k-approval

We show that Possible Winner for k-approval becomes NP-complete for input in-stances consisting of at least two partial votes. Herein, we assume that k can be chosen appropriately within the given many-one reductions. The case that k as well as the number of votes are bounded will be considered in Chapter 9. The following NP-hardness results rely on many-one reductions from the NP-complete Independent Set (IS) problem. Given an undirected graph G = (U, E) and a positive integer t, it asks whether there is a size-t vertex subset U^′⊆ U such that there is no edge between any two vertices of U^′.

Theorem 8.2. For k-approval, Possible Winner is NP-complete for a partial profile that consists of at least two partial votes when k is part of the input.

Proof. The NP-membership of Possible Winner for k-approval is obvious. To show the NP-hardness we give a many-one reduction from Independent Set. Consider an IS-instance (G = (U, E), t) with n-vertex set U = {u¹, . . . , un}². We assume that t < (n− 1)/2. This case clearly still leads to NP-completeness. We construct a partial profile P over a set C of candidates in which the distinguished candidate c∈ C is a possible unique winner according to k-approval with

k := n +

if and only if G has an independent set of size at most t. Observe that the assump-tion t < (n− 1)/2 implies that k > 0, and thus k-approval is well-defined. We first give a construction to show the hardness for a 2-voter profile and then explain how to extend this construction to work for an arbitrary fixed number of votes. The set of candidates is otherwise, there is one candidate e_{i,j} in CE.³

• The set D consists of dummy candidates with |D| := n − t + ⁿ2

+|E| − tn + 2^t

. Since t < (n− 1)/2, this gives a positive integer.

The basic idea to construct the two partial votes (given in Figure 8.1) can be described as follows. The distinguished candidate is fixed such that it makes two

2Since the number of votes is constant, we use the variable n for the number of vertices.

3Note that e_{i,j}= e_{j,i}and e^′_{i,j}= e^′_{j,i}.

v1: c≻ D ≻ C^V ≻ C^E.

v2: c≻ C^V ∪ C^E≻ D, and, for 1 ≤ i < j ≤ n, if{vⁱ, vj} ∈ E, then cⁱ≻ e{i,j} and cj ≻ e^′_{i,j}; otherwise, ci≻ e{i,j} and cj≻ e{i,j}.

Figure 8.1: Partial votes of a Possible Winner instance resulting from a many-one reduction from Independent Set.

points in total. Thus, in a winning extension every other candidate must be assigned to a zero-position in at least one of the two votes. The first vote is used to select a set C_V^′ of t “vertex candidates”. More specifically, we discuss below that k is adjusted such that exactly t vertex candidates from CV (forming C_V^′ ) must be assigned to one-positions whereas the remaining candidates from CV and all candidates from CE

must be assigned to zero-positions and thus are beaten by c. It follows that the t candidates from C_V^′ must take a zero-position in the second vote. The second vote is constructed such that assigning a candidate from CV to a zero-position implies that n− 1 candidates from C^E are shifted to a zero-position (this is redundant in the sense that these candidates are already beaten by c since they assume zero-positions in the first vote). Now, the crucial part is that in a winning extension the t candidates from C_V^′ cannot shift pairwisely disjoint subsets of candidates from CE since then the number of zero-positions would not be sufficiently large and a candidate from C_V^′ would be assigned to a one-position and thus would not be beaten by c. Clearly, if two candidates from C_V^′ shift a “common” candidate from CE, then this “saves”

one zero-position. The construction of the second vote ensures that two candidates from CV can shift a common candidate from CE to a zero-position only if there is no edge between them. Due to an appropriate bound on the number of zero-positions, this enforces that in a winning extension every pair of candidates from C_V^′ must shift a common candidate from CE and thus there cannot be an edge between the corresponding vertices in the Independent Set-instance.

Claim: There is a winning extension of P if and only if G has a independent set of size t.

“⇐”: Let CV^I ⊂ C^V denote the subset of candidates corresponding to an independent set of G. Then, the “edge candidates” that must be shifted to the right by the candidates from C_V^I are

C_E^I :={e{i,j}∈ C^E| cⁱ∈ CV^I and {vⁱ, vj} /∈ E} (8.3)

∪{e{i,j}∈ C^E| cⁱ∈ CV^I, i < j and{vⁱ, vj} ∈ E} (8.4)

∪{e^′{i,j}∈ C^E| cⁱ∈ CV^I, i > j and{vⁱ, vj} ∈ E}. (8.5) Before giving a winning extension of the partial votes, we verify that

|CE^I| = t · (n − 1) −

t 2



. (8.6)

For every candidate ci ∈ CV^I, C_E^I contains either the candidate e_{i,j} or the candi-date e^′_{i,j}, for all j ∈ {1, . . . , n} \ {i}, leading to the term t · (n − 1). From this term we need to subtract the number of candidates that are counted twice. Since C_V^I

8.2 Number of votes 131 v1: c > D > C_V^I > CV \ CV^I > CE

v2: c > CV \ CV^I > CE\ CE^I > C_V^I > C_E^I > D

Figure 8.2: Winning extension of the partial votes with C_V^I denoting the candidates corresponding to an independent set and C_E^I ⊆ C^E the candidates shifted to the right by candidates from C_V^I. The zero-positions are highlighted.

corresponds to an independent set, there are no edges between any pair of vertices corresponding to {cⁱ, cj} ⊂ CV^′ and thus there is no candidate e^′_ij. Hence, for ev-ery unordered pair of candidates{cⁱ, cj} ⊂ CV^′ , the candidate e_{i,j} is counted twice, giving the term ₂^t

.

Extend the partial votes as described in Figure 8.2. To see that c wins in this extension, we verify that the highlighted positions are the zero-positions. To this end, we first compute the number of zero-positions per vote which is the number of candidates minus k and thus

Since every candidate except c takes a highlighted position and thus a zero-position in at least one vote, it directly follows that c wins.

“⇒”: Consider an extension in which c wins. By construction, in every extension of v¹, the last|C^E| = ⁿ2

+|E| zero-positions must be assumed by the candidates from C^E. Due to Equation (8.7) there remain n− t zero-positions which can only be assigned to candidates from CV. Let C_V^⋆ be the set of the remaining t candidates from CV with a one-position in v1. Let C_E^⋆ ⊆ C^E be defined analogously to C_E^I by replacing C_V^I through C_V^⋆ in (8.5). In a winning extension, all candidates from C_V^⋆ must assume a zero-position in v2. This shifts all candidates from C_E^⋆ to zero-positions in v2 as well.

Since all candidates from D must assume zero-positions in every extension of v2, it is easy to verify that there are exactly tn− 2^t

zero-positions left over for the candidates from C_V^⋆ and C_E^⋆ (and c beats all candidates of D). In addition, a candidate from C_E^⋆ can be shifted to a zero-position by at most two candidates from C_V^⋆. Hence, there must be ₂^t

“shared” candidates in C_E^′ , that is, candidates e_{i,j} with ci ≻ e{i,j}

and cj ≻ e{i,j} in the partial vote v2. Since|CV^⋆| = t, there are ^t2

unordered pairs of candidates and each pair {cⁱ, cj} can share the candidate e^ij only if there is no edge between vi and vj in G (otherwise, the one of them with smaller index would

additionally shift e^′_{i,j}). Hence the vertices corresponding to C_V^⋆ form an independent set in G. This finishes the proof of the Claim.

Finally, let us consider the problem for s > 2 votes. To show NP-hardness, one pads the construction for two votes as follows. Add (s− 2) · k new dummy candidates and fix them at the first k positions in s−2 of the votes such that every new candidate takes a one-position exactly once. The remaining two votes are constructed as in the 2-voter profile (given in Figure 8.1) and the new candidates are appended at the end.

Clearly, every new candidate makes exactly one point and thus is always beaten by c.

All old candidates make zero points in the newly added votes and thus the situation in the two “old” votes is still the same. Hence, c is a possible winner in the new s-voter profile if and only if it is a possible winner in the 2-voter profile.

The construction given in the above NP-hardness proof can be adapted in a straightforward way to work for the co-winner case. Finally, combining Observa-tion 8.1 with Theorem 8.2 one directly obtains the following dichotomy result.

Corollary 8.1. For k-approval, Possible Winner with k being part of the input is NP-complete if the input profile consists of at least two partial votes and it is solvable in polynomial time otherwise.