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 of agents and the task is to identify a socalled socially qualified subgroup of the agents [3]. To do so, we are given a qualification profile that denotes for each agent which of the other agents deems qualified, i.e., agent qualifies if and disqualifies if . For an agent , let denote the set of agents qualifying and the set of agents disqualifying . 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 and with is the consent rule, denoted [4]. Under the consent rule, an agent with is socially qualified if and only if at least agents (including itself) qualify . Similarly, an agent with is socially disqualified if and only if at least agents (including itself) disqualify .
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 Agent Bribery Input: Set of agents, qualification profile , subset of agents to be made socially qualified, and budget . Question: Is it possible to modify the opinion of at most agents such that after the modifications all agents from are socially qualified under the consent rule ?Erdélyi et al. [2] proved that this problem is NPhard even for and . Subsequently, Boehmer et al. [1] conducted a detailed study of the parameterized complexity of this question considering the parameters , , , and . Among others, they proved that Constructive Agent Bribery is W[1]hard with respect to even if .
Let be the maximum number of agents from that an agent qualifies, i.e., . We now prove that our algorithm for Set Cover with Demands can be used to prove that Constructive Agent Bribery with is fixedparameter tractable with respect to .
Theorem 2.
Constructive Agent Bribery is solvable in time for .
Proof.
First of all, as implies that every agent that disqualifies itself is also socially disqualified, we bribe all agents in who do not qualify themselves to qualify everyone and adjust the budget accordingly. We delete from all agents who are already socially qualified after this bribery, while keeping them in the set of agents.
If we have for the resulting budget, we are done as we can simply pick agents and make them qualify everyone, which results in all agents from being socially qualified.
Consequently, we are left with the situation where . We now reduce the problem to an instance of Set Cover with Demands as follows. We set the universe and for each its demand to (the number of additional qualification needs to get by the bribery to become socially qualified). For each agent who does not qualify all other agents, we add a set to our covering system containing all agents from which does not qualify, i.e., . Finally, we set . Bribing an agent , which results in all agents from getting an additional qualification, corresponds to including 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 is bounded by . Moreover, for each , it holds that is equal to the number of agents qualifies before the bribery and, thus, . Lastly, note that as each agent can only be approved by at most agents before the bribery, needs to appear in all but at most sets. Thus, applying the algorithm from Theorem 1, we can solve the problem in time. ∎
It is even possible to extend this result to a fixedparameter tractable algorithm for the parameters (note that as proven by Boehmer et al. [1] Constructive Agent Bribery is W[1]hard with respect to ):
Corollary 4.
Constructive Agent Bribery is solvable in time.
Proof.
If we bribe an agent, we always make him qualify all agents. For all agents with and , must be bribed; so we bribe . Thus, we can assume for all with . Now, as long as there exists an with , we branch on bribing or bribing agents from and update , , and accordingly (we delete agents from if they became socially qualified). We reject the current branch if . For each nonrejected branch, it remains to consider agents from who qualify themselves. This problem is similar to the case when and we can apply Theorem 2.
As the branching factor in each step is bounded by and the depth is bounded by , the algorithm from Theorem 2 is employed at most 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”.
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