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

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

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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

Related Research

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Application: Bribery in Group Identification

Theorem 2.

Proof.

Corollary 4.

Proof.

Acknowledgments

References