Parameterized Complexity of Voter Control in Multi-Peaked Elections

Abstract

We study the parameterized complexity of voter control problems in κ-peaked elections, where κ is a positive integer. In particular, we focus on the constructive/destructive control by adding/deleting votes for Condorcet, Maximin and Copeland^α. It is known that in general elections all these problems are NP-hard, except for the destructive control by adding/deleting votes for Condorcet which is polynomial-time solvable. We strengthen these results by showing that, when restricted to κ-peaked elections where κ = 3,4, the above NP-hard problems not only remain NP-hard but also are W[1]-hard with respect to the number of added/deleted votes.

Appendices

Appendix: Missing Proof in Theorem 2

In the appendix, we prove the W[1]-hardness of constructive/destructive control by adding votes for Copeland^α where 0 < α ≤ 1 in 3-peaked elections. Consider first the case where 0 < α < 1. Let α be a fixed rational number with 0 < α < 1. We first consider DCAV-Copeland^α-UNI. The W[1]-hardness reduction is adapted from that for DCAV-Copeland⁰-UNI by creating polynomially many dummy candidates. The role of these dummy candidates is to enlarge the score gap between p and q to a certain extent, when two unregistered votes corresponding to two intersected 2-intervals are added to the registered votes; hence guarantees that q would still be the unique winner when a multiset of at most k unregistered votes corresponding to a non-independent set are added. To this end, we also adopt the same restriction on 2-intervals here as we did for DCAV-Copeland⁰-UNI: every two 2-intervals either do not intersect or they intersect in a non-trivial interval. We create these dummy candidates in a way so that any nonempty intersection of two 2-intervals corresponds to at least $\lceil \frac {1}{1-\alpha }\rceil + 2$ candidates. To this end, we do the following. We call an intersection [x₁, x₂] with x₁ < x₂ of two 2-intervals a minimal intersection if there is no other 2-interval which has at least one of its endpoints in [x₁, x₂]. Clearly, given a 2-interval representation of a 2-interval graph, all minimal intersections can be found in polynomial time. Apart from creating all the candidates as in the reduction for DCAV-Copeland⁰-UNI, we create for each minimal intersection [x₁, x₂] a set of $\lceil \frac {1}{1-\alpha }\rceil$ dummy candidates which lie in distinguished places between x₁ and x₂. See Fig. 10 for an illustration. This construction ensures that p can prevent q from being the unique winner only if p beats every other candidate in the final election. The observation is that once p ties or is beaten by some candidate x ∈Γ, then p actually ties or is beaten by at least $\lceil \frac {1}{1-\alpha }\rceil + 2$ candidates. The amount $\lceil \frac {1}{1-\alpha }\rceil + 2$ is large enough to make p have a strictly less score than that of q, and hence cannot prevent q from being the unique winner. However, p beats every other candidate if and only if there is an independent set of size k for $\mathcal {F}$, implying the correctness of the reduction. The proofs for other three problems DCAV-Copeland^α-NON, CCAV-Copeland^α-UNI and CCAV-Copeland^α-NON are adapted from DCAV-Copeland⁰-NON, CCAV-Copeland⁰-UNI and CCAV-Copeland⁰-NON, respectively. The constructions are analogous to the above reduction.

Fig. 10

An illustration of the dummy candidates in the W[1]-hardness reduction for DCAV-Copeland^α-UNI in Theorem 2. The figure on the left-hand side shows two 2-intervals (the black one above and the gray one below) which intersect in a non-trivial interval [x₂, x₃]. Moreover, there is no other 2-interval whose endpoint is in [x₂, x₃]. We create exactly $\lceil \frac {1}{1-\alpha }\rceil$ dummy candidates between x₂ and x₃. The figure on the right-hand side shows four 2-intervals, with only one interval of each 2-intervals is shown. There are $\lceil \frac {1}{1-\alpha }\rceil + 2$ candidates corresponding to the intersection (this is a minimal intersection since no other 2-interval has any of its endpoints in this intersection) of the second 2-interval and the fourth 2-interval, and $\lceil \frac {1}{1-\alpha }\rceil + 2$ candidates corresponding to the intersection (a minimal intersection) of the first 2-interval and the third 2-interval

Finally we consider Copeland¹. The reductions for the four problems DCAV-Copeland¹-UNI, DCAV-Copeland¹-NON, CCAV-Copeland¹-UNI, CCAV-Copeland¹-NON are adapted from DCAV-Copeland⁰-UNI, DCAV-Copeland⁰-NON, CCAV-Copeland⁰-UNI, CCAV-Copeland⁰-NON, respectively. Precisely, each reduction is different from the corresponding one in the way that we set R = k − 1. Moreover, instead of searching for an independent set of size k, we search for an independent set of size k − 1. $\square$

Missing Proof in Theorem 5

We consider the missing proof for DCDV-Copeland^α-UNI in Theorem 5. The reduction is similar to the one for CCDV-Copeland^α-UNI with the difference that we create one more dummy candidate q^′ which lies immediately on the right side of q in $\mathcal {L}$. That is, the candidate set is Γ ∪{p,q,q^′} with q being the distinguished candidate, and we define $\mathcal {L}=(q,q^{\prime },\vec {\Gamma },p)$. The role of the dummy candidate q^′ is to guarantee that, in the final election, every candidate in Γ is beaten by both q and q^′. This excludes the possibility that some x_i would have a higher score than that of q in the final election. To achieve this goal, we rank q^′ immediately after q in every vote and remains the order of other candidates unchanged. Precisely, we create n − 1 votes defined as $(\mathcal {L}[x_{1},p],q,q^{\prime })$, and n − k + 2 votes defined as $(p,\mathcal {L}[q, x_{|{\Gamma }|}])$. Besides, for every 2-interval $I_{i}=\{{I_{i}^{1}}=[x_{\alpha }, x_{\beta }], {I_{i}^{2}}=[x_{\gamma },x_{\delta }]\}$ with x_β < x_γ, we create two votes as follows.

$$(\mathcal{L}[q,x_{\alpha}),\mathcal{L}(x_{\beta},x_{\gamma}),\mathcal{L}(x_{\delta},p],\mathcal{L}[x_{\alpha},x_{\beta}],\mathcal{L}[x_{\gamma},x_{\delta}]); \text{and}$$

$$(q,q^{\prime},\mathcal{L}[x_{\alpha},x_{\beta}],\mathcal{L}[x_{\gamma},x_{\delta}],p,\mathcal{L}[x_{1},x_{\alpha}),\mathcal{L}(x_{\beta},x_{\gamma}),\mathcal{L}(x_{\delta},x_{|{\Gamma}|}]).$$

The comparisons are shown in Table 7.

Table 7 Comparisons between candidates in the W[1]-hardness proof for DCDV-Copeland^α-UNI in Theorem 5

Using the same argument for CCDV-Copeland^α-UNI, we can conclude that if there is an independent set of size k, the candidate p can prevent q from being the unique winner by deleting k votes. For the other direction, observe first that no candidate x_i ∈Γ has a chance to have a higher score than that of q since every x_i is beaten by both q and q^′ in the final election. Clearly, q^′ also has no chance to prevent q from being the unique winner since every vote ranks q above q^′. Therefore, the only candidate which can prevent q from being the unique winner is p and, moreover, this happens only if p beats every candidate in Γ. The remaining argument is the same as for CCDV-Copeland^α-UNI.