Finding Small Multi-Demand Set Covers with Ubiquitous Elements and Large Sets is Fixed-Parameter Tractable

04/20/2021 ∙ by Niclas Boehmer, et al. ∙ Czech Technical University in Prague ∙ Berlin Institute of Technology (Technische Universität Berlin) ∙ Nanyang Technological University ∙ Humboldt-Universität zu Berlin ∙ 0 ∙ share

We study a variant of Set Cover where each element of the universe has some demand that determines how many times the element needs to be covered. Moreover, we examine two generalizations of this problem when a set can be included multiple times and when sets have different prices. We prove that all three problems are fixed-parameter tractable with respect to the combined parameter budget, the maximum number of elements missing in one of the sets, and the maximum number of sets in which one of the elements does not occur. Lastly, we point out how our fixed-parameter tractability algorithm can be applied in the context of bribery for the (collective-decision) group identification problem.

READ FULL TEXT VIEW PDF

Application: Bribery in Group Identification

We now describe an application where the Set Cover with Capacities problem naturally arises and our FPT algorithm is directly applicable: In group identification, we are given a set $A = {a_{1}, \dots, a_{n}}$ of agents and the task is to identify a so-called socially qualified subgroup of the agents [3]. To do so, we are given a qualification profile $φ : A \times A \to {- 1, 1}$ that denotes for each agent $a$ which of the other agents $a$ deems qualified, i.e., agent $a$ qualifies $a^{'}$ if $φ (a, a^{'}) = 1$ and disqualifies $a^{'}$ if $φ (a, a^{'}) = - 1$ . For an agent $a \in A$ , let $Q_{φ}^{+} (a) = {a^{'} \in A ∣ φ (a^{'}, a) = 1}$ denote the set of agents qualifying $a$ and $Q_{φ}^{-} (a) = {a^{'} \in A ∣ φ (a^{'}, a) = - 1}$ the set of agents disqualifying $a$ . To decide given a set of agents and a qualification profile which agents are socially qualified, different social rules have been proposed. One popular rule parameterized by two integers $s$ and $t$ with $s + t \leq n + 2$ is the consent rule, denoted $f^{(s, t)}$ [4]. Under the consent rule, an agent $a \in A$ with $φ (a, a) = 1$ is socially qualified if and only if at least $s$ agents (including $a$ itself) qualify $a$ . Similarly, an agent $a \in A$ with $φ (a, a) = - 1$ is socially disqualified if and only if at least $t$ agents (including $a$ itself) disqualify $a$ .

Recently, Erdélyi et al. [2] initiated the study of the computational complexity of bribery in the context of group identification, among others, asking the following question:

Constructive-

f^{(s, t)}

Agent Bribery Input: Set

A

of agents, qualification profile

φ

, subset

A^{+} \subseteq A

of agents to be made socially qualified, and budget

ℓ

. Question: Is it possible to modify the opinion of at most

ℓ

agents such that after the modifications all agents from

A^{+}

are socially qualified under the consent rule

f^{(s, t)}

Erdélyi et al. [2] proved that this problem is NP-hard even for $s = 1$ and $t = 2$ . Subsequently, Boehmer et al. [1] conducted a detailed study of the parameterized complexity of this question considering the parameters $ℓ$ , $s$ , $t$ , and $| A^{+} |$ . Among others, they proved that Constructive- $f^{(s, t)}$ Agent Bribery is W[1]-hard with respect to $s + ℓ$ even if $t = 1$ .

Let $Δ$ be the maximum number of agents from $A^{+}$ that an agent $a \in A$ qualifies, i.e., $Δ := {max}_{a \in A} | {a^{'} \in A^{+} ∣ φ (a, a^{'}) = 1} |$ . We now prove that our algorithm for Set Cover with Demands can be used to prove that Constructive- $f^{(s, t)}$ Agent Bribery with $t = 1$ is fixed-parameter tractable with respect to $Δ + s$ .

Theorem 2.

Constructive- $f^{(s, t)}$ Agent Bribery is solvable in $O (n^{2} \cdot (Δ \cdot s)^{s})$ time for $t = 1$ .

Proof.

First of all, as $t = 1$ implies that every agent that disqualifies itself is also socially disqualified, we bribe all agents in $A^{+}$ who do not qualify themselves to qualify everyone and adjust the budget $ℓ$ accordingly. We delete from $A^{+}$ all agents who are already socially qualified after this bribery, while keeping them in the set of agents.

If we have $ℓ \geq s$ for the resulting budget, we are done as we can simply pick $ℓ$ agents and make them qualify everyone, which results in all agents from $A^{+}$ being socially qualified.

Consequently, we are left with the situation where $ℓ < s$ . We now reduce the problem to an instance of Set Cover with Demands as follows. We set the universe $U = A^{+}$ and for each $a \in A^{+}$ its demand to $s - | Q^{+} (a) |$ (the number of additional qualification $a$ needs to get by the bribery to become socially qualified). For each agent $a \in A$ who does not qualify all other agents, we add a set $F_{a}$ to our covering system $F$ containing all agents from $A^{+}$ which $a$ does not qualify, i.e., $F_{a} := {a^{'} \in A^{+} ∣ φ (a, a^{'}) = - 1}$ . Finally, we set $k := ℓ$ . Bribing an agent $a \in A$ , which results in all agents from $F_{a}$ getting an additional qualification, corresponds to including $F_{a}$ in the cover. It is easy to see that there exists a successful bribery if and only if there exists a solution to the constructed Set Cover with Demands instance. Note that in the constructed instance $k$ is bounded by $s$ . Moreover, for each $a \in A$ , it holds that $n - | F_{a} |$ is equal to the number of agents $a$ qualifies before the bribery and, thus, $Δ \geq | A^{+} | - {min}_{F \in F} | F |$ . Lastly, note that as each agent $a \in A^{+}$ can only be approved by at most $s - 1$ agents before the bribery, $a$ needs to appear in all but at most $s - 1$ sets. Thus, applying the algorithm from Theorem 1, we can solve the problem in $O (n \cdot (Δ \cdot s)^{s})$ time. ∎

It is even possible to extend this result to a fixed-parameter tractable algorithm for the parameters $s + t + ℓ + Δ$ (note that as proven by Boehmer et al. [1] Constructive- $f^{(s, t)}$ Agent Bribery is W[1]-hard with respect to $s + t + ℓ$ ):

Corollary 4.

Constructive- $f^{(s, t)}$ Agent Bribery is solvable in time.

Proof.

If we bribe an agent, we always make him qualify all agents. For all agents $a \in A^{+}$ with $φ (a, a) = - 1$ and $| Q^{-} (a) | \geq ℓ + t$ , $a$ must be bribed; so we bribe $a$ . Thus, we can assume $| Q^{-} (a) | < ℓ + t$ for all $a \in A^{+}$ with $φ (a, a) = - 1$ . Now, as long as there exists an $a \in A^{+}$ with $φ (a, a) = - 1$ , we branch on bribing $a$ or bribing $| Q^{-} (a) | - (t - 1)$ agents from $Q^{-} (a)$ and update $φ$ , $ℓ$ , and $A^{+}$ accordingly (we delete agents from $A^{+}$ if they became socially qualified). We reject the current branch if $ℓ < 0$ . For each non-rejected branch, it remains to consider agents from $A^{+}$ who qualify themselves. This problem is similar to the case when $t = 1$ and we can apply Theorem 2.

As the branching factor in each step is bounded by $ℓ + t$ and the depth is bounded by $ℓ$ , the algorithm from Theorem 2 is employed at most $(ℓ + t)^{ℓ}$ times, which results in an overall running time of ∎

Acknowledgments

NB is supported by the DFG project MaMu (NI 369/19). DK is partly supported by the OP VVV MEYS funded project CZ.02.1.01/0.0/0.0/16_019/0000765 “Research Center for Informatics”.

References

Boehmer et al. [2020] Niclas Boehmer, Robert Bredereck, Dusan Knop, and Junjie Luo. Fine-grained view on bribery for group identification. In Proceedings of the Twenty-Ninth International Joint Conference on Artificial Intelligence (IJCAI 2020)
, pages 67–73. ijcai.org, 2020.
Erdélyi et al. [2020] Gábor Erdélyi, Christian Reger, and Yongjie Yang. The complexity of bribery and control in group identification. Autonomous Agents and Multi-Agent Systems, 34(1):8, 2020.
Kasher and Rubinstein [1997] Asa Kasher and Ariel Rubinstein. On the question ”who is a j?” a social choice approach. Logique et Analyse, 40(160):385–395, 1997.
Samet and Schmeidler [2003] Dov Samet and David Schmeidler. Between liberalism and democracy. J. Economic Theory, 110(2):213–233, 2003.

Finding Small Multi-Demand Set Covers with Ubiquitous Elements and Large Sets is Fixed-Parameter Tractable

Authors

Approximation Algorithm for the Partial Set Multi-Cover Problem

Maximum n-times Coverage for COVID-19 Vaccine Design

The Only Undoable CRDTs are Counters

Levelable Sets and the Algebraic Structure of Parameterizations

Set Cover and Vertex Cover with Delay

Computing L(p,1)-Labeling with Combined Parameters

Multiparty Selection

Application: Bribery in Group Identification

Theorem 2.

Proof.

Corollary 4.

Proof.

Acknowledgments

References

Finding Small Multi-Demand Set Covers with Ubiquitous Elements and Large Sets is Fixed-Parameter Tractable

Authors

Approximation Algorithm for the Partial Set Multi-Cover Problem

Maximum n-times Coverage for COVID-19 Vaccine Design

The Only Undoable CRDTs are Counters

Levelable Sets and the Algebraic Structure of Parameterizations

Set Cover and Vertex Cover with Delay

Computing L(p,1)-Labeling with Combined Parameters

Multiparty Selection

This week in AI

Application: Bribery in Group Identification

Theorem 2.

Proof.

Corollary 4.

Proof.

Acknowledgments

References