those excluded from the solution are in T 0 . Call a pair (T , T 0 ) a feasible solution for an instance of Budgeted Unique Coverage if T 0 = S − T , c(T ) ≤ B and p(T ) ≥ k. A pair (T , T 0 ) is a partial solution if T , T 0 ⊆ S and T ∩ T 0 = ∅. A partial solution (T , T 0 ) can be extended to a feasible solution if there exist X , X 0 ⊆ S − (T ∪ T 0 ) such that X ∩ X 0 = ∅ and (T ∪ X , T 0 ∪ X 0 ) is a feasible solution. A partial solution (T , T 0 ) is isolated if for each set Si ∈ T , N [Si ] ⊆ T ∪ T 0 . Note that if a partial solution (T , T 0 ) is isolated then for all S ∈ S − (T ∪ T 0 ) we S have S ∩ T ∈T T = ∅. If a partial solution does not satisfy this property we call it non-isolated. Given an isolated partial solution (T , T 0 ), let U1 , . . . , Ut be a partition of S − (T ∪ T 0 ) according to costs. That is, all members in any set Ui have the same cost ci and for all i 6= j, ci 6= cj . Note that t ≤ B. For 13
each Ui , let Uimax denote a member of Ui with maximum total profit. Lemma 22 Let (T , T 0 ) be an isolated partial solution and let U1 , U2 , . . . , Ut be a partition of S − (T ∪ T 0 ) according to costs. Suppose (T , T 0 ) can be extended to a feasible solution by adding a member of Ui to T . Then there exists an extension of (T , T 0 ) into a feasible solution (T ∪ X , T 0 ∪ X 0 ) such that X ∩ N [Uimax ] 6= ∅. PROOF. Suppose (T , T 0 ) can be extended to a feasible solution (T ∪X , T 0 ∪ X 0 ) by adding a member U ∈ Ui to T and that X ∩ N [Uimax ] = ∅. This means that N [Uimax ] ⊆ X 0 . Remove U from X and replace it by Uimax . Note that every element of Uimax is uniquely covered and that the total profit of these newly uniquely covered elements is at least as that of those covered by U . Since c(U ) = c(Uimax ), the new solution continues to be feasible. 2 One can use Lemma 22 to develop an FPT algorithm with time complexity O∗ ((B 2 · f (k) · 2f (k) )min{B,k} ). This algorithm works in stages; each stage consists of two phases: the building phase and the isolation phase. We will argue that at the end of each stage, the cost of the solution being constructed and the profit of the elements uniquely covered are strictly greater than what they were at the end of the previous stage. The number of stages required is therefore at most min{B, k}. We will then show that the algorithm makes at most B 2 · f (k) · 2f (k) branches at each stage which would then prove that the overall running time of the algorithm is O∗ ((B 2 · f (k) · 2f (k) )min{B,k} ). At the beginning of each stage, the algorithm starts with an isolated partial solution. At the beginning of the first stage, this isolated partial solution is (T = ∅, T 0 = ∅). The algorithm then partitions the family S − (T ∪ T 0 ) according to costs into the subfamilies U1 , . . . , Ut . If we assume that there exists a feasible solution, it has to include a set from one of the subfamilies Ui . For each choice of a subfamily, Lemma 22 assures us that it is sufficient to consider a set S ∈ N [Smax ] where Smax is a set in the subfamily that maximizes profit. Since t ≤ B and |N [Smax ]| ≤ f (k) there are at most B · f (k) ways of choosing this set S. The algorithm makes at most B · f (k) branches and chooses which set to include in the solution. If there is no subfamily Ui such that for U ∈ Ui , c(U ) ≤ B − c(T ), the algorithm aborts this branch. This ends the building-phase. Adding a set to an isolated solution may cause it to become non-isolated. In the isolation phase, the algorithm makes the current non-isolated partial solution isolated. It chooses a set T ∈ T such that N [T ] − (T ∪ T 0 ) 6= ∅ and considers all bipartitions (A, A0 ) of N [T ] − (T ∪ T 0 ) satisfying the property that each set in T uniquely covers at least one element in the family T ∪ A. It branches on each such bipartition and sets T ← T ∪ A and T 0 ← T 0 ∪ A0 . If there is no bipartition satisfying this property the algorithm aborts this branch. Since |T | can never exceed B and |N [T ]| ≤ f (k) for any T ∈ T , the number 14
of branches that the algorithm creates in an isolation phase at any stage is at most B · 2f (k) . Since the building phase produces at most B · f (k) branches and the isolation phase works on each branch generated by the building phase, the total number of branches at any stage is at most B 2 · f (k) · 2f (k) . To prove the claimed time complexity of our algorithm, we have to show that at the end of each stage both the cost and profit of the solution being constructed increase. To argue this it is helpful to introduce some terminology. Call a set S ∈ T distinguished if N [S] ⊆ T ∪ T 0 . Thus if (T , T 0 ) is an isolated partial solution then every set in T is distinguished. At the beginning of each building stage, the algorithm adds a set S to T and this increases the cost of the solution. Since the algorithm considers an isolated partial solution at the beginning of each building stage, no set in N [S] has a nonempty intersection with a set in T . In the isolation phase S becomes distinguished. The algorithm considers all bipartitions (A, A0 ) of N [S] that satisfy the property that each set in T (which includes the newly added set S as well) uniquely covers at least one element in the family T ∪A. This ensures that the profit of the uniquely covered elements increases at least in this first step of the isolation phase. Subsequent additions to T do not change the elements uniquely covered by the distinguished sets of T . Thus the profit does not decrease during the isolation phase. Theorem 23 Let (U, S, c, p, B, k) be an instance of the Budgeted Unique Coverage problem such that for every set S ∈ S we have |N [S]| ≤ f (k). Then there is an algorithm with time complexity O∗ ((B 2 · f (k) · 2f (k) )min{B,k} ) for this problem. The Budgeted Max Cut problem is a special case where |N [S]| ≤ k − 1 for all S ∈ S, and the following corollary is immediate. Corollary 24 The Budgeted Max Cut problem with positive integer costs and profits is fixed-parameter tractable when parameterized by B and k. There is an algorithm with time complexity O∗ ((B 2 · k · 2k )min{B,k} ) for this problem. This algorithm can be used for the special case of the (unweighted) Unique Coverage problem where every set in the input family has a nonempty intersection with at most f (k) sets. Note that B ≤ k and for the unweighted case we have the following result. Corollary 25 Let (U, S, k) be an instance of (unweighted) Unique Coverage where every set in S intersects with at most f (k) distinct sets of the family, where f is a function of k alone. Then this instance can be solved in time O∗ ((k 2 · f (k) · 2f (k) )k ). Hence this algorithm is better than that developed in Section 4.3 (Corollary 14) whenever log f (k) + f (k) < 2(k − log k). 15
6
Concluding Remarks
In this paper, we considered the parameterized complexity of the Unique Coverage problem. We note that our FPT algorithms are applicable even under the generalized objective function defined in terms of satisfactions s0 = 0, s1 ≥ s2 ≥ . . . ≥ 0 [4,6], where an element u of the ground set yields a satisfaction modulated profit si · p(u) if it is covered by exactly i sets of the solution. For the (unweighted) Unique Coverage problem, the objective P function is then i si · ni where ni elements are covered exactly i times and P for the Budgeted Unique Coverage problem it is u si · p(u). It is not difficult to see that our kernel result for the (unweighted) Unique Coverage problem shows that the (unweighted) version with this generalized objective function is FPT too. For the budgeted version, if we assume that every set in the input family has a nonempty intersection with at most f (k) sets in the family, then Lemma 22 holds and the algorithm outlined in Section 5.2 works here as well. There are several directions in which to proceed. Firstly, the reduction rules that we give for the unweighted case are simple and the kernel that we obtain is exponential in k. Kernelization is a very important topic in the design of FPT algorithms and the challenge is to devise reduction rules to obtain a polynomial (linear?) kernel or prove that no such kernel exists under some plausible complexity-theoretic assumption. Are there reduction rules that lead to a better problem kernel? In particular, is there a polynomial kernel for the Unique Coverage problem? What about the variant Unique Coverage on Unit Disks introduced by Erlebach et al. [6]? We can show that with respect to a broader setqof reduction rules, which we state below, the kernel size is at least Ω(2k / k/2). The input instance is (U = {1, 2, . . . , n}, S = {S1 , S2 , . . . , Sm }, k). R3 If S covers at most k−1 elements, then the given instance is a no-instance. R4 Suppose the input satisfies the following: (1) S covers exactly k elements, and (2) any two sets in the family have a non-empty intersection. Then the given instance is a no-instance. R5 If there exists distinct sets Si , Sj ∈ S such that |Si ∆Sj | ≥ k, then the given instance is a yes-instance. Here ∆ denotes the symmetric difference. R6 If there exists S ∈ S such that for all S 0 ∈ S \ S, we have S ∩ S 0 = ∅, then delete S and set k := k − |S|. R7 If an element e ∈ U appears in each set of S, then delete e from U and all the sets in S. This element cannot be uniquely covered. The following example illustrates the situation where the kernel-size can be q k Ω(2 / k/2) under reduction rules R1 through R7. Let U = {1, 2, . . . , k}, S = S1 ∪ S2 , where S1 consists of all subsets of U of size exactly dk/2e + 1 and S2 is some collection of subsets of U of size at most k/4. Note that |S1 | = 16
k dk/2e+1
q
, which by Stirling’s approximation is, Ω(2k / k/2). If S2 = ∅ then the given instance is a no-instance by Rule R4. But if we take S2 = {{dk/2e+ 2}, . . . , {k}}, then this is a yes-instance and one can show that the reduction rules do not change theqinput size. Thus there are yes and no instances of the problem with Ω(2k / k/2) sets unaltered by the reduction rules. Another important question is whether there exists a good branching algorithm for Unique Coverage and, in particular, for Unique Coverage on Unit Disks. The algorithm that we gave for Unique Coverage runs in time 2 O∗ (4k ). Finally, is the Budgeted Unique Coverage problem with positive integer costs/profits, with parameters B and k, fixed-parameter tractable? Acknowledgements We thank Saket Saurabh for pointing out the connection between Unique Coverage and strong systems of distinct representatives and for several fruitful discussions.
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