Computational aspects of strategic behaviour in elections with top-truncated ballots

Abstract

Understanding when and how computational complexity can be used to protect elections against different manipulative actions has been a highly active research area over the past two decades. Much of this literature, however, makes the assumption that the voters or agents specify a complete preference ordering over the set of candidates. There are many multiagent systems applications, and even real-world elections, where this assumption is not warranted, and this in turn raises a series of questions on the impact of partial voting on the complexity of manipulative actions. In this paper, we focus on two of these questions. First, we address the question of how hard it is to manipulate elections when the agents specify only top-truncated ballots. Here, in particular, we look at the weighted manipulation problem—both constructive and destructive manipulation—when the voters are allowed to specify top-truncated ballots, and we provide general results for all scoring rules, for elimination versions of all scoring rules, for the plurality with runoff rule, for a family of election systems known as Copeland$^{\alpha }$, and for the maximin protocol. The second question we address is the impact of top-truncated voting on the complexity of manipulative actions in electorates with structured preference profiles. In particular, we consider electorates that are single-peaked and we show how, for many voting protocols, allowing top-truncated voting reimposes the $\mathcal {NP}$-hardness shields that normally vanish in such electorates.

Appendix: Additional proofs

1.1 Proofs for Sect. 3.3

Lemma 1

Partition$'$ is NP-complete.

Proof

It is easy to see that Partition$'$ is in $\mathcal {NP}$. To prove $\mathcal {NP}$-hardness, we show a reduction from the Partition (P1) problem. Let $\{a_1, \ldots , a_n\}$, where $\sum _i a_i = 2M$, be an arbitrary instance of the Partition problem. Next, construct the following instance of Partition$'$(P2):

• 1. For each $a_i$, construct $b_i = a_i + 2M^2$ and $b'_i = 2M^2$.

• 2. Construct $X_1 = X_2 = 2M^2n + 4M^2 - M$.

Now, it can be seen that $\sum _i (b_i + b'_i) + X_1 + X_2 = 2\cdot (2n+2)\cdot 2M^2$ and that each of $b_i, b'_i, X_1, X_2 \ge 2M^2$.

Suppose there exists a partition $S_1, S_2$ for P1. Then it is easy to see that P2 also has a partition $T_1$, $T_2$ such that $\sum T_1 = \sum T_2 = (2n+2)2M^2$, where $T_1 = \{b_i\}_{i | a_i \in S_1} \cup \{b'_i\}_{i | a_i \in S_2} \cup \{X_1\}$ and $T_2 = \{b_i\}_{i | a_i \in S_2} \cup \{b'_i\}_{i | a_i \in S_1} \cup \{X_2\}$.

Conversely, suppose there exists a partition for P2. This implies that there exists two subsets $T_1, T_2$ such that $\sum T_1 = \sum T_2 = (2n+2)2M^2$. Now, the first thing to observe is that both $X_1, X_2$ cannot belong to the same set, because if so then the sum of the rest of the elements $= 4nM^2 + 2M < (2n+2)2M^2$. So let us assume that $X_1 \in T_1$ and $X_2 \in T_2$. Also, since $X_1 > b_i, \forall i$, we can remove $X_1$ and $X_2$ from the two subsets. Removing them we have $\sum T_1 - X_1 = \sum T_2 - X_2 = M + 2nM^2$. So now we can construct $S_1$, and $S_2$ as follows:

$$\begin{aligned} S_1 = \{a_i\}_{i | b_i \in T_1} \\ S_2 = \{a_i\}_{i | b_i \in T_2} \end{aligned}$$

This in turn implies that P1 has a partition. $\square$

Lemma 2

Fixed-Difference Subset Sum is $\mathcal {NP}$-complete.

Proof

It is easy to see that Fixed-Difference Subset Sum is in $\mathcal {NP}$. To prove $\mathcal {NP}$-hardness, we show a reduction from the Partition probem. Let $\{k_1, \ldots , k_t\}$, where $\sum _i k_i = 2K$, be an arbitrary instance of Partition. Next, construct the following instance $\{\ell _1, \ldots , \ell _t, \ell '_1, \ldots , \ell '_t\}$ of Fixed-Difference Subset Sum(P2), where $\ell _i = k_i + 2^{n + i}$, $\ell '_i = 2^{n + i}$, and $n = \lceil \log {2K} \rceil$.

Suppose there exists a partition $S_1, S_2$ for P1. Then P2 has subsets $T_1$, $T_2$ such that $\sum T_1 - \sum T_2 = K + \sum _{i=1}^{t} 2^{i+n}$, where $T_1 = \{\ell _i\}_{i | k_i \in S_1} \cup \{\ell '_i\}_{i | k_i \in S_2}$ and $T_2 = \emptyset$.

Conversely, suppose there exists subsets $T_1$, $T_2$ in P2 such that $\sum T_1 - \sum T_2 = K + \sum _{i=1}^{t} 2^{i+n}$. Now, it is easy to argue that none of $\ell _i$ or $\ell '_i$ for $i = 1, \ldots , t$ can belong to $T_2$, because if so then the second term ($\sum _{i=1}^{t} 2^{i+n}$) of $\sum T_1 - \sum T_2$ will not be attainable. Therefore, we can construct $S_1$ and $S_2$ such that $S_1 = \{k_i\}_{i | \ell _i \in T_1}$ and $S_2 = \{k_i\}_{i | \ell _i \notin T_1}$, and this in turn implies that P1 has a partition. $\square$

1.2 Additional proofs for Sect. 5.2

Theorem 27

For any 3-candidate scoring rule X that is not isomorphic to plurality or veto, in single-peaked electorates, CWCM with top-truncated votes in $X_{\downarrow }$ is NP-complete.

Proof (sketch)

The proof here is similar to Theorem 2 and hence we only sketch the same. To prove $\mathcal {NP}$-hardness, we can proceed like in Theorem 2 by considering the following three cases. In all the cases we can use a reduction from the Partition problem.

• Case 1 $\alpha _2< \alpha _1 < 2\alpha _2$. Given an arbitrary instance of Partition, construct the following instance of CWCM with $p\ L\ a\ L\ b$ as the linear ordering over the candidates. This linear ordering in turn restricts the set of allowed votes to $\{(p> a> b), (a> p> b), (a> b> p), (b> a> p), (a> b \sim p), (b> a \sim p), (p > a \sim b)\}$. In S let there be a vote of weight K voting $(a> b > p)$, $(b> a > p)$, $(b > a \sim p)$, and $(p > a \sim b)$ each, and let each $k_i$ in T have a vote of weight $k_i$. Now, proceed like in Theorem 2.

• Case 2 $\alpha _1 > 2\alpha _2$. In this case, construct the following instance of CWCM with $a\ L\ p\ L\ b$ as the linear ordering over the candidates. This linear ordering in turn restricts the set of allowed votes to $\{(a> p> b), (p> a> b), (p> b> a), (b> p> a), (a> b \sim p), (b> a \sim p), (p > a \sim b)\}$. In S let there be a vote of weight $(2\alpha _1 - \alpha _2)K$ voting $(a> p > b)$ and $(b> p > a)$ each, and let each $k_i$ in T have a vote of weight $(\alpha _1 - 2\alpha _2)k_i$. Now, proceed like in Theorem 2.

• Case 3 $\alpha _1 = 2\alpha _2$. In this case, construct the following instance of CWCM with $a\ L\ p\ L\ b$ be the linear ordering over the candidates. In S let there be a vote of weight 3K voting $(a > b \sim p)$ and $(b > a \sim p)$ each, and let each $k_i$ in T have a vote of weight $k_i$. Now, proceed like in Theorem 2. $\square$

Theorem 28

For any 3-candidate scoring rule X that is not isomorphic to plurality, in single-peaked electorates, CWCM with top-truncated votes in $X_{av}$ is NP-complete.

Proof (sketch)

The proof here is similar to Theorem 3 and hence we only sketch the same. To prove $\mathcal {NP}$-hardness, we use a reduction from Fixed-difference Subset Sum. Given an arbitrary instance of Fixed-difference Subset Sum, construct the following instance of CWCM with $a\ L\ p\ L\ b$ as the linear ordering over the candidates. This linear ordering in turn restricts the set of allowed votes to $\{(a> p> b), (p> a> b), (p> b> a), (b> p> a), (a> b \sim p), (b> a \sim p), (p > a \sim b)\}$. In S let there be a vote of weight 8K voting $(a > b \sim p)$ and $(b > a \sim p)$ each, vote of weight 3K voting $(p> b > a)$, and a vote of weight K voting $(p> a > b)$. In T let each $k_i$ have a vote of weight $2k_i$. Now, proceed like in Theorem 3. $\square$

Lemma 3

During each round of the eliminate(Veto), only the rightmost or the leftmost candidate in the linear ordering L is eliminated.

Proof

Let $c_1\ L\ c_2\ L\ \cdots \ L\ c_m$ be any arbitrary linear ordering. It is clear that in the first round of eliminate(Veto) only $c_1$ or $c_m$ will be eliminated since all the votes only have either of them placed at the end. Also, since the votes are single-peaked, for any candidate $c_j$, the only candidates which can be immediately on top of it in a vote where $c_j$ is placed last are its right and left neighbours, where the right neighbour of the rightmost candidate is the leftmost one and vice-versa. Therefore, in particular, if $c_1$ is the candidate eliminated in the first round then all its last votes gets transferred to either $c_2$ or $c_m$. Hence in the subsequent round only one of $c_2$ or $c_m$ will be eliminated. Continuing this way we can see that at every round r only one of the corner candidates in L (which now is the linear ordering over the remaining $m -r + 1$ candidates) will be eliminated. $\square$

Lemma 4

The reverse of e is a single-peaked order with respect to the given linear order L.

Proof

Our claim is that $O: (p = c_m)> c_{m-1}> \cdots > c_1$ is a valid single-peaked order with respect to L. To prove this, let us assume the contrary. This implies that there exists three candidates $c_i$, $c_j$, $c_k$ such that $c_j>_{O} c_i >_{O} c_k$, but according to the linear ordering L it is only possible that $c_i\ L\ \cdots \ L\ c_k\ L\ \cdots \ L\ c_j$ or $c_j\ L\ \cdots \ L\ c_k\ L\ \cdots \ L\ c_i$. Now, $c_j>_{O} c_i >_{O} c_k$ implies that, among $c_i, c_j,$ and $c_k$, $c_k$ is eliminated first, followed by $c_i$, and then $c_j$. But then, since at any round r only the corner elements can be eliminated (Lemma 3), $c_k$ can never be eliminated before both $c_i$ and $c_j$ in either of the two linear orderings. Hence e cannot be a valid elimination order if its reverse is not single-peaked. $\square$

1.3 Additional proofs for Sect. 5.3

Lemma 5

For the linear order $a\ L\ p\ L\ b$, when the input elimination order is (a, p, b), if there exists a successful bribery, then it can be induced by the elimination order (a, b, p) (or in other words, we do not need to consider any other elimination orders).

Proof

Since the linear order is $a\ L\ p\ L\ b$, only the following votes are allowed: $\{(a> p> b), (p> a> b), (p> b> a), (b> p > a)\}$. Let us assume that the total weight of all the voters voting $(a> p > b)$ is $W_1$, those voting $(p> a > b)$ is $W_2$, those voting $(p> b > a)$ is $W_3$, and those voting $(b> p > a)$ is $W_4$. Now, since the input elimination order is (a, p, b), we have the following inequalities: (1) $W_1 + W_2 \le W_3 + W_4$ (since a is getting eliminated in the first round, $LV(a) \ge LV(b)$, where LV(a) is the total weight of all the voters who place a at the end), and (2) $W_1 + W_2 + W_3 < W_4$ (since p gets eliminated in the second round). A successful bribery for p can proceed in two ways. First is if the bribery induces the elimination order (a, b, p) and the other way is to induce the elimination order (b, a, p). The task here is to prove that inducing (a, b, p) is better than inducing (b, a, p), or in other words, prove that inducing (b, a, p) requires bribing at least one more voter than what is required to induce (a, b, p).

To prove this, let us consider the case when the elimination order (a, b, p) is induced. Since the input already has a getting eliminated in the first round, we only need to look at the second round and see how much total weight should be bribed so that p is saved from elimination. Once a is eliminated, $LV(b) = W_1 + W_2 + W_3$, and $LV(p) = W_4$. Therefore, for p to win in the second round, the weight of votes we need to bribe is at least $D_{pb} = \left\lceil \frac{W_4 - (W_1 + W_2 + W_3)}{2} \right\rceil$. Let this bribery require $k_1$ voters to be bribed. Moreover, note that all the bribed votes are of the form $(b> p > a)$ and they will be bribed to vote $(p> b > a)$.

Now, let us consider the second elimination order (b, a, p). To induce this, that is to make b get eliminated in the first round, the total weight of votes we need to bribe is at least $D_{ab} = \left\lceil \frac{(W_3 + W_4) - (W_1 + W_2)}{2} \right\rceil = \left\lceil \frac{2W_3 + W_4 - (W_1 + W_2 + W_3)}{2} \right\rceil = W_3 + \left\lceil \frac{ W_4 - (W_1 + W_2 + W_3)}{2} \right\rceil = W_3 + D_{pb}$. As a result, among voters voting $(b> p > a)$ we need to bribe a weight of at least $D_{pb}$ so as to get b eliminated in the first round. But then, this bribery requires at least $k_1$ voters (from above), which in turn implies that we need at least one more voter to be bribed here than in the case where the elimination order (a, b, p) was induced. Hence, when the input elimination order is (a, p, b) and the linear order aLpLb, for a successful bribery it is enough to induce the elimination order (a, b, p). $\square$

Theorem 37

In single-peaked electorates, weighted-bribery with top-truncated votes is NP-complete for 3-candidate Baldwin’s rule.

Proof (sketch)

The proof here is similar to that of Theorem 36 and hence we only sketch the same. To prove $\mathcal {NP}$-hardness, we show a reduction from Partition$'$. Given an arbitrary instance $\{a_1, \ldots , a_n\}$ of Partition$'$, construct an instance of weighted-bribery (C, V, p, k), where $C = \{a, b, p\}$ is the set of candidates, $a\ L\ p\ L\ b$ is the linear order of the candidates, and V is the set of the following voters.

• 1. For each $a_i$, construct a voter $v_i$ whose weight is $2a_i$ and who votes for $(b > a \sim p)$. Let T be the set of all these voters.

• 2. Construct the following set of voters S: n voters of weight 2K each voting $(b> p > a)$, n voters of weight 2K each voting $(b > a \sim p)$, n voters of weight K each voting $(a > b \sim p)$, 2 voters of weight 1 each voting $(b> p > a)$, 2 voters of weight 1 each voting $(a > b \sim p)$, and 1 voter of weight 1 voting $(p > a \sim b)$.

Set the bribe limit $k = n$ and $V = S \cup T$. Now, proceed like in Theorem 36. $\square$