Control in the presence of manipulators: cooperative and competitive cases

Abstract

Control and manipulation are two of the most studied types of attacks on elections. In this paper, we study the complexity of control attacks on elections in which there are manipulators. We study both the case where the “chair” who is seeking to control the election is allied with the manipulators, and the case where the manipulators seek to thwart the chair. In the latter case, we see that the order of play substantially influences the complexity. We prove upper bounds, holding over every election system with a polynomial-time winner problem, for all standard control cases, and some of these bounds are at the second or third level of the polynomial hierarchy, and we provide matching lower bounds to prove these tight. Nonetheless, for important natural systems the complexity can be much lower. We prove that for approval and plurality elections, the complexity of even competitive clashes between a controller and manipulators falls far below those high bounds, even as low as polynomial time. Yet for a Borda-voting case we show that such clashes raise the complexity unless NP = coNP.

Specific systems

In some of the proofs in this section, we use the notation ${{\text{ score}_{(C,V)}(a)}}$ to denote the score of candidate a in election (C, V). When it is clear from context, we may leave out C, V, or both.

1.1 Plurality

In this subsection we prove the remaining plurality cases of Theorem 4.5.

Theorem A.1

For plurality elections, the following hold.

• 1. M+$\left[ \begin{array}{c}\mathrm{C}\\ \mathrm{D}\end{array}\right]$C$\left[ \begin{array}{c}\mathrm{A}\\ \mathrm{D}\end{array}\right]$V are each in $\mathrm{P}$.

• 2. $\left[ \begin{array}{c}\mathrm{C}\\ \mathrm{D}\end{array}\right]$C$\left[ \begin{array}{c}\mathrm{A}\\ \mathrm{D}\end{array}\right]$V-$\left[ \begin{array}{c}\mathrm{CF}\\ \mathrm{MF}\end{array}\right]$ are each in $\mathrm{P}$.

Proof

For the constructive cooperative and the destructive competitive cases it is clear that the manipulators should all vote for p.

For the destructive cooperative and the constructive competitive cases the optimal action for the manipulators is to all vote for the same highest-scoring candidate in $C - \{p\}$.

In all cases we can determine if the chair can be successful by assuming the manipulators vote as above and using the corresponding polynomial-time algorithm for control from Bartholdi et al. [3] (for the constructive cases) or from Hemaspaandra et al. [42] (for the destructive cases), modified in the obvious way for the nonunique-winner case (see Observation 4.6). $\square$

For the remaining proofs in this section, given an election (C, V) containing k manipulators, we say that a candidate r is a rival of p if r can beat p pairwise, i.e., if ${{\text{ score}_{\{p,r\}}(r)}} + k > {{\text{ score}_{\{p,r\}}(p)}}$. Recall that the manipulators come in without votes (i.e., as blank slates), and so ${{\text{ score}_{\{p,r\}}(r)}}$ and ${{\text{ score}_{\{p,r\}}(p)}}$ denote the scores of r and p in the election comprised of the candidates p and r, and the nonmanipulators in V.

Lemma A.2

If there exists a partition of voters such that p is an overall winner in the “TE” model when all manipulators vote for the same highest-scoring rival r, then there exists a partition such that p is always (i.e., regardless of the votes of the manipulators) an overall winner.

Proof

Given an election (C, V) where V contains k manipulators, a candidate $p \in C$, and a highest-scoring rival r, we do the following.

Let $(V_1,V_2)$ be a partition such that p is an overall winner when all manipulators vote for r. Let $k_1$ be the number of manipulators in $V_1$, let $k_2$ be the number of manipulators in $V_2$, let $\ell _1$ be the number of nonmanipulator votes for r in $V_1$, and let $\ell _2$ be the number of nonmanipulator votes for r in $V_2$. Without loss of generality assume that p is the unique winner of $(C,V_1)$ when all manipulators vote for r. Since we can exchange a manipulator in $V_2$ that votes for r with a nonmanipulator in $V_1$ that votes for r without changing the outcome of the election, we can assume that there are no manipulators in $V_2$ (i.e., $k_2 = 0$) or that there are no nonmanipulators voting for r in $V_1$ (i.e., $\ell _1 = 0$).

We first consider the case where $k_2 = 0$. We will construct a new partition $(\widehat{V}_1,\widehat{V}_2)$ that will work regardless of how the manipulators vote. Let $\widehat{V}_2$ consist of $\ell _2$ nonmanipulator votes for r, ${{\text{ score}_{V_2}(p)}}$ nonmanipulator votes for p, for every rival $\widehat{r} \ne r$, $min(\ell _2, score(\widehat{r}))$ votes for $\widehat{r}$, and for every nonrival $c \ne p$ all the nonmanipulator votes for c. Let $\widehat{V}_1 = V-\widehat{V}_2$.

We first show that p is always the unique winner of $(C,\widehat{V}_1)$. We know that ${{\text{ score}_{V_1}(r)}} + k_1 = \ell _1 + k_1 < {{\text{ score}_{V_1}(p)}} = {{\text{ score}_{\widehat{V}_1}(p)}}$. Since there are $k_1$ manipulators in $\widehat{V}_1$, it suffices to show that for all $c \ne p$, ${{\text{ score}_{\widehat{V}_1}(c)}} \le \ell _1$. This is immediate from the construction: ${{\text{ score}_{\widehat{V}_1}(r)}} = \ell _1$, for every nonrival $c \ne p$, ${{\text{ score}_{\widehat{V}_1}(c)}} = 0$, and for every rival $\widehat{r} \ne r$, $\textit{score}(\,{\widehat{r}}\,) \le {{\text{ score }(r)}} = \ell _1 + \ell _2$, and so ${{\text{ score}_{\widehat{V}_1}(\,\widehat{r}\,)}} \le \ell _1$.

So the only way in which p can be precluded from winning the runoff is if there is a rival $\widehat{r}$ of p such that $\widehat{r}$ is the unique winner of $(C,\widehat{V}_2)$ (recall that there are no manipulators in $\widehat{V}_2$). Since ${{\text{ score}_{\widehat{V}_2}(r)}} = \ell _2$ and ${{\text{ score}_{\widehat{V}_2}({\,\widehat{r}\,})}} \le \ell _2$ for every rival $\widehat{r}$, it then follows that r is the unique winner of $(C,\widehat{V}_2)$. But then $\textit{score}{(\,{\widehat{r}}\,)} < \ell _2$ for every rival $\widehat{r} \ne r$ and ${{\text{ score }(c)}} < \ell _2$ for every nonrival $c \ne p$, and ${{\text{ score}_{V_2}(p)}} = {{\text{ score}_{\widehat{V}_2}(p)}} < \ell _2$. It follows that r is the unique winner of $(C,V_2)$. Then p is never an overall winner of (C, V) using the partition $(V_1,V_2)$, which contradicts our assumption. It follows that p is always a winner of (C, V) using the partition $(\widehat{V}_1, \widehat{V}_2)$.

It remains to show the claim for $k_2 > 0$ and $\ell _1 = 0$. We will again construct a new partition $(\widehat{V}_1,\widehat{V}_2)$ that will work regardless of how the manipulators vote. Let $\widehat{V}_1$ consist of ${{\text{ score}_{V_1}(p)}}$ nonmanipulator votes for p and $k_1$ manipulators. Then $\widehat{V}_2$ consists of ${{\text{ score}_{V_2}(p)}}$ nonmanipulator votes for p, all nonmanipulator votes for c, $c \ne p$, and $k_2$ manipulators.

Since p is the unique winner of $(C,V_1)$ when all manipulators vote for r, it follows that $k_1 < {{\text{ score}_{V_1}(p)}} = {{\text{ score}_{\widehat{V}_1}(p)}}$, and so p is always the unique winner of $(C,\widehat{V}_1)$.

So the only way in which p can be precluded from winning the runoff is if there is manipulation and a rival $\widehat{r}$ of p such that $\widehat{r}$ is the unique winner of $(C,\widehat{V}_2)$. Then ${{\text{ score }({\,\widehat{r}\,})}} + k_2 = {{\text{ score}_{\widehat{V}_2}(\,\widehat{r}\,)}} > {{\text{ score}_{\widehat{V}_2}(c)}}$ for all $c \ne \widehat{r}$. In particular, ${{\text{ score }({\,\widehat{r}\,})}} + k_2 > {{\text{ score }(c)}}$ for every nonrival $c \ne p$ and ${{\text{ score }(\,{\widehat{r}\,})}}+ k_2 > {{\text{ score}_{{V}_2}(p)}}$. Since ${{\text{ score }({\,\widehat{r}\,})}} \le {{\text{ score }(r)}} = \ell _2$, it follows that r is the unique winner of $(C,V_2)$ when all manipulators vote for r. This contradicts our assumption. It follows that p is always a winner of (C, V) using the partition $(\widehat{V}_1, \widehat{V}_2)$. $\square$

Theorem A.3

For plurality elections, CCPV-TE-CF is in P.

Proof

Given an election (C, V) and a preferred candidate of the chair $p \in C$, p can be made a winner if and only if there exists a partition $(V_1,V_2)$ such that p is always an overall winner.

If no rivals of p exist, then clearly control is possible if and only if $C = \{p\}$ or there is at least one vote for p (in the latter case, let $V_1$ consist of one voter for p).

Otherwise, let r be a highest-scoring rival of p. It is immediate from Lemma A.2 that control is possible if and only if there exists a partition such that p wins when all manipulators vote for r and rank the remaining candidates in lexicographic order. This can be determined by running the polynomial-time algorithm for plurality-CCPV-TE from [42], modified in the obvious way for the nonunique-winner case (see Observation 4.6). $\square$

Theorem A.4

It holds that .

Proof

It immediately follows from the definition that .

Now suppose that “MF” control is possible. Then for all manipulations there exists a partition such that the preferred candidate p wins. Then either no rival to p exists, in which case “CF” control is possible since either p is the only candidate or there exists at least one vote for p. When a rival r to p exists, control is certainly possible when all the manipulators vote for r. By Lemma A.2 we know that then there exists a partition where p is always a winner, so “CF” control is possible. $\square$

Corollary A.5

For plurality elections, CCPV-TE-MF is in P.

Theorem A.6

For plurality elections, M+DCPV-TE is in P.

Proof

Given an election (C, V) and a despised candidate of the chair $p \in C$, we can determine in polynomial time if p can be precluded from winning by partitioning voters as follows. If there are no manipulators, run the polynomial-time algorithm for plurality-DCPV-TE from [42], modified in the obvious way for the nonunique-winner case (see Observation 4.6).

So, let $k > 0$ denote the number of manipulators in V. If there exists a rival r to p (i.e., a candidate that can beat p pairwise, i.e., a candidate for which ${{\text{ score}_{\{p,r\}}(p)}} < {{\text{ score}_{\{p,r\}}(r)}} + k$), then control is possible: Let $V_2$ consist of one manipulator and let all manipulators vote for r.

If there are no rivals, we must ensure that p doesn’t make it to the runoff, i.e., we need to ensure there is a candidate $c \ne p$ such that c ties-or-beats p in both subelections or there are two distinct candidates $c, d \ne p$ such that c ties-or-beats p in the first subelection and d ties-or-beats p in the second subelection. It is easy to see that this can be done if and only if we are in one of the following two cases.

• 1. There are at least two candidates, c is a highest-scoring candidate in $C - \{p\}$, and ${{\text{ score }(p)}} \le {{\text{ score }(c)}} + k$. (Have all manipulators vote for c and use partition $(V,\emptyset )$.)

• 2. There are at least three candidates, c and d are two highest-scoring candidates in $C - \{p\}$, and ${{\text{ score }(p)}} \le {{\text{ score }(c)}}+{{\text{ score }(d)}}+k$. (Have $V_1$ consist of $\min ({{\text{ score }(p)}},{{\text{ score }(c)}})$ votes for p and all votes for c. The remaining votes, including all manipulators, who will vote for d, will be in $V_2$.) $\square$

Lemma A.7

If there exists a partition of voters such that p is not a plurality winner in the “TE” model when all manipulators vote for p, then there exists a partition such that p can never be made a plurality winner by the manipulators.

Proof

Given an election (C, V) and a candidate $p \in C$, we do the following.

Let $(V_1,V_2)$ be a partition such that p is not a winner when all manipulators vote for p. If p can never be made a winner by the manipulators in this partition then we are done. So, suppose there exists a manipulation such that p is an overall winner (with the partition $(V_1,V_2)$). Without loss of generality assume that p is the unique winner of $(C,V_1)$. Then p is also the unique winner in $(C,V_1)$ if all manipulators vote for p. However, since p is not an overall winner if all manipulators vote for p there is a candidate $c \ne p$ such that if all manipulators vote for p, c is the unique winner of $(C,V_2)$ and c is the unique winner of the runoff $(\{p,c\},V)$.

Now move all manipulators from $V_2$ to $V_1$. Note that c remains the unique winner of $(C,V_2)$ and that c is always the unique winner of $(\{p,c\},V)$. It follows that in this new partition, p is never a winner, no matter what the manipulators do. $\square$

Lemma A.7 implies that is in P, since control is possible if and only if control is possible when all manipulators vote for p. This can be checked using the polynomial-time algorithm for plurality - DCPV-TE from Hemaspaandra et al. [42], modified in the obvious way for the nonunique-winner case (see Observation 4.6).

Theorem A.8

For plurality elections, DCPV-TE-CF is in P.

We will now show that Lemma A.7 also implies that plurality - DCPV-TE-MF is in P.

Theorem A.9

For plurality elections, DCPV-TE-MF is in P.

Proof

Given an election (C, V) and a despised candidate of the chair $p \in C$, we will show that we can determine in polynomial time if p can be precluded from winning by partitioning voters.

As in the “CF” case we will use Lemma A.7 to show that control is possible if and only if there exists a partition such that p is precluded from winning when all manipulators vote for p. This also implies that = .

It immediately follows from the definition that if the instance of plurality - DCPV-TE-MF is positive, then there exists a partition such that p is not a winner when all manipulators vote for p.

For the other direction, by Lemma A.7 if there exists a partition such that p is not a winner when all the manipulators vote for p, then there exists a partition $(V_1,V_2)$ such that p can never be made a winner by the manipulators. This implies that no matter what the manipulators do, there exists a partition (in fact, always the same partition) such that p is not a winner. This then implies that the instance of plurality - DCPV-TE-MF is positive. $\square$

1.2 Condorcet

In this subsection we prove the Condorcet cases of Theorem 4.5.

Theorem A.10

For Condorcet elections, the following hold.

• 1. M+$\left[ \begin{array}{c}\mathrm{C}\\ \mathrm{D}\end{array}\right]$C$\left[ \begin{array}{c}\mathrm{A}\\ \mathrm{D}\end{array}\right]$C are each in $\mathrm{P}$.

• 2. M+DC$\left[ \begin{array}{c}\mathrm{A}\\ \mathrm{D}\end{array}\right]$V are both in $\mathrm{P}$.

• 3. $\left[ \begin{array}{c}\mathrm{C}\\ \mathrm{D}\end{array}\right]$C$\left[ \begin{array}{c}\mathrm{A}\\ \mathrm{D}\end{array}\right]$C-$\left[ \begin{array}{c}\mathrm{CF}\\ \mathrm{MF}\end{array}\right]$ are each in $\mathrm{P}$.

• 4. DC$\left[ \begin{array}{c}\mathrm{A}\\ \mathrm{D}\end{array}\right]$V-$\left[ \begin{array}{c}\mathrm{CF}\\ \mathrm{MF}\end{array}\right]$ are both in $\mathrm{P}$.

Proof

For the constructive cooperative and the destructive competitive cases it is clear that the manipulators should all rank p first. The remaining candidates can be ranked in any order since this does not affect whether or not p is a Condorcet winner.

For the destructive cooperative and the constructive competitive cases the optimal action for the manipulators is to rank p last.

In all cases we can determine if the chair can be successful by assuming the manipulators vote as above and using the corresponding polynomial-time algorithm for control from Bartholdi et al. [3] (for the constructive cases) or from Hemaspaandra et al. [42] (for the destructive cases), modified in the obvious way for the nonunique-winner case (see Observation 4.6). $\square$

We now prove the Condorcet partition cases. Since Condorcet winners are always unique, the “TE” and “TP” cases coincide and so we will leave out this notation, following [42].

Theorem A.11

For Condorcet elections, M+$\left[ \begin{array}{c}\mathrm{C}\\ \mathrm{D}\end{array}\right]$C$\left[ \begin{array}{c}\mathrm{PC}\\ \mathrm{RPC}\end{array}\right]$ are each in P.

Proof

Given an election (C, V) and a preferred candidate of the chair $p \in C$, we can determine in polynomial time if p can be made a winner (in the presence of manipulators) by partitioning of candidates and by runoff partitioning of candidates as follows.

The use we’re about to make of the notion of control being possible is as described immediately before Theorem 4.7. For the constructive cases we do the following. Since Condorcet elections satisfy both WARP and unique-WARP, we know from Theorems 4.7 and 4.9 that control is possible if and only if control is possible using partition $(C-\{p\},\{p\})$. Set all manipulators to rank p first. Rank the candidates that do not beat p pairwise next in all manipulator votes (in any order). Then, as long as there exists an unranked candidate c that can never be a Condorcet winner in $(C-\{p\}, V)$, rank c next in all manipulator votes.

Let $\widehat{C}$ be the set of candidates not yet ranked by the manipulators. Notice that every $c \in \widehat{C}$ beats p pairwise, and every $c \in \widehat{C}$ can become a Condorcet winner in $(\widehat{C},V)$ (and thus also in (C, V)).

So, to determine if control is possible, we must determine if the manipulators can vote in such a way that $(\widehat{C},V)$ has no Condorcet winner, i.e., for each $c \in \widehat{C}$ there exists a $c' \in \widehat{C}-\{c\}$ such that $c'$ ties-or-beats c pairwise.

For $\Vert V\Vert$ even, assume that there are at least two candidates in $\widehat{C}$ and for $\Vert V\Vert$ odd, assume there are at least three candidates in $\widehat{C}$ (otherwise there will always be Condorcet winners). We have the following cases, depending on whether or not there is a Condorcet winner in $(\widehat{C},V)$ before the manipulators vote and depending on the parity of $\Vert V\Vert$. Let $k \ge 1$ denote the number of manipulators in V.

• 1. If there exists a Condorcet winner and $\Vert V\Vert$ is even, then let c be the Condorcet winner, and let $d \in \widehat{C}-\{c\}$. It is easy to see that each of the manipulators can vote $c> d > \widehat{C}-\{c,d\}$ or $d> c > \widehat{C}-\{c,d\}$ in such a way that c ties d pairwise. So, c is no longer a Condorcet winner and no other candidate becomes a Condorcet winner, since c ties-or-beats every other candidate pairwise.

• 2. If there exists a Condorcet winner and $\Vert V\Vert$ is odd, then let c be the Condorcet winner, and let $a,b \in \widehat{C}-\{c\}$ be such that a ties-or-beats b pairwise. Have $\lceil k/2 \rceil$ manipulators vote $a> b> c > \widehat{C}-\{a,b,c\}$ and $\lfloor k/2 \rfloor$ manipulators vote $b> c> a > \widehat{C}-\{a,b,c\}$. After this manipulation, b beats c pairwise, a beats b pairwise, and c beats every candidate in $\widehat{C} - \{b,c\}$ pairwise.

• 3. If there is no Condorcet winner and $\Vert V\Vert$ is even, then have $\lfloor k/2 \rfloor$ manipulators vote $\widehat{C}$ (i.e., the candidates in $\widehat{C}$ in some fixed order) and $\lfloor k/2 \rfloor$ manipulators vote $\overleftarrow{\widehat{C}}$ (i.e., the candidates in $\widehat{C}$ in reverse order). When k is odd, let the remaining manipulator vote arbitrarily. It is clear that no Condorcet winners are created by the manipulators.

• 4. If there is no Condorcet winner and $\Vert V\Vert$ is odd, then we have the following cases.

  • (a) If k is even, then have k/2 manipulators vote $\widehat{C}$ and the remaining k/2 manipulators vote $\overleftarrow{\widehat{C}}$.

  • (b) If k is odd and there is no weak Condorcet winner (a weak Condorcet winner is a candidate that ties-or-beats every other candidate pairwise), then have $\lfloor k/2 \rfloor$ manipulators vote $\widehat{C}$ and $\lfloor k/2 \rfloor$ manipulators vote $\overleftarrow{\widehat{C}}$. Let the remaining manipulator vote arbitrarily. It is clear that no Condorcet winner is created by the manipulators.

  • (c) If k is odd and there exists a weak Condorcet winner, then let c be a weak Condorcet winner and let a be a candidate such that a ties c pairwise. We have the following two cases.

    • i. If for all $b \in \widehat{C}-\{a,c\}$, a beats b pairwise and c beats b pairwise, then have $\lceil k/2 \rceil$ manipulators vote $\widehat{C}-\{a,c\}> a > c$ and have the remaining $\lfloor k/2 \rfloor$ manipulators vote $c> \widehat{C}-\{a,c\} > a$. So, now a beats c pairwise, and for all $b \in \widehat{C}-\{a,c\}$, c beats b pairwise and b beats a pairwise, and thus there is still no Condorcet winner.

    • ii. Otherwise, there exists a candidate $b \in C-\{a,c\}$ such that it is not the case that a and c both beat b pairwise. Suppose there are at least three manipulators, and set their votes in the following way. (If there is only one manipulator, then since each candidate in $\widehat{C}$ can become a Condorcet winner, all candidates in $\widehat{C}$ tie pairwise. And so there is always a Condorcet winner after manipulation.)

      • A. If a does not beat b pairwise, then let $\lfloor k/3 \rfloor$ manipulators vote $c> b> a > \widehat{C}-\{a,b,c\}$, $\lfloor k/3 \rfloor$ manipulators vote $b> a> c > \widehat{C}-\{a,b,c\}$, and $\lfloor k/3 \rfloor$ manipulators vote $a> c> b > \widehat{C}-\{a,b,c\}$. Note that a beats c pairwise, b beats a pairwise, and c beats every candidate in $\widehat{C} - \{a,c\}$ pairwise, so there is no Condorcet winner. If two manipulators remain, then have one vote $\widehat{C}$ and the other vote $\overleftarrow{\widehat{C}}$. Otherwise, if a single manipulator remains, since a beats c pairwise after the manipulators act as above, when the one remaining manipulator votes $c > \cdots$, no Condorcet winner is created.

      • B. If a beats b pairwise, then c does not beat b pairwise. It follows that c ties b pairwise. Now switch candidates a and b, and we are in the previous case.

For the destructive cases, since Condorcet elections satisfy unique-WARP, the chair cannot, by partitioning of candidates or by runoff partitioning of candidates, cause a candidate that is a unique winner to no longer be a unique winner [42]. This implies that control is possible if and only if the manipulators can vote so that p is not a winner in (C, V). It is immediate that the optimal action for the manipulators is to put p last. $\square$

Theorem A.12

For Condorcet elections, $\left[ \begin{array}{c}\mathrm{C}\\ \mathrm{D}\end{array}\right]$C$\left[ \begin{array}{c}\mathrm{PC}\\ \mathrm{RPC}\end{array}\right]$-$\left[ \begin{array}{c}\mathrm{CF}\\ \mathrm{MF}\end{array}\right]$ are each in P.

Proof

Given an election (C, V) and a preferred candidate of the chair $p \in C$, we can determine in polynomial time if p can be made a winner by partitioning candidates and by runoff partitioning of candidates as follows.

For the constructive cases, since Condorcet elections satisfy both WARP and unique-WARP, we know from Theorems 4.7 and 4.9 (which each apply only to the TE model, but since the Condorcet election system never has more than one winner, for Condorcet elections TE and TP are in effect identical) that control is possible if and only if control is possible using partition $(C-\{p\},\{p\})$. The manipulators can preclude p from winning if and only if there is a candidate $c \ne p$ that can be made to uniquely win $(C-\{p\},V)$ and ties-or-beats p pairwise. This can easily be checked by having all manipulators vote for c.

For the destructive cases, since Condorcet elections satisfy unique-WARP, the chair cannot, by partitioning of candidates or by runoff partitioning of candidates, cause a candidate that is a unique winner to no longer be a unique winner [42]. This implies that control is possible if and only if the manipulators cannot vote so that p becomes a winner in (C, V). It is immediate that the optimal action for the manipulators is to vote for p. $\square$

Theorem A.13

For Condorcet elections, M+DCPV is in P.

Proof

Given an election (C, V) and a despised candidate of the chair $p \in C$, we can determine in polynomial time if p can be precluded from winning by partitioning voters as follows.

If there exists a candidate $r \in C-\{p\}$ such that when all manipulators rank p last (and ranking the remaining candidates in any order), r ties-or-beats p pairwise, then control is possible by having all manipulators rank p last and using partition $(V,\emptyset )$.

If no such candidate exists, the only way to ensure that p is not a winner is to ensure that p does not participate in the runoff. Suppose there exists a partition and a manipulation such that p is not a unique winner of either subelection. If in this partition we set all manipulators to rank p last, p still does not win either subelection. So, we can check whether we are in this case by having all manipulators rank p last, and then use the polynomial-time algorithm for Condorcet - DCPV from [42], modified in the obvious way for the nonunique-winner case (see Observation 4.6). $\square$

Below we state a lemma analogous to Lemma A.7, but for Condorcet elections.

Lemma A.14

If there exists a partition of voters such that p is not a Condorcet winner when all manipulators vote for p, then there exists a partition such that p can never be made a winner by the manipulators.

Proof

Given an election (C, V) and a candidate $p \in C$, we do the following.

Let $(V_1,V_2)$ be a partition such that p is not a winner when all manipulators vote for p. So, either there exists a candidate $r \in C-\{p\}$ such that r ties-or-beats p pairwise when all manipulators vote for p, or p is not a unique winner of either subelection.

In the former case the partition $(V,\emptyset )$ will always work, and in the latter case it is clear to see that there is no way for the manipulators to make p a unique winner of either subelection, so we are done. $\square$

Lemma A.14 implies that is in P, since control is possible if and only if control is possible when all manipulators vote for p. This can be checked using the polynomial-time algorithm from [42], modified in the obvious way for the nonunique-winner case (see Observation 4.6).

Theorem A.15

For Condorcet elections, DCPV-CF is in P.

A similar argument as in the proof of Theorem A.9 shows that Lemma A.14 above also implies that the corresponding “MF” case is also in P.

Theorem A.16

For Condorcet elections, DCPV-MF is in P.

1.3 Approval

In this subsection we prove the Approval cases of Theorem 4.5.

Theorem A.17

For approval elections, the following hold.

• 1. M+$\left[ \begin{array}{c}\mathrm{C}\\ \mathrm{D}\end{array}\right]$C$\left[ \begin{array}{c}\mathrm{A}\\ \mathrm{D}\end{array}\right]$C are each in $\mathrm{P}$.

• 2. M+DC$\left[ \begin{array}{c}\mathrm{A}\\ \mathrm{D}\end{array}\right]$V are both in $\mathrm{P}$.

• 3. $\left[ \begin{array}{c}\mathrm{C}\\ \mathrm{D}\end{array}\right]$C$\left[ \begin{array}{c}\mathrm{A}\\ \mathrm{D}\end{array}\right]$C-$\left[ \begin{array}{c}\mathrm{CF}\\ \mathrm{MF}\end{array}\right]$ are each in $\mathrm{P}$.

• 4. DC$\left[ \begin{array}{c}\mathrm{A}\\ \mathrm{D}\end{array}\right]$V-$\left[ \begin{array}{c}\mathrm{CF}\\ \mathrm{MF}\end{array}\right]$ are each in $\mathrm{P}$.

Proof

For the constructive cooperative and the destructive competitive cases it is clear that the manipulators should all approve of only p.

For the destructive cooperative and the constructive competitive cases the optimal action for the manipulators is approve of all candidates except p.

In all cases we can determine if the chair can be successful by assuming the manipulators vote as above and using the corresponding polynomial-time algorithm for control from Bartholdi et al. [3] (for the constructive cases) or from Hemaspaandra et al. [42] (for the destructive cases), modified in the obvious way for the nonunique-winner case (see Observation 4.6). $\square$

Theorem A.18

For approval elections, M+$\left[ \begin{array}{c}\mathrm{C}\\ \mathrm{D}\end{array}\right]$C$\left[ \begin{array}{c}\mathrm{PC}\\ \mathrm{RPC}\end{array}\right]$-$\left[ \begin{array}{c}\mathrm{TE}\\ \mathrm{TP}\end{array}\right]$ are each in P.

Proof

Given an election (C, V) and a preferred candidate of the chair $p \in C$, we can determine in polynomial time if p can be made a winner (in the presence of manipulators) by partitioning of candidates and by runoff partitioning of candidates as follows. Let k denote the number of manipulators in V.

For the constructive “TE” cases we do the following. Since approval elections satisfy both WARP and unique-WARP, we know from Theorems 4.7 and 4.9 that control is possible if and only if control is possible using partition $(C-\{p\},\{p\})$. Set all manipulators to approve of p. If that makes p an overall winner of the election, we are done. If not, let c be the unique winner of subelection $(C-\{p\},V)$ (since p will participate in the runoff, the only way p can fail to then win overall is if there is a unique winner of $(C-\{p\}, V)$ who beats p in the runoff). As just mentioned parenthetically, note that after manipulation, c’s score in this case must be greater than p’s score. If for all $d \in C - \{p,c\}$, ${{\text{ score }(c)}} > {{\text{ score }(d)}} + k$, c will always be the unique winner of $(C-\{p\},V)$ and so p will never be an overall winner. If there exists a candidate d in $C - \{p,c\}$ such that ${{\text{ score }(c)}} \le {{\text{ score }(d)}} + k$, let ${{\text{ score }(c)}} - {{\text{ score }(d)}}$ voters approve of d (in addition to p). In this case, $(C-\{p\},V)$ does not have a unique winner and so p is an overall winner.

For the constructive “TP” cases, note that control is possible if and only if the manipulators can vote so that p becomes a winner in (C, V). So the optimal action for the manipulators is to approve of only p. Similarly, for the destructive cases, control is possible if and only if the manipulators can vote so that p does not win (for the “TP” cases) or does not uniquely win (for the “TE” cases) in (C, V). So the optimal action for the manipulators is to approve of all candidates except p. $\square$

Theorem A.19

For approval elections, $\left[ \begin{array}{c}\mathrm{C}\\ \mathrm{D}\end{array}\right]$C$\left[ \begin{array}{c}\mathrm{PC}\\ \mathrm{RPC}\end{array}\right]$-$\left[ \begin{array}{c}\mathrm{TE}\\ \mathrm{TP}\end{array}\right]$-$\left[ \begin{array}{c}\mathrm{CF}\\ \mathrm{MF}\end{array}\right]$ are each in P.

Proof

Given an election (C, V) and a preferred candidate of the chair $p \in C$, we can determine in polynomial time if p can be made a winner (in the presence of manipulators) by partitioning candidates and by runoff partitioning of candidates as follows.

For the constructive “TE” cases, since approval elections satisfy both WARP and unique-WARP, we know from Theorems 4.7 and 4.9 that control is possible if and only if control is possible using partition $(C-\{p\},\{p\})$. The manipulators can preclude p from winning if and only if there is a candidate $c \ne p$ that can be made to uniquely win using partition $(C-\{p\},\{p\})$. This can easily be checked by having all manipulators approve of only c.

For the constructive “TP” cases, note that control is possible if and only if the manipulators cannot vote so that p does not become a winner in (C, V). So the optimal action for the manipulators, regardless of who goes first, is to approve of all candidates except p. Similarly, for the destructive cases, control is possible if and only if the manipulators cannot vote so that p becomes a winner (for the “TP” cases) or a unique winner (for the “TE” cases) in (C, V). So the optimal action for the manipulators, regardless of who goes first, is to approve of only p. $\square$

Theorem A.20

For approval elections, M+DCPV-$\left[ \begin{array}{c}\mathrm{TE}\\ \mathrm{TP}\end{array}\right]$ are both in P.

Proof

Given an election (C, V) containing k manipulators, and a despised candidate of the chair $p \in C$, we can determine in polynomial time if p can be precluded from winning by partitioning voters for the “TE” case as follows. Let k denote the number of manipulators in V.

• 1. If there is a candidate, $c \ne p$ such that ${{\text{ score }(p)}} \le {{\text{ score }(c)}}+k$, then control is possible by having all manipulators disapprove of only p and using partition $(V,\emptyset )$.

• 2. If we are not in Case 1, the only way to preclude p from being a winner is if p doesn’t make it to the runoff, i.e., if there exist a partition and a manipulation such that p is not a unique winner of either subelection. If in this partition we make all manipulators vote to disapprove of only p, p is still not a unique winner of either subelection. So, we can check whether we are in this case by having all manipulators vote to disapprove of only p, and then using the polynomial-time algorithm for approval-DCPV-TE from [42], modified in the obvious way for the nonunique-winner case (see Observation 4.6).

For the “TP” case, replace “$\le$” by “<” in Case 1, and “unique winner” by “winner” and “approval-DCPV-TE” by “approval-DCPV-TP” in Case 2. $\square$

Below we state a lemma analogous to Lemma A.7, but for approval elections.

Lemma A.21

If there exists a partition of voters such that p is not an approval winner in the “TE” (“TP”) model when all manipulators approve of only p, then there exists a partition such that p can never be made an approval winner by the manipulators in the same tie-breaking model.

Proof

The proof for the “TE” case follows similarly to the proof of Lemma A.7, so we just provide the proof of the “TP” case.

Given an election (C, V) and a candidate $p \in C$, we do the following.

Let $(V_1,V_2)$ be a partition such that p is not a winner when all manipulators approve of only p. If p can never be made a winner by the manipulators in this partition then we are done. So, suppose there exists a manipulation such that p is an overall winner (with the partition $(V_1,V_2)$). Without loss of generality p is a winner of the subelection $(C,V_1)$. Then if all manipulators in $V_1$ approve of only p, we know that p remains a winner of $(C,V_1)$. Note we don’t get any new winners in $(C,V_1)$. Since p is not an overall winner if all manipulators approve of only p there is a candidate $c \ne p$ such that if all manipulators vote for p, c is a winner of $(C,V_2)$ and ${{\text{ score }(c)}} > {{\text{ score }(p)}}$.

Now move all manipulators from $V_2$ to $V_1$. Note that c remains a winner of $(C,V_2)$ and that c will always beat p in the runoff. It follows that in this new partition, p is never a winner, no matter what the manipulators do. $\square$

Lemma A.21 implies that and are both in P, since control is possible if and only if (nonmanipulator) control is possible when all manipulators approve of only p. This can be checked using the corresponding polynomial-time algorithms from Hemaspaandra et al. [42], modified in the obvious way for the nonunique-winner case (see Observation 4.6).

Theorem A.22

For approval elections, DCPV-$\left[ \begin{array}{c}\mathrm{TE}\\ \mathrm{TP}\end{array}\right]$-CF are both in P.

Lemma A.21 above also implies that the corresponding manipulators-first cases are both in P. The proof of the following theorem follows from a similar argument as the proof of Theorem A.9.

Theorem A.23

For approval elections, DCPV-$\left[ \begin{array}{c}\mathrm{TE}\\ \mathrm{TP}\end{array}\right]$-MF are both in P.