# On the Computational Complexity of Variants of Combinatorial Voter Control in Elections

## TL;DR

This paper investigates the computational complexity of combinatorial voter control problems in elections, revealing that destructive control is often easier than constructive control, with certain cases solvable in polynomial time and others remaining NP-hard.

## Contribution

It introduces and analyzes the complexity of combinatorial voter control variants, highlighting conditions under which problems are tractable or NP-hard, and discusses approximation limitations.

## Key findings

- Destructive control is generally easier than constructive control.
- Symmetric bundling with limited associations allows polynomial-time solutions for destructive control.
- Constructive control remains NP-hard even with disjoint bundles.

## Abstract

Voter control problems model situations in which an external agent tries toaffect the result of an election by adding or deleting the fewest number of voters. The goal of the agent is to make a specific candidate either win (\emph{constructive} control) or lose (\emph{destructive} control) the election. We study the constructive and destructive voter control problems whenadding and deleting voters have a \emph{combinatorial flavor}: If we add (resp.\ delete) a voter~$v$, we also add (resp.\ delete) a bundle~$\kappa(v) $ of voters that are associated with~$v$. While the bundle~$\kappa(v)$ may have more than one voter, a voter may also be associated with more than one voter. We analyze the computational complexity of the four voter control problems for the Plurality rule. We obtain that, in general, making a candidate lose is computationally easier than making her win. In particular, if the bundling relation is symmetric (i.e.\ $\forall w\colon w \in \kappa(v) \Leftrightarrow v \in \kappa(w) $), and if each voter has at most two voters associated with him, then destructive control is polynomial-time solvable while the constructive variant remains $\NP$-hard. Even if the bundles are disjoint (i.e.\ $\forall w\colon w \in \kappa(v) \Leftrightarrow \kappa(v) = \kappa(w) $), the constructive problem variants remain intractable. Finally, the minimization variant of constructive control by adding voters does not admit an efficient approximation algorithm, unless P=NP.