The Complexity of Online Bribery in Sequential Elections

TL;DR

This paper explores the complexity of bribery in sequential elections, revealing that online, sequential decision-making can significantly increase computational difficulty, though some systems remain manageable.

Contribution

It introduces a new model for online, sequential bribery and analyzes its complexity, highlighting cases where complexity increases or remains manageable.

Findings

Sequential bribery can be PSPACE-complete for some systems.

For certain election systems, bribery complexity does not increase dramatically.

The model captures realistic voting scenarios with limited information and timing constraints.

Abstract

Prior work on the complexity of bribery assumes that the bribery happens simultaneously, and that the briber has full knowledge of all votes. However, in many real-world settings votes come in sequentially, and the briber may have a use-it-or-lose-it moment to decide whether to alter a given vote, and when making that decision the briber may not know what votes remaining voters will cast. We introduce a model for, and initiate the study of, bribery in such an online, sequential setting. We show that even for election systems whose winner-determination problem is polynomial-time computable, an online, sequential setting may vastly increase the complexity of bribery, jumping the problem up to completeness for high levels of the polynomial hierarchy or even PSPACE. But we also show that for some natural, important election systems, such a dramatic complexity increase does not occur, and we pinpoint the complexity of their bribery problems.

Tables

Table 1. Table 1: Comparing the complexity of online bribery and nonsequential bribery for natural election systems

Problem	Online bribery complexity	Nonsequential bribery complexity
Plurality
$\text{-}\mathrm{Bribery}$	P (Thm. 5.3)	P [FHH09, Thm. 3.1]
$\text{-}\$\mathrm{Bribery}$	P (Thm. 5.3)	P [FHH09, Thm. 3.3]
$\text{-}\mathrm{Weighted}\text{-}\mathrm{Bribery}$	P (Thm. 5.3)	P [FHH09, Thm. 3.3]
$\text{-}\mathrm{Weighted}\text{-}\$\mathrm{Bribery}$	NP-complete (Thm. 5.4)	NP-complete [FHH09, Thm. 3.2]
3-candidate-Veto
$\text{-}\mathrm{Bribery}$	P (Thm. 5.6)	P [FHH09, Thm. 4.13]
$\text{-}\$\mathrm{Bribery}$	P (Thm. 5.6)	P [FHH09, Thm. 4.13]
$\text{-}\mathrm{Weighted}\text{-}\mathrm{Bribery}$	$P^{\mathrm{NP}\left[1\right]}$-complete (Thm. 5.6)	NP-complete [FHH09, Thm. 4.9]
$\text{-}\mathrm{Weighted}\text{-}\$\mathrm{Bribery}$	$P^{\mathrm{NP}\left[1\right]}$-hard and in ${\Delta }_2^p$ (Thm. 5.6)	NP-complete [FHH09, Thm. 4.8]
Approval
$\text{-}\mathrm{Bribery}$	P (Thm. 5.7)	NP-complete [FHH09, Thm. 4.2]
$\text{-}\$\mathrm{Bribery}$	P (Thm. 5.7)	NP-complete [FHH09, Thm. 4.2]
$\text{-}\mathrm{Weighted}\text{-}\mathrm{Bribery}$	P (Thm. 5.7)	NP-complete [FHH09, Thm. 4.2]
$\text{-}\mathrm{Weighted}\text{-}\$\mathrm{Bribery}$	NP-complete (Thm. 5.7)	NP-complete [FHH09, Thm. 4.2]

Equations

$$(\forall \overrightarrow{x_1})(\exists \overrightarrow{x_2})(\forall \overrightarrow{x_3})[F(\overrightarrow{x_1},\overrightarrow{x_2},\overrightarrow{x_3})],$$

$$(\exists \overrightarrow{x_1})(\forall \overrightarrow{x_2})(\forall \overrightarrow{x_3})[F(\overrightarrow{x_1},\overrightarrow{x_2},\overrightarrow{x_3})]\vee$$

$$(\forall \overrightarrow{x_1})(\exists \overrightarrow{x_2})(\forall \overrightarrow{x_3})[F(\overrightarrow{x_1},\overrightarrow{x_2},\overrightarrow{x_3})]\vee$$

$$(\forall \overrightarrow{x_1})(\forall \overrightarrow{x_2})(\exists \overrightarrow{x_3})[F(\overrightarrow{x_1},\overrightarrow{x_2},\overrightarrow{x_3})]\vee$$

$$(\forall \overrightarrow{x_1})(\forall \overrightarrow{x_2})(\forall \overrightarrow{x_3})[F(\overrightarrow{x_1}.\overrightarrow{x_2},\overrightarrow{x_3})].$$

$$(Q_1\overrightarrow{x_1})(Q_2\overrightarrow{x_2})\cdots (Q_{\ell }\overrightarrow{x_{\ell }})[F(\overrightarrow{x_1},\overrightarrow{x_2},\dots ,\overrightarrow{x_{\ell }})],$$

$$(Q_{1}\overrightarrow{x_{1}},b_{1})(Q_{2}\overrightarrow{x_{2}},b_{2})\cdots(Q_{\ell}\overrightarrow{x_{\ell}},b_{\ell})\Big{[}\Big{(}F(\overrightarrow{x_{1}},\overrightarrow{x_{2}},\dots,\overrightarrow{x_{\ell}})\Big{)}~{}{\;\wedge\;}~{}\Big{(}\bigwedge_{\{i\,\mid\>Q_{i}=\exists\}}b_{i}\Big{)}\Big{]}.$$

$$(\forall \overrightarrow{x_1})(\exists \overrightarrow{x_2})(\forall \overrightarrow{x_3})(\exists \overrightarrow{x_4})\cdots (\forall \overrightarrow{x_{2k+1}})[F(\overrightarrow{x_1},\overrightarrow{x_2},\dots ,\overrightarrow{x_{2k+1}})]$$

$$(\forall \overrightarrow{x_1},b_1)(\exists \overrightarrow{x_2},b_2)(\forall \overrightarrow{x_3},b_3)(\exists \overrightarrow{x_4},b_4)\cdots (\forall \overrightarrow{x_{2k+1}},b_{2k+1})$$

$$\displaystyle\Big{[}\Big{(}F(\overrightarrow{x_{1}},\overrightarrow{x_{2}},\dots,\overrightarrow{x_{2k+1}})\Big{)}~{}{\;\wedge\;}~{}\Big{(}b_{2}{\;\wedge\;}b_{4}\wedge\dots\wedge b_{2k}\Big{)}\Big{]}.$$

$$(\forall x_{1,1},x_{1,2},\dots ,x_{1,k_1})(\exists x_{2,1},x_{2,2},\dots ,x_{2,k_2})\cdots ({\forall }_{2k+1}{x}_{2k+1,1},x_{2k+1,2},\dots ,x_{2k+1,k_{2k+1}})$$

$$[\Phi (x_{1,1},x_{1,2},\dots ,x_{1,k_1},x_{2,1},x_{2,2},\dots ,x_{2,k_2},\dots ,x_{2k_1,1},x_{2k+1,2},\dots ,x_{2k+1,k_{2k+1}})],$$

$$({\forall }_1\overrightarrow{x_1})({\exists }_2\overrightarrow{x_2})({\forall }_3\overrightarrow{x_3})(\exists \overrightarrow{x_4})\cdots ({\forall }_z\overrightarrow{x_z})[F(\overrightarrow{x_1},\overrightarrow{x_2},\dots ,\overrightarrow{x_z})]$$

Full text

We introduce a model for, and initiate the study of, bribery in such an online, sequential setting. We show that even for election systems whose winner-determination problem is polynomial-time computable, an online, sequential setting may vastly increase the complexity of bribery, jumping the problem up to completeness for high levels of the polynomial hierarchy or even $\mathrm{PSPACE}$. But we also show that for some natural, important election systems, such a dramatic complexity increase does not occur, and we pinpoint the complexity of their bribery problems.

Key words: bribery, computational complexity, computational social choice, logic, quantifier assignment, sequential elections.

1 Introduction

In computational social choice theory, the three most studied types of attacks on elections are bribery, control, and manipulation, and the models of those that are studied seek to model the analogous real-world actions. Informally speaking, bribery means that an external agent, the briber, by bribing some voters without exceeding a given budget seeks to influence the outcome of an election; electoral control refers to an external agent, commonly called the chair, who seeks to influence the outcome of an election by structural changes such as adding, deleting, or partitioning candidates or voters; and manipulation (see Footnote 7 for a formal definition) means that some voter or coalition of voters may vote strategically. These strategic actions and their applications in artificial intelligence and multiagent systems have been surveyed in many book chapters and articles [FHH10, CW16, FR16, BR16].

Such studies are typically carried out for the model in which all the voters vote simultaneously. That sometimes is the case in the real world. But it also is sometimes the case that the voters vote in sequence—in what is sometimes called a roll-call election (see Section 2 for some related work); in political settings such as American political-party presidential primary conventions, one may actually have a verbal roll call for votes going, for example, from Alabama to Alaska to Arizona and so on through Wyoming. That type of setting, i.e., sequential, has been relatively recently introduced and studied for control and manipulation—in particular, studies have been done of both control and manipulation in the so-called online, sequential setting [HHR14, HHR17a, HHR17b].

In the present paper, we study the complexity of, and algorithms for, the online, sequential case of bribery. Briefly put, we are studying the case where the voting order (and the voter weights and cost of bribing each voter) is known ahead of time to the briber. But at the moment a voter seeks to vote, the voter’s planned vote is revealed to the briber, who then has a use-it-or-lose-it chance to bribe the voter, by paying the voter’s bribe-price (and doing so allows that vote to be changed to any vote the briber desires).

The problem we are studying is the complexity of that decision. In particular, how hard is it to decide whether under optimal play on the part of the briber there is an action for the briber regarding the current voter such that under continued future optimal play by the briber (in the face of all future revelations of unknown information being pessimal), the briber can reach a certain goal (e.g., having one of his or her two favorite candidates win; or not having any of his or her three most hated candidates win; those two types of goals are examples of what are known respectively as constructive and destructive goals). We mention that the text “there is an action for the briber regarding the current voter such that under continued future optimal play by the briber (in the face of all future revelations of unknown information being pessimal)” in the previous sentence is in fact, as will be made clear in Section 3.2, about alternating existential and universal quantifiers. Section 3.2, and for some issues Section 2, provide a detailed discussion of issues regarding the model, the varying forms the costs in bribery can take (from actual dollars to time or effort spent to risk accepted), and the fact that, despite the typical associations with the word “bribery,” in many settings bribery is not modeling illegal, immoral, or evil acts.

The following list presents the section structure of our results.

Sections 4.1 and 4.2 establish our upper bounds—of $\mathrm{PSPACE}$ and the ${\Pi^{p}_{2k+1}}$ level of the polynomial hierarchy—on online bribery (i.e., online, sequential bribery, but we will for the rest of the paper and especially in our problem names often omit the word “sequential” when the word “online” is present) in the general case and in the case of being restricted to at most $k$ bribes. In Section 3.3, we will briefly discuss why our ${\Pi^{p}_{2k+1}}$ upper-bound proofs are far from trivial and how we meet their challenges through establishing a new result, that may be of interest in its own right, about alternating Turing machines whose accepting paths are “weight-bounded.” 2. 2.

Section 4.3 proves that there are election systems, with simple winner problems, such that each of the abovementioned upper-bounds is tight, i.e., that PSPACE-completeness holds or ${\Pi^{p}_{2k+1}}$-completeness holds. Again, we will briefly discuss in Section 3.3 some substantial, novel challenges in our lower-bound proofs and how we surmount them. 3. 3.

In Section 5, we look at the complexity of online bribery for various natural systems. We show that for both Plurality and Approval, it holds that priced, weighted online bribery is $\mathrm{NP}$-complete, whereas all other problem variants of online bribery are in $\mathrm{P}$. Since these other problem variants in the case of traditional (i.e., nonsequential) bribery are known to be $\mathrm{NP}$-complete [FHH09], this also shows that nonsequential bribery can be harder than online bribery for natural systems. In addition, we provide complete dichotomy theorems that distinguish NP-hard from easy cases for all our online bribery problems for scoring protocols and additionally we show that Veto elections, even with three candidates, have even higher lower bounds for weighted online bribery, namely $\mathrm{P}^{\mathrm{NP}[1]}$-hardness, where $\mathrm{P}^{\mathrm{NP}[1]}$ denotes the class of problems that can be solved by a P machine querying its NP oracle at most once on each input.

We handle weighted election systems throughout this paper in the standard way that one would naturally expect. In Appendix A, we discuss the strengths and weaknesses of using this approach.

2 Related Work

Our paper’s general area is computational social choice, in which studying the complexity of election and preference aggregation problems and manipulative attacks on them is a central theme. There are many excellent surveys and book chapters on computational social choice [BCE13, Rot16, BCE*+*16], and computational social choice and computational complexity have a long history of close, mutually beneficial interaction (see the survey [Hem18]).

The prior papers most related to our work are the papers that defined and studied the complexity of online control [HHR17a, HHR17b], of online manipulation [HHR14], and of traditional (i.e., nonsequential) bribery [FHH09]. Particularly important among those is online manipulation, as we will show connections/inheritance between online manipulation and our problems. We also will show connections/inheritance between nonsequential manipulation and our problems. Traditional (i.e., nonsequential) manipulation was introduced by Bartholdi, Tovey, and Trick [BTT89] in the unweighted case and by Conitzer, Sandholm, and Lang [CSL07] in the weighted case.

As alluded to in Section 1, in our sequential problems the briber’s goal in the so-called constructive case is, loosely speaking, that at least one of a collection of “liked” candidates be a winner, and in the so-called destructive case is, loosely speaking, that none of a set of “disliked” candidates is a winner. This approach to framing the goal is the same as is used throughout the line of papers mentioned above on online attacks on elections, and supports connections and comparisons between this work and that earlier work. This model differs from the single-candidate focused model used in papers on nonsequential bribery, and in Footnote 3 we will discuss the model choice, and why we feel this goal is natural, and will present some complexity inheritances (and a noninheritance) between the two approaches. But we mention here that this approach to framing the goal, in addition to being the settled one in papers on online electoral attack problems, simply seems natural. For example, agents often do come into an election system focused not on a particular candidate winning, but having a collection of candidates from which they hope that at least one wins; and the agents often act accordingly. And similarly, agents often come into an election with some set of candidates they view as so dangerous or wrong-headed that the agent wants to ensure that none of them win. More broadly, we view the problems studied here as natural and interesting. Admittedly, the problems are formal problems, and so are not perfectly capturing the noise of reality. However, formalizing and studying a crisp model of a problem is an important step, even if further steps are needed. Relatedly, we mention that the worst-case nature of our analysis is itself creating a quite hostile environment for the briber—and so is modeling conservative bribers who want to handle the case in which unrevealed things all come out against them—and in our open questions collection at the end of this paper we urge the study of online bribery in contexts that go beyond the worst-case setting.

The existing work most closely related to our work on the effect on alternating Turing machines and formulas of limits on existential actions is the work on online voter control [HHR17b], though the issues tackled here are different and harder.

The work of Xia and Conitzer [XC10] (see also [Slo93, DP01, DE10, BDT19]) that defines and explores the Stackelberg voting game is also about sequential voting, although unlike this paper their analysis is game-theoretic and is about manipulation rather than bribery. Sequential (and related types of) voting have also been studied in an axiomatic way [Ten04] and using Markov decision processes [PP13], though neither of those works focuses on issues of bribery. Poole and Rosenthal [PR97] provide a history of roll-call voting. This is used, for instance, in the US Senate; Thomas discusses the strategic behavior of US senators who, seeking to be re-elected, “deliberately change the ideological tenor of their roll-call voting during the course of their terms” [Tho85, p. 96]. Another real-life example of a roll-call election can be observed when in a department meeting the chair goes around the table, asking each faculty member—one after the other—about their preferences regarding some important departmental matter (and perhaps about their reasons for these preferences).

In the original paper on nonsequential bribery there were other types of bribery, e.g., bribery*′*, unary-coding, and succinct variants [FHH09]. Many other types have been studied since, e.g., microbribery [FHHR09], nonuniform bribery [Fal08], swap- (and its special case shift-) bribery [EFS09] (see also [EF10, BCF*+*16, MNRS18]), and extension bribery [BFLR12]. However, for compactness and since they are very natural, this paper focuses completely on bribery in its eight typical versions (as to prices, weights, and constructive/destructive), except now in an online, sequential setting. It would be interesting to see in future work whether our model of online bribery in sequential elections (to be formally described in Section 3.2) can also be applied to these other variants of bribery.

Pulling back to the bigger picture, it is very important to stress that bribery can be about bribing in the “natural” sense of the word: paying people to change their votes. But bribery more generally models the situation where for each of a number of voters there is a cost associated with changing that voter’s vote. The cost indeed could be cash given to bribe them. But it could also be the “shoe leather” cost of sending campaign workers to the voters’ doors to spend the time to change the voters’ minds so that the voters actually sincerely believe in, and thus vote in, a given way. A variant on this that is more explicitly sequential would be a local political candidate canvassing door-to-door through a neighborhood along a fixed path, before an election, and after an initial few moments of chatting forming an assessment of the voter’s preferences and then deciding whether to spend the time needed to change the voter’s mind. Or it could be that the cost is measuring the danger to the briber of corrupting without the voter’s knowledge the given vote as it passes through the briber’s hands (and the briber is operating within a limit of how much total danger he or she is willing to risk).

That is, bribery provides a relatively broad framework for allocating a limited resource (framed as “cost”) to change the votes of some among a number of agents. In fact, the original bribery paper of Faliszewski, Hemaspaandra, Hemaspaandra [FHH09] itself already allowed both prices and weights, and also studied the case where the cost of the bribe varied based on “how far” from the original preference of the voter the briber wanted to move the vote via bribery (see that paper’s coverage of so-called bribery*′*, and see also the related notion of microbribery from Faliszewski, Hemaspaandra, Hemaspaandra, and Rothe [FHHR09]); and thus that paper itself was quite flexible in what its framework encompassed. That paper correctly stressed that the “bribery” being modeled was not necessarily an illegal, immoral, or evil act ([FHH09, p. 490], see also [FHHR09, p. 280] and those papers, from the citation list later in this paragraph, that are on “campaign management”). For example, the bribery could simply be a transaction in some broader optimization seeking to find the lowest cost to reach a certain type of outcome. Papers since the work of Faliszewski, Hemaspaandra, Hemaspaandra [FHH09] have proposed a wide range of new variants of the cost structure or the allowed bribery moves (or even the vote types or vote ensembles), depending on the situation being studied (as just a few examples, [EFS09, EF10, BBHH15, FRRS15, BFNT16, SFE17, MNRS18]).

This paper’s approach to the briber’s goal, which is assuming worst-case revelations of information, is inspired by the approach used in the area known as online algorithms [BE98]. However, our goal notion is not of the competitive-ratio type often used there, since here, as in general is true in computational social choice, we are not dealing with numerically-valued notions of degree of preference for a candidate. Rather, as mentioned earlier in this section, we adopt the goal model used by the existing line of work on online attacks on elections.

Interesting work that is related—though somewhat distantly—in flavor to our study is the paper of Chevaleyre et al. [CLM*+*12] on the addition of candidates. They also focus on the moment at which one has to make a key decision, in their case whether all of a group of potential additional candidates should be added.

3 Preliminaries

In this section, we first provide some basic notions from complexity theory and social choice theory. Then we formally define and discuss our model of online bribery. Finally, we briefly discuss some technical points of the proofs to come. We mention that during a first reading of this paper, the reader may wish to skip most or all of the footnotes, especially the two long ones in Section 3.2, since our footnotes are mostly used to give extra information, context, contrasts, and in some cases claims.

3.1 Basics

$\mathrm{P}$ is the class of decision problems in deterministic polynomial time. $\mathrm{NP}$ is the class of decision problems in nondeterministic polynomial time. For each $k\geq 0$, ${\Sigma^{p}_{k}}$ is the class of decision problems in the $k$th $\Sigma$ level of the polynomial hierarchy [MS72, Sto76], e.g., ${\Sigma^{p}_{0}}=\mathrm{P}$, ${\Sigma^{p}_{1}}=\mathrm{NP}$, and ${\Sigma^{p}_{2}}=\mathrm{NP}^{\mathrm{NP}}$ (i.e., the class of sets accepted by nondeterministic polynomial-time oracle Turing machines given unit-cost access to an NP oracle). For each $k\geq 0$, ${\Pi^{p}_{k}}=\{L\mid\overline{L}\in{\Sigma^{p}_{k}}\}$, e.g., ${\Pi^{p}_{0}}={\Sigma^{p}_{0}}=\mathrm{P}$, ${\Pi^{p}_{1}}=\mathrm{coNP}$, and ${\Pi^{p}_{3}}=\mathrm{coNP}^{\mathrm{NP}^{\mathrm{NP}}}$. The polynomial-hierarchy level ${\Delta_{2}^{{p}}}=\mathrm{P}^{\mathrm{NP}}$ is the class of sets accepted by deterministic polynomial-time oracle Turing machines given unit-cost access to an $\mathrm{NP}$ oracle, and $\mathrm{P}^{\mathrm{NP}[1]}$ is the same class restricted to one oracle query per input. Chandra, Kozen, and Stockmeyer [CKS81] defined alternating polynomial-time Turing machines and showed that the set of languages accepted by alternating polynomial-time Turing machines is exactly PSPACE (i.e., the class of problems that can be solved in polynomial space). We will not go into detail about alternating Turing machines, but simply put, they are Turing machines that can make both universal and existential moves.

We say that $A\leq_{m}^{p}B$ ($A$ polynomial-time many-one reduces to $B$) exactly if there is a polynomial-time computable function $f$ such that $(\forall x)[x\in A\iff f(x)\in B]$.

Fact 3.1**.**

For each complexity class ${\cal C}\in\{{\Sigma^{p}_{0}},\allowbreak{\Sigma^{p}_{1}},\allowbreak{\Pi^{p}_{1}},\allowbreak{\Sigma^{p}_{2}},\allowbreak{\Pi^{p}_{2}},\dots\}$, ${\cal C}$ is closed downwards under polynomial-time many-one reductions, i.e., $(B\in{\cal C}{\;\wedge\;}A\leq_{m}^{p}B)\implies A\in{\cal C}$.

Each of the classes mentioned in Fact 3.1 is even closed downwards under what is known as polynomial-time disjunctive truth-table reducibility [LLS75]. Disjunctive truth-table reducibility can be defined as follows. We say that $A\leq_{dtt}^{p}B$ ($A$ polynomial-time disjunctive truth-table reduces to $B$) exactly if there is a polynomial-time computable function $f$ such that, for each $x$, it holds that (a) $f(x)$ outputs a list of 0 or more strings, and (b) $x\in A$ if and only if at least one string output by $f(x)$ is a member of $B$. (Polynomial-time many-one reductions are simply the special case of polynomial-time disjunctive truth-table reductions where the polynomial-time disjunctive truth-table reduction’s output-list function is required to always contain exactly one element.)

Fact 3.2**.**

For each complexity class ${\cal C}\in\{{\Sigma^{p}_{0}},\allowbreak{\Sigma^{p}_{1}},\allowbreak{\Pi^{p}_{1}},\allowbreak{\Sigma^{p}_{2}},\allowbreak{\Pi^{p}_{2}},\dots\}$, ${\cal C}$ is closed downwards under polynomial-time disjunctive truth-table reductions, i.e., $(B\in{\cal C}{\;\wedge\;}A\leq_{dtt}^{p}B)\implies A\in{\cal C}$.

The above fact is obvious for P. It is also easy to see and well known for NP and $\mathrm{coNP}$ (for example, the results follow immediately from the result of Selman [Sel82] that NP is closed downwards under so-called positive Turing reductions). The results for the NP and coNP cases relativize (as Selman’s mentioned result’s proof clearly relativizes). That gives (namely, by relativizing the NP and coNP cases by complete sets for NP, $\mathrm{NP}^{\mathrm{NP}}$, etc.) the claims for the higher levels of the polynomial hierarchy (in fact, it gives something even stronger, since it gives downward closure under disjunctive truth-table reductions that themselves are relativized, but we won’t need that stronger version in this paper).

All the many-one and disjunctive truth-table reductions discussed in the paper will be polynomial-time ones. So we henceforth will sometimes skip the words “polynomial-time” when speaking of a polynomial-time many-one or disjunctive truth-table reduction.

A set $L$ is said to be (polynomial-time many-one) hard for a class ${\cal C}$ (for short, “$L$ is ${\cal C}$-hard”) exactly if $(\forall B\in{\cal C})[B\leq_{m}^{p}L]$. If in addition $L\in{\cal C}$, we say that $L$ is polynomial-time many-one complete for ${\cal C}$, or simply that $L$ is ${\cal C}$-complete.

An (unweighted) election system ${\cal E}$ takes as input a voter collection $V$ and a candidate set $C$, such that each element of $V$ contains a voter name and a preference order over the candidates in $C$; and for us in this paper preference orders are always total orders, except when we are speaking of approval voting where the preference orders are bit-vectors from $\{0,1\}^{\|C\|}$. (For the rest of the preliminaries, we’ll always speak of total orders as the preference orders’ type, with it being implicit that when later in the paper we speak of and prove results about approval voting, all such places will tacitly be viewed as speaking of bit-vectors.) The election system maps from that to a (possibly nonproper) subset of $C$, often called the winner set. We often will call each element of $V$ a vote, though as is common sometimes we will use the term vote to refer just to the preference order. We often will use the variable names $\sigma$, $\sigma_{1}$, $\sigma_{2}$, …, $\sigma_{i}$ for total orders. We allow election systems to, on some inputs, have no winners.111Although in social choice this is often disallowed, as has been discussed previously, see, e.g., [FHH16, Footnote 3], artificially excluding the case of no winners is unnatural, and many papers in computational social choice allow this case. A typical real-world motivating example is that in Baseball Hall of Fame votes, having no inductees in a given year is a natural outcome that has at times occurred.

For a given (unweighted, simultaneous) election system, ${\cal E}$, the (unweighted) winner (aka the winner-determination) problem (in the unweighted case) is the set $\{\mathopen{\langle}{C,V,c}\mathclose{\rangle}\mid c\in C{\;\wedge\;}$ $c$ is a winner of the election $(C,V)$ under election system ${\cal E}\}$.

For a given (unweighted, simultaneous) election system, ${\cal E}$, the winner problem in the weighted case will be the set of all strings $\mathopen{\langle}{C,V,c}\mathclose{\rangle}$ such that $C$ is a candidate set, $V$ is a set of weighted (via binary nonnegative integers as weights) votes (each consisting of a voter name and a total order over $C$), $c\in C$, and in the unweighted election created from this by replacing each $w$-weighted vote in $V$ with $w$ unweighted copies of that same vote, $c$ is a winner in that election under the (unweighted) election system ${\cal E}$. For an election system ${\cal E}$, it is clear that if the winner problem in the weighted case is in $\mathrm{P}$, then so is the winner problem in the unweighted case. However, there are election systems ${\cal E}$ for which the converse fails. The above approach to defining the weighted winner problem is natural and appropriate for the election systems discussed in this paper. However, see Appendix A for a discussion of the strengths and weaknesses of using this approach to the weighted winner problem in other settings, and for more discussion of the claims in this paragraph.

3.2 Online Bribery in Sequential Elections

This paper is about the study of online bribery in sequential elections. In this setting, we are—this is the sequential part—assuming that the voters vote in a well-known order, sequentially, with each casting a ballot that expresses preferences over all the candidates. And we are assuming—this is the online part—that the attacker, called “the briber,” as each new vote comes in has his or her one and only chance to bribe that voter, i.e., to alter that vote to any vote of the briber’s choice.

Bribery has aspects of both the other standard types of electoral attacks: bribery is like manipulation in that one changes votes and it is like (voter) control in that one is deciding on a set of voters (in the case of bribery, which ones to bribe). Reflecting this, our model follows as closely as possible the relevant parts of the existing models that study manipulation and control in online settings [HHR14, HHR17b, HHR17a]. In particular, we will follow insofar as possible both the model of, and the notation of the model of, the paper by Hemaspaandra, Hemaspaandra, and Rothe [HHR17b] that introduced the study of online voter control in sequential elections. In particular, we will follow the flavor of their model of control by deleting voters, except here the key decision is not whether to delete a given voter, but rather is whether a given voter should be bribed, i.e., whether the voter’s vote should be erased and replaced with a vote supplied by the briber. That “replace[ment]” part is more similar to what happens in the study of online manipulation, which was modeled and studied by Hemaspaandra, Hemaspaandra, and Rothe [HHR14]. We will, as both those papers do, focus on a key moment—a moment of decision—and in particular on the complexity of deciding whether there exists an action the briber can take, at that moment, such that doing so will ensure, even under the most hostile of conditions regarding the information that has not yet been revealed, that the briber will be able to meet his or her goal.

If $u$ is a voter and $C$ is a candidate set, an election snapshot for $C$ and $u$ is specified by a triple $V=(V_{<u},u,V_{>u})$, which loosely put (and in the paragraph after this one we will provide close coverage of the precise details) is made up of all voters in the order they vote, each accompanied in models where there are prices and/or weights with their prices and/or weights (which in this paper are assumed to be nonnegative integers coded in binary).222Why do we feel it natural in most situations for the prices and weights to be in binary rather than unary? A TARK referee, for example, asked whether it was not natural to assume that weights and prices would always be small, or if not, would always be multiples of some integer that when divided out would make the remaining numbers small. Our answer is that both prices and weights in many settings tend to be large, and without any large, shared-by-all divisor. To see this clearly regarding weights, consider for example the number of shares of stock the various stockholders hold in some large corporation or the number of residents in each of the states of a country. Prices too are potentially as rich and varied as are individuals and objects, e.g., in some settings each person’s bribe-price might be the exact fair market value of his or her house, or might be closely related to the number of visits a web site they own has had in the past year. Pulling back, we note that requiring prices and weights to be in unary is often tremendously (and arguably inappropriately) helping the algorithms as to what their complexity is, since in effect one is “padding” the many inputs’ sizes as much as exponentially. But if weighted votes are viewed as indivisible objects—and that is indeed how they are typically treated in the literature—the right approach indeed is to code the weights in binary, and not to give algorithm designers the potentially vastly lowered bar created by the padding effect of coding the weights in unary. Indeed, it is known in the study of nonsequential bribery that changing prices or weights to unary can shift problems’ complexities from NP-hardness to being in deterministic polynomial time [FHH09, pp. 500–504]. Also, people typically do code natural numbers in binary, not unary. In addition, for each voter voting before $u$ (namely, the voters in $V_{<u}$), also included in this listing will be the vote they cast (or if they were bribed, what vote was cast for them by the briber) and whether they were bribed; and for $u$ the listing will also include the vote $u$ will cast unless bribed to cast a different vote. So $V_{>u}$ is simply a list, in the order they will vote, of the voters, if any, who come after $u$, each also including the voter’s price and/or weight data if we are in a priced and/or weighted setting. Further, the vote for $u$ and all the votes in $V_{<u}$ must be votes over the candidate set $C$ (and in particular, in this paper votes are total orderings of the candidates, e.g., $a>b>c$).

There is a slight overloading of notation above, in that we have not explicitly listed in the structure the location of the mentioned extra data. In fact, our actual definition is that the first and last components of the 3-tuple $V$ are lists of tuples, and the middle component is a single tuple. Each of these contain the appropriate information, as mentioned above. For example, for priced, weighted bribery:

the elements of the list $V_{<u}$ will be 5-tuples $(v_{i},p_{i},w_{i},\sigma_{i},b_{i})$ whose components respectively are the voter’s name, the voter’s price, the voter’s weight, the voter’s cast ballot (which is the voter’s original preference order if the voter was not bribed and is whatever the voter was bribed into casting if the voter was bribed), and a bit specifying whether that vote resulted from being bribed, 2. 2.

the middle component of $V$ will be a tuple that contains the first four of those five components, and 3. 3.

the elements of the list $V_{>u}$ will contain the first three of the above-mentioned five components.

Similarly, for example, for unpriced, unweighted bribery, the three tuple types would respectively have three components, two components, and one component.

As a remaining tidbit of notational overloading, in some places we will speak of $u$ when we in fact mean the voter name that is the first component of the tuple that makes up the middle tuple of $V$. That is, we will use $u$ both for a tuple that names $u$ and gives some of its properties, and as a stand-in for the voter him- or herself. Which use we mean will always be clear from context. (We mention in passing that the fact that our voters and candidates have names is consistent with the existing line of work on online elections, see, e.g., [HHR14, HHR17b], though it also is in keeping with that fact that voters and candidates typically truly do have names. We will at times use this in proofs.)

Let us, with the above in hand, define our notions of online bribery for sequential elections. Settings can independently allow or not allow prices and weights, and so we have four basic types of bribery in our online, sequential model, each having both constructive and destructive versions.

Our specification of these problems as languages is centered around what Hemaspaandra, Hemaspaandra, and Rothe [HHR14] called a magnifying-glass moment. This is a moment of decision as to a particular voter. To capture precisely what information the briber does and does not have at that moment, and to thus allow us to define our problems, we define a structure that we will call an OBS, which stands for online bribery setting. An OBS is defined as a 5-tuple $(C,V,\sigma,d,k)$, where $C$ is a set of candidates; $V=(V_{<u},u,V_{u<})$ is an election snapshot for $C$ and $u$ as discussed earlier; $\sigma$ is the preference order of the briber; $d\in C$ is a distinguished candidate; and $k$ is a nonnegative integer (representing for unpriced cases the maximum number of voters that can be bribed, and for priced cases the maximum total cost, i.e., the sum of the prices of all the bribed voters).

Given an election system ${\cal E}$, we define the online unpriced, unweighted bribery problem, abbreviated by $\mathrm{online}\hbox{-}\allowbreak\mathrm{{{\cal E}}}\hbox{-}\allowbreak\mathrm{Bribery}$, as the following decision problem. The input is an OBS. And the question is: Does there exist a legal (i.e., not violating whatever bribe limit holds) choice by the briber on whether to bribe $u$ (recall that $u$ is specified in the OBS, namely, via the middle component of $V$) and, if the choice is to bribe, of what vote to bribe $u$ into casting, such that if the briber makes that choice then no matter what votes the remaining voters after $u$ are (later) revealed to have, the briber’s goal (the meeting of which itself depends on ${\cal E}$ and will be defined explicitly two paragraphs from now) can be reached by the current decision regarding $u$ and by using the briber’s future (legal-only, of course) decisions (if any), each being made using the briber’s then-in-hand knowledge about what votes have been cast by then?

Note that this approach is about alternating quantifiers. It is asking whether there is a current choice by the briber such that for all potential revealed vote values for the next voter there exists a choice by the briber such that for all potential revealed vote values for the next-still voter there exists a choice by the briber such that… and so on… such that the resulting winner set under election system ${\cal E}$ meets the briber’s goal. This is a bit more subtle than it might at first seem. The briber is acting somewhat powerfully, since the briber is represented by existential quantifiers. But the briber is not all-powerful in this model. In particular, the briber can’t see and act on future revelations of vote values; after all, those are handled by a universal quantifier that occurs downstream from an existential quantifier that commits the briber to a particular choice.

In the above we have not defined what “the briber’s goal” is, so let us do that now. $W_{{\cal E}}(C,U)$ will denote the winner set, according to election system ${\cal E}$, of the (nonsequential) election $(C,U)$, where $C$ is the candidate set and $U$ is the set of votes. By the briber’s goal we mean, in the constructive case, that if at the end of the above process $U^{\prime}$ is the set of votes (some may be the original ones and some may be the result of bribes), it holds that $W_{{\cal E}}(C,U^{\prime})\cap\{c\mid c\geq_{\sigma}d\}\neq\emptyset$, i.e., the winner set includes some candidate (possibly itself being $d$) that the briber likes at least as much as the briber likes $d$. In the destructive case, the goal is to ensure that no candidate that the briber hates as much or more than the briber hates $d$ belongs to the winner set, i.e., the briber’s goal is to ensure that $W_{{\cal E}}(C,U^{\prime})\cap\{c\mid d\geq_{\sigma}c\}=\emptyset$.333In each of these two cases, although we have in the problem statement used an order $\sigma$, that order $\sigma$ really is merely being used to determine what set of candidates to (constructive case) try to get one of into the winner set, or to (destructive case) try to keep all of out of the winner set. So one might wonder why we don’t simply pass in such a set, rather than passing in an ordering. The answer is that for the version of the problem that this paper is studying, namely the decision version, either way would be fine. However, allowing an order to be passed in keeps our approach in harmony with the approach of earlier work [HHR14, HHR17b, HHR17a], which made that choice because if one studies the optimization version of the problem—e.g., in the constructive case trying to find the most preferred candidate within $\sigma$ that the briber can ensure will be a winner—a set does not bring in enough information to frame the problem but an order does. Our approach is also following—and thus better allowing comparisons and connections to—the earlier papers on online attacks on elections, in that our paper is using an upper segment of the order for the constructive case and a lower segment of the order for the destructive case (see [HHR17b, second paragraph of Footnote 4] for more discussion of why that choice is most natural). That model intentionally does not limit one to—as is the case in the nonsequential case of bribery—specifying in the constructive case a single candidate whom one wants to win, or in the destructive case a single candidate whom one wants to keep from being a winner. There is so much uncertainty in the online setting that it seems natural to allow the attacker to have such flexible goals. Of course, the attacker is free to specify just one candidate, and so the just-one-focus-candidate cases are special cases of our model. And so, for each problem here, the version that is limited to a single focus-candidate certainly many-one polynomial-time reduces to the same problem in our more-flexible-goal model. So all our upper-bound complexity results in our model immediately are inherited by the one-focus-candidate sequential versions. On the other hand, it follows from the present paper’s work that, unless $\mathrm{NP}=\mathrm{coNP}$, there are cases when the one-focus-candidate sequential model falls into a simpler complexity class than the analogous problem in our model. In particular, the second part of Theorem 5.6 proves that $\mathrm{online}\hbox{-}\allowbreak\mathrm{{\mbox{\rm 3-candidate-Veto}}}\hbox{-}\allowbreak\mathrm{Weighted}\hbox{-}\allowbreak\mathrm{Bribery}$ is $\mathrm{P}^{\mathrm{NP}[1]}$-complete in our goal model, but the second bullet point of the proof of that theorem notes that in the one-focus-candidate sequential model the problem is NP-complete.

In a moment, we will go on to define online bribery variants that have prices and/or weights. However, let us first pause for a moment to give a toy example of $\mathrm{online}\hbox{-}\allowbreak\mathrm{{{\cal E}}}\hbox{-}\allowbreak\mathrm{Bribery}$, simply to give some concreteness to the problem. (Note: Later in the paper, Section 5 will build efficient algorithms for many important online bribery problems.) In our example here, we will not focus very much on the actual formal structures that we have defined to capture the problem, but will give the example informally. In this example, the election system will be 2-Approval, in which the top two candidates in each voter’s vote get one point each from that voter and the other candidates get zero points from that voter. And whichever candidate (or candidates if there is a tie for most points) gets the most points wins the election. In our example, the candidate set $C$ will be $a$, $b$, and $c$, the briber’s preference order $\sigma$ will be $a<b<c$, and the briber’s $d$ will be $b$. So the briber’s goal is to have at least one of $b$ or $c$ be a winner. Suppose that our magnifying-glass moment is the moment when the second-to-last voter’s vote is revealed, and suppose that as we enter that moment—we will not focus on the details of what came before—the current scores are 8 points for $a$, and $7$ points each for $b$ and $c$. Suppose that at this magnifying-glass moment we are in the extreme case in which the briber’s budget has already been completely expended, and so the briber is helpless to do any further bribing. Let us completely cover whether this instance is a Yes instance of $\mathrm{online}\hbox{-}\allowbreak\mathrm{{\mbox{\rm 2-Approval}}}\hbox{-}\allowbreak\mathrm{Bribery}$ (i.e., belongs to that set) or is a No instance. If the revealed vote of the second-to-last voter is $a<b<c$ or is $a<c<b$—i.e., the points of that voter will go to $b$ and $c$—then this is a Yes instance of the problem. Why? After that second-to-last vote—and keep in mind that the briber has no bribing budget left—all three candidates would be tied at 8 points, and so since the final voter has to give one point to exactly two candidates, whichever of $b$ or $c$ gets one of those points will be a winner. So at least one of $b$ or $c$ will be a winner, though we don’t know which. On the other hand, if the revealed vote of the second-to-last voter is any of the other four preference orders (which are the orders that will cause that voter to give one of its points to $a$), then this is a No instance of $\mathrm{online}\hbox{-}\allowbreak\mathrm{{\mbox{\rm 2-Approval}}}\hbox{-}\allowbreak\mathrm{Bribery}$. That is so simply because, since the briber has no bribing budget left, if the second-to-last voter’s points go to $a$ and $b$ (respectively, $a$ and $c$), then if the final voter gives his or her points to $a$ and $c$ (respectively, $a$ and $b$), then $a$ is the sole winner of the election, and so the briber’s goal has not been achieved.

Turning back to definitions, we already have defined both $\mathrm{online}\hbox{-}\allowbreak\mathrm{{{\cal E}}}\hbox{-}\allowbreak\mathrm{Bribery}$ and $\mathrm{online}\hbox{-}\allowbreak\mathrm{{{\cal E}}}\hbox{-}\allowbreak\mathrm{Destructive}\hbox{-}\allowbreak\mathrm{Bribery}$. Those both are in the unpriced, unweighted setting. And so as per our definitions, the voters passed in as part of the problem statement do not come with or need price or weight information.

In contrast, for the priced and/or weighted settings, our definitions, naturally enough, require that the OBS includes those prices and/or weights. And so the same definition text that was used above defines all the other cases, except that one must keep in mind for the priced cases that when the “bribery limit” is mentioned one must instead speak of the “bribery budget,” and in the weighted cases the winner set $W$ is of course defined in terms of the weighted version of the given voting system (which must, for that to be meaningful, have a well-defined notion of what its weighted version is; Section 3.1 provides that notion for all systems in this paper, see also Appendix A for further discussion). Thus, we also have tacitly defined the six problems \mathrm{online}\hbox{-}\allowbreak\mathrm{{{\cal E}}}\hbox{-}\allowbreak\mathrm{\Bribery}$,$\mathrm{online}\hbox{-}\allowbreak\mathrm{{{\cal E}}}\hbox{-}\allowbreak\mathrm{Destructive}\hbox{-}\allowbreak\mathrm{$Bribery}$,$\mathrm{online}\hbox{-}\allowbreak\mathrm{{{\cal E}}}\hbox{-}\allowbreak\mathrm{Weighted}\hbox{-}\allowbreak\mathrm{Bribery}$,$\mathrm{online}\hbox{-}\allowbreak\mathrm{{{\cal E}}}\hbox{-}\allowbreak\mathrm{Destructive}\hbox{-}\allowbreak\mathrm{Weighted}\hbox{-}\allowbreak\mathrm{Bribery}$,$\mathrm{online}\hbox{-}\allowbreak\mathrm{{{\cal E}}}\hbox{-}\allowbreak\mathrm{Weighted}\hbox{-}\allowbreak\mathrm{$Bribery}$, and$\mathrm{online}\hbox{-}\allowbreak\mathrm{{{\cal E}}}\hbox{-}\allowbreak\mathrm{Destructive}\hbox{-}\allowbreak\mathrm{Weighted}\hbox{-}\allowbreak\mathrm{$Bribery}$.

For an unpriced online bribery problem, we will postpend the problem name with a “[$k$]” to define the version where as part of the problem definition itself the bribery limit is—in contrast with the above unpriced problem—not part of the input but rather is fixed to be the value $k$. For example, $\mathrm{online}\hbox{-}\allowbreak\mathrm{{{\cal E}}}\hbox{-}\allowbreak\mathrm{Bribery}[k]$ denotes the unpriced, unweighted bribery problem where the number of voters who can be bribed is set not by the problem input but rather is limited to be at most $k$.

Why might these “[$k$]” cases be interesting and worthwhile to study? For many of our problems, we will show that the versions without a bound on the number of bribes are extraordinarily complex, namely, PSPACE-complete. It is natural to wonder if such simpler versions might drop to less extremely high levels of complexity, and indeed we will see that that is the case. So this is a type of parameterized investigation, though a somewhat unusual one. Also, in various situations one might naturally have both a budget bound and a fixed upper bound on the number of bribes that one can do. For example, suppose a kingdom, to form each valuable alliance, will need to always offer in marriage a child of the monarch, plus some amount of treasure that varies depending on the country being bribed into an alliance. One might well suppose that the number of children of the monarch is limited by some relatively small number, such as ten, and that problem might better model the problem if such bounds reduce the complexity, as we will see that they seem to.

Note that in each of the “[$k$]” variants, we tacitly are altering the definition of OBS from its standard 5-tuple, $(C,V,\sigma,d,k)$, to instead the 4-tuple $(C,V,\sigma,d)$; that is because for these cases, the $k$ is fixed as part of the general problem itself, rather than being a variable part of the individual instances. For priced “[$k$]” variants, there will be both a limit (being a variable part of the input) on the total price of the bribes and a fixed as part of the general problem itself limit on the number of voters who can be bribed.

Of course, there are some immediate relationships that hold between these eight problems. One has to be slightly careful since there is a technical hitch here. We cannot for example simply claim that $\mathrm{online}\hbox{-}\allowbreak\mathrm{{{\cal E}}}\hbox{-}\allowbreak\mathrm{Bribery}[k]$ is a subcase of \mathrm{online}\hbox{-}\allowbreak\mathrm{{{\cal E}}}\hbox{-}\allowbreak\mathrm{\Bribery}[k]$. If we had implemented the unpriced case by still including prices in the input but requiring them all to be 1, then it would be a subcase. But regarding both prices (weights), our definitions simply omit them completely from problems that are not about prices (weights). In spirit, it is a subcase, but formally it is not. Nonetheless, we can still reflect the relationship between these problems, namely, by stating how they are related via polynomial-time many-one reductions. (We could even make claims regarding more restrictive reduction types, but since this paper is concerned with complexity classes that are closed downwards under polynomial-time many-one reductions, there is no reason to do so.) The following proposition (and the connections that follow from it by the transitivity of polynomial-time many-one reductions) captures this.

Proposition 3.3**.**

For each $k>0$ and for each election system ${\cal E}$,

1. (a)

\mathrm{online}\hbox{-}\allowbreak\mathrm{{{\cal E}}}\hbox{-}\allowbreak\mathrm{Bribery}[k]\leq_{m}^{p}\mathrm{online}\hbox{-}\allowbreak\mathrm{{{\cal E}}}\hbox{-}\allowbreak\mathrm{\Bribery}[k]$, 2. (b)

$\mathrm{online}\hbox{-}\allowbreak\mathrm{{{\cal E}}}\hbox{-}\allowbreak\mathrm{Bribery}[k]\leq_{m}^{p}\mathrm{online}\hbox{-}\allowbreak\mathrm{{{\cal E}}}\hbox{-}\allowbreak\mathrm{Weighted}\hbox{-}\allowbreak\mathrm{Bribery}[k]$, 3. (c)

\mathrm{online}\hbox{-}\allowbreak\mathrm{{{\cal E}}}\hbox{-}\allowbreak\mathrm{\Bribery}[k]\leq_{m}^{p}\mathrm{online}\hbox{-}\allowbreak\mathrm{{{\cal E}}}\hbox{-}\allowbreak\mathrm{Weighted}\hbox{-}\allowbreak\mathrm{$Bribery}[k]$, and 4. (d)

\mathrm{online}\hbox{-}\allowbreak\mathrm{{{\cal E}}}\hbox{-}\allowbreak\mathrm{Weighted}\hbox{-}\allowbreak\mathrm{Bribery}[k]\leq_{m}^{p}\mathrm{online}\hbox{-}\allowbreak\mathrm{{{\cal E}}}\hbox{-}\allowbreak\mathrm{Weighted}\hbox{-}\allowbreak\mathrm{\Bribery}[k]$. 2. 2.

The above item also holds for the case when all of its problems are changed to their destructive versions. 3. 3.

The above two items also hold for the case when all the “$[k]$”s are removed (e.g., we have \mathrm{online}\hbox{-}\allowbreak\mathrm{{{\cal E}}}\hbox{-}\allowbreak\mathrm{Bribery}\leq_{m}^{p}\mathrm{online}\hbox{-}\allowbreak\mathrm{{{\cal E}}}\hbox{-}\allowbreak\mathrm{\Bribery}$).

One might ask why our model reveals the votes only as the voters vote, but makes the voters’ prices/weights known in advance. The answer is that in most natural situations of weighted voting, the weights are no secret. For example, it is well known how many votes each state has at a political nominating convention. And though people generally do not publicize how easily they can be bribed, in our model for the priced cases, we are assuming that the briber is acting with knowledge of each voter’s price, perhaps due to familiarity with the voter. Similarly, we are taking the order that voters vote as being known in advance, as also is the case in many central situations, from roll-call votes by state at political nominating conventions to the briber—a CS department chairperson—meeting with faculty members one at a time in his or her office to solicit (and perhaps bribe) their votes on some central issue.

3.3 A Brief Discussion of Some Upcoming Proofs

Finally, let us mention and briefly discuss some of the novel proof approaches needed to obtain our results.

The ${\Pi^{p}_{2k+1}}$ upper bounds that we will show in Sections 4.1 and 4.2 are far less straightforward than upper bounds in the polynomial hierarchy typically are. Since bribes can occur on any voter (until one runs out of allowed bribes), and so a yes-no decision has to be made, even for the case of at most $k$ bribes, there can be long strings of alternating existential and universal choices in the natural alternating Turing machine programs for the problems. And so there is the threat that one can prove merely a PSPACE upper bound.

However, in Section 4.2.1 we prove a more general result about alternating Turing machines that, while perhaps making polynomially many alternations between existential and universal choices, make most of the existential choices in a boring way (the exact restriction will be defined rigorously in that section). Basically, we show that in the relevant setting one can pull much of the existential guessing upstream and make it external to the alternating Turing machine, and indeed one can do so in such a way that one transforms the problem into the disjunction of a polynomial number of uniformly generated questions about actions of alternating Turing machines each of which itself has at most $2k+1$ alternation blocks. From that, we establish the needed upper bound, both for the relevant abstract case of alternating Turing machines and for our online bribery problems.

Regarding our result in Section 4.3 that there are election systems, with simple winner problems, such that each of the abovementioned upper-bounds is tight (so PSPACE-completeness or ${\Pi^{p}_{2k+1}}$-completeness holds), there is a substantial, novel challenge that the proof here has to overcome. Namely, to prove for example ${\Pi^{p}_{2k+1}}$-hardness, we generally need to reduce from quantified boolean formulas with particular quantifiers applying to particular variables. However, in online bribery, the briber is allowed to choose where to do the bribing. This in effect corresponds to having a formula with clusters of quantified variables, yet such that we, as we attempt to prove theorems related to these structures, don’t have control over which quantifiers are existential and which are universal. Rather, in effect what the online bribery setting will test is whether there exists an assignment (consistent with the number of bribes allowed—which limits the number of existential quantifiers one can set) of each quantifier to be either existential or universal, such that for that quantifier-assignment the formula evaluates as true. (This is not at all the same as quantifier exchange. In quantifier exchange, the exchanged quantifiers move around together with their associated variables.)

However, we handle this by showing how to construct a new formula that builds in protection against this setting. In particular, we note that one can take a quantified boolean formula and turn it into one such that, in this Wild West setting of quantifier assignment, the new formula can be made true by a legal (i.e., having at most as many $\exists$ quantifiers as the original formula) quantifier assignment exactly if the original formula is true.

4 General Upper Bounds and Matching Lower Bounds

Even for election systems with simple winner problems, the best general upper bounds that we can prove for our problems reflect an extremely high level of complexity.

One might wonder whether that merely is a weakness in our upper-bound proofs. However, in each case, we provide a matching completeness result proving that these really are the hardest problems in the classes their upper bounds put them in.

However, in Section 5, we will see that for many specific natural, important systems, the complexity is tremendously lower than the upper bounds, despite the fact that the present section shows that there exist systems that meet the upper bounds.

4.1 The General Upper Bound, Without Limits on the Number of Bribes

This section covers upper bounds for the case when any bribe limit/bribery budget is passed in through the input—not hardwired into the problem itself.

Theorem 4.1**.**

For each election system ${\cal E}$ whose winner problem in the unweighted case is in polynomial time (or even in polynomial space), each of the problems $\mathrm{online}\hbox{-}\allowbreak\mathrm{{{\cal E}}}\hbox{-}\allowbreak\mathrm{Bribery}$, $\mathrm{online}\hbox{-}\allowbreak\mathrm{{{\cal E}}}\hbox{-}\allowbreak\mathrm{Destructive}\hbox{-}\allowbreak\mathrm{Bribery}$, \mathrm{online}\hbox{-}\allowbreak\mathrm{{{\cal E}}}\hbox{-}\allowbreak\mathrm{\Bribery}$*, and *$\mathrm{online}\hbox{-}\allowbreak\mathrm{{{\cal E}}}\hbox{-}\allowbreak\mathrm{Destructive}\hbox{-}\allowbreak\mathrm{$Bribery}$* is in *$\mathrm{PSPACE}$. 2. 2.

For each election system ${\cal E}$ whose winner problem in the weighted case is in polynomial time (or even in polynomial space), each of the problems $\mathrm{online}\hbox{-}\allowbreak\mathrm{{{\cal E}}}\hbox{-}\allowbreak\mathrm{Weighted}\hbox{-}\allowbreak\mathrm{Bribery}$, $\mathrm{online}\hbox{-}\allowbreak\mathrm{{{\cal E}}}\hbox{-}\allowbreak\mathrm{Destructive}\hbox{-}\allowbreak\mathrm{Weighted}\hbox{-}\allowbreak\mathrm{Bribery}$, \mathrm{online}\hbox{-}\allowbreak\mathrm{{{\cal E}}}\hbox{-}\allowbreak\mathrm{Weighted}\hbox{-}\allowbreak\mathrm{\Bribery}$*, and *$\mathrm{online}\hbox{-}\allowbreak\mathrm{{{\cal E}}}\hbox{-}\allowbreak\mathrm{Destructive}\hbox{-}\allowbreak\mathrm{Weighted}\hbox{-}\allowbreak\mathrm{$Bribery}$* is in *$\mathrm{PSPACE}$.

Proof.

Consider first the case in which the winner problem is in polynomial time. For that problem, each of the eight problems can clearly be solved by (what is known as) an alternating polynomial-time Turing machine. It follows from the same paper of Chandra, Kozen, and Stockmeyer that defined alternating polynomial-time Turing machines [CKS81] that each is in $\mathrm{PSPACE}$, simply from the problems’ definitions.

One can see in various ways that the eight problems remain in $\mathrm{PSPACE}$ even if their winner problem is merely assumed to be in $\mathrm{PSPACE}$. Perhaps the simplest way to see that is that it follows from the above combined with the fact that $\mathrm{PSPACE}^{\rm PSPACE}=\mathrm{PSPACE}$, in the model in which oracle queries are themselves polynomially length-bounded, since all eight of our problems when generalized to allowing $\mathrm{PSPACE}$ winner problems are, in the model in which oracle queries are polynomially length-bounded, in $\mathrm{PSPACE}^{\rm PSPACE}$.   ❑

4.2 The General Upper Bound, With Limits on the Number of Bribes

Turning to the case where in the problem the number of bribes has a fixed bound of $k$, these problems fall into the ${\Pi^{p}_{2k+1}}$ level of the polynomial hierarchy. That is not immediately obvious. After all, even when one can bribe at most $k$ times, one still for each of the current and future voters seems to need to explore the one-bit-per-voter decision of whether to bribe the voters (plus in those cases where one does decide to bribe, one potentially has to explore the exponential—in the number of candidates—possible votes to which the voter can be bribed). On its surface, for our problems, that would seem to say that the number of alternations between universal and existential moves that the natural polynomial-time alternating Turing program for our problem would have to make is about the number of voters (since for each voter we are asking whether there exists a legal bribing decision such that for all possible votes of the remaining voters, the briber’s goal can be met)—a bound that would not leave the problem in any fixed level of the polynomial hierarchy, but would merely seem to put the problem in $\mathrm{PSPACE}$.

So these problems are cases where even obtaining the stated upper bound is interesting and requires a twist to prove. The twist is as follows. On the surface the exploration of these problems has an unbounded number of alternations between universal and existential states in the natural, brute-force alternating Turing machine program. But for all but $k$ of the existential guesses on each accepting path, the guess is a boring one, namely, we guess that regarding that voter we don’t bribe. We will show, by proving a more general result about alternating Turing machines and restrictions on the structure of their maximal existential move segments (i.e., maximal sequences of existential guesses) along accepting paths, a ${\Pi^{p}_{2k+1}}$ upper bound on our sets of interest. In some sense, in terms of being charged as to levels of the polynomial hierarchy, we will be showing that if for a certain collection of 0-or-1 existential decisions one on each accepting path chooses 0 all but a fixed number of times (although for the other times one may then make many more nondeterministic choices), one can manage to in effect not be charged at all for the guessing acts that guessed 0.

We know of only one result in the literature that is anything like this. That result, which also came up in the complexity of online attacks on elections, is a result of Hemaspaandra, Hemaspaandra, and Rothe [HHR17b], where in the context not of bribery but of voter control they showed that for each fixed $k>0$ it holds that, for each polynomial-time alternating Turing machine $M$ whose alternation blocks are each one bit long and that for at most $k$ of the existential blocks guess a zero, the language accepted by $M$ is in $\mathrm{coNP}$.

In contrast, in the present paper’s case we are in a far more complicated situation, since in bribery our existential blocks are burdened not just by 1-bit bribe-or-not decisions, but for the cases when we decide to try bribing, we need to existentially guess what bribe to do. And so we do not stay in $\mathrm{coNP}$ regardless of how large $k$ is, as held in that earlier case. But we show that we can at least limit the growth to at most $2k+1$ alternating quantifiers—in particular, to the class ${\Pi^{p}_{2k+1}}$. And since we later provide problems of this sort that are complete for ${\Pi^{p}_{2k+1}}$, our $2k+1$ is optimal unless the polynomial hierarchy collapses.

We will approach this in two steps. First, as Section 4.2.1, we will prove the result about alternating Turing machines. And then, as Section 4.2.2, we will apply that to online bribery in the case of only globally fixed numbers of bribers being allowed.

4.2.1 A Result about Alternating Turing Machines

Briefly put, an alternating Turing machine [CKS81] (aka ATM) is a generalization of nondeterministic and conondeterministic computation. We will now briefly review the basics (see [CKS81] for a more complete treatment). An ATM can make both universal and existential choices. For a universal “node” of the machine’s action to evaluate to true, all its child nodes (one each for each of its possible choices) must evaluate to true. For an existential “node” of the machine’s action to evaluate to true, at least one of its child nodes (it has one child node for each of its possible choices) must evaluate to true. A leaf of the computation tree (a path, at its end) is said to evaluate to true if the path halted in an accepting state and is said to evaluate to false if the path halted in a nonaccepting state. (As our machines are time-bounded, all paths halt.) Without loss of generality, in this paper we assume that each universal or existential node has either two children (namely, does a universal or existential split over the choices 0 and 1; we will often call this a “1-bit move”) or has exactly one child (it does a trivial/degenerate universal or existential choice of an element from the one-element set $\{0\}$; we will often call this a “0-bit move”). The latter case is in effect a deterministic move, except allowing degenerate $\forall$ steps of that sort will let us put a “separator” between otherwise contiguous $\exists$ computation segments. Of course, long existential guesses can be done in this model, for example by guessing a number of bits sequentially. An ATM accepts or rejects based on what its root node evaluates to (which is determined inductively in the way described above).

Definition 4.2**.**

Consider a path $\rho$ in the tree of an ATM. The weight of that path is as follows. Consider all maximal segments of existential nodes with their guesses along the path. (As mentioned above, we may without loss of generality, and do, assume that each nonleaf node is $\exists$ or $\forall$, although perhaps a degenerate such node in the way mentioned above.) The weight of path $\rho$ is its number of maximal existential segments such that the concatenation of the bits guessed in that segment is not the 1-bit string 0 (i.e., the number of maximal existential segments of length at least 2 plus the number of maximal existential segments of length 1 with guess 1).

Let us illustrate this, as Figure 1. In the figure, the illustrated path (the leftmost one at the left edge of the tree) has weight 0; it has three maximal existential segments, but each is of length one and makes the guess 0. If we change the last $\forall$ to $\exists$ we have a path of weight 1; it has two maximal existential segments, the first one is of length 1 and makes the guess 0, and so does not count, but the second one is of length 3, and so does count.

With this definition in hand, we can now state our key theorem showing that limited weight on accepting paths for ATMs simplifies the complexity of the languages accepted. The result is one where one may go back and forth between thinking it is obvious and thinking it is not obvious. In particular, note that even on accepting paths of weight at most $k$, it is completely possible that the number of alternations between existential and universal nodes may be far greater than $k$ and may be far greater than $2k+1$, and indeed may grow unboundedly as the input’s size increases (and this might naturally lead one to worry that perhaps even a machine with bounded accepting-path weight potentially might accept a $\mathrm{PSPACE}$-complete language). What the theorem below is saying is that despite that, machines with bounded weight on their accepting paths still accept only ${\Pi^{p}_{2k+1}}$ sets.

Theorem 4.3**.**

Let $k\geq 0$ be fixed. Each polynomial-time ATM $M$ such that on no input does $M$ have an accepting path of weight strictly greater than $k$ accepts a language in ${\Pi^{p}_{2k+1}}$.

Proof.

Let $k\geq 0$ be fixed. Let $L$ be the language accepted by polynomial-time ATM $M$ that has the property that each of its accepting paths has weight at most $k$. Our goal is to prove that $L\in{\Pi^{p}_{2k+1}}$.

We will do so by proving that there is a set $G\in{\Pi^{p}_{2k+1}}$ such that $L\leq_{dtt}^{p}G$, i.e., $L$ polynomial-time disjunctively truth-table reduces to $G$. By Fact 3.2, it follows that $L\in{\Pi^{p}_{2k+1}}$.

Let $q$ be a nondecreasing polynomial that upper-bounds the running time of $M$. Let $\mathopen{\langle}{\cdot,\cdot}\mathclose{\rangle}$ be a standard pairing function, i.e., a polynomial-time computable, polynomial-time invertible bijection between $\Sigma^{\ast}\times\Sigma^{\ast}$ and $\Sigma^{\ast}$. Recall that every step of our ATM $M$ involves either a 0-bit existential move (which we’ll think of basically as existentially choosing 0 from the choice palette set $\{0\}$) or a 1-bit existential move (which, recall, involves choosing one element from the choice palette set $\{0,1\}$ with the machine enforcing an “or” over the two children thus reached) or a 0-bit universal move (which we’ll think of basically as universally choosing 0 from the choice palette set $\{0\}$) or a 1-bit universal move. And the 1-bit moves involve successor states hinged on whether the move-choice is a 0 or a 1. (As Turing machines are standardly defined, there can be (one or multiple) successor states to a given state, hinged on a (degenerate or nondegenerate) choice.)

$G$ will be the set of all $\mathopen{\langle}{x,s}\mathclose{\rangle}$ such that all of the list of conditions that we will give below hold relative to $x$ and $s$. The intuition here is that $s$ is a bit-vector whose $i$th bit controls how the $i$th maximal existential segment is handled. In particular, if that $i$th bit is a 1, then the segment moves forward unrestrained. But if that $i$th bit is a 0, then we expect and require (and cut off that part of the tree otherwise) the maximal existential segment to be a single existential step (either guessing a bit from $\{0,1\}$ or the allowed but superfluous existential step of guessing a bit from the one-element set $\{0\}$) and we basically will (as described in part of step 3 below) cut that step out of the tree by replacing it by a trivially universal step. Returning to our defining of $G$, the set $G$ will be all $\mathopen{\langle}{x,s}\mathclose{\rangle}$ such that all the following claims hold.

$x\in\Sigma^{\ast}$ and $s\in\{0,1\}^{q(|x|)}$. 2. 2.

The number of “1”s in the bit-string $s$ is at most $k$. 3. 3.

$M$ accepts when we simulate it on input $x$ but with the following changes in the machine’s action.

As one simulates $M$ on a given path, consider the first existential node (if any) that one encounters.

If the first bit of $s$ is a 1, then for that node we will directly simulate it, and on all paths that follow from this one, on all the following existential nodes (if any) that are in an unbroken segment of existential nodes from this one, we will similarly directly simulate them. On the other hand, if the first bit of $s$ is a 0, then (a) if it is the case that if the current node makes the choice 0 then the node that follows it is an existential node, then the current path halts and rejects (because something that $s$ is specifying as being a maximal existential segment consisting of a single 0 clearly is not); and (b) if (a) does not hold (and so the node that follows if we make the choice 0 is either universal or a leaf), then do not take an existential action at the current node but rather implement it as a degenerate universal step (namely, a “$\forall$” guess over one option, namely, 0, matching as to next state and so on whatever the existential node would have done on the choice of 0).

If the path we are simulating didn’t already end or get cut off during the above-described handling of its first, if any, maximal existential segment, then continue on until we hit the start of its second existential segment. We handle that exactly as described above, except now our actions are controlled not by the first bit of $s$ but by the second.

And similarly for the third maximal existential segment, the fourth, and so on.

All other aspects of this simulation are unchanged from $M$’s own native behavior.

Note that $G\in{\Pi^{p}_{2k+1}}$. Why? Even in the worst of cases for us, the modified computation of $M$ starts with a $\forall$ block and then has $k$ $\exists$ blocks each separated by a $\forall$ block; and then we finish with a $\forall$ block. But then our ATM as it does that simulation starts with $\forall$ and has $2k$ alternations of quantifier type, and thus has $2k+1$ alternation blocks with $\forall$ as the leading one. And so by Chandra, Kozen, and Stockmeyer’s [CKS81] characterization of the languages accepted by ATMs with that leading quantifier and that number of alternations, this set is in ${\Pi^{p}_{2k+1}}$.

Finally, we argue that $L\leq_{dtt}^{p}G$. In particular, note that $L\leq_{dtt}^{p}G$ via the reduction that on input $x$ generates every length $q(|x|)$ bit-string having less than or equal to $k$ occurrences of the bit 1, and as its $\leq_{dtt}^{p}$ output list outputs each of those paired with $x$. This list is easily generated and is polynomial in size. In particular, the number of pairs in the list is clearly at most $\sum_{0\leq j\leq k}{q(|x|)\choose j}\leq\sum_{0\leq j\leq k}(q(|x|))^{j}\leq(k+1)(q(|x|))^{k}$. That completes the proof.   ❑

In the above, we focused on maximal existential guess sequences, and limiting the number of those, on accepting paths, whose bit-sequence-guessed was other than the string 0. So we barred from accepting paths any maximal existential guess sequences that contain a 1 and any that have two or more bits. We mention in passing that we could have framed things more generally in various ways. For example, we could have made each maximal existential guess be of a fixed polynomial length and could have defined our notion of a “boring” guess sequence not as the string “0” but as a string of 0’s of exactly that length. The ${\Pi^{p}_{2k+1}}$ upper bound holds also in that setting, via only slight modifications to the proof.

4.2.2 The Upper-Bound Results Obtained by Applying

the Previous Section’s Result about Alternating Turing Machines

We now establish our upper bounds on online bribery.

Theorem 4.4**.**

For each $k\in\{0,1,2,\dots\}$, and for each election system ${\cal E}$ whose winner problem in the unweighted case is in polynomial time,444Unlike Theorem 4.1, we cannot allow the winner problem here to be in $\mathrm{PSPACE}$ and argue that the rest of the theorem holds unchanged. However, we can allow the winner problem here to even be in $\mathrm{NP}\cap\mathrm{coNP}$, and then the rest of the theorem holds unchanged. The key point to notice to see that that holds is—as follows immediately from the fact that $\mathrm{NP}^{\mathrm{NP}\cap\mathrm{coNP}}=\mathrm{NP}$ [Sch83]—that for each $k\geq 0$, $\mathrm{NP}\cap\mathrm{coNP}$ is ${\Pi^{p}_{2k+1}}$-low, i.e., that $\left({{\Pi^{p}_{2k+1}}}\right)^{\mathrm{NP}\cap\mathrm{coNP}}={\Pi^{p}_{2k+1}}$. each of the problems $\mathrm{online}\hbox{-}\allowbreak\mathrm{{{\cal E}}}\hbox{-}\allowbreak\mathrm{Bribery}[k]$, $\mathrm{online}\hbox{-}\allowbreak\mathrm{{{\cal E}}}\hbox{-}\allowbreak\mathrm{Destructive}\hbox{-}\allowbreak\mathrm{Bribery}[k]$, \mathrm{online}\hbox{-}\allowbreak\mathrm{{{\cal E}}}\hbox{-}\allowbreak\mathrm{\Bribery}[k]$*, and *$\mathrm{online}\hbox{-}\allowbreak\mathrm{{{\cal E}}}\hbox{-}\allowbreak\mathrm{Destructive}\hbox{-}\allowbreak\mathrm{$Bribery}[k]$* is in *${\Pi^{p}_{2k+1}}$. 2. 2.

For each $k\in\{0,1,2,\ldots\}$, and for each election system ${\cal E}$ whose winner problem in the weighted case is in polynomial time,555As in the case of Footnote 4, the rest of the theorem remains unchanged even if we relax the “polynomial time” to instead be “$\mathrm{NP}\cap\mathrm{coNP}$.” each of the problems $\mathrm{online}\hbox{-}\allowbreak\mathrm{{{\cal E}}}\hbox{-}\allowbreak\mathrm{Weighted}\hbox{-}\allowbreak\mathrm{Bribery}[k]$, $\mathrm{online}\hbox{-}\allowbreak\mathrm{{{\cal E}}}\hbox{-}\allowbreak\mathrm{Destructive}\hbox{-}\allowbreak\mathrm{Weighted}\hbox{-}\allowbreak\mathrm{Bribery}[k]$, \mathrm{online}\hbox{-}\allowbreak\mathrm{{{\cal E}}}\hbox{-}\allowbreak\mathrm{Weighted}\hbox{-}\allowbreak\mathrm{\Bribery}[k]$*, and *$\mathrm{online}\hbox{-}\allowbreak\mathrm{{{\cal E}}}\hbox{-}\allowbreak\mathrm{Destructive}\hbox{-}\allowbreak\mathrm{Weighted}\hbox{-}\allowbreak\mathrm{$Bribery}[k]$* is in *${\Pi^{p}_{2k+1}}$.

Proof.

Let $k\geq 0$ be fixed. Let us start by arguing that \mathrm{online}\hbox{-}\allowbreak\mathrm{{{\cal E}}}\hbox{-}\allowbreak\mathrm{Weighted}\hbox{-}\allowbreak\mathrm{\Bribery}[k]\in{\Pi^{p}_{2k+1}}$. After that, we will show the remaining cases of part 2 and then we will show part 1.

As noted (for the case without the bound of $k$) in Section 3.2, what is really going on here is about alternating quantifiers. Consider a given input to the problem \mathrm{online}\hbox{-}\allowbreak\mathrm{{{\cal E}}}\hbox{-}\allowbreak\mathrm{Weighted}\hbox{-}\allowbreak\mathrm{\Bribery}[k]$. Let the voter under consideration (i.e.,$u$) in the focus moment of that problem just for this paragraph be referred to as$u_{1}$, and let the ones coming after it be called, in the order they occur,$u_{2}$,$u_{3}$,$\dots$,$u_{\ell}$. What the membership problem is in essence asking is whether there *exists* an allowable (within both the price budget and the global limit of$k$allowed bribes) choice as to whether to bribe$u_{1}$(and if the decision is to bribe, then whether there *exists* a vote to which to bribe$u_{1}$) such that, *for each* vote that$u_{2}$may then be revealed to have, there *exists* an allowable (within both the price budget and the global limit of$k$allowed bribes) choice as to whether to bribe$u_{2}$(and if the decision is to bribe, then whether there *exists* a vote to which to bribe$u_{2}$) such that, *for each* vote that$u_{3}$may then be revealed to have,$\dots$, such that there *exists* an allowable (within both the price budget and the global limit of$k$allowed bribes) choice as to whether to bribe$u_{\ell}$(and if the decision is to bribe, then whether there *exists* a vote to which to bribe$u_{\ell}$) such that$W_{{\cal E}}(C,U^{\prime})\cap{c\mid c\geq_{\sigma}d}\neq\emptyset$(recall that that inequality says that the winner set includes some candidate that the briber likes at least as much as the briber likes$d$;$U^{\prime}$ is here representing the vote set after all the voting/bribing, as per Section 3.2’s definitions).

Note that for at most $k$ of the choice blocks associated with $u_{1},u_{2},\dots$ can we make the choice to bribe. (In fact, if we have already done bribing of one or more voters in $V_{<u}$, then our remaining number of allowed bribes will be less than $k$.) Keeping that in mind, imagine implementing the above paragraph’s alternating-quantifier-based algorithm on a polynomial-time ATM. In our model, every step is either a universal or an existential one, and let us program up all deterministic computations that are part of the above via degenerate universal steps. (We do that rather than using degenerate existential steps since those degenerate existential steps would interact fatally with our definition of maximal existential sequence; we really need those places where one guesses that one will not bribe to be captured as a maximal existential sequence of length one with guess bit 0; this comment is quietly using the fact that when making a 1-bit choice as to whether to bribe we associate the choice 1 with “yes bribe this voter” and 0 with “do not bribe this voter.”) In light of that and the fact that we know that the weighted winner problem of election system ${\cal E}$ is in $\mathrm{P}$, the limit of $k$ ensures that no accepting path will have weight greater than $k$. And so by Theorem 4.3, we have that \mathrm{online}\hbox{-}\allowbreak\mathrm{{{\cal E}}}\hbox{-}\allowbreak\mathrm{Weighted}\hbox{-}\allowbreak\mathrm{\Bribery}[k]\in{\Pi^{p}_{2k+1}}$.

By the exact same argument, except changing the test at the end to $W_{{\cal E}}(C,U^{\prime})\cap\{c\mid d\geq_{\sigma}c\}=\emptyset$, we have that \mathrm{online}\hbox{-}\allowbreak\mathrm{{{\cal E}}}\hbox{-}\allowbreak\mathrm{Destructive}\hbox{-}\allowbreak\mathrm{Weighted}\hbox{-}\allowbreak\mathrm{\Bribery}[k]\in{\Pi^{p}_{2k+1}}$.

From these two results, it follows by Proposition 3.3 that $\mathrm{online}\hbox{-}\allowbreak\mathrm{{{\cal E}}}\hbox{-}\allowbreak\mathrm{Weighted}\hbox{-}\allowbreak\mathrm{Bribery}[k]\in{\Pi^{p}_{2k+1}}$ and $\mathrm{online}\hbox{-}\allowbreak\mathrm{{{\cal E}}}\hbox{-}\allowbreak\mathrm{Destructive}\hbox{-}\allowbreak\mathrm{Weighted}\hbox{-}\allowbreak\mathrm{Bribery}[k]\in{\Pi^{p}_{2k+1}}$.

That completes the proof of part 2 of the theorem. Now, we cannot simply invoke Proposition 3.3 to claim that part 1 holds. The reason is that part 1’s hypothesis about the winner problem merely puts the unweighted winner problem in $\mathrm{P}$, but the proof we just gave of part 2 used the fact that for that part we could assume that the weighted winner problem is in $\mathrm{P}$.

However, the entire construction of this proof works perfectly well in the unweighted case, namely, we are only given that the unweighted winner problem is in $\mathrm{P}$, but the four problems we are studying are the four unweighted problems of part 1 of the theorem statement. So we have that each of the problems $\mathrm{online}\hbox{-}\allowbreak\mathrm{{{\cal E}}}\hbox{-}\allowbreak\mathrm{Bribery}[k]$, $\mathrm{online}\hbox{-}\allowbreak\mathrm{{{\cal E}}}\hbox{-}\allowbreak\mathrm{Destructive}\hbox{-}\allowbreak\mathrm{Bribery}[k]$, \mathrm{online}\hbox{-}\allowbreak\mathrm{{{\cal E}}}\hbox{-}\allowbreak\mathrm{\Bribery}[k]$, and$\mathrm{online}\hbox{-}\allowbreak\mathrm{{{\cal E}}}\hbox{-}\allowbreak\mathrm{Destructive}\hbox{-}\allowbreak\mathrm{$Bribery}[k]$is in${\Pi^{p}_{2k+1}}$. That completes the proof of the theorem.   ❑

4.3 Matching Lower Bounds

For each of the $\mathrm{PSPACE}$ and ${\Pi^{p}_{2k+1}}$ upper bounds established so far in this section, we can in fact establish a matching lower bound. We show that by, for each, proving that there is an election system, with a polynomial-time winner problem, such that the given problem is polynomial-time many-one hard for the relevant class (and so, in light of the upper-bound results, is polynomial-time many-one complete for the relevant class).

Theorem 4.5**.**

For each problem $I$ from this list of problems: $\mathrm{online}\hbox{-}\allowbreak\mathrm{{{\cal E}}}\hbox{-}\allowbreak\mathrm{Bribery}$, $\mathrm{online}\hbox{-}\allowbreak\mathrm{{{\cal E}}}\hbox{-}\allowbreak\mathrm{Destructive}\hbox{-}\allowbreak\mathrm{Bribery}$, \mathrm{online}\hbox{-}\allowbreak\mathrm{{{\cal E}}}\hbox{-}\allowbreak\mathrm{\Bribery}$*, *$\mathrm{online}\hbox{-}\allowbreak\mathrm{{{\cal E}}}\hbox{-}\allowbreak\mathrm{Destructive}\hbox{-}\allowbreak\mathrm{$Bribery}$*, *$\mathrm{online}\hbox{-}\allowbreak\mathrm{{{\cal E}}}\hbox{-}\allowbreak\mathrm{Weighted}\hbox{-}\allowbreak\mathrm{Bribery}$*, *$\mathrm{online}\hbox{-}\allowbreak\mathrm{{{\cal E}}}\hbox{-}\allowbreak\mathrm{Destructive}\hbox{-}\allowbreak\mathrm{Weighted}\hbox{-}\allowbreak\mathrm{Bribery}$*, *$\mathrm{online}\hbox{-}\allowbreak\mathrm{{{\cal E}}}\hbox{-}\allowbreak\mathrm{Weighted}\hbox{-}\allowbreak\mathrm{$Bribery}$*, and *$\mathrm{online}\hbox{-}\allowbreak\mathrm{{{\cal E}}}\hbox{-}\allowbreak\mathrm{Destructive}\hbox{-}\allowbreak\mathrm{Weighted}\hbox{-}\allowbreak\mathrm{$Bribery}$*, there exists an (unweighted) election system *${\cal E}$*, whose winner problem in both the unweighted case and the weighted case is in polynomial time, such that *$I$* is *$\mathrm{PSPACE}$-complete. 2. 2.

For each $k\in\{0,1,2,\ldots\}$, and for each problem $I$ from this list of problems: $\mathrm{online}\hbox{-}\allowbreak\mathrm{{{\cal E}}}\hbox{-}\allowbreak\mathrm{Bribery}[k]$, $\mathrm{online}\hbox{-}\allowbreak\mathrm{{{\cal E}}}\hbox{-}\allowbreak\mathrm{Destructive}\hbox{-}\allowbreak\mathrm{Bribery}[k]$, \mathrm{online}\hbox{-}\allowbreak\mathrm{{{\cal E}}}\hbox{-}\allowbreak\mathrm{\Bribery}[k]$*, *$\mathrm{online}\hbox{-}\allowbreak\mathrm{{{\cal E}}}\hbox{-}\allowbreak\mathrm{Destructive}\hbox{-}\allowbreak\mathrm{$Bribery}[k]$*, *$\mathrm{online}\hbox{-}\allowbreak\mathrm{{{\cal E}}}\hbox{-}\allowbreak\mathrm{Weighted}\hbox{-}\allowbreak\mathrm{Bribery}[k]$*, *$\mathrm{online}\hbox{-}\allowbreak\mathrm{{{\cal E}}}\hbox{-}\allowbreak\mathrm{Destructive}\hbox{-}\allowbreak\mathrm{Weighted}\hbox{-}\allowbreak\mathrm{Bribery}[k]$*, *$\mathrm{online}\hbox{-}\allowbreak\mathrm{{{\cal E}}}\hbox{-}\allowbreak\mathrm{Weighted}\hbox{-}\allowbreak\mathrm{$Bribery}[k]$*, and *$\mathrm{online}\hbox{-}\allowbreak\mathrm{{{\cal E}}}\hbox{-}\allowbreak\mathrm{Destructive}\hbox{-}\allowbreak\mathrm{Weighted}\hbox{-}\allowbreak\mathrm{$Bribery}[k]$*, there exists an (unweighted) election system *${\cal E}$*, whose winner problem in both the unweighted case and the weighted case is in polynomial time, such that *$I$* is *${\Pi^{p}_{2k+1}}$-complete.

Proof.

All sixteen upper bounds follow from Theorems 4.1 or 4.4. So we need only prove the matching lower bound in each case.

We focus on the problem $\mathrm{online}\hbox{-}\allowbreak\mathrm{{{\cal E}}}\hbox{-}\allowbreak\mathrm{Bribery}[k]$ and will discuss the other fifteen cases in Section 4.3.4. We will show that there is an election system ${\cal E}$, whose unweighted and weighted winner problems are in polynomial time, for which this problem is ${\Pi^{p}_{2k+1}}$-hard.

To do this, we will insofar as possible be inspired by Hemaspaandra, Hemaspaandra, and Rothe’s [HHR14] work on online manipulation, and will insofar as possible adopt its notations and phrasings. This will help the novel challenge here to better stand out. That novel challenge is the one mentioned in list item 2 on page 2 of the current paper. In the manipulation case, we can by shaping the output of the reduction (from a ${\Pi^{p}_{2k+1}}$-complete set) control which voters are manipulators, and so can match and control the quantifier structure of the quantified boolean formula whose truth we are trying to determine via an online manipulation question. However, in online bribery, the briber can bribe any voters the briber wants (within the other constraints). This gives the briber quite a lot of freedom—enough to in effect shuffle which quantifiers are which in the boolean formula we are trying to in effect simulate; but that would cause chaos and would break our simulation’s connection to the formula.

We will handle this by doing our simulation in effect on a transformed version of the formula that is immune to quantifier shuffling, thanks to the transformed formula being rigged so that any assignment as to which quantifiers are chosen to be $\exists$ and which are chosen to be $\forall$ cannot possibly lead to a successful bribe unless the assignment is precisely the one that causes the online bribery to in effect correctly model and study the underlying formula we care about as to where the quantifiers are. (Put somewhat differently, we create an online bribery problem where we in effect are forcing the hand of the briber as to whom to bribe.)

We will show that this can be done quite directly, due to the power of boolean formulas. Let us first show how to do this, and then we will use this in our proof.

4.3.1 Transforming the Formula

Consider any quantified boolean formula, i.e., $(Q_{1}\overrightarrow{x_{1}})(Q_{2}\overrightarrow{x_{2}})\cdots(Q_{\ell}\overrightarrow{x_{\ell}})[F(\overrightarrow{x_{1}},\overrightarrow{x_{2}},\dots,\allowbreak\overrightarrow{x_{\ell}})]$, where each $Q_{i}$ is either an $\exists$ or a $\forall$, each $\overrightarrow{x_{i}}$ is shorthand for a list of boolean variables (with no boolean variable appearing in more than one of the $\overrightarrow{x_{i}}$’s), and $F$ is a propositional boolean formula. Let $j$ be the number of $Q_{i}$ that are $\exists$. Now, suppose we want to put that formula into a setting where that same interior formula $F$ is used, but the $\ell$ quantifiers can be reassigned in every possible way that has at most $j$ of them being $\exists$; and then we do an “OR” over all the thus-generated quantified boolean formulas. For example, if our original formula is

[TABLE]

then what we will in fact be assessing in that setting is this statement:

[TABLE]

(Since $\forall$ is never less demanding than $\exists$, the final disjunct above doesn’t affect the formula’s truth value, and more generally, we could just always consider just the quantifier assignments that have exactly $j$ $\exists$ quantifiers, rather than those that have $j$ or fewer $\exists$ quantifiers. But there is no need to use that, so we will just ignore it.)

Notice that our giant disjunctive formula may very well not be satisfied on the same set of inputs as the original one, namely, it may be satisfied on additional inputs. What we’ll need in our proof is to make an easily-computed cousin of the original formula that somehow is “preinsulated” from having its truth value changed by the above type of quantifier (re)assignment, yet that evaluates to true if and only if the original formula does. Let us give that cousin. If the original formula is (as above)

[TABLE]

then our preinsulated cousin for it is

[TABLE]

Note that this cousin’s propositional statement \Big{(}F(\overrightarrow{x_{1}},\overrightarrow{x_{2}},\dots,\overrightarrow{x_{\ell}})\Big{)}~{}{\;\wedge\;}~{}\Big{(}\bigwedge_{\{i\,\mid\>Q_{i}=\exists\}}b_{i}\Big{)} can never evaluate to true unless each $Q_{i}$ that is an $\exists$ in the original formula is an $\exists$ here. Of course, that is true in this naked version of the cousin, since the quantifiers are the same and have not been shuffled yet. However, the key point is that if we take this cousin, and as above take the disjunction of every possible assignment of its quantifiers that assigns at most $j$ of them to be $\exists$, then that new quantified boolean formula evaluates to true if and only if the original formula does. Further, the transformation from the original formula to the cousin (and here we really mean the cousin, not the larger disjunctive item) is clearly a polynomial-time transformation and does not change the quantifiers, and so in particular if the original quantifiers have each $Q_{i}$ being $\forall$ when $i$ is odd and $\exists$ when $i$ is even, then so will the cousin formula (and of course the $\ell$ and the $F$ of the cousin formula are the same as those of the original formula).

Basically, we’ve made a propositional formula that nails down its existential quantifiers as to how the formula can possibly be made true, even if it has to weather a disjunction over all possible shufflings of its quantifiers (even if that shuffling is allowed to—pointlessly—decrease the number of $\exists$ quantifiers).

Fix any $k\geq 0$. With the above “(preinsulated) cousin” work in hand, we can now give our proof that there exists an election system ${\cal E}$, whose unweighted and weighted winner problems are in polynomial time, such that $\mathrm{online}\hbox{-}\allowbreak\mathrm{{{\cal E}}}\hbox{-}\allowbreak\mathrm{Bribery}[k]$ is ${\Pi^{p}_{2k+1}}$-hard. We will do so by giving such an election system ${\cal E}$ (which depends on $k$) and giving a polynomial-time many-one reduction to $\mathrm{online}\hbox{-}\allowbreak\mathrm{{{\cal E}}}\hbox{-}\allowbreak\mathrm{Bribery}[k]$ from the ${\Pi^{p}_{2k+1}}$-complete problem we will denote $A_{2k+1}$, where $A_{2k+1}$ is all formulas of the form

[TABLE]

that evaluate to true (here, $F$ is required to be a propositional boolean formula and as above, the $\overrightarrow{x_{i}}$’s are pairwise disjoint variable collections) and such that (this additional nonstandard requirement clearly can be made without loss of generality, in the sense that even with this the set we are defining is clearly ${\Pi^{p}_{2k+1}}$-complete) at least one variable from within each of the $\overrightarrow{x_{i}}$’s occurs in the formula $F(\overrightarrow{x_{1}},\overrightarrow{x_{2}},\dots,\overrightarrow{x_{2k+1}})$.

4.3.2 The Election System ${\cal E}$

Let us give the election system ${\cal E}$ and the polynomial-time many-one reduction from $A_{2k+1}$ to $\mathrm{online}\hbox{-}\allowbreak\mathrm{{{\cal E}}}\hbox{-}\allowbreak\mathrm{Bribery}[k]$. As mentioned before, we will stay as close as possible to the argument line used in the hardness arguments for online manipulation.

Let us first specify our (unweighted) election system ${\cal E}$. If the input to ${\cal E}$ is $(C,V)$, the election system will do the following. The system will first look at the candidate set and determine which candidate, let us call it $c$, has the lexicographically smallest among the candidate names in $C$. Next, ${\cal E}$ will look at the bit-string $c$ to determine whether it is (i.e., whether it encodes) a tiered boolean formula. A tiered boolean formula [HHR14] is a formula whose variable names are each of the form $x_{i,m}$ (which really means a direct encoding of a string, such as “$x_{14,92}$”); the $i,m$ fields must be positive integers. If $c$ cannot be validly parsed in this way, then we declare no candidates to be the winner of the election. But otherwise, we have in hand a tiered formula represented by $c$. For a given $i$, we will think of all the variables $x_{i,m}$ that occur in $c$ as being a “block” of variables (as they will all be falling under a particular quantifier). Let $\mathit{SubscriptOneMax}$ denote the maximum value that occurs in the first component of any of our $x_{i,m}$ variables that occur in $c$ (technically, in the formula encoded by $c$, but we will henceforward just refer to that as $c$); this will tell us the number of blocks. Let $0pt$ denote the maximum value that occurs in the second component of any of our $x_{i,m}$ variables that occur in $c$. Now, ${\cal E}$ will immediately declare that everyone loses unless all of the following things hold:

The number of voter names in $V$ is at least $\mathit{SubscriptOneMax}+1$. (That is, each vote consists of a name and an order. It is possible that the same name appears multiple times, perhaps some times with the same order—as happens when an unweighted winner problem is created from a weighted election that has at least one weight that is 2 or more—and perhaps sometimes with different orders. But what this condition is saying is that if one considers the set of all names that occur in at least one of the votes in $V$, that set of names is of cardinality at least $\mathit{SubscriptOneMax}+1$.) 2. 2.

$\mathit{SubscriptOneMax}=2k+1$. 3. 3.

The number of candidates in $C$ is greater than or equal to $1+2\cdot 0pt$. (This condition is to ensure that each vote’s preference order is about a large enough number of candidates that it can be used to assign all the variables in one quantifier block.) 4. 4.

No block is unpopulated; that is, for each $b$, $1\leq b\leq\mathit{SubscriptOneMax}$, $c$ contains at least one variable $x_{i,m}$ whose first component is $b$.

Now, ${\cal E}$ will make a list, which we will refer to as the “special list,” of $\mathit{SubscriptOneMax}$ votes from $V$. ${\cal E}$ will make the list in the following somewhat involved fashion. Let $\mathit{Names}$ be the set of all voter names that occur in $V$, i.e., it is the set of all names that are the first component of at least one vote in $V$. Recall that if we have reached this point, we know that $\|\mathit{Names}\|\geq\mathit{SubscriptOneMax}+1$. Our list will not include any vote whose first component is the lexicographically smallest string in $\mathit{Names}$. (This feature will be used later in the proof to keep the pathological voter $u$ from breaking the proof.) For each $i$, $1\leq i\leq\mathit{SubscriptOneMax}$, the $i$th vote in our special list will be, among all votes whose first component is lexicographically the $(i+1)$st smallest string in $\mathit{Names}$, the vote with the lexicographically smallest second component. Note that, for example, even if “Alice” appears many times as the first-component field of elements in the list $V$, at most one Alice vote will appear in our special list of $\mathit{SubscriptOneMax}$ votes.

Now (recall, we are still defining the election system ${\cal E}$’s action; we are not here speaking of our online bribery problem), ${\cal E}$ will use the vote of the first voter in this special list to assign truth values to all variables of the form $x_{1,m}$, will use the vote of the second voter in this special list to assign truth values to all variables of the form $x_{2,m}$, and so on up to the vote of the $\mathit{SubscriptOneMax}$th voter, which will assign truth values to all variables of the form $x_{\mathit{SubscriptOneMax},m}$.

Let us now set out how votes create those assignments. For this, we will use the coding scheme from Hemaspaandra, Hemaspaandra, and Rothe [HHR14], which is as follows. Consider a vote whose total order over $C$ is $\sigma^{\prime}$ (recall that we have that $\|C\|\geq 1+2\cdot 0pt$). Remove $c$ (the candidate encoding the formula) from the order $\sigma^{\prime}$, yielding $\sigma^{\prime\prime}$. Let $c_{1}<_{\sigma^{\prime\prime}}c_{2}<_{\sigma^{\prime\prime}}\cdots<_{\sigma^{\prime\prime}}c_{2\cdot 0pt}$ be the $2\cdot 0pt$ least preferred candidates in $\sigma^{\prime\prime}$. Then ${\cal E}$ will build a vector in $\{0,1\}^{0pt}$ as follows: The $d$th bit of the vector is [math] if the string that names candidate $c_{2d-1}$ (e.g., “Eilonwy”) is lexicographically less than the string that names candidate $c_{2d}$ (e.g., “Rowella”), and this bit is $1$ otherwise. Let $\overrightarrow{b_{i}}$ denote the vector thus built from the $i$th vote (in the above special list), $1\leq i\leq\mathit{SubscriptOneMax}$. For each variable $x_{i,m}$ occurring in $c$, assign to that variable the value of the $m$th bit of $\overrightarrow{b_{i}}$, where [math] represents false and $1$ represents true. This process has assigned all the variables of $c$, so $c$ evaluates to either true or false. If $c$ evaluates to true, everyone wins under system ${\cal E}$, and otherwise everyone loses under system ${\cal E}$. This completes our definition of election system ${\cal E}$. Note that the election system ${\cal E}$ that we just defined has a polynomial-time winner problem in both the unweighted and the weighted cases. This is because our special list was simply constructed and each boolean formula, given an assignment for all its variables, can easily be evaluated in polynomial time.

4.3.3 The Reduction

With our ${\cal E}$ in hand, let us now show that $\mathrm{online}\hbox{-}\allowbreak\mathrm{{{\cal E}}}\hbox{-}\allowbreak\mathrm{Bribery}[k]$ is ${\Pi^{p}_{2k+1}}$-hard, via giving a polynomial-time many-one reduction to $\mathrm{online}\hbox{-}\allowbreak\mathrm{{{\cal E}}}\hbox{-}\allowbreak\mathrm{Bribery}[k]$ from the ${\Pi^{p}_{2k+1}}$-complete problem $A_{2k+1}$ that was defined above.

Here is how the reduction works. Let $y$ be an instance of $A_{2k+1}$, i.e., $y$ is of the form:

[TABLE]

and at least one variable from within each of the $\overrightarrow{x_{i}}$’s occurs in $F$. The above formula may evaluate to true or may evaluate to false; that is the question we want answered. (We here are ignoring syntactically illegal inputs, as they obviously are not in $A_{2k+1}$ and so can be easily handled.) Let us transform $y$ into its cousin formula, as per the above discussion, i.e., we transform $y$ into the following formula $y^{\prime}$:

[TABLE]

(For the $k=0$ case, simply skip the second conjunct. The odd $b_{i}$ are superfluous, but do no harm.) Note that we can view the above $y^{\prime}$ as being the formula (here $\Phi$ will be the entire above propositional part):

[TABLE]

where the quantifier applied to the $x_{i,m}$ variables is $\forall$ if $i$ is odd and is $\exists$ if $i$ is even, the $x_{i,m}$ are boolean variables, $\Phi$ is the abovementioned propositional boolean formula, and note that for each $i$, $1\leq i\leq 2k+1$, $\Phi$ will contain at least one variable of the form $x_{i,m}$.

Let us, with $y^{\prime}$ in hand, and viewed in terms of the just-given variable names and formula $\Phi$, continue with specifying the reduction, keeping in mind that this is a preinsulated formula.

Our many-one reduction will map to the instance $(C,V,\sigma,d,B)$ of $\mathrm{online}\hbox{-}\allowbreak\mathrm{{{\cal E}}}\hbox{-}\allowbreak\mathrm{Bribery}[k]$, specified by the following:

$C$ contains a candidate whose name, $c$, encodes $\Phi$, and in addition $C$ contains $2\cdot\max(k_{1},\ldots,k_{2k+1})$ other candidates, all with names lexicographically greater than $c$—for specificity, let us say their names are the $2\cdot\max(k_{1},\ldots,k_{2k+1})$ strings that immediately follow $c$ in lexicographic order. 2. 2.

$V$ contains $2k+2$ voters, $0,1,2,\ldots,2k+1$, who vote in that order, where $u=0$ is the distinguished voter (so there are no voters in $V_{<u}$ and there are $2k+1$ voters in $V_{>u}$). The voter names will be lexicographically ordered by their number, so [math] is least and $2k+1$ is greatest. $u$’s preference order $\sigma$ will actually be irrelevant, because since $u$ will have the lexicographically smallest name among all voters, $u$’s vote will be ignored by our election system ${\cal E}$; but for specificity, let us say $u$’s preference order is to simply rank the candidates in lexicographic order. 3. 3.

The briber’s preference order $\sigma$ is to like candidates in the opposite of their lexicographic order. In particular, $c$ is the briber’s most preferred candidate. And we set $d$ to be $c$ (so the goal of the online bribery problem becomes simply to make $c$ be a winner). 4. 4.

$B$, the limit on the number of allowed bribes, is $k$.

Note that this is a polynomial-time reduction. And it follows from this reduction’s construction and the definition of ${\cal E}$ that $y^{\prime}$ is in $A_{2k+1}$ if and only if the thus-constructed $(C,V\sigma,d,B)$ is in $\mathrm{online}\hbox{-}\allowbreak\mathrm{{{\cal E}}}\hbox{-}\allowbreak\mathrm{Bribery}[k]$.

Why? The online bribery problem will be asking whether there is some way of bribing at most $k$ of the voters so as to make $c$ be a winner. Since this is an online bribery problem, we thus are existentially quantifying for each of those at most $k$ bribed voters as to his or her cast preference order, and for all the other voters we are universally quantifying as to their preference orders. The preference order of the voter named $i$ (who will on this input be the $(i+1)$st voter on our “special list” defined earlier) will be controlling the setting of the variables in the $i$th block of $\Phi$.

Now, all possible choices of where to bribe are allowed, as long as the total number of bribes is at most $B$ (i.e., at most $k$). However, the election system is routing these assigned variables to our preinsulated formula $\Phi$, and so by the properties of our preinsulated formula, the only possible case that can result in the formula being true (and thus all candidates—and so in particular candidate $c$—winning, as opposed to all candidates losing) is when every even-numbered voter other than voter [math] is bribed. But as there are only $k$ allowed bribes and there are $k$ even-numbered voters other than [math] (namely, the voters named $2,4,\dots,2k$), those bribes already use every allowed bribe, and so we know that the only way $c$ can be a winner is if every odd-numbered voter and voter [math] remain unbribed and every even-numbered voter other than 0 is bribed. And so our online-bribery problem is in fact testing whether $y\in A_{2k+1}$; $d$ is a winner (in fact, all candidates are winners) in the constructed instance $(C,V\sigma,d,B)$ if and only if $y\in A_{2k+1}$.

We thus have shown that $\mathrm{online}\hbox{-}\allowbreak\mathrm{{{\cal E}}}\hbox{-}\allowbreak\mathrm{Bribery}[k]$ is ${\Pi^{p}_{2k+1}}$-hard, and thus in light of the earlier upper bound, have shown that $\mathrm{online}\hbox{-}\allowbreak\mathrm{{{\cal E}}}\hbox{-}\allowbreak\mathrm{Bribery}[k]$ is ${\Pi^{p}_{2k+1}}$-complete.

4.3.4 The Other Fifteen Cases

It follows immediately from Proposition 3.3 that \mathrm{online}\hbox{-}\allowbreak\mathrm{{{\cal E}}}\hbox{-}\allowbreak\mathrm{\Bribery}[k]$is${\Pi^{p}{2k+1}}$-hard, and thus in light of the earlier upper bound is${\Pi^{p}{2k+1}}$-complete.

As to $\mathrm{online}\hbox{-}\allowbreak\mathrm{{{\cal E}}}\hbox{-}\allowbreak\mathrm{Weighted}\hbox{-}\allowbreak\mathrm{Bribery}[k]$, if we use the entire above construction and assign to each voter weight 1, we have (keeping in mind that for ${\cal E}$ both the unweighted and the weighted winner problems are in polynomial time), the thus-altered construction shows that $\mathrm{online}\hbox{-}\allowbreak\mathrm{{{\cal E}}}\hbox{-}\allowbreak\mathrm{Weighted}\hbox{-}\allowbreak\mathrm{Bribery}[k]$ is ${\Pi^{p}_{2k+1}}$-hard, and thus in light of the earlier upper bound is ${\Pi^{p}_{2k+1}}$-complete. And from that, the earlier upper bound, and Proposition 3.3, we have that \mathrm{online}\hbox{-}\allowbreak\mathrm{{{\cal E}}}\hbox{-}\allowbreak\mathrm{Weighted}\hbox{-}\allowbreak\mathrm{\Bribery}[k]$is${\Pi^{p}_{2k+1}}$-complete.

That covers the four constructive (i.e., not destructive) cases of part 2 of the theorem.

But the same construction, easily modified for the case of unbounded numbers of quantifiers, analogously yields PSPACE-hardness results, and thus by the earlier upper bounds PSPACE-completeness results, for the four constructive cases from part 1 of the theorem. In particular, to show that $\mathrm{online}\hbox{-}\allowbreak\mathrm{{{\cal E}}}\hbox{-}\allowbreak\mathrm{Bribery}$ is $\mathrm{PSPACE}$-hard, and thus in light of the earlier upper bound PSPACE-complete, we now map to $\mathrm{online}\hbox{-}\allowbreak\mathrm{{{\cal E}}}\hbox{-}\allowbreak\mathrm{Bribery}$ from the PSPACE-complete set $A_{\mathrm{QBF}}$, defined here as all formulas of the form

[TABLE]

that evaluate to true (here, $F$ is required to be a propositional boolean formula, and the $\overrightarrow{x_{i}}$’s are required to be pairwise disjoint variable collections, and $z$ is an odd integer) and such that at least one variable from within each of the $\overrightarrow{x_{i}}$’s occurs in the formula $F(\overrightarrow{x_{1}},\overrightarrow{x_{2}},\dots,\overrightarrow{x_{z}})$. Specifying that the leading quantifier is a $\forall$ and that the number of quantifiers in our alternating quantifier sequence is odd and that at least one variable from each block occurs in $F$ is not the standard version of QBF, but clearly also yields a PSPACE-complete set. And having it be of this form makes it clear how to specify ${\cal E}$ (namely, for inputs whose $\mathit{SubscriptOneMax}$ value is $2k+1$, we use the version of the above ${\cal E}$ that assumes and enforces that $\mathit{SubscriptOneMax}=2k+1$) and what reduction to use (namely, given a formula with the syntax and properties (except perhaps truth) of $A_{\mathrm{QBF}}$, having $2k+1$ alternating quantifiers, we use the actions of the reduction for the ${\Pi^{p}_{2k+1}}$ case above). (The reduction’s actions are sufficiently uniform and simple that what was just mentioned can itself be done in a single polynomial-time many-one reduction that handles all odd sequence lengths of alternating quantifiers whose first quantifier is a $\forall$.)

So we have handled all eight constructive cases.666For (no pun intended) completeness, we mention that we could have alternatively established the lower bounds for \mathrm{online}\hbox{-}\allowbreak\mathrm{{{\cal E}}}\hbox{-}\allowbreak\mathrm{\Bribery}[k]$and$\mathrm{online}\hbox{-}\allowbreak\mathrm{{{\cal E}}}\hbox{-}\allowbreak\mathrm{$Bribery}via results proved or stated in Hemaspaandra, Hemaspaandra, and Rothe’s [[HHR14](#bib.bibx39)] work on online manipulation, in light of the fact that one can simulate online manipulation by priced online bribery (via setting the budget to 1, the price of the manipulators each to 0, and the price of the nonmanipulators each to 2). That paper handles weights differently than this paper, and it doesn’t provide lower-bound matching results for any of the general-case destructive settings. However, from what that paper does do one can, using the gateway we just mentioned, claim (from that paper’s stated-without-proof result regarding the “freeform online manipulation problem” [[HHR14](#bib.bibx39), p. 702]) the{\Pi^{p}_{2k+1}}$-hardness of$\mathrm{online}\hbox{-}\allowbreak\mathrm{{{\cal E}}}\hbox{-}\allowbreak\mathrm{$Bribery}[k]$for some${\cal E}$whose unweighted winner problem is in polynomial time, and also one can claim (regarding the other part of our theorem) the$\mathrm{PSPACE}$-hardness of$\mathrm{online}\hbox{-}\allowbreak\mathrm{{{\cal E}}}\hbox{-}\allowbreak\mathrm{$Bribery}$for some${\cal E}$ whose unweighted winner problem is in polynomial time.

The eight destructive cases are analogous. We won’t do this in detail, but basically as to ${\cal E}$ one does everything as above, except every place that in the definition of an election system ${\cal E}$ we said everyone wins one changes that to saying that everyone loses, and everywhere we above in the definition of an election system ${\cal E}$ said everyone loses one changes that to saying that everyone wins. And as to the reductions, one uses the same reductions as above.

This completes the proof of the theorem.    ❑

5 Online Bribery for Specific Election Systems

In this section, we look at the complexity of online bribery for various natural systems. For both Plurality and Approval, we show that priced, weighted online bribery is $\mathrm{NP}$-complete but that the election system’s other three online bribery variants are in $\mathrm{P}$. Since these other three problem variants of nonsequential bribery are known to be $\mathrm{NP}$-complete [FHH09], this also shows that nonsequential bribery can be harder than online bribery for natural systems. In addition, we provide complete dichotomy theorems that distinguish NP-hard from easy cases for all our online bribery problems for scoring protocols and additionally we show that Veto elections, even with three candidates, have even higher lower bounds for weighted online bribery, namely $\mathrm{P}^{\mathrm{NP}[1]}$-hardness.

The following theorem is useful for proving lower bounds for online bribery for specific systems.777For an election system $\mathcal{E}$, ${\mathrm{{\cal E}}\hbox{-}\mathrm{UCM}}$ denotes the unweighted coalitional manipulation problem: Given a set $C$ of candidates, a collection $V$ of nonmanipulative voters over $C$, a collection $W$ of manipulative voters (who will come in without specified preferences), and a designated candidate $c\in C$, can we assign preference orders over $C$ to the members of $W$ in such a way that $c$ is a winner of the election $(C,V\cup W)$? If the voters are weighted, we obtain the weighted coalitional manipulation problem ${\mathrm{{\cal E}}\hbox{-}\mathrm{WCM}}$; note that the manipulators’ weights but not their preferences are given in the problem instance. The destructive variants of these two problems (where the goal is to prevent $c$ from being a winner in the manipulated election) are denoted by ${\mathrm{{\cal E}}\hbox{-}\mathrm{DUCM}}$ and ${\mathrm{{\cal E}}\hbox{-}\mathrm{DWCM}}$, respectively.

Theorem 5.1**.**

Nonsequential manipulation reduces to corresponding online bribery. (So, ${\mathrm{{\cal E}}\hbox{-}\mathrm{UCM}}$ reduces to $\mathrm{online}\hbox{-}\allowbreak\mathrm{{\cal E}}\hbox{-}\allowbreak\mathrm{Bribery}$, ${\mathrm{{\cal E}}\hbox{-}\mathrm{DUCM}}$ reduces to $\mathrm{online}\hbox{-}\allowbreak\mathrm{{\cal E}}\hbox{-}\allowbreak\mathrm{Destructive}\hbox{-}\allowbreak\mathrm{Bribery}$, ${\mathrm{{\cal E}}\hbox{-}\mathrm{WCM}}$ reduces to $\mathrm{online}\hbox{-}\allowbreak\mathrm{{\cal E}}\hbox{-}\allowbreak\mathrm{Weighted}\hbox{-}\allowbreak\mathrm{Bribery}$, and ${\mathrm{{\cal E}}\hbox{-}\mathrm{DWCM}}$ reduces to $\mathrm{online}\hbox{-}\allowbreak\mathrm{{\cal E}}\hbox{-}\allowbreak\mathrm{Destructive}\hbox{-}\allowbreak\mathrm{Weighted}\hbox{-}\allowbreak\mathrm{Bribery}$.) 2. 2.

Constructive manipulation in the unique winner model888In the unique winner model, the goal of a constructive manipulation action is to have the designated candidate be the only winner of the manipulated election. In the destructive case, the goal is to ensure that the designated candidate is not a unique winner of the manipulated election.* reduces to corresponding online destructive bribery (so, ${\mathrm{{\cal E}}\hbox{-}\mathrm{UCM}}$ in the unique winner model reduces to $\mathrm{online}\hbox{-}\allowbreak\mathrm{{\cal E}}\hbox{-}\allowbreak\mathrm{Destructive}\hbox{-}\allowbreak\mathrm{Bribery}$ and ${\mathrm{{\cal E}}\hbox{-}\mathrm{WCM}}$ in the unique winner model reduces to $\mathrm{online}\hbox{-}\allowbreak\mathrm{{\cal E}}\hbox{-}\allowbreak\mathrm{Destructive}\hbox{-}\allowbreak\mathrm{Weighted}\hbox{-}\allowbreak\mathrm{Bribery}$) for election systems that always have winners (if there are candidates).* 3. 3.

Online manipulation reduces to corresponding online priced bribery. (So, ${\mathrm{online}\hbox{-}\mathrm{{\cal E}}\hbox{-}\mathrm{UCM}}$ reduces to \mathrm{online}\hbox{-}\allowbreak\mathrm{{\cal E}}\hbox{-}\allowbreak\mathrm{\Bribery}$*, *${\mathrm{online}\hbox{-}\mathrm{{\cal E}}\hbox{-}\mathrm{DUCM}}$* reduces to *$\mathrm{online}\hbox{-}\allowbreak\mathrm{{\cal E}}\hbox{-}\allowbreak\mathrm{Destructive}\hbox{-}\allowbreak\mathrm{$Bribery}$*, *${\mathrm{online}\hbox{-}\mathrm{{\cal E}}\hbox{-}\mathrm{WCM}}$* reduces to *$\mathrm{online}\hbox{-}\allowbreak\mathrm{{\cal E}}\hbox{-}\allowbreak\mathrm{Weighted}\hbox{-}\allowbreak\mathrm{$Bribery}$*, and *${\mathrm{online}\hbox{-}\mathrm{{\cal E}}\hbox{-}\mathrm{DWCM}}$* reduces to *$\mathrm{online}\hbox{-}\allowbreak\mathrm{{\cal E}}\hbox{-}\allowbreak\mathrm{Destructive}\hbox{-}\allowbreak\mathrm{Weighted}\hbox{-}\allowbreak\mathrm{$Bribery}$.)

Proof.

For the first part, let $V_{<u}$ be the nonmanipulators and let $\{u\}\cup V_{>u}$ be the manipulators. The vote of $u$ is irrelevant. Let $k$ be the number of manipulators (meaning all voters in $\{u\}\cup V_{>u}$ can be bribed). In the constructive case, the preferred candidate $p$ becomes the designated candidate, which is ranked first in $\sigma$. In the destructive case, the despised candidate $d$ becomes the designated candidate, which is ranked last in $\sigma$.

For the second part, let again be $V_{<u}$ the nonmanipulators and $\{u\}\cup V_{>u}$ the manipulators. The vote of $u$ is irrelevant. Let $k$ be the number of manipulators (meaning again that all voters in $\{u\}\cup V_{>u}$ can be bribed). The ranking $\sigma$ puts the preferred candidate $p$ first. The other candidates are ranked in lexicographic order, and the designated candidate is the lexicographically smallest candidate in $C-\{p\}$, i.e., the candidate ranked second in $\sigma$.

For the last part, set the price of the manipulators to 0, the price of the nonmanipulators to 1, and set $k$ to 0.   ❑

It is interesting to note that, assuming $\mathrm{P}\neq\mathrm{NP}$, bribery does not reduce to corresponding online bribery, not even for natural systems. For example, Approval-Bribery is NP-complete [FHH09, Theorem 4.2], but we will show below in Theorem 5.7 that $\mathrm{online}\hbox{-}\allowbreak\mathrm{{Approval}}\hbox{-}\allowbreak\mathrm{Bribery}$ (and even $\mathrm{online}\hbox{-}\allowbreak\mathrm{{Approval}}\hbox{-}\allowbreak\mathrm{Weighted}\hbox{-}\allowbreak\mathrm{Bribery}$ and \mathrm{online}\hbox{-}\allowbreak\mathrm{{Approval}}\hbox{-}\allowbreak\mathrm{\Bribery}$) are in$\mathrm{P}$.

We end this section with a simple observation about unpriced, unweighted online bribery.

Observation 5.2**.**

For unpriced, unweighted online bribery, it is always optimal (meaning that if the briber can reach his or her goal, it can be reached in this way) to bribe the last $k$ voters (we don’t even have to handle $u$ in a special way). This implies that unpriced, unweighted online bribery is certainly reducible to unweighted online manipulation, and so we inherit those upper bounds.

5.1 Plurality

In this section, we completely classify the complexity of all our versions of online bribery for the most important natural system, Plurality. In this system, each candidate scores a point when it is ranked first in a vote and the candidates with the most points are the winners. We show that these problems are NP-complete if we have both prices and weights, and in P in all other cases.

The following observation is crucial in our upper bound proofs: For \mathrm{online}\hbox{-}\allowbreak\mathrm{{Plurality}}\hbox{-}\allowbreak\mathrm{Weighted}\hbox{-}\allowbreak\mathrm{\Bribery}$and$\mathrm{online}\hbox{-}\allowbreak\mathrm{{Plurality}}\hbox{-}\allowbreak\mathrm{Destructive}\hbox{-}\allowbreak\mathrm{Weighted}\hbox{-}\allowbreak\mathrm{$Bribery}$, there is a successful bribery if and only if there is a successful bribery where all bribed voters from$u$onward vote for the same highest-scoring desired candidate999In the constructive case we call all members of${c\ |\ c\geq_{\sigma}d}$—and in the destructive case we call all members of${c\ |\ c>_{\sigma}d}$—*desired* candidates, where$\sigma$is the briber’s ideal ranking and$d$the designated candidate. and all nonbribed voters after$u$vote for the same so-far highest-scoring undesired candidate. If$u$is bribed, we do not count$u$’s original vote to compute the highest score. If$u$is not bribed, then we count$u$’s vote.

Theorem 5.3**.**

$\mathrm{online}\hbox{-}\allowbreak\mathrm{{Plurality}}\hbox{-}\allowbreak\mathrm{Bribery}$, $\mathrm{online}\hbox{-}\allowbreak\mathrm{{Plurality}}\hbox{-}\allowbreak\mathrm{Destructive}\hbox{-}\allowbreak\mathrm{Bribery}$, $\mathrm{online}\hbox{-}\allowbreak\mathrm{{Plurality}}\hbox{-}\allowbreak\mathrm{Weighted}\hbox{-}\allowbreak\mathrm{Bribery}$, $\mathrm{online}\hbox{-}\allowbreak\mathrm{{Plurality}}\hbox{-}\allowbreak\mathrm{Destructive}\hbox{-}\allowbreak\mathrm{Weighted}\hbox{-}\allowbreak\mathrm{Bribery}$, \mathrm{online}\hbox{-}\allowbreak\mathrm{{Plurality}}\hbox{-}\allowbreak\mathrm{\Bribery}$, and$\mathrm{online}\hbox{-}\allowbreak\mathrm{{Plurality}}\hbox{-}\allowbreak\mathrm{Destructive}\hbox{-}\allowbreak\mathrm{$Bribery}$are in$\mathrm{P}$.

Proof.

First look at $\mathrm{online}\hbox{-}\allowbreak\mathrm{{Plurality}}\hbox{-}\allowbreak\mathrm{Weighted}\hbox{-}\allowbreak\mathrm{Bribery}$. We are given an OBS $(C,V,\sigma,d,k)$, where $V=(V_{<u},u,V_{>u})$. We can bribe successfully if and only if we can bribe successfully and we bribe $u$ or we can bribe successfully and we do not bribe $u$. Let $\Gamma_{d}=\{c\ |\ c\geq_{\sigma}d\}$ be the desired candidates and let $k^{\prime}$ be the number of voters in $V_{<u}$ that have already been bribed. If $k^{\prime}>k$, our instance is illegal and we reject. If $\Gamma_{d}=C$, we can successfully bribe, and we accept.

To check whether we can bribe successfully and bribe $u$, let $c$ be a candidate in $\Gamma_{d}$ with highest score in $V_{<u}$. If $k^{\prime}\geq k$, we can’t bribe $u$ and we reject. Otherwise, bribe $u$ to vote for $c$ and bribe $\min(k-k^{\prime}-1,\|V_{>u}\|)$ highest-weight voters in $V_{>u}$ to vote for $c$. This will give us the score of $c$ after bribery. Now let $h$ be a candidate in $C-\Gamma_{d}$ with highest score in $V_{<u}$. Assume that all the nonbribed voters in $V_{>u}$ vote for $h$. Then we can successfully bribe if and only if the score of $c$ is at least the score of $h$.

To check whether we can bribe successfully and not bribe $u$, let $c$ be a candidate in $\Gamma_{d}$ with highest score in $V_{<u}\cup\{u\}$. Bribe $\min(k-k^{\prime},\|V_{>u}\|)$ highest-weight voters in $V_{>u}$ to vote for $c$. This will give us the score of $c$ after bribery. Now let $h$ be a candidate in $C-\Gamma_{d}$ with highest score in $V_{<u}\cup\{u\}$. Assume that all the nonbribed voters in $V_{>u}$ vote for $h$. Then we can successfully bribe if and only if the score of $c$ is at least the score of $h$.

For the destructive case, we argue similarly, except that we let $\Gamma_{d}=\{c\ |\ c>_{\sigma}d\}$ and we are successful if the score of $c$ is greater than the score of $h$.

Next look at \mathrm{online}\hbox{-}\allowbreak\mathrm{{Plurality}}\hbox{-}\allowbreak\mathrm{\Bribery}$. We argue similarly as in the weighted case, except that we are now looking at bribery budgets instead of bribery limits, and so$k^{\prime}$is the price of the voters in$V_{<u}$that are bribed, we make sure that the price of the bribed voters is not higher than$k^{\prime}$, and we bribe the lowest-priced voters.

For completeness, we give the complete proof. We are given an OBS $(C,V,\sigma,d,k)$, where $V=(V_{<u},u,V_{>u})$. We can bribe successfully if and only if we can bribe successfully and we bribe $u$ or we can bribe successfully and we do not bribe $u$. Let $\Gamma_{d}=\{c\ |\ c\geq_{\sigma}d\}$ and let $k^{\prime}$ be the price of the voters in $V_{<u}$ that are bribed. If $k^{\prime}>k$, our instance is illegal and we reject. If $\Gamma_{d}=C$, we can successfully bribe, and we accept.

To check whether we can bribe successfully and bribe $u$, let $c$ be a candidate in $\Gamma_{d}$ with highest score in $V_{<u}$. If $k^{\prime}+\pi(u)>k$, we can’t bribe $u$ and we reject. Otherwise, bribe $u$ to vote for $c$ and bribe a lowest-priced voter in $V_{>u}$ to vote for $c$, as long as the price of the bribed voters is at most $k$. This will give us the score of $c$ after bribery. Now let $h$ be a candidate in $C-\Gamma_{d}$ with highest score in $V_{<u}$. Assume that all the nonbribed voters in $V_{>u}$ vote for $h$. Then we can successfully bribe if and only if the score of $c$ is at least the score of $h$.

To check whether we can bribe successfully and not bribe $u$, let $c$ be a candidate in $\Gamma_{d}$ with highest score in $V_{<u}\cup\{u\}$. Bribe a lowest-priced voter in $V_{>u}$ to vote for $c$, as long as the price of the bribed voters is at most $k$. This will give us the score of $c$ after bribery. Now let $h$ be a candidate in $C-\Gamma_{d}$ with highest score in $V_{<u}\cup\{u\}$. Assume that all the nonbribed voters in $V_{>u}$ vote for $h$. Then we can successfully bribe if and only if the score of $c$ is at least the score of $h$.

For the destructive case, we again argue similarly, except that we let $\Gamma_{d}=\{c\ |\ c>_{\sigma}d\}$ and we are successful if the score of $c$ is greater than the score of $h$.   ❑

Theorem 5.4**.**

\mathrm{online}\hbox{-}\allowbreak\mathrm{{Plurality}}\hbox{-}\allowbreak\mathrm{Weighted}\hbox{-}\allowbreak\mathrm{\Bribery}$* and$\mathrm{online}\hbox{-}\allowbreak\mathrm{{Plurality}}\hbox{-}\allowbreak\mathrm{Destructive}\hbox{-}\allowbreak\mathrm{Weighted}\hbox{-}\allowbreak\mathrm{$Bribery}$are$\mathrm{NP}$-complete, even when restricted to two candidates.*

Proof.

These problems are in NP, using the observation at the start of this section: Guess a set of voters to bribe, check that their price is within the budget, let all these bribed voters vote for the same highest-scoring desired candidate, and let all nonbribed voters vote for the same highest-scoring undesired candidate. Accept if a desired candidate wins in the constructive case and accept if no undesired candidate wins in the destructive case.

To show NP-hardness for the constructive case, we use the same construction as for nonsequential bribery with prices and weights. We reduce from (the standard NP-complete problem) Partition. Let $s_{1},\ldots,s_{n}$ be a sequence of nonnegative integers such that $\sum_{i=1}^{n}=2S$. We map to OBS $(C,V,\sigma,d,k)$, where $C=\{d,c\}$, $d>_{\sigma}c$, the price and weight of the $i$th voter are both $s_{i}$, $u$ is the first voter and votes for $c$, and $k=S$.

For the destructive case, our designated candidate will be $c$, and $V_{<u}$ consists of one unbribed weight-1 voter who votes for $d$.   ❑

5.2 Beyond Plurality

A scoring protocol is a vector $\alpha=(\alpha_{1},\ldots,\alpha_{m})$ of integers $\alpha_{1}\geq\alpha_{2}\geq\cdots\geq\alpha_{m}\geq 0$. This defines an election system on $m$ candidates where each candidate earns $\alpha_{i}$ points for each vote that ranks it in the $i$th position and the candidates with the most points are the winners.

Theorem 5.5**.**

For each scoring vector $\alpha=(\alpha_{1},\ldots,\alpha_{m})$,

\mathrm{online}\hbox{-}\allowbreak\mathrm{{\alpha}}\hbox{-}\allowbreak\mathrm{Weighted}\hbox{-}\allowbreak\mathrm{\Bribery}$* and *$\mathrm{online}\hbox{-}\allowbreak\mathrm{{\alpha}}\hbox{-}\allowbreak\mathrm{Destructive}\hbox{-}\allowbreak\mathrm{Weighted}\hbox{-}\allowbreak\mathrm{$Bribery}$* are in *$\mathrm{P}$* if *$\alpha_{1}=\alpha_{m}$* and *$\mathrm{NP}$-hard otherwise; 2. 2.

$\mathrm{online}\hbox{-}\allowbreak\mathrm{{\alpha}}\hbox{-}\allowbreak\mathrm{Weighted}\hbox{-}\allowbreak\mathrm{Bribery}$* and $\mathrm{online}\hbox{-}\allowbreak\mathrm{{\alpha}}\hbox{-}\allowbreak\mathrm{Destructive}\hbox{-}\allowbreak\mathrm{Weighted}\hbox{-}\allowbreak\mathrm{Bribery}$ are in $\mathrm{P}$ if $\alpha_{2}=\alpha_{m}$ and $\mathrm{NP}$-hard otherwise; and* 3. 3.

\mathrm{online}\hbox{-}\allowbreak\mathrm{{\alpha}}\hbox{-}\allowbreak\mathrm{Bribery},\mathrm{online}\hbox{-}\allowbreak\mathrm{{\alpha}}\hbox{-}\allowbreak\mathrm{Destructive}\hbox{-}\allowbreak\mathrm{Bribery},\mathrm{online}\hbox{-}\allowbreak\mathrm{{\alpha}}\hbox{-}\allowbreak\mathrm{\Bribery}$*, and *$\mathrm{online}\hbox{-}\allowbreak\mathrm{{\alpha}}\hbox{-}\allowbreak\mathrm{Destructive}\hbox{-}\allowbreak\mathrm{$Bribery}$* are in *$\mathrm{P}$.

Note that Theorem 5.5 implies the lower bound of Theorem 5.4, but does not imply Theorem 5.3, since in that theorem the number of candidates is not fixed. Theorem 5.5 does apply to 3-candidate Veto, for which we will additionally prove higher lower bounds in Theorem 5.6.

Proof.

If $\alpha_{1}=\alpha_{m}$, all candidates are always winners. If $\alpha_{1}>\alpha_{2}=\alpha_{m}$, this is in essence Plurality, which is handled in the theorems above. In all other constructive cases, the hardness follows from the hardness for ${\mathrm{{\alpha}}\hbox{-}\mathrm{WCM}}$ from Hemaspaandra and Hemaspaandra [HH07] and Theorem 5.1. In all other destructive cases, the hardness follows from the hardness for ${\mathrm{{\alpha}}\hbox{-}\mathrm{WCM}}$ in the unique winner model from Hemaspaandra and Hemaspaandra [HH07] and Theorem 5.1. (Note that ${\mathrm{{\alpha}}\hbox{-}\mathrm{DWCM}}$ is easily seen to be in $\mathrm{P}$. Basically, to make $d$ not a winner, we need an $a$ such that $a$’s score is higher than $d$’s score. So, make all manipulators vote $a>\cdots>d$, and compute the scores. So, this does not help us with the lower bound for \mathrm{online}\hbox{-}\allowbreak\mathrm{{\alpha}}\hbox{-}\allowbreak\mathrm{Destructive}\hbox{-}\allowbreak\mathrm{Weighted}\hbox{-}\allowbreak\mathrm{\Bribery}$.)

For the last case, we first look at the unpriced, unweighted case, even though the result follows from the priced case. We do this because the algorithm for this case is much simpler. Note that it follows from Observation 5.2 that we can assume that all bribed voters go last and so we have $k^{\prime}$ nonbribed voters followed by $k$ bribed voters. Since there are only a constant number of different votes (since $m$ is a constant), simply brute-force to determine whether it is the case that for all (polynomially many) possible $k^{\prime}$ votes, there are $k$ votes such that a candidate that is preferred to $d$ is a winner (for constructive) or such that no candidate that $d$ is preferred to is a winner.

Note that if we have prices, we can not assume that the bribed voters come last. We also can not assume that we bribe the cheapest voters, since later voters have more power than earlier voters, and so a more expensive later voter could be a better choice to bribe than a cheaper earlier voter. Still, we can solve the priced cases in polynomial time, by using dynamic programming. It is crucial that the scores for unweighted elections are $O(\log n)$.

We want to compute $\pi(s_{1},\ldots,s_{m},u,k)$ to be the minimum budget such that if the score of $c_{i}$ before $u$ is $s_{i}$ and there are $k$ voters after $u$, the briber can accomplish their goal. Note that there are a constant number of votes for $u$ and that all other numbers are $O(\log n)$. Also note that $\pi(s_{1},\ldots,s_{m},u,k)$ is the minimum of the minimum budget needed when $u$ is bribed and the minimum budget needed when $u$ is not bribed. To compute the minimum budget needed when $u$ is bribed, compute the minimum over all votes $v$ and $v^{\prime}$ of $\pi(u)+\pi(s^{\prime}_{1},\ldots,s_{m}^{\prime},v^{\prime},k-1)$, where $s_{i}^{\prime}$ is the score of $c_{i}$ from all but the last $k$ voters when $u$ is bribed to vote $v$. To compute the minimum budget needed when $u$ is not bribed, compute the minimum over all votes $v^{\prime}$ of $\pi(s^{\prime}_{1},\ldots,s_{m}^{\prime},v^{\prime},k-1)$, where $s_{i}^{\prime}$ is the score of $c_{i}$ from all but the last $k$ voters when $u$ is not bribed.   ❑

In Veto, each candidate scores a point when it is not ranked last in a vote and the candidates with the most points are the winners. Now let’s look at 3-candidate-Veto.

Theorem 5.6**.**

$\mathrm{online}\hbox{-}\allowbreak\mathrm{{\mbox{\rm 3-candidate-Veto}}}\hbox{-}\allowbreak\mathrm{Bribery}$, $\mathrm{online}\hbox{-}\allowbreak\mathrm{{\mbox{\rm 3-candidate-Veto}}}\hbox{-}\allowbreak\mathrm{Destructive}\hbox{-}\allowbreak\mathrm{Bribery}$, \mathrm{online}\hbox{-}\allowbreak\mathrm{{\mbox{\rm 3-candidate-Veto}}}\hbox{-}\allowbreak\mathrm{\Bribery}$*, and *$\mathrm{online}\hbox{-}\allowbreak\mathrm{{\mbox{\rm 3-candidate-Veto}}}\hbox{-}\allowbreak\mathrm{Destructive}\hbox{-}\allowbreak\mathrm{$Bribery}$* are in *$\mathrm{P}$. 2. 2.

$\mathrm{online}\hbox{-}\allowbreak\mathrm{{\mbox{\rm 3-candidate-Veto}}}\hbox{-}\allowbreak\mathrm{Weighted}\hbox{-}\allowbreak\mathrm{Bribery}$* and $\mathrm{online}\hbox{-}\allowbreak\mathrm{{\mbox{\rm 3-candidate-Veto}}}\hbox{-}\allowbreak\mathrm{Destructive}\hbox{-}\allowbreak\mathrm{Weighted}\hbox{-}\allowbreak\mathrm{Bribery}$ are $\mathrm{P}^{\mathrm{NP}[1]}$-complete.* 3. 3.

\mathrm{online}\hbox{-}\allowbreak\mathrm{{\mbox{\rm 3-candidate-Veto}}}\hbox{-}\allowbreak\mathrm{Weighted}\hbox{-}\allowbreak\mathrm{\Bribery}$* and *$\mathrm{online}\hbox{-}\allowbreak\mathrm{{\mbox{\rm 3-candidate-Veto}}}\hbox{-}\allowbreak\mathrm{Destructive}\hbox{-}\allowbreak\mathrm{Weighted}\hbox{-}\allowbreak\mathrm{$Bribery}$* are *$\mathrm{P}^{\mathrm{NP}[1]}$*-hard and in *${\Delta_{2}^{{p}}}$* (and we conjecture that they are *${\Delta_{2}^{{p}}}$-complete).

Proof.

The first part follows immediately from Theorem 5.5.

For the second part, we look at different cases for the placement of the designated candidate in the preference order $a>_{\sigma}b>_{\sigma}c$.

• •

If $d=c$ in the constructive case or $d=a$ in the destructive case, the problem is trivial.

• •

If $d=a$ in the constructive case, we need to ensure that $a$ is a winner, and if $d=b$ in the destructive case, we need to ensure that $a$ is the unique winner. In both these cases, the nonbribed voters veto $a$, no matter what. This means that the location of the bribed voters doesn’t matter (though their weights will). This is $\mathrm{NP}$-complete. For the upper bound, guess a set of voters to bribe, check that they are within the bribery limit, guess votes for the bribed voters, and have all nonbribed voters veto $a$. NP-hardness follows from the NP-hardness for weighted manipulation [HH07] plus the proof of Theorem 5.1.

• •

If $d=b$ in the constructive case, the goal is to not have $c$ win uniquely, and if $d=c$ in the destructive case, the goal is to have $c$ not win. In this case, all bribed voters veto $c$ (no matter what).

This is coNP-complete. To show that the complement is in NP, pick the $k$ heaviest voters to bribe. Then check if you can partition the remaining voters to veto $a$ or $b$ in such a way that $c$ wins uniquely in the constructive case or that $c$ wins in the destructive case. Note that it is always best for the briber to bribe the $k$ heaviest voters: Swapping the weights of a lighter voter to be bribed with a heavier voter not to be bribed will never make things worse for the briber. To show hardness, note that the complement is basically (the standard NP-complete problem) Partition.

Putting the three cases together, the unpriced, weighted case is $\mathrm{P}^{\mathrm{NP}[1]}$-complete, since it can be written as the union of a NP-complete set and a coNP-complete set that are P-separable.101010Two sets $A$ and $B$ are P-separable if there exists a set $X$ computable in polynomial time such that $A\subseteq X\subseteq\overline{B}$ (see [GS88]).

It remains to show the third part, i.e., the priced, weighted case. Note that this case inherits the $\mathrm{P}^{\mathrm{NP}[1]}$-hardness from the weighted case (in fact, it already inherited this from online manipulation, using Theorem 5.1). For the ${\Delta_{2}^{{p}}}$ upper bound, we again look at different cases for the placement of the designated candidate in the preference order $a>_{\sigma}b>_{\sigma}c$.

• •

If $d=c$ in the constructive case or $d=a$ in the destructive case, the problem is trivial.

• •

If $d=a$ in the constructive case, we need to ensure that $a$ is a winner, and if $d=b$ in the destructive case, we need to ensure that $a$ is the unique winner. In both these cases, the nonbribed voters veto $a$, no matter what. This means that the location of the bribed voters doesn’t matter (though their prices and weights will). All cases are in $\mathrm{NP}$: Guess a set of voters to bribe, check that they are within the budget, guess votes for the bribed voters, and have all nonbribed voters veto $a$.

• •

If $d=b$ in the constructive case, the goal is to not have $c$ win uniquely, and if $d=c$ in the destructive case, the goal is to have $c$ not win. In this case, all bribed voters veto $c$ (no matter what).

To show the upper bound for the priced, weighted case, note that we need to check that there exists a set of voters that can be bribed within the budget such that if all bribed voters veto $c$, then for all votes for the nonbribed voters, $c$ is not the unique winner (in the constructive case) or not a winner (in the destructive case). This is clearly in ${\Sigma_{2}^{{p}}}$. With some care, we can show that it is in fact in ${\Delta_{2}^{{p}}}$. First use an NP oracle to determine the largest possible total weight (within the budget) of bribed voters. Then determine whether we should bribe $u$, by using the oracle again to determine the largest possible total weight (within the budget) of bribed voters, assuming we bribe $u$. If that weight is the same as the previous weight, bribe $u$. Otherwise, do not bribe $u$. Repeating this will give us a set of voters to bribe of maximum weight. All these voters will veto $c$. It remains to check that for all votes for the nonbribed voters, $c$ is not the unique winner (in the constructive case) or not a winner (in the destructive case). This takes one more query to an NP oracle.  ❑

We end this section by looking at approval voting. In approval voting, each candidate scores a point when it is approved in a vote and the candidates with the most points are the winners. Note that approval sets have arbitrary sizes. Though Approval-Bribery is NP-complete [FHH09, Theorem 4.2], we show that $\mathrm{online}\hbox{-}\allowbreak\mathrm{{Approval}}\hbox{-}\allowbreak\mathrm{Bribery}$ (and even $\mathrm{online}\hbox{-}\allowbreak\mathrm{{Approval}}\hbox{-}\allowbreak\mathrm{Weighted}\hbox{-}\allowbreak\mathrm{Bribery}$ and \mathrm{online}\hbox{-}\allowbreak\mathrm{{Approval}}\hbox{-}\allowbreak\mathrm{\Bribery}$) are in$\mathrm{P}$. This implies that even for natural systems, bribery can be harder than online bribery (assuming$\mathrm{P}\neq\mathrm{NP}$).

Theorem 5.7**.**

$\mathrm{online}\hbox{-}\allowbreak\mathrm{{Approval}}\hbox{-}\allowbreak\mathrm{Bribery}$, $\mathrm{online}\hbox{-}\allowbreak\mathrm{{Approval}}\hbox{-}\allowbreak\mathrm{Destructive}\hbox{-}\allowbreak\mathrm{Bribery}$, \mathrm{online}\hbox{-}\allowbreak\mathrm{{Approval}}\hbox{-}\allowbreak\mathrm{\Bribery}$*, *$\mathrm{online}\hbox{-}\allowbreak\mathrm{{Approval}}\hbox{-}\allowbreak\mathrm{Destructive}\hbox{-}\allowbreak\mathrm{$Bribery}$*, *$\mathrm{online}\hbox{-}\allowbreak\mathrm{{Approval}}\hbox{-}\allowbreak\mathrm{Weighted}\hbox{-}\allowbreak\mathrm{Bribery}$*, and *$\mathrm{online}\hbox{-}\allowbreak\mathrm{{Approval}}\hbox{-}\allowbreak\mathrm{Destructive}\hbox{-}\allowbreak\mathrm{Weighted}\hbox{-}\allowbreak\mathrm{Bribery}$* are each in *$\mathrm{P}$. 2. 2.

\mathrm{online}\hbox{-}\allowbreak\mathrm{{Approval}}\hbox{-}\allowbreak\mathrm{Weighted}\hbox{-}\allowbreak\mathrm{\Bribery}$* and *$\mathrm{online}\hbox{-}\allowbreak\mathrm{{Approval}}\hbox{-}\allowbreak\mathrm{Destructive}\hbox{-}\allowbreak\mathrm{Weighted}\hbox{-}\allowbreak\mathrm{$Bribery}$* are each *$\mathrm{NP}$-complete.

Proof.

The upper bounds are immediate from the observation that it is optimal for the briber to have the bribed voters approve all desired candidates and to have the nonbribed voters approve all other candidates. The NP-hardness for the priced, weighted cases follows with the same reduction as for the priced, weighted cases of online bribery for Plurality from Theorem 5.4.   ❑

6 Conclusions

We have introduced a model of online, sequential bribery in voting and have initiated the study of the complexity of the most natural problems in this setting. In particular, we have shown that even for election systems whose winners can be determined in polynomial time, in an online, sequential setting these bribery problems can be complete for PSPACE or, when restricted to at most $k$ bribes, for ${\Pi^{p}_{2k+1}}$.

On the other hand, we have also shown that for some natural, important election systems, namely Plurality, Approval, and $3$-candidate-Veto, such a dramatic complexity jump does not occur, and we pinpoint the complexity of their bribery problems. Table 1 compares our complexity results on online bribery for these specific natural election systems to the known complexity results for nonsequential bribery.

A very natural direction for further research is to investigate the complexity of online bribery in natural election systems other than those studied in Section 5 of this paper. A more narrow, focused challenge would be to close the gap (see Theorem 5.6), for online-3-candidate-Veto-Weighted-$Bribery and online-3-candidate-Veto-Destructive-Weighted-$Bribery, between the $\mathrm{P}^{\mathrm{NP}[1]}$-hardness lower bounds and the ${\Delta_{2}^{{p}}}$ upper bounds; we conjecture that both problems in fact are ${\Delta_{2}^{{p}}}$-complete. The argument made in Footnote 2 notwithstanding, another potential direction for additional study would be to investigate the complexity of online bribery when prices and/or weights are encoded in unary rather than binary. A broad area for further research would be to investigate online bribery in a probabilistic setting, perhaps with probabilistic distribution information about future votes and their correlations; that would have to be done keeping in mind that later votes could be influenced by earlier cast votes. Finally, it could be interesting to extend our model of online, sequential bribery to other bribery variants such as swap bribery or shift bribery.

Acknowledgments

A preliminary version of this paper appeared in the Seventeenth Conference on Theoretical Aspects of Rationality and Knowledge (TARK 2019) [HHR19]. We are very grateful to the anonymous conference and journal reviewers and Eric Pacuit for many helpful comments and suggestions.

Appendix A Appendix: A Discussion of the

Weighted Version of a Given Election System

Let us discuss the issue of weighted versions of election systems, since the issue is not as straightforward as it might at first seem. This is a somewhat rarefied, model-focused discussion, and the reader can safely skip this section unless interested in the issue of models—and can rest assured that the outcome of the section is that throughout this paper we use the notion of weighted versions that is a common and intuitive one.

For natural systems, the weighted versions typically are already defined. In particular, usually typical and natural is to treat each weight-$w$ vote as $w$ unweighted copies of that same vote, and to do whatever the unweighted system would do given that particular collection of votes. Let us call this notion of an election system’s weighted version multiplicity expansion. This is simply interpreting weights as what called the “succinct” (e.g., [FHH09, FHHR09, HHR09, HH09] and many more)—as opposed to the standard, aka “nonsuccinct”—version of an election’s votes, which is using binary numbers to represent the number of copies.

The unweighted winner problem is in a strictly formal sense not a special case of the defined-by-multiplicity-expansion weighted winner problem, because the types of the voters differ (the former problem has no weights and the latter problem has weights). However, the unweighted winner problem polynomial-time reduces (and indeed reduces even via far more restrictive reductions) to the defined-by-multiplicity-expansion weighted winner problem simply by the near-trivial action of setting the weight of each voter to be one. So the unweighted winner problem is in effect never of greater complexity than the defined-by-multiplicity-expansion weighted winner problem. In particular, if for an election system ${\cal E}$ the defined-by-multiplicity-expansion weighted winner problem is in $\mathrm{P}$, then so is the unweighted winner problem for ${\cal E}$.

Note, however, that since the weights in our problems are in binary (see Footnote 2), the multiplicity-expansion approach means that we may be simulating the original system on an exponentially long input. And so in this notion of weighted elections it does not in general hold that if an (unweighted) election system has a polynomial-time winner problem, then its weighted version—defined by multiplicity expansion in the way just mentioned—has a polynomial-time winner problem. Indeed, it is very easy to make artificial examples of election systems with polynomial-time winner problems where the winner problem of the weighted version (as defined via multiplicity expansion) is complete for exponential time, i.e., is complete for ${\rm E}=\cup_{c>0}\mathrm{DTIME}[2^{cn}]$.111111For completeness, let us give an (admittedly artificial) construction of such a case. Let $L_{\rm E}$ be any problem that is complete for E with respect to linear-time many-one reductions; it is well-known that E has such complete sets, e.g., the natural universal set $\{\langle M,x,0^{k}\rangle\mid M$ accepts input $x$ within $k$ steps$\}$ (see [Har78]). Let ${\cal E}$ be the election system such that if there is exactly one candidate—$c$—in the election, and all voters in the election cast the same vote, and the number of voters is at least $2^{|c|}$, then $c$ wins if and only if $c\in L_{\rm E}$. Here, we are viewing $c$ as the bit-string naming that candidate, and $|c|$ denotes the number of bits in that string. In all other cases, all candidates win. The unweighted winner problem for this election system ${\cal E}$ is clearly in $\mathrm{P}$. However, note that the defined-by-multiplicity-expansion weighted winner problem of ${\cal E}$ is linear-time many-one complete for $\rm E$. Why? It clearly is in $\rm E$, since (though this is overkill) one can expand all weighted votes, that expansion transforms problem instances of size $n$ to problem instances of size at most $2^{n}$, and then one can accept if and only if the expanded instance is a member of the (polynomial-time solvable) unweighted winner problem of ${\cal E}$. But the defined-by-multiplicity-expansion weighted winner problem of ${\cal E}$ is also many-one linear-time hard for $\rm E$, since $L_{\rm E}$ reduces to it by the many-one linear-time reduction that maps from $x$ to an election with one candidate, whose name is $x$, and with one voter, whose weight is $c^{\prime}|x|$, where $c^{\prime}$ is the smallest natural number such that $L_{\rm E}\in\mathrm{DTIME}[2^{c^{\prime}n}]$ (such a $c^{\prime}$ exists by the definition of $\rm E$ and the fact that $L_{\rm E}\in{\rm E}$). Thus, the defined-by-multiplicity-expansion weighted winner problem of ${\cal E}$ is many-one linear-time complete for $\rm E$, despite the fact that the unweighted winner problem for ${\cal E}$ belongs to $\mathrm{P}$. Despite that, natural systems often have nice properties, and in particular, beautifully, for a wide variety of natural systems, their weighted versions are defined by multiplicity expansion yet those versions still have polynomial-time winner problems. For example, the family of election systems known as Copeland and Llull elections [Cop51, MLC08] (see also, e.g., [FHHR09]) has this property. Even more importantly, so-called scoring systems have this property (basically because, in their case, instead of doing an exponential number of additions one at a time, one can simply multiply). And, for example, Section 5 of this paper discusses a number of such real-world systems, and when speaking of their weighted versions is indeed doing so via multiplicity expansion, and in doing so, will in each case need to note that the thus-defined weighted version’s winner problem is in polynomial time.

In light of the previous paragraph, one still is left with the question: Should we (a) simply define the weighted version of every election system to be what is defined by the multiplicity expansion approach above? Or (b) for each election system, should each author, using his or her human taste, hand-tailor, possibly in a quirky nonuniform way, what he or she feels is a natural notion if its weighted version?

Both those approaches have advantages and downsides. We will discuss those, and then—having made clear the downsides we are embracing—will in this paper adopt approach (a), namely, multiplicity expansion. One disadvantage of multiplicity expansion is that as mentioned above it in some cases takes unweighted election systems whose winner problems are in polynomial time and boosts the winner-problem complexity of their weighted versions as high as exponential time. But that is more a feature to be aware of—and to avoid being bitten by due to forgetting that the feature might exist—than a disadvantage. The only true disadvantages we know of regarding using multiplicity expansion as providing a general notion of interpreting weights in elections are (i) the approach is conflating the issue of weights with the issue of succinct representations, and in particular (ii) for a few natural election systems, the approach arguably gets things wrong (and we will soon come back and discuss this “gets things wrong” in more detail). The advantage of the multiplicity-expansion approach is that it is usually highly natural, it is what almost anyone would think of when asked what weight should be interpreted as, and it gives an across-the-board, uniform approach to weight, rather than having the notion be a system-by-system, ad hoc, argued-and-debated construct. Turning to alternative (b), since human tastes differ, alternative (b)’s worst downside is that in the extreme one might be just basically hand in the weights to an election system’s computation function and let it do whatever it feels like with them—making them just information/bits it can use, perhaps in utterly unnatural ways. That the definition of weight may not match what people broadly feel that weight means, and also (aside from the promise that the humans hand-defining the weighted versions of each system are using their taste) this approach does nothing to ensure that the weighted version of an election systems has any connection whatsoever to the system’s unweighted version. In light of this, throughout this paper we outright define, for any (unweighted) election system ${\cal E}$, the weighted version of ${\cal E}$ (which via overloading we will also sometimes describe as ${\cal E}$—the inputs to the system though will make clear which version we are speaking of, since in one case there are no weights and in the other case there are) via multiplicity expansion.

It might seem strange for any paper to take a different approach to what “weighted” means. Indeed, the literature’s papers are generally taking the multiplicity-expansion approach to weight as so natural and compelling that they simply are employing it, generally without mentioning it or mentioning why they are using it. However, to further discuss this, let us briefly come back to the worst weakness of the choice we made, namely, that there are cases where this approach arguably “gets things wrong.” In particular, the fact that the approach conflates succinctness with weight means that the indivisibility of a vote is being shattered. Yet this is a severe problem for election systems that are very focused on actions affecting individual votes. For example, a famous election system known as Dodgson elections [Dod76] is based on the number of sequential exchanges of adjacent preferences within voters needed to make a given candidate become “a Condorcet winner” [Con85], i.e., a candidate who beats each other candidate in pairwise head-on-head elections. The multiplicity-expansion approach would allow, for a example, some unweighted copies of a given weighted vote to have an exchange made in them while other unweighted copies of that same weighted vote didn’t have the exchange made in them; yet that seems not to respect the spirit of Dodgson’s system. (If one were hand-defining a notion of weight for Dodgson’s system, it is simply not clear whether for a weight $i$ voter, now viewed as utterly indivisible, one would want to view an adjacent-candidates exchange as one exchange or $w_{i}$ exchanges. Each could be argued for, and so there are at least two quite different, quite reasonable notions of hand-defined weightedness for Dodgson election.) In Young elections [You77], which are defined based on how few deletions of voters are needed to make a given candidate a (“weak,” but let us not here worry about that distinction) Condorcet winner, a similar issue occurs—similar both in the difficulty and in the fact that there are, in a hand-built notion of weight for the problem, two quite different notions that reasonable people could disagree on as to which is more appropriate. To the best of our knowledge, the issue of how to frame weighted Dodgson and weighted Young has been touched on only once in the literature, namely, it is discussed in the technical-report version of a paper of Fitzsimmons and Hemaspaandra [FH16], although not in the later versions of that same paper. That paper provides a valuable discussion of many of the issues discussed in this section, and we highly commend it to the reader. Interestingly, that paper finds only one natural election system for which the weighted (in the sense of multiplicity expansion) winner problem can even conditionally be shown to be of greater complexity than the unweighted winner problem, namely, the paper shows that to be the case for the system known as Kemeny elections [Kem59]—which are a case where the natural interpretation of weights is via multiplicity expansion (see [FH16])—if a certain complexity-theoretic conjecture holds (namely, that there are sets acceptable via sequential access to $\mathrm{NP}$ that are not acceptable via parallel access to $\mathrm{NP}$). (In contrast, footnote 11 of the present paper constructs a case where the weighted version of a problem is, unconditionally, of greater complexity than its unweighted version.) The system Single Transferable Vote (see, e.g., [BO91]) has similar issues to those of Dodgson and Young elections, as to how to define weight for it.

To be clear, we are not suggesting that multiplicity expansion captures the right notion of weightedness for such systems as Dodgson’s, Young’s, or Single Transferable Vote. It does not, and papers studying those in weighted contexts should not employ multiplicity expansion as their notion of weight. However, what we are saying is that multiplicity expansion is the best uniform, general approach to weight, and so we use it here—while carefully making sure not to apply it to cases such as Dodgson elections where it is not a good fit, and making sure to always be aware that multiplicity expansion can distort the complexity of winner problems and that one must for whatever cases one covers make sure not to casually assume otherwise. We refer the reader to the work of Fitzsimmons and Hemaspaandra [FH16, FH19] for further discussion of, and results on, weights/succinctness/multiply in elections.

As a final comment, we stress that our definition regarding how the winner problem is defined for the weighted version of an (unweighted) election system in no way is something that binds the definitions of manipulative actions within those systems. Those actions and their costs are defined not by the winner-problem handling but rather by the manipulative actions themselves, e.g., in bribery of priced, weighted elections, the price of a weighted voter is the cost of bribing that particular voter (i.e., the weighted vote bribes or fails to be bribed as a single unit).