A Control Dichotomy for Pure Scoring Rules
aa r X i v : . [ c s . G T ] A p r A Control Dichotomy for Pure Scoring Rules ∗ ∼ Abstract
Scoring systems are an extremely important class of election systems. A length- m (so-called) scoring vector applies only to m -candidate elections. To handle generalelections, one must use a family of vectors, one per length. The most elegant ap-proach to making sure such families are “family-like” is the recently introduced notionof (polynomial-time uniform) pure scoring rules (Betzler and Dorn, 2010), where eachscoring vector is obtained from its precursor by adding one new coefficient. We obtainthe first dichotomy theorem for pure scoring rules for a control problem. In particular,for constructive control by adding voters (CCAV), we show that CCAV is solvable inpolynomial time for k -approval with k ≤ k -veto with k ≤
2, every pure scoring rulein which only the two top-rated candidates gain nonzero scores, and a particular rulethat is a “hybrid” of 1-approval and 1-veto. For all other pure scoring rules, CCAV isNP-complete. We also investigate the descriptive richness of different models for defin-ing pure scoring rules, proving how more rule-generation time gives more rules, provingthat rationals give more rules than do the natural numbers, and proving that somerestrictions previously thought to be “w.l.o.g.” in fact do lose generality.
Elections give rise to a plethora of interesting questions in the social and political sciences,and have been extensively studied from a computer-science point of view in the last two ∗ Supported in part by NSF grants CCF- { } and by COST Action IC1205. Workdone in part while H. Schnoor visited RIT supported by an STSM grant of Cost Action IC1205. control problem , in which the chair of an election (ab)uses her powerto try to affect the election outcome. In this paper we focus on constructive control byadding voters (CCAV), i.e., where the chair tries to make her favorite candidate win byadding voters. Constructive control by adding voters is an extremely important controltype, since it occurs in (political) practice very often. A standard example is the “Get-out-the-Vote” efforts of political parties, aimed at (supposed) supporters of those parties.However, we should also mention that, as has been pointed out for example in books byRiker (1986) and Taylor (2005), in modern politics issues of candidate introduction orremoval (CCAC/CCDC) have become highly important; the case of Ralph Nader in tworecent American elections vividly supports this point.The computational complexity of CCAV and other forms of control was first studiedby Bartholdi, Tovey, and Trick (1992), for plurality and (so-called) Condorcet elections. Inthis paper, we study the complexity of CCAV for pure scoring rules , an attractive classintroduced by Betzler and Dorn (2010) that contains many important voting systems. Ascoring rule for an election with m candidates is defined by m coefficients α ≥ α ≥ · · · ≥ α m . Each voter ranks the m candidates from her most favorite to her least favorite; acandidate gains α i points from being in position i on that voter’s ballot.Well-known examples of families of scoring rules include the following. Borda Countfor m candidates uses coefficients m − , m − , . . . , , k -approval uses coefficients1 , . . . , | {z } k , , . . . , | {z } m − k ; k -veto uses 1 , . . . , | {z } m − k , , . . . , | {z } k . Dowdall voting, used for Nauru’s parliament,uses 1 , , , . . . , m .The construction of the scoring vector for a specific number of candidates usually followsa natural pattern, as in the above examples. This leads to the definition of a “pure scoringrule.” We discuss the notion of “purity” in detail in this paper. (Basically, it means thatat each length we insert one entry into the previous length’s vector; all our above examplesare pure.)There is a rich literature on computational aspects of scoring rules, e.g., dichotomytheorems on weighted manipulation (Hemaspaandra and Hemaspaandra, 2007), the pos-sible winner problem (Betzler and Dorn, 2010; Baumeister and Rothe, 2012), and bribery(Faliszewski et al., 2009), as well as results about specific voting systems (Betzler et al.,2011; Davies et al., 2011; Faliszewski et al., 2013).In this paper, we provide the first complete investigation of the complexity of the un-weighted CCAV problem for pure scoring rules. We prove a dichotomy theorem that gives acomplete complexity-theoretic classification of that control problem for pure scoring rules.Our result is as follows.It turns out that there are only 4 types of pure scoring rules for which CCAV is solvablein polynomial time:1. k -approval for k ≤ k -veto for k ≤
2, 2. every pure scoring rule in which only the two top-rated candidates receive a nonzeroscore,4. a particular rule which is a “hybrid” of 1-approval and 1-veto: each voter awards herfavorite candidate 1 point, and her least favorite candidate − < -order of the values: While thiselection system is equal to the rule generated by coefficients 2 , , . . . , ,
0, the rule generatedby using coefficients 3 , , . . . , , As is standard, given an m -component (so-called “scoring”) vector α = ( α , . . . , α m ) suchthat α ≥ . . . ≥ α m , we define a so-called ( m -candidate) “scoring system” election basedon this by, for each voter (the voters vote by strict, linear orders over the candidates),giving α i points to the voter’s i th-most-favorite candidate. Whichever candidate(s) get thehighest number of points, summed over all voters, are the winner(s). We refer to the α i s as“coefficients.”We will use the following very pure definition of pure scoring rules, which removes someof the additional assumptions that have been used in earlier work. Indeed, earlier workasserted that (at least some of) these assumptions were not restrictions; we will look atthat issue anew below. 3n election system E is a T -GSR (generalized scoring rule) if there is a function f thaton each input 0 m , m ≥
1, outputs an m -component scoring vector α m = ( α m , . . . , α mm )such that α m ≥ . . . ≥ α mm and each α ij belongs to T , and for each m , the winner setunder E is exactly the winner set given by the scoring system using the scoring vector α m .The notation 0 m denotes a string of m “0”s. Using this as the argument ensures thatthe generator’s computation time is measured as a function of m , since its input lengthis exactly m . Since the votes are of size at least m each, this provides the natural, fairapproach to framing FP-uniformity (as we will do below). We call f a generator for E .While one can consider election systems based on an f that is not computable, in practicewe want there to be an efficient algorithm computing f . This is expressed with different uniformity conditions. If E is a T -GSR via some generator f that can be computed in acomplexity class F , then E is an F -uniform- T -GSR and f is an F -uniform T -generator.The values of T we will be interested in are the naturals N , the nonnegative rationals Q ≥ ,the rationals Q , and the integers Z . Our most important value of F will be the polynomial-time functions, FP. When we do not state “nonuniform” or some specific uniformity, wealways mean FP-uniform. When we put a name in boldface, it indicates all the electionsthat can be generated by a generator of the named sort, e.g., FP-uniform- Z -GSR is theclass of all polynomial-time-uniform generalized scoring rules with integer coefficients, aclass first defined and discussed by Hemaspaandra and Hemaspaandra (2007).A T -GSR (of whatever uniformity) is a T -PSR (pure scoring rule) if it has a generator(of the same uniformity) f satisfying the following “purity” constraint: For each m ≥ α m that when deleted leaves exactly the vector α m − .Throughout the paper we use the following observation, which notes that two scoringvectors, after being “normalized,” differ if and only if they are capturing distinct electionsystems: An m -position scoring vector α = ( α , . . . , α m ) (over N ) is normalized if α m = 0and the greatest common divisor of its nonzero α i ’s is 1. Normalizing a given scoring vector(over N or Z ) is easily achievable in polynomial time: subtract α m from each coefficientand then divide each coefficient by the gcd (computed using Euclid’s gcd algorithm) of thenonzero thus-altered coefficients. The normalization of a scoring vector over Q (resp., Q ≥ )is done by multiplying through by the lcm of the denominators of the nonzero coefficients,and then viewing that as a vector over Z (resp., N ) and normalizing it as above. Proposition 2.1.
Let m ≥ and let α and α ′ be m -position scoring vectors over T ⊆ Q .Then α and α ′ have the same winner sets on each m -candidate election if and only if α and α ′ both have the same normalized version. The if direction basically follows from Observation 2.2 of Hemaspaandra and Hema-spaandra (2007), as noted by Betzler and Dorn (2010). The only if follows by giving aconstruction that for any two unequal normalized scoring vectors constructs a vote set onwhich their winner sets differ. The construction works by “aligning” the vectors by multi-plying each so that their first coefficients are equal, and then using a padding constructionto ensure that only two candidates are crucial and that the winner sets can be distinguishedby appropriately exploiting the first position at which the aligned vectors differ.4enerators f and f are equivalent if they generate the same election system. Due toProposition 2.1, this is the case if and only if, for each length m , the normalized scoringvectors generated by f and f for m candidates are identical. We now examine how amount of time used to generate PSRs, and the universe the PSR’scoefficients are drawn from, affect the family of election rules that can be obtained. We alsolook at whether such a seemingly innocuous and standard assumption as having the lastcoefficient always being zero in fact loses generality; we’ll see that it does lose generality,but in a way that can in part be papered over.
Does more generation time give a richer class of pure scoring rules? A very tight timehierarchy can be achieved by a legal form of cheating. In particular, consider any (nice) timeclass that can for some m ≥ N of the form ( α , , . . . , | {z } m − , α is so big that some other time class cannot generate this scoring vector (forexample, because it simply doesn’t have enough time to write down enough bits to get anumber as large as α ). Our vector is normalized, and by Proposition 2.1 this already isenough to allow us to argue that the two time classes differ in their winner sets. (This proofworks for GSRs. And it works for PSRs if the “nice”ness of the class allows it to obey puritywhile employing the above approach—hardly an onerous requirement.) However, that claimsimply uses the fact that more time can write more bits. A truly fair and far more interestingseparation would show that one can with more time obtain more pure scoring rules in away that does not depend on using coefficient lengths that simply cannot be produced bythe weaker time class. In the dream case, all coefficients in fact would simply be 0 or 1,so there are no long coefficients in play at all. We in fact have achieved such a hierarchytheorem. Full time constructibility is a standard notion that most natural time functionssatisfy and, as is standard, when we speak of a time function T ( m ) by convention that isshorthand for max( m + 1 , ⌈ T ( m ) ⌉ ) (Hopcroft and Ullman, 1979). FDTIME[ g ( · )] denotesthe functions computable in the given amount of deterministic time. Theorem 3.1. If T ( m ) is a fully time-constructible function and lim sup n →∞ T ( m ) log T ( m ) T ( m ) = 0 ,then there is an election rule in FDTIME[ T ( m )] -uniform- { , } - PSR that is not in
FDTIME[ T ( m )] -uniform- Q - GSR . The log factor here is not surprising; this is the standard overhead it takes for a 2-tapeTuring machine to simulate a multitape TM. What is surprising is that no additional factorof m is needed. Why might we expect such a factor? (We caution that the rest of this para-graph is intended mostly for those having familiarity with the diagonalization techniquesused to prove hierarchy theorems.) In the context of a diagonalization construction (which5s the basic technique used in the proof of Theorem 3.1), one might expect to (all countingagainst the overall time T limit) at vector-length m have to “recreate” all the shorter vec-tors used in earlier diagonalizations to ensure that the length- m and length-( m −
1) vectorsare related in a “pure” way. We however sidestep the need for that overhead by a “purity”-inducing trick: For each odd m our scoring vector will be of the form 1 ⌊ m/ ⌋ ⌊ m/ ⌋ +1 , and ateach even length, we purely extend that to whichever of 1 ⌊ m/ ⌋ +1 ⌊ m/ ⌋ +1 or 1 ⌊ m/ ⌋ ⌊ m/ ⌋ +2 diagonalizes against the T -time machine that is being diagonalized against (if this is an m when we have time to so diagonalize). Briefly, we lurch back to fixed, safe way-stations atevery second length, and this removes the need to recompute our own history. The richness and structure of the coefficient set for PSRs affects how broad a class ofelection rules can be captured, as shown by the following result. (Of course, trivially N -PSR ⊆ Q ≥ -PSR ∩ Z -PSR ⊆ Q ≥ -PSR ∪ Z -PSR ⊆ Q -PSR. ) Theorem 3.2. Q ≥ - PSR nonuniform- Z - PSR , Z - PSR nonuniform- Q ≥ - PSR , and Q - PSR nonuniform- Q ≥ - PSR ∪ nonuniform- Z - PSR . So for example, in pure scoring rules, (polynomial-time uniformly) using positive ratio-nals cannot be simulated by naturals or integers, even nonuniformly. And in pure scoringrules, (polynomial-time uniformly) using integers cannot be simulated by naturals or pos-itive rationals, even nonuniformly. One might think these claims are impossible, and thatby normalizing one can go back and forth, but it is precisely the purity requirement that ismaking that sort of manipulation impossible—there is a price to purity, and it is showingitself here. (The final part of the theorem does not follow automatically from the first twoparts plus the trivial observation before the theorem; just the weaker variant of that part inwhich ∪ is replaced with ∩ follows from those.) Note that enlarging the universe does notnecessarily lead to a larger class of election systems: For example, requiring that coefficientsare odd natural numbers gives the same set of election systems as merely requiring them tobe natural numbers.We mention a more flexible and highly attractive notion of purity that erases the dif-ferences just discussed. T -FPSRs (flexible pure scoring rules), of whatever uniformity ornonuniformity, will be defined exactly the same way as T -PSRs were defined, except thepurity condition is changed to: For each m ≥
2, there is a component of α m whose removalgives a scoring vector equivalent to α m − . Due to Proposition 2.1, this means that therelevant scoring vectors have the same normalization. We call such generators flexible . For this notion we have for the nonuniform case and the FP-uniform case (and most other nicecases), the following equality. Theorem 3.3. N - FPSR = Q - FPSR = Q ≥ - FPSR = Z - FPSR = Q - PSR . .3 Having Smallest Coefficient of Zero Loses Generality—Slightly Betzler and Dorn (2010) in their definition of scoring rules require that at each m , we have α mm = 0 (let us call this condition norm-0), and comment that this is not a restriction.We note that there are PSRs that cannot be generated by any pure scoring rule meetingthat constraint. What is at issue here is a bit subtle: At each fixed length, the restrictionis innocuous. But in the context of families that are bound by the purity constraint, therestriction loses generality. On the other hand, we will also note that each pure scoring rulehas a generator that is “close” to meeting that constraint—it meets it at all but finitelymany lengths.Betzler and Dorn (2010) also have a “gcd is 1” condition (although the phrasing is notcrystal clear as to whether the gcd constraint applies to the union of all nonzero coefficientsthat occur over all lengths, or whether it must in fact apply separately at each length; thelatter (which we call norm-gcd) would block the vector (2 ,
0) but the former would not ifthe next step were, for example, (3 , , Theorem 3.4.
There is an FP -uniform PSR that is not generated even by any nonuniform
PSR norm-0 generator. On the other hand, every FP -uniform (respectively, nonuniform) PSR is generated by a FP -uniform (respectively, nonuniform) PSR generator that for all but afinite number of m has the property that the last coefficient, α m,m , is zero and the gcd ofthe nonzero coefficients in the length- m vector is one. Does the second sentence of the theorem imply that each PSR has all its vectors thesame at each length (except for a finite number of exceptional lengths) as the vectors ofsome PSR that satisfies norm-0 and norm-gcd? The answer is actually “no.” The somewhatsubtle issue at play is that PSRs can generate vectors that no generator satisfying norm-0and norm-gcd can ever generate, such as the family (3 , , . . . , , theorem of Betzlerand Dorn (2010), as completed by Baumeister and Rothe (2012), that we can read offthe complexity (of the possible winner problem) even in our slightly more flexible case.However, our above theorem does—in light of the actual proof case decomposition used inthose papers (which is based on issues such as whether one has an unbounded number ofpositions that differ and so on) and some additional argumentation to connect to that andin particular to note that Betzler and Dorn (and Baumeister and Rothe) are in effect quietlycovering well even those cases that do not satisfy gcd constraints—connect so well to theirwork that each of our cases is settled by their proofs. We study the following problem for an election system E : When R is a set of registeredvoters, is there some subset of the unregistered voters U of size at most k that we can addto the election to ensure that p is the winner?7 efinition 4.1. Let E be an election system. The constructive control problem for E byadding voters , E -CCAV, is the following problem: Given two multisets sets of votes R and U , a candidate p and a number k , is there a set A ⊆ U with k A k ≤ k such that p is awinner of the election if the votes in the multiset R ∪ A are evaluated using the system E ? We often use a generator, f , as a shorthand for the election system (scoring rule family)it generates, e.g., we write f -CCAV. For a generator f , we use α f,m = ( α f,m , . . . , α f,mm ) todenote the scoring vector generated by f for m candidates and its individual coefficients.To simplify presentation, we only consider FP-uniform generators. However our resultscontinue to hold as long as we can solve the following question in polynomial time: Given m , i , and j in unary , does α mi > α mj hold, where f ( m ) = ( α m , . . . , α mm )?Our main result is a complexity dichotomy for f -CCAV when f is an FP-uniform pure Q -generator (or, equivalently due to Theorem 3.3, a FP-uniform flexible N -generator). Recallthat equivalent generators result in the same election system, hence, due to Proposition 2.1,Theorem 4.2 implies polynomial-time results for all generators with the same normalizationas one below. We state our main result. Theorem 4.2.
Let f be a FP -uniform pure Q -generator. Then f -CCAV is solvable inpolynomial time if f is equivalent to one of the following generators: • f = (1 , , , , . . . , (this generates -approval), • f = (1 , . . . , , or f = (1 , . . . , , , ( / -veto), • for some α ≥ β , f = ( α, β, , . . . , , • f = (2 , , . . . , , .In all other cases, f -CCAV is NP-complete. Note that 1- and 2-approval are covered by the generator f . The remainder of the papercontains the proof of Theorem 4.2: Section 4.1 contains the algorithms for all polynomial-time solvable cases, Section 4.2 contains our hardness results (with some proofs deferred tothe appendix), and the proof of Theorem 4.2 follows in Section 4.3. The following result is proven by Lin (2012).
Theorem 4.3. E -CCAV is solvable in polynomial time if E is k -approval with k ≤ or k -veto with k ≤ . Due to Proposition 2.1, Theorem 4.3 implies that CCAV remains polynomial-time solv-able for “scaled” versions of k -approval with k ≤ k -veto with k ≤
2, i.e., generatorsof the form ( α, β, γ, δ, . . . , δ ) with β, γ ∈ { α, δ } or ( α, . . . , α, β, γ ) with β ∈ { α, γ } . We nowlook at a generalization of 2-approval: Voters approve of 2 candidates, and can distinguish8etween their first and second choice. CCAV for this generalization remains efficientlysolvable. In contrast, Theorem 4.11 shows that our control problem for the correspondinggeneralization of 3-approval is NP-hard. Theorem 4.4.
Let α ≥ β be fixed. Then f -CCAV is polynomial-time solvable for f =( α, β, , . . . , .Proof. Due to Theorem 4.3, assume α > β . Let ℓ be such that ℓ ( α − β ) ≥ β . Let aninstance with candidates C , favorite candidate p , registered voters R , and potential voters U be given; let k be the number of voters that can be added. Assume ℓ ≤ k , otherwise brute-force. Let V ( V ) be the set of voters in U that put p in the first (second) spot. Clearly, weadd voters only from V ∪ V . Assume w.l.o.g. that two voters who vote identically in thefirst two positions also rank the remaining candidates identically. In particular, two votersin V ( V ) are different if and only if they vote different candidates in the second (first)place. We use the following facts. Fact 1.
For i ∈ { , } , given a set S ⊆ V ∪ V , it can be checked in polynomial time whether S can be extended, by adding at most k − k S k voters from V i to make p win. Fact 2.
To make p win, it is never better to add ℓ voters from V than adding ℓ pairwisedifferent voters from V . Due to Fact 2, we do not have to consider solutions that use ℓ or more voters from V and leave ℓ or more distinct voters from V unused. So if a solution exists, we can find oneusing fewer than ℓ voters from V , or leaving fewer than ℓ pairwise different voters from V unused. For both cases we will test whether there is a corresponding solution.We start with the first case. Since ℓ is constant, we can test every subset S ⊆ V with k S k < ℓ . For each of these S , we use Fact 1 to check in polynomial time whether S can beextended to a solution by adding voters only from V .For the second case, we determine in polynomial time whether there is a solution thatdoes not leave ℓ pairwise different voters from V unused as follows: We encode the choiceof the unused voters from V as a function u : C → { , . . . , k V k} that states, for eachcandidate c , the number of unused V -voters placing c second. Since we look for solutionssatisfying the second case, we only consider functions u for which u ( c ) > ℓ − S ⊆ C with k S k ≤ ℓ −
1. Since ℓ is a constant,there are only polynomially many of these sets.2. For each such S , we test all functions u as above for which u ( c ) > c ∈ S . Sucha function can be regarded as a function u : S → { , . . . , k V k} . There are k V k k S k ≤k V k ℓ − many of these functions. Since k V k is bound by the input size and ℓ is aconstant, this number is polynomial in the input size.9o we can polynomially go through all possibilities of potentially unused V -voters, whichis the same as going through all possible sets S ′ of used V -voters. For each of these sets S ′ ,we again use Fact 1 to check in polynomial time whether S ′ can be extended to a solution.This completes the proof.Our final polynomial-time case is the generator (2 , , . . . , , > -order. Namely, for all α > β > α = 2 β , f -CCAV with f = ( α, β, . . . , β,
0) isNP-complete (Theorem 4.12.2). Second, this case was the only one left open in Betzler andDorn’s (2010) possible winner dichotomy; the question was eventually settled by Baumeisterand Rothe (2012), who proved NP-completeness.
Theorem 4.5. f -CCAV is solvable in polynomial time for the generator f = (2 , , . . . , , .Proof. Let C , R , and U be the set of candidates, registered voters, and unregistered voters, p the preferred candidate, and k the number of voters we can add. We add no voter voting p last, and it is never better to add a voter voting p second than to add one voting p first.So we first add all voters from U that place p in the first position. If there are more than k of these voters, we choose the ones to add with the obvious greedy strategy that alwayspicks, among all available votes of the form p > · · · > c , the one where c currently hasthe highest score. After this preprocessing, all relevant voters in U vote c > · · · > c with p / ∈ { c , c } . To simplify presentation, we use Proposition 2.1 and consider f as thegenerator (1 , , . . . , , − p is determined by the votes in R .We reduce the problem to min-cost (network) flow, which can be solved in polynomialtime. Let S = P c ∈ C −{ p } score ( c ). We use the following nodes and edges: • For each c ∈ C − { p } , there is a node c , additionally, there are source and target nodes s and t . • There is an edge from candidate c to candidate c with cost 1 and with capacity equalto the number of voters in U voting c > · · · > c . • For each candidate-node c , there is an edge from s to c with cost 0 and capacity score ( c )and an edge from c to t with cost 0 and capacity score ( p ).Now p can be made winner with at most k additional voters if and only if there isa flow from s to t with value S and cost at most k : Clearly, network flows with cost atmost k correspond to subsets of U with size at most k , and using an edge ( c , c ) r timescorresponds to adding r voters voting c > · · · > c , since this vote transfers one point from c to c . The capacity of the outgoing edges of s ensure that each candidate initially getsthe correct number of points (since S points must be distributed), the edges to t ensurethat in the end, no candidate may have more points than p .The above results cover all polynomial-time cases of Theorem 4.2. We now turn to theNP-complete cases. 10 .2 Hardness Results We use the standard NP-complete problem 3DM (3-dimensional matching).
Definition 4.6. X , Y , and Z with k X k = k Y k = k Z k , and a set M ⊆ X × Y × Z .Question Is there a set C ⊆ M with k C k = k X k that covers X , Y , and Z ? We say that C covers X (resp., Y , Z ) if every element from X (resp., Y , Z ) appears in atuple of C . Since X , Y , and Z are pairwise disjoint, in this case every element from X ( Y , Z )appears in the first (second, third) component of a tuple from C . Since k X k = k Y k = k Z k ,a set C ⊆ M with k C k = k X k covers X ( Y , Z ) if and only if no two tuples from C agreein the first (second, third) component. A set C covering X , Y , and Z is called a cover . In our hardness proofs, we often need to set up the registered voters to ensure specific scoresfor the candidates. The following lemma shows that, if there is a “dummy” candidate towhom any surplus points can be “shifted,” we can obtain every set of relative scores thatcan be expressed as a polynomial-size linear combination of the coefficients in the scoringvector.
Lemma 4.7.
Given a scoring vector ( α , . . . , α m ) , and for each c ∈ { , . . . , m − } , numbers a c , . . . , a cm in signed unary, and a number k in unary, we can compute, in polynomialtime, votes such that the scores of the candidates when evaluating these votes accordingto the scoring vector ( α , . . . , α m ) are as follows: There is some o such that for each c ∈{ , . . . , m − } , score ( c ) = o + P mi =1 a ci α i , and score ( c ) > score ( m ) + kα . The value o in Lemma 4.7 is the common offset for all relevant scores. The actual valueof o is irrelevant, since the winner of the election is determined by the relative scores. Thevalue k is given so that the computed votes ensure that the dummy candidate m cannotwin the election with the addition of at most k voters. We now show that the CCAV-problem is NP-complete for generators using “many” differentcoefficients. Consider any generator f using (at least) 7 different coefficients for somelength m . Then with α f,m = ( α f,m , α f,m , α f,m , α f,m , . . . , α f,mm − , α f,mm − , α f,mm ) we know that α f,m > α f,mm − . This condition in fact suffices for the CCAV problem to be NP-hard; theresult applies to, e.g., Borda, 3-veto, and 4-approval (the latter two use just two differentcoefficients, but satisfy α f,m > α f,mm − for m ≥ M from a 3DM-instance, along with p , in the 4 top positions of an unregistered 4-approval voteor (without p ) in the last 3 positions of an unregistered 3-veto vote. In our cases, we can11lways “simulate” one of these systems: If α f,m > α f,mm − , then being ranked in one of thefirst 4 positions is strictly better than being ranked in one of the last 3 positions. Roughlyspeaking, if “many” intermediate coefficients are larger than the last 3, then the last 3 arethe “exception,” and we can use them to “simulate” 3-veto. On the other hand, if “many”intermediate coefficients are smaller than the first 4, then the first 4 are the “exception”and we “simulate” 4-approval. NP-hardness for both 3-veto and 4-approval is proved byLin (2012); however we use a direct reduction from 3DM in our generalization.We start with the “simulation” of 3-veto. The statement of the following result is a bitunusual. It indeed gives a reduction for generators meeting the condition α f,m k +1 > α f,mm − forall m . But beyond that the function g gives what we call a “partial” reduction from 3DMto f -CCAV for f ’s that meet the condition for some values of m . In the proof, the size ofthe 3DM instance is artificially enlarged to ensure that this “partial reduction” meets ananalogue counterpart in such a way that for every generator f that satisfies α f,m > α f,mm − for some m , we know that for each large enough m , one of the two reductions can be applied.(Appendix Section B.3 has more on this.) Theorem 4.8.
Let f be an FP -uniform Q -generator. Then there exists an FP -computablefunction g such that • g takes as input an instance I of 3DM and produces an instance I CCAV of f -CCAVwith m = 6 k candidates, where k = k X k = k Y k = k Z k . • If α f,m k +1 > α f,mm − , then: I is a positive instance of 3DM iff I CCAV is a positive instanceof f -CCAV.Proof. We write α i for α f,mi . W.l.o.g., let X = { s , . . . , s k } , Y = { s k +1 , . . . , s k } , and Z = { s k +1 , . . . , s k } . We use the following candidates: • Each s i ∈ { s , . . . , s k } is a candidate. • p is the preferred candidate. • There are dummy candidates d , . . . , d m − k − . We assume there are at least 3 dummycandidates, i.e., k ≥ R of registered voters such that the scores ofthe candidates are as follows. (In the following, we “normalize” the scores of all candidatesusing the score of p as a base. So we pretend that the number o from the application ofLemma 4.7 is zero in order to simplify the presentation, clearly the absolute points of allcandidates must be positive and are shifted by the actual number o from the lemma.) • score ( p ) = 0. For generators where both cases apply such as f = (2 , , , , , , , . . . , , , , score ( s i ) = kα − ( k − α i − α m − r ( i ) , where r ( i ) = 2 , , s i ∈ X, Y, Z , respectively. (So α m − r ( i ) is exactly the amount of points that s i gains froma vote that “vetoes” s i , see below) • score ( d i ) < − kα for i ∈ { , . . . , d m − k − } .Let M ⊆ X × Y × Z be the set from I . For each ( x, y, z ) = ( s h , s i , s j ) ∈ M (soclearly h < i < j ), we add an available voter to U voting as follows: p > s > · · · > s h − > d > s h +1 > · · · > s i − >d > s i +1 > · · · > s j − > d > s j +1 > · · · > s k >d > · · · > d m − k − > x > y > z. We say that such a vote vetoes the candidates x , y , and z , and identify elements of M and the corresponding votes.We show that the reduction is correct. First assume that the instance of 3DM is positive,and let C ⊆ M be the cover with k C k = k . We add the voters corresponding to the elementsof C in the obvious way and show that p indeed wins the resulting election.To see this, it suffices to show that p has at least as many points as each candidate s i ,since by construction, the dummy candidates cannot win the election with adding at most k votes. So let i ∈ { , . . . , k } . The final score for p and s i are as follows: • p gains α points in each of the k additional votes, so p ends up with exactly kα points. • s i gains ( k − α i points from the ( k −
1) votes corresponding to elements ( x, y, z ) ∈ C with s i / ∈ { x, y, z } , and α m − r ( i ) points from the single vote vetoing s i . So s i ends up witha final score of kα − ( k − α i − α m − r ( i ) + ( k − α i + α m − r ( i ) = kα as well.For the converse, let C ⊆ M be a set of at most k votes whose addition lets p win. Ifthis is not a cover, then there is some s i that is vetoed in none of the added votes. We nowcompare the points of p and s i . • p gains α points in each of the k C k additional votes, so p ends up with exactly k C k α points. • Since s i is not vetoed in any new vote, s i gains α i points in each added vote and thusends up with kα − ( k − α i − α m − r ( i ) + k C k α i points.Since k C k ≤ k , α ≥ α i +1 and α i +1 ≥ α k +1 > α m − ≥ α m − r ( i ) , it follows that kα − ( k − α i − α m − r ( i ) + k C k α i − k C k α = ( k − k C k )( α − α i ) | {z } ≥ + α i − α m − r ( i ) | {z } > >
0. So s i beats p if C is not a cover; since byassumption adding C makes p win, C must be a cover.In a similar way, we can prove an analogous result for all scoring rules that “can im-plement” 4-approval in the sense that being voted in one of the first 4 positions is strictlybetter than being voted in most “later” positions. The proof of the following result is verysimilar to the proof of Theorem 4.8, except that an additional argument is needed to ensurethat the favorite candidate cannot be made a winner with less than k additional voters.13 heorem 4.9. Theorem 4.8 also holds when the condition α f,mk +1 > α f,mm − is replaced with α f,m > α f,mm − k +1 . As mentioned above, we now put the two reductions above together to obtain the NP-hardness result of this section, i.e., to prove that f -CCAV is NP-complete as soon as thereis a number m where the coefficients of f satisfy α f,m > α f,mn − . If this condition is true, thenwe know that one of the inequalities α f,m ≥ α f,m ≥ · · · ≥ α f,mm − ≥ α f,mm − is in fact strict.Depending on the position of this strict inequality, we choose which reduction to apply: Ifthe strict inequality appears “close” to the first candidate, then the first “few” positions arestrictly better than “most,” and the system can “simulate” k -approval for some k ≥
4. Onthe other hand, if the strict inequality appears “close” to the last candidate, then the last“few” positions are worse than “most,” and we can similarly “simulate” k -veto for some k ≥ Theorem 4.10. f -CCAV is NP-complete for every FP -uniform pure Q -generator f with α f,m > α f,mm − for some m . We now study pure generators f not covered by Theorem 4.10, i.e., where α f,m ≤ α f,mm − forall m . Then for m ≥ α f,m is of the form ( α f,m , α f,m , α f,m , α f,m , . . . , α f,m , α f,m , α f,m ).The reductions above cannot work in this case, since there are no 3 positions “worse thanmost” and no 4 positions “better than most.”Due to Theorem 3.3, we can regard f equivalently as flexible N -generators or as pure Q -generators. For the latter representation, purity requires that all coefficients from α f,m also appear in α f,m +1 . So the above numbers α , . . . , α do not depend on m . We can usea fixed affine transformation for these finitely many coefficients and, using Proposition 2.1,rewrite all coefficients as natural numbers.Our next hardness result concerns a generalization of 3-approval. Recall fromTheorem 4.3 that CCAV for 3-approval itself, i.e., the generator ( α, α, α, , . . . , Theorem 4.11.
Let α ≥ β ≥ γ > and α = γ . Let f be the generator giving ( α, β, γ, , . . . , . Then f -CCAV is NP-complete.Proof. Let M be the set from an instance of 3DM with k M k = n , and let k = k X k = k Y k = k Z k (recall X , Y , and Z must be pairwise disjoint). We use the candidates X ∪ Y ∪ Z ∪ { p } ∪ { S i , S ′ i | S i ∈ M } and a dummy candidate d to be able to apply Lemma 4.7.We use the lemma to set up the registered votes such that the resulting relative scoresare as follows: score ( p ) = α + 2 γ , score ( c ) = ( n + 2 k ) β + 2 γ for all c ∈ X ∪ Y ∪ Z , score ( S i ) = ( n + 2 k ) β + min( α, γ ), and score ( S ′ i ) = ( n + 2 k ) β + α + γ for each S i ∈ M .Further, score ( d ) < − ( n + 2 k ) α . For each S i = ( x, y, z ), we introduce four unregisteredvoters voting as follows: 14 > p > S i > . . . . y > p > S i > . . . . z > p > S ′ i > . . . . S i > p > S ′ i > . . . .We show that p can be made a winner of the election by adding at most n + 2 k votersif and only if the 3DM-instance is positive, i.e., there is a set C ⊆ M with k C k = k and for S i = S j ∈ C , S i and S j differ in all three components.First assume that there is such a cover. In this case, p can be made a winner of theelection by adding the following voters: For each S i = ( x, y, z ) ∈ C , we add the votes x > p > S i , y > p > S i , and z > p > S ′ i . For each S i = ( x, y, z ) / ∈ I , we add the vote S i > p > S ′ i . Note that this adds exactly 3 k + ( n − k ) = n + 2 k votes. Adding these votesresults in the following scores: • p gains β points in each added vote, so p gains ( n + 2 k ) β points and p ’s final score is α + 2 γ + ( n + 2 k ) β , • each candidate in X ∪ Y ∪ Z gains α points, leading to a final score of α + ( n + 2 k ) β + 2 γ | {z } previous score as well, • for each S i ∈ C , we have that score ( S i ) = ( n + 2 k ) β + min( α, γ ) | {z } previous score +2 γ ≤ ( n + 2 k ) β + α +2 γ , which again is the score of p . • for each S i / ∈ C , we have score ( S i ) = ( n + 2 k ) β + min( α, γ ) + α ≤ ( n + 2 k ) β + 2 γ + α ,equal to the score of p . • for each S ′ i (independent of whether S ′ i ∈ C or S ′ i / ∈ C ), we have score ( S ′ i ) =( n + 2 k ) β + α + γ | {z } previous score + γ = ( n + 2 k ) β + α + 2 γ , again this is the score of p .Thus all candidates tie and so in particular, p is a winner of the election.For the converse, assume that p can be made a winner by adding at most n + 2 k voters.Since each S ′ i initially beats p , at least one vote is added. Thus there is a candidate c ∈ X ∪ Y ∪ Z with score ( c ) ≥ ( n +2 k ) β +2 γ + α , or some S ′ i with score ( S ′ i ) ≥ ( n +2 k ) β + α +2 γ . Inboth cases, we need to add at least n +2 k voters to ensure that p has at least α +2 γ +( n +2 k ) β points as well.Since n + 2 k votes are added, and each of these votes gives points to p and 2 othercandidates, there are 2 n + 4 k positions awarding points in the added votes that are filledwith (not necessarily different) candidates other than p . Each of the 3 k candidates from X ∪ Y ∪ Z can only gain α points without beating p in the election, so each of these can15ll at most one of these 2 n + 4 k positions. So at least 2 n + k positions must be filled by(again, not necessarily different) candidates from { S i , S ′ i | ≤ i ≤ n } . Each S ′ i can appearat most once in the third position without beating p . Since there are n candidates of theform S ′ i , it follows that there must be n + k occurrences of candidates S i in the first threepositions of the added votes. Since no S i can gain α + γ points without beating p , each S i can either appear in a vote S i > p > S ′ i , or in up to two votes of the form c > p > S i with c ∈ X ∪ Y . ( S i = ( x, y, z ) cannot appear in three of these, since then one of x and y wouldgain too many points.) So the only way to fill n + k positions with candidates of the form S i is having 2 k occurrences of S i in the third place, and n − k occurrences of S i in the firstplace. In order to fill all positions, each S ′ i has to appear once in the final position, and dueto the above, n − k of these occurrences are in a vote of the form S i > p > S ′ i . Thus thereare k votes of the form z > p > S ′ i . It follows that there are 3 k votes added that vote acandidate from X ∪ Y ∪ Z in the first position, and n − k voters are added that vote some S i first. Since no S i may appear both in first and in last position, and each S ′ i may appearonly once, and each x i , y i , and z i may gain only α points, it follows that the added votescorrespond to a cover.We also have proved the following cases NP-complete. Theorem 4.12.
The problem f -CCAV is NP-complete if f is one of the following puregenerators:1. f = ( α , α , α , α , . . . , α , α , with α > α > .2. f = ( α , α , . . . , α , with α / ∈ { α , α } , α > .3. f = ( α , α , . . . , α , α , with α > α > α .4. f = ( α , . . . , α , α , with α > α > . We now use the individual results from Sections 4.1 and 4.2 to prove our main dichotomyresult, Theorem 4.2:
Proof.
The polynomial cases follow from Theorems 4.3, 4.4, and 4.5, we prove hardness.If α f,m > α f,mm − for some m , hardness follows from Theorem 4.10. So assume α f,m = · · · = α f,mm − for all m ≥
6. As argued in the discussion after Theorem 4.10, we assume α m,f = ( α , α , α , α , . . . , α , α , α ) for each m ≥
6. Due to Proposition 2.1, we canassume α = 0. We reduce the number of relevant coefficients from 5 to 3: • If α = 0, then, since f does not generate 3-approval and is not equivalent to f , α >α >
0. Hardness follows from Theorem 4.11. To see this, we compute the difference between the score of S i after gaining α + γ points and that of p after gaining ( n +2 k ) β points. This value is ( n +2 k ) β +min( α, γ )+ α + γ − α − γ − ( n +2 k ) β = min( α, γ ) − γ .Since α > γ and γ >
0, this value is strictly positive, so S i indeed beats p if S i gains α + γ points. If α > α >
0, hardness follows from Theorem 4.12.1.So assume α = α = α >
0, i.e., f is of the form ( α , α , . . . , α , α , • If α = α , then since f does not generate 1-veto, we know that α = α = α . Since f is not equivalent to (2 , , . . . , , α = 2 α . Thus NP-hardness followsfrom Theorem 4.12.2. • If α > α , then depending on whether α > α > α or α = α > α , hardnessfollows from Theorem 4.12.3 or Theorem 4.12.4 (note that in the latter case, we knowthat α = 0, since f does not generate 2-veto). Acknowledgments
We thank the AAAI 2014 reviewers for helpful comments and sug-gestions.
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Appendix
The appendix is structured as follows: • In Section A, we provide additional discussion and full proofs for the results on descrip-tional richness and pure scoring rules contained in Sections 2 and 3 of this paper. • Section B contains the proofs that were omitted from Section 4, i.e., the dichotomy result,and also provides additional discussion of “partial reductions.”
A Omitted Proofs and Discussion from Sections 2 and 3
A.1 Proof of Proposition 2.1
Proposition 2.1.
Let m ≥ and let α and α ′ be m -position scoring vectors over T ⊆ Q .Then α and α ′ have the same winner sets on each m -candidate election if and only if α and α ′ both have the same normalized version.Proof. The “if” direction follows from Observation 2.2 of Hemaspaandra and Hemaspaan-dra (2007), as noted in Betzler and Dorn (2010).Let us prove the “only if” direction. This result seems so natural and important thatit feels as if it should be a folk theorem, although we don’t know of it as such; but in anycase, since the result is crucial to this paper, we include a construction that clearly andexplicitly establishes this claim. Let D and D ′ be arbitrary, fixed length- m scoring vectorsover N such that their normalizations, A and A ′ , differ. We will construct an m -candidateelection in which the winner sets under A and A ′ differ.Since by the “if” direction of the present proposition A and D always yield the samewinner set, and also by the “if” direction of the present proposition A ′ and D ′ always yieldthe same winner set, we may conclude that our constructed election has different winnersets under D and D ′ .Let the components of the normalized scoring vector A be ( α , α , . . . , A ′ be ( α ′ , α ′ , . . . , m = 1, A = A ′ is impossible. If m = 2, A = A ′ exactly if one of them is (0 ,
0) andone is (1 , m ≥ , , A and A ′ are both trivial, then A = A ′ , and if exactly one is trivial then building an input separatingthem is easy. Thus we from now on in this proof assume that both A and A ′ are nontrivial.Recall that both A and A ′ are normalized. We will in polynomial time “align” them, i.e.,we will scale them so that the first components of A and A ′ become identical. In particular,to align them we will multiply A by α ′ and we will multiply A ′ by α . So the first coefficientof each will now be α α ′ . Let us rename the thus-scaled vectors (each equivalent to itsoriginal vector as to what winner sets it gives, due to Observation 2.2 of Hemaspaandraand Hemaspaandra (2007)) as B = ( β , β , . . . , β m − ,
0) and B ′ = ( β ′ , β ′ , . . . , β ′ m − , γ be the least i such that β i = β ′ i . W.l.o.g., assume β i > β ′ i . γ = 1 is impossiblesince β = β ′ = α α ′ . So 2 ≤ i ≤ m − B and B ′ have different winner sets.Our candidates will be a , b , and “dummy” candidates d , . . . , d m − . We will ensure thatonly a and b are serious contenders for winning.Let s be a shorthand for “ d > d > · · · > d m − ,” let s be a shorthand for “ d > d > · · · > d m − > d ,” and so on, up to s m − being a shorthand for “ d m − > d > · · · > d m − .”Our vote set is as follows:1. For each i , 1 ≤ i ≤ m −
2, we will have H votes “ a > b > s i ” and we will have H votes“ b > a > s i .” H ’s exact value will be specified later in the proof, but will be chosen tobe so large that a and b are the only serious contenders.2. β γ votes of the form “ a is the top choice and b is the last choice (and the other choices willbe irrelevant to our proofs, but for specificity let us say they are filled in in lexicographicalorder).”3. β votes of the form “ b is the γ th choice and a is the last choice (and the other choices willbe irrelevant to our proofs, but for specificity let us say they are filled in in lexicographicalorder).”That ends our specification of the votes.Let us tally up the points that each candidate gets under B and under B ′ . Clearly,and keeping in mind that β = β ′ , we have: score B ( a ) = ( m − H ( β + β ) + β β γ + 0, score B ′ ( a ) = ( m − H ( β ′ + β ′ ) + β γ β ′ + 0, score B ( b ) = ( m − H ( β + β ) + β β γ + 0, score B ′ ( b ) = ( m − H ( β ′ + β ′ ) + β β ′ γ + 0, score B ( d i ) ≤ ( m − β ) + β γ β + ( β ) ,and score B ′ ( d i ) ≤ ( m − β ′ ) + β γ β ′ + β β ′ . We now set the value of H , namely tobe H = 2 β . Keeping in mind that β = β ′ ≥
1, it is easy to see that this choice of H ensures that for each i , 1 ≤ i ≤ m −
2, min( score B ( a ) , score B ( b )) > score B ( d i ) andmin( score B ′ ( a ) , score B ′ ( b )) > score B ′ ( d i ). So we have ensured that, under B and under B ′ , a and b have more points than any d i . 19nder B ′ , note that a is the one and only winner (recall β = β ′ and β γ > β ′ γ ). But ourvotes ensure that under B , a and b tie as the (sole) winners.So the votes we gave show that B and B ′ (and thus D and D ′ ) have different winnersets on some example, namely, the above example. A.2 Proof of Theorem 3.1
Theorem 3.1. If T ( m ) is a fully time-constructible function and lim sup n →∞ T ( m ) log T ( m ) T ( m ) = 0 ,then there is an election rule in FDTIME[ T ( m )] -uniform- { , } - PSR that is not in
FDTIME[ T ( m )] -uniform- Q - GSR .Proof.
Rather than proving Theorem 3.1, we will prove a slightly weaker result,“Theorem X,” and then will explain how to modify that proof to establish Theorem 3.1.
Theorem X
There is an integer constant k > T ( m ) is a fullytime-constructible function, and lim sup n →∞ T ( m ) log T ( m ) T ( m ) = 0, and ( ∀ m )[ T ( m ) ≥ k ( m + 1)], then there is an election rule in FDTIME[ T ( m )]-uniform- { , } -PSR that is not in FDTIME[ T ( m )]-uniform- Z -GSR .We will prove Theorem X by describing how to appropriately adjust the classic presenta-tion by Hopcroft and Ullman (1979), henceforward “HU,” of the proof of the deterministictime hierarchy theorem (their Theorem 12.9); that proof of HU can be found as pages297–298 of that book, and that proof itself draws also on the framework of that book’searlier proof of the deterministic space hierarchy theorem. We will assume that the readeris familiar with those deep, classic proof presentations; anyone not expert in complexity willwant to either first master that proof framework or just skip the present proof.So, suppose we are given T and T satisfying the conditions of Theorem X. We’ll laterspecify k , but k must not and will not depend on T or T .Our goal is to show that there is an FDTIME[ T ( m )]-uniform- { , } -PSR generator, callit f , whose rule’s winner set is not obtained by any FDTIME[ T ( m )]-uniform- Z -GSR.Generally, we just follow the overall architecture of the HU proof, but the differencesare as follows. First, we will be computing and diagonalizing against functions. So ourenumeration of machines will be an enumeration of all function-computing Turing machines(i.e., machines with an output tape such that when they halt, whatever is on the outputtape is viewed as being the output). Like HU, we will assume that our enumeration ofsuch machines has the property that if w encodes a machine, then 1 k w encodes exactly thesame machine; “padding” by adding 1 ∗ as a prefix to a machine’s coding does not changethe machine encoded. (Unlike HU, who truly need this due to their making a liminf claim,we merely need each machine to appear infinitely often in the enumeration, since for otherreasons we are simply making a limsup claim. However, this padding property certainly isa fine way of achieving what we need.)Another difference is that we need to be a pure scoring rule. So each length rule mustappropriately link to and extend the rule from the previous length. As mentioned in the20ext immediately after Theorem 3.1, doing so in the obvious fashion would seem to adda multiplicative factor of m , but instead we use the trick mentioned there to avoid this.That is, for each odd natural number m , f (0 m ) will be 1 ⌊ m/ ⌋ ⌊ m/ ⌋ +1 (e.g., (0), (1 , , , , , , m we might not even have time to realizethat m was odd and write the appropriate output; that worry is because one might worrythat for some small m we may not have time greater than m + 1 and that is not enoughtime to both see what m is, and that it is odd, and to already just before halfway throughit have known that we are just before halfway through it and to have switched what we areoutputting from 1s to 0s. However, the ( ∀ m )[ T ( m ) ≥ k ( m + 1)] assumption of Theorem Xwill ensures us that we do have that time.)Now, at each even length m , we’ll try to do a diagonalization, if we have time. Ourframework is that we will always have as our vector at this length either 1 ⌊ m/ ⌋ +1 ⌊ m/ ⌋ +1 or 1 ⌊ m/ ⌋ ⌊ m/ ⌋ +2 (e.g., at length 4, either 1100 or 1000; note that in this proof we willquietly go back and forth notationally between, for example, (1 , , ,
0) and 1100 and 1 ).Basically, we want to ensure that if whatever FDTIME[ T ( m )]-uniform- Z -GSR generatorwe (“we” are the function f ) at stage m are trying to diagonalize against outputs a vectorwhose winner set is the same as that given by 1 ⌊ m/ ⌋ +1 ⌊ m/ ⌋ +1 then we will output thevector 1 ⌊ m/ ⌋ ⌊ m/ ⌋ +2 and otherwise we will output the vector 1 ⌊ m/ ⌋ +1 ⌊ m/ ⌋ +1 . To do this,we immediately write onto our output tape the vector 1 ⌊ m/ ⌋ +1 ⌊ m/ ⌋ +1 . We do so so thatif we run out of time in our diagonalization on this input, we at least have one of the twolegal vectors that keep our rule pure within our framework. And we do have time to writethis vector, thanks to the ( ∀ m )[ T ( m ) ≥ k ( m + 1)] assumption of Theorem X.Now, as to the diagonalization, it goes as follows. If m ≥ m in binary and the trailing 0 of m in binary, and call the string that remains w . (We strip the trailing 0 because we diagonalize only at even m , but we want the entireset of such strings created to exactly equal { , } ∗ .) We will view w as the encoding of aTuring machine from our enumeration of function-computing TMs. We’ll run it (unless werun out of time: as per the entire HU framework, we’ll be using a separate tape and thefully time-constructible nature of T to enforce a time cutoff; actually, we’ll use two separatetapes, since our time cutoff is max( T ( m ) , k ( m + 1))) on input 0 m to get the length- m vectorit outputs, call it v (if it halts and has the wrong length output, then it clearly isn’t evena valid GSR). We must then efficiently evaluate whether that vector has the same winnerset as would 1 ⌊ m/ ⌋ +1 ⌊ m/ ⌋ +1 . One might think to do so we would have to normalize v ,which includes gcd’s and other time-eating computations, but that is not so. To tell if v has the same winner set as 1 ⌊ m/ ⌋ +1 ⌊ m/ ⌋ +1 , we need only test whether v is of the form a ⌊ m/ ⌋ +1 b ⌊ m/ ⌋ +1 , where a and b are members of Z and a > b . If so they have the samewinner set, and if not they don’t. Ignoring for the moment the cost of getting v , this testis an easy linear-time test on a multihead, multitape TM. If we find that v does have thesame winner set as 1 ⌊ m/ ⌋ +1 ⌊ m/ ⌋ +1 , then we on our output tape just overwrite the ⌊ m/ ⌋ thcharacter, changing it from a 1 to a 0 (for example, if m = 4, we’d overwrite the second bitto change our placeholder 1100 into 1000). So, if we have time for our simulation of w to21omplete and to evaluate whether the v has the same winner set as 1 ⌊ m/ ⌋ +1 ⌊ m/ ⌋ +1 , andif it is, have time to overwrite the one bit, then we have successfully diagonalized againstthe machine w .On the other hand, what if we run out of time? No problem. If the machine associatedwith w happens to run in time T ( m ) for all m (and we’re doing all w , so some will havethat time behavior and some won’t), then—since w will appear infinitely often again in ourconstruction, as it will also appear as 1 w , as 11 w , and so on—due to the limsup assumptionwe eventually will have the time to fully run the diagonalization (the simulation and ouron-the-cheap comparison of winner sets and our bit-fixing). The reason is that the logmultiplicative overhead in the limsup is (see HU) enough to do the machine simulation,and the limsup ensures that for any constant c our T for every m beyond some point willsatisfy T ( m ) > cT ( m ) log T ( m ). Let us discuss the constant k . k is simply there because for small values of m , we maynot have enough time to see the input, and write the appropriate starting string to ouroutput tape. (All we know is we have at least m + 1 steps. That isn’t enough, as out inputis of the form 0 m so until we hit the right-end marker, all we see is 0s, but the starting stringwe need to put on the output tape as a placeholder is roughly half 1’s and half 0’s, so if wejust get m + 1 steps, we don’t know roughly halfway through the 0’s that we are roughlyhalfway through and that we have to switch from outputting 1’s to outputting 0’s. Thefact that asymptotically we have at least T ( m ) log T ( m ) time, and thus at least m log m time, doesn’t help us for small values of m , since if we run out of placeholder time at evenone value of m we’ve violated purity of our alleged PSR.) However, we can set a value of k that allows us to do the placeholder writing (and on any natural TM framework, it will bea very small constant k ; go to the end of the string 0 m , see if it is even, and then output theright placeholder string, for example by having on an extra dummy tape basically build astring of just under m/ k depends just on the TM framework,not on T or T .All that remains is how to move from our proof of Theorem X to a proof of Theorem 3.1. Readers familiar with the time hierarchy theorem may wonder why we here use a limsup but the HUtheorem uses a liminf. The reason is subtle. Briefly put, in HU, once a machine w is being diagonalizedagainst at some length, it will be (in its sibling forms 1 w , 11 w , etc.) diagonalized against at every greaterlength. So a liminf suffices, as liminf ensures that at some (indeed, infinitely many, though they don’t reallyneed that) greater lengths we get enough time overhead to diagonalize, and that is all HU needs. (Warning:The HU proof itself sets up the right machinery and construction for liminf to suffice, but then inside itsproofs, seems to forget this and by such phrases as “infinitely often” on its p. 298 line 8 and “has arbitrarilylong encodings” on page 298 and 299, say things that would require limsup to support. However, the HUconstruction supports liminf, and one can change the two mentioned phrases to “almost everywhere” and“at every length onward once it first occurs” and the thus adjusted proof becomes correct.) In contrast,PSRs take as input 0 m , and so at each length are diagonalizing against just one Turing machine. And sowe do know that once we face off against w (say, on input 0 (1 w ) binary ), we’ll infinitely often see its identicalsiblings at longer lengths (such as on input 0 (11 w ) binary , but note that that is about twice as long, not onelonger), but we don’t know we’ll have them at every length. Basically, since our input is 0 m , we get toattempt just one diagonalization at each length. Happily, under a limsup assumption, we’re safe and fine,since at every sufficiently long length, we have the headroom to diagonalize. ∀ m )[ T ( m ) ≥ k ( m + 1)]. Our salvation is the limsup claim, which ensures that even-tually, T becomes very big. So all we need to do is find a way to mark time until thatkicks in, and by marking time, we have to make do with m + 1 steps as that is all we knowwe always have. We can’t such marking time within the framework of the particular PSRwe’ve been using above, because as noted above, one would seem to have to be able toknow the midpoint of 0 m in real time as one was passing it, and that is impossible. Theworkaround is as follows. We change the PSR that we’ll use. In particular, for each T and T , there exists some odd value m ′ beyond which the limsup ensures that we do haveeasily enough time headroom. So what we’ll do is our PSR will simply be 0 m at each length m ≤ m ′ (note that we can do this in time m + 1, at length m ), and from then on, it willmimic our trick except with those extra 0’s, namely, we will fixing our vector at each oddlength, m > m ′ , to be 1 ⌊ ( m − m ′ ) / ⌋ ⌊ ( m − m ′ ) / ⌋ +1+ m ′ , and at even lengths m > m ′ , our vectorwill be either 1 ⌊ ( m − m ′ ) / ⌋ +1 ⌊ ( m − m ′ ) / ⌋ +1+ m ′ or 1 ⌊ ( m − m ′ ) / ⌋ ⌊ ( m − m ′ ) / ⌋ +2+ m ′ . This allowsus to remove the requirement that ( ∀ m )[ T ( m ) ≥ k ( m + 1)], and so we have establishedTheorem 3.1. Note that m ′ will depend on T and T , but that is perfectly legal; it doesnot undermine the proof.We finish with some brief, technical comments about possible extensions and alternateproofs. First, our diagonalization is carried out the world that it is in, namely, the world offunctions. We mention in passing that since the PSR we build is variable only as to whichof two possible vectors it has at each even length, the information behind its choice is ineffect a set—indeed, a set of the form A ⊆ (11) ∗ , and so a particular type of tally set. Onecould use this to make the proof more set-focused. But one would still have to accountfor overhead time and so on, and we feel it is more natural to simply have the proof bein the world the items in question are inhabiting. Second, we mention that it is possiblethat one could turn Theorem 3.1’s limsup into a liminf, thus making the theorem stronger.However, for most natural time functions T and T , the liminf and the limsup are equal,and so we do not consider this an important direction, especially as it would likely makethe proof far more complex if attempted. We do mention that the natural path to use toattempt such an improvement would be, once one first was trying to diagonalize against agiven machine, to attempt diagonalizing against that machine at every length from thenon until one had successfully diagonalized against it. But we warn that to even know whatone was currently trying to diagonalize against would require recomputing one’s history,which itself takes time and can interfere with the proof. Worse, our current setup doesn’tdo any diagonalizations at even lengths, but the liminf could go to zero just due to evenlengths, and that would not at all help us ever have time to diagonalize; so our entire zigzagframework becomes poisonous, yet abandoning it loses its advantage (already lost anywayin the type of new construction we are speculating about) of avoiding history rebuilding.23 .3 Proof of Theorem 3.2 In the following theorem, nonuniform simply means that the corresponding generators arenot required to be computable in any specific complexity class. In fact, the generators maybe uncomputable.
Theorem 3.2. Q ≥ - PSR nonuniform- Z - PSR , Z - PSR nonuniform- Q ≥ - PSR , and Q - PSR nonuniform- Q ≥ - PSR ∪ nonuniform- Z - PSR .Proof.
Let us first show that Q ≥ -PSR nonuniform- Z -PSR . Consider the followingFP-uniform Q ≥ -PSR generator f . f (0 ) = (1), f (0 ) = ( , f (0 ) = ( , , ), f (0 ) = ( , , , ), f (0 ) = ( , , , , ), f (0 ) = ( , , , , , ), f (0 ) = ( , , , , , , ),etc.Consider a nonuniform Z -PSR type generator g that claims to have the same winnerset as this on all instances. By the (clear) extension of Proposition 2.1 mentioned at theend of the paragraph that follows that result, the 5-candidate vector of g must normalize to(6 , , , , Z ) of g satisfies 2( α − α ) = ( α − α ) = ( α − α ) = 2( α − α ). Similarly, the 7-candidate vector of g must normalize to a vector having gap pattern: 1,2,4,4,2,1. Since thatvector retains all 5 coefficients from the 5-candidate vector, with two additional coefficientsappropriately added, by inspection of the possibilities, it is clear that the only possibleextension that achieves this is one that adds exactly the coefficients α + ( α − α ) and α − ( α − α ) (and due to the 6-candidate vector, they must be added in the order juststated). That is, the vector extends to the right and left, but an amount half the “base”gap. However, note that if α − α is not a multiple of 2, this extension is already not legalover Z , as it would have coefficients not in Z . If α − α is a multiple of 2, then we indeeddo have this valid extension of the 5-candidate vector. But by the same argument, the onlypossible 9-candidate vector again will have to be formed from the 7-candidate vector byextending to the right and left ends by half the then-current “base” gap. Note that thatwill not be possible unless α − α is a multiple of 2 = 4. And so on. Since there is somepower of 2 that is the maximum power of two that divides α − α , this process will eventualend in failure, i.e., there will be no valid vector over Z that properly extends the pattern g has trapped itself into. Thus, contrary to the claim made for g , g in fact cannot match thewinner sets of f .Let us now show that Z -PSR nonuniform- Q ′ -PSR . We will give the argumentfor this case more compactly, since the reader by now will be familiar with the flavor ofarguments of this sort, from the above case. Consider the following FP-uniform Z -PSRgenerator f . 24 (0 ) = (0), f (0 ) = (1 , f (0 ) = (1 , , − f (0 ) = (2 , , , − f (0 ) = (2 , , , − , − f (0 ) = (4 , , , , − , − f (0 ) = (4 , , , , − , − , − Q ≥ -PSR type generator g that claims to have the same winnerset as this on all instances. Note that the gap pattern, at each length starting at 5, is thatwe have four identical, adjacent smallest-size (among the gaps that exist) gaps in the center,and then surrounding that the gaps grow by repeatedly doubling (with the extra one beingon the “upper” side on even-length cases). Note that no single internal insertion anywheremaintains a four, identical, adjacent, smallest-size (among the gaps that exist) gap pattern.So we can only expand by going doubly-far to the right and left in each pair of extensions.But that means the values α m +12 m +1 will be decreasing at an exponentially increasing rate.And so since α was finite, at some m the value α m +12 m +1 will necessarily have to be less thanzero, and so will not be an element of Q ≥ .We’ll sketch Q -PSR nonuniform- Q ≥ -PSR ∪ nonuniform- Z -PSR even morebriefly. It does not automatically follow from the earlier parts; claiming that would be likeclaiming that since not all integers are even and not all integers are odd, it follows that notall integers are odd integers or even integers. Nonetheless, by building a construction thatuses both of the weaknesses exploited by the constructions of the preceding two parts, wecan easily establish that Q -PSR nonuniform- Q ≥ -PSR ∪ nonuniform- Z -PSR . Inparticular, it is easy to prove that the following FP-uniform Q -PSR has the desired property.After setting down 4 equally spaced gaps, it in term makes re-doublingly large gaps goingin the negative direction and re-halvingly large gaps going in the positive direction. f (0 ) = (2), f (0 ) = (2 , f (0 ) = (2 , , f (0 ) = (2 , , , − f (0 ) = (2 , , , − , − f (0 ) = (2 , , , − , − , − f (0 ) = (2 , , , , − , − , − f (0 ) = (2 , , , , − , − , − , − f (0 ) = (2 , , , , , − , − , − , − f (0 ) = (2 , , , , , − , − , − , − , − f (0 ) = (2 , , , , , , − , − , − , − Q ≥ -typegenerator nor a Z -type generator. 25 .4 Proof of Theorem 3.3 Theorem 3.3. N - FPSR = Q - FPSR = Q ≥ - FPSR = Z - FPSR = Q - PSR .Proof.
Theorem 3.3 is immediately clear, in light of Proposition 2.1 and the end of theparagraph following it, since our normalizations shifted each of these types to equivalentvectors over N , which is the most restrictive of all these types.The equality with Q -PSR can easily be seen by inductively constructing, for everyflexible generator f , an equivalent pure generator over the rationals. A.5 Proof of Theorem 3.4
Theorem 3.4.
There is an FP -uniform PSR that is not generated even by any nonuniform
PSR norm-0 generator. On the other hand, every FP -uniform (respectively, nonuniform) PSR is generated by a FP -uniform (respectively, nonuniform) PSR generator that for all but afinite number of m has the property that the last coefficient, α m,m , is zero and the gcd ofthe nonzero coefficients in the length- m vector is one.Proof. Let us prove the first part of the theorem. Consider this FP-uniform N -PSR, whichwe mention in passing is even an FP-uniform N -PSR norm-gcd . f (0 ) = (5), f (0 ) = (6 , f (0 ) = (7 , , f (0 ) = (8 , , , f (0 ) = (8 , , , , f (0 ) = (8 , , , , , f (0 ) = (8 , , , , , , N -PSR norm-0 type generator g ′ that claims to have the same winner set asthis on all instances. So by Proposition 2.1, it is clear that the 4-candidate vector of g ′ mustbe of the form (3 k, k, k, k ∈ N − { } .But clearly addition of one coefficient to this vector cannot yield anything that normal-izes to (8 , , , , g ′ differs from the original rule on some5-candidate examples, by Proposition 2.1.(We mention in passing that the exact same f we just used can also be easily argued tobe accepted by no PSR generator that has the property that at each length, the gcd of itsnonzero coefficients (if any) is one.)Let us turn to the second claim of the theorem, namely, that every N -PSR is generatedby an N -PSR generator that on all but at most a finite number of lengths m has the propertythat the last coefficient is zero and the gcd of the nonzero coefficients in the length- m vectoris one.Given E , an FP-uniform N -PSR, fix an FP generator for that rule. Let Pool = { α mj | m ≥ ∧ ≤ j ≤ m } , i.e., it is the set of all coefficients.If k Pool k = 1 we are done easily using the scoring vector sequence (0), (0 , , , k Pool k ≥
2. Let min(
Pool ) = k . Now, alter the generatorfrom now on (and thus each α ij , and so indirectly we also have altered Pool ) to subtract k α ij . So it is clear that, after the alteration, at all but a finite number ofvalues of m we have α mm = 0.Let j = gcd( Pool ), for the (altered)
Pool .Let us keep in mind that with PSRs, coefficients that occur at a length must also occur ateach greater length. If j = 1, then there is some finite length at which the set of coefficientsat that length has a gcd of one, and so at all greater lengths the coefficient set also has agcd of one. If j ≥
2, then further re-alter our generator to, after subtracting k , divide by j .Applying the j = 1 argument to this, we again have that the gcd will be one at all but afinite number of lengths, and as dividing by j does not change the set of lengths at which α mm = 0, we also have that for our re-altered generator α mm is zero for all but a finite numberof m ’s. The re-altered generator is clearly polynomial-time computable if the original oneis, so we have completed the proof of the theorem’s second part for the FP-uniform case,and the result clearly also holds for the nonuniform case, by the same argument with thediscussion of runtimes removed. B Omitted Proofs and Discussion from Section 4
B.1 Missing Pieces of the Proof of Theorem 4.4
Here we present the proofs of the two facts used in the proof of Theorem 4.4. We start withFact 1:
Fact 1.
For i ∈ { , } , given a set S ⊆ V ∪ V , it can be checked in polynomial time whether S can be extended, by adding at most k − k S k voters from V i to make p win.Proof. For each candidate c ∈ C , let s c be the score of that candidate after the voters in V and the voters in R have been counted. For the remainder of the proof of Fact 1, weassume that i = 1, i.e., we are adding voters with p in the first position. Since the proof ofFact 1 does not use the fact that α ≥ β , the proof works in the same way for i = 2, i.e.,adding voters voting p in the second position.We can add at most r = k − k S k voters from V . Clearly, k is bound by the size of theinstance. Thus we can brute-force over every j ∈ { , . . . , r } and check whether S can beextended with exactly j voters from V . To do this for a given j , we proceed as follows: • First, we compute the number of points that p will have after adding j voters from V ,this is s p + jα . We denote this number with s finalp . • For each candidate c ∈ C , we compute the maximal number m c of votes giving points to c that can be added. This is the maximal number m c with s c + m c β ≤ s finalp . If m c isnegative for some c , then S cannot be extended with exactly j voters from V . • For each candidate c ∈ C , we add voters voting c in the second place while still possible,i.e., while all of the following hold: – the number of added voters up to now does not exceed j ,27 the number of added voters voting c second does not exceed m c , – there are still voters available voting c second. • If the resulting set is a solution to the control problem, accept. Otherwise, S cannot beextended with exactly j voters from V .The proof for Fact 2 follows: Fact 2.
To make p win, it is never better to add ℓ voters from V than adding ℓ pairwisedifferent voters from V .Proof. Let S ( S ) be the corresponding subset of V ( V ). Let c , . . . , c ℓ be the candidatesvoted in the second place by the voters in S . We compare the effect that adding S or S has on the relationship between p and the c i (clearly, for the relationship between p andcandidates c / ∈ { c , . . . , c ℓ } , adding S is always better). • When adding S , p gains ( ℓ − α + ( α − β ) = ℓα − β points against each c i . • When adding S , p gains at most ℓβ points against each of the c i .Since ℓ ( α − β ) ≥ β , we have that ℓα − β ≥ ℓβ , and so adding S is at least as good asadding S . B.2 Proof of Lemma 4.7
The proof Lemma 4.7 makes extensive use of the following construction:
Lemma B.1.
Given a scoring vector ( α , . . . , α m ) , and numbers i = j , k = l ∈ { , . . . , m } ,there is a set V of m votes such that when counting the votes in V , the scores of all candidatesis A := P nm =1 α m , except: • candidate c i has A + α k − α l points, • candidate c j has A + α l − α k points.Proof. Without loss of generality, assume i ≤ j and k ≤ l , otherwise we swap.Let v be a vote having c i in position l and c j in position k , the remaining candidatesare ordered arbitrarily. For any ℓ ≥
1, let v ℓ +1 be the vote obtained from v ℓ by moving eachcandidate one position to the right, i.e., the candidate at position t in the vote v ℓ ends atposition t + 1 in the vote v ℓ +1 if t + 1 ≤ m , and at position 1 if t + 1 = m . Now let V contain the votes v , . . . , v m . Clearly, each candidate gains exactly A points in the votes V .Now, let V be obtained from V by exchanging, in the vote v , the positions of candidate c i and c j . Then, relatively to the votes V , the points of c i and c j change as follows: • c i loses α l points and gains α k points, • c j loses α k points and gains α l points. 28o the relative score is as required.The actual proof of Lemma 4.7 follows: Lemma 4.7.
Given a scoring vector ( α , . . . , α m ) , and for each c ∈ { , . . . , m − } , numbers a c , . . . , a cm in signed unary, and a number k in unary, we can compute, in polynomialtime, votes such that the scores of the candidates when evaluating these votes accordingto the scoring vector ( α , . . . , α m ) are as follows: There is some o such that for each c ∈{ , . . . , m − } , score ( c ) = o + P mi =1 a ci α i , and score ( c ) > score ( m ) + kα .Proof. The algorithm produces the set V by consecutive applications of Lemma B.1. Wesay that an application of the Lemma transfers ( α k − α l ) points from c j to c i , since therelative score (in relation to all candidates except c i and c j ) of c i increases by α k − α l , whilethe score of c j decreases by the same amount. Note that in order for such a “transfer” to bepossible, neither candidate is required to actually have a nonzero score. We start with theempty set V , and then, for each a ci = 0 for some c ∈ { , . . . , m − } , we proceed as follows: • if a ci >
0, we apply Lemma B.1 a ci times to transfer α i points from the candidate m tothe candidate c , • if a ci <
0, we apply Lemma B.1 ( m − a ci times to transfer α i points from the candidate m to each candidate c ′ = { c, m } .In the above steps, clearly the score of m is the lowest among all candidates. So to ensurethat score ( i ) > score ( m ) + kα , it suffices to transfer ( k + 1) α points from the candidate m to each candidate c ∈ { , . . . , m − } . Clearly, the resulting scores are as required. B.3 Proof of Theorem 4.9 and Discussion of “Partial Reductions”
In the discussion surrounding Theorems 4.8 and 4.10, we described Theorem 4.8 and its“dual”—which we now state formallyas Theorem B.2—as “partial reductions.” To clar-ify this unusual terminology, we now explain why we use this term. For a generator f that is covered in Theorem 4.10, it is not necessarily the case that hardness follows fromTheorem 4.8 alone (or from Theorem B.2 alone). Rather, for each value of m result-ing from a 3DM-instance, the reduction establishing NP-hardness (which is given in theproof of Theorem 4.10) needs to check which of the two cases (as to meeting the differing“ α ... < α ... ” conditions listed in Theorem 4.8 and Theorem B.2) is satisfied. Depending onwhich condition applies, either the reduction g from Theorem 4.8 or the reduction g fromTheorem B.2 is used for this particular 3DM-instance. In this sense, Theorem 4.8 (and alsoits “dual,” Theorem B.2) are “partial” reductions: For a given f , each of these theorems mayonly be “half” of the reduction needed to prove the NP-hardness of f -CCAV. Theorems 4.8and B.2 together, with a simple additional observation, then give Theorem 4.10, which isthe “complete” reduction to obtain NP-completeness of f -CCAV.Let us turn from the above discussion to the formal statement of Theorem 4.9 and itsproof: We recall the statement of the Theorem:29 heorem 4.9. Theorem 4.8 also holds when the condition α f,mk +1 > α f,mm − is replaced with α f,m > α f,mm − k +1 . The complete statement of Theorem 4.9 is as follows. Note that again, we do not requirethe generator f to respect any purity conditions. Theorem B.2.
Let f be an FP -uniform Q -generator. Then there exists an FP -computablefunction g such that • g takes as input an instance I of 3DM where k = k X k = k Y k = k Z k , • g produces an instance I CCAV of f -CCAV, • If for m = 6 k we have that α f,m > α f,mm − k +1 , then: I is a positive instance of 3DM ifand only if I CCAV is a positive instance of f -CCAV.Proof. The proof is very similar to the proof of Theorem 4.8 above, in particular we usethe same set of candidates. Again, let M ⊆ X × Y × Z be the set from I , and let X = { s , . . . , s k } , Y = { s k +1 , . . . , s k } , Z = { s k +1 , . . . , s k } . Similarly to the proof oftheorem 4.8, we use Lemma 4.7 to construct an election with relative points as follows: • score ( p ) = 0, • for each i ∈ { , . . . , k } , score ( s i ) = kα − ( k − α m − k + i − α r ( i ) , • all dummy candidates have points such that adding at most k votes does not let themwin the election.For each ( x, y, z ) = ( s h , s i , s j ) ∈ M with h < i < j , we add an available voter vot-ing as follows: p > x > y > z > d > · · · > d m − k − > s > · · · > s h − > d >s h +1 > · · · > s i − > d > s i +1 > · · · > s j − > d > s j +1 > . . . , s k .We say that x , y , and z are approved in this vote. Note that a candidate s i gets α r ( i ) (where r ( i ) = 1, 2 or 3 depending whether s i ∈ X , Y or Z ) points in a vote approving s i ,and gets α m − k + i in a vote not approving s i .We claim that p can be made a winner by adding at most k of the available votes if andonly if the 3DM instance is positive.First assume that the 3DM instance is positive, and let C ⊆ M be a cover with k C k = k .We add the votes corresponding to the elements of C , i.e., for each candidate s i , we addone vote that approves s i and ( k −
1) votes that do not approve s i . The final score of thenondummy candidates is as follows: • Candidate p gains α points in each of the k added votes, so the final score of p is kα , • each candidate s i gains α r ( i ) + ( k − α m − k + i points from the one vote approving s i and the k − s i . Thus the final score of s i is kα − ( k − α m − k + i − α r ( i ) + α r ( i ) + ( k − α m − k + i = kα .30hus p and s i tie, and so p is a winner of the election.For the converse, assume that p can be made a winner by adding at most k voters. Let C be the elements of M corresponding to the added votes, then k C k ≤ k .Since α ≥ α > α m − k +1 , we know that the score of s before adding any votes ispositive. Thus C = ∅ , i.e., at least one vote is added. Let z be a candidate from Z who isapproved in at least one of the added votes. We show that k C k = k . For this, indirectlyassume that k C k < k . We show that z strictly beats p in this case: Consider the finalscores of p and z . Let i be the index of z , i.e., the i with s i = z . Note that r ( i ) = 3, since s i = z ∈ Z . • candidate p gains k C k α points, • candidate z gains at least α + ( k C k − α m − k + i points (even more if z is approved inmore than one of the additional votes). Thus the final score of z is at least kα − ( k − α m − k + i − α + α + ( k C k − α m − k + i = kα + ( k C k − k ) α m − k + i To see that z strictly beats p , we compute the difference between their scores, which is kα + ( k C k − k ) α m − k + i | {z } score ( s i ) − k C k α | {z } score ( p ) = ( k − k C k ) α − ( k − k C k ) α m − k + i = ( k − k C k )( α − α m − k + i ) . Since, by assumption, k − k C k >
0, and α ≥ α > α m − k +1 ≥ α m − k + i , this differenceis positive and so s i strictly beats p if k C k < k . Thus we know that k C k = k .We now show that C is indeed a cover. Assume that this is not the case, so, since k C k = k , without loss of generality, there is some z ′ ∈ Z which is covered twice, i.e., whichis approved in at least two of the added votes. We show that this z ′ strictly beats p in theelection after the addition of votes by comparing the score of p and z ′ . As above, let i bethe index of z ′ , i.e., chose i such that z ′ = s i . The scores are as follows (again, r ( i ) = 3since s i ∈ Z ): • The candidate p again gets kα points, and so its final score is kα . • The candidate z ′ is approved in at least 2 of the k additional votes. Thus z ′ gains at least2 α + ( k − α m − k + i points. The final score of z ′ is thus at least kα − ( k − α m − k + i − α r ( i ) | {z } = α +2 α + ( k − α m − k + i = kα − α m − k + i + α .Since α > α m − k +1 ≥ α m − k + i , it follows that the final score of z ′ exceeds the score of p , and so z ′ indeed strictly beats p as claimed if C is not a cover. Thus C is indeed a coveras required. 31 .4 Proof of Theorem 4.10 Theorem 4.10. f -CCAV is NP-complete for every FP -uniform pure Q -generator f with α f,m > α f,mm − for some m .Proof. Clearly, for pure scoring rules, if the condition α f,m > α f,mm − is true for some m , thenit remains true for all m ′ ≥ m , since it is easy to see that α f,m +14 ≥ α f,m and α f,m +1 m +1 − ≤ α f,mm − .We prove NP-hardness by a reduction from 3DM. So let I be a 3DM-instance, andlet k be the cardinality of the set X in I . Without loss of generality, we can assumethat k is large enough such that the number m = 6 k satisfies the condition α f,m > α f,mm − .Due to the monotonicity of the coefficients and since m − k + 1 = 3 k + 1, we have that α f,m ≥ α f,mm − k +1 = α f,m k +1 ≥ α f,mm − , and since α f,m > α f,mm − , we know that one of the cases α f,m k +1 > α f,mm − , or α f,m > α f,mm − k +1 occurs. In the first case, we use the reduction from Theorem 4.8, in the second case, theone from Theorem B.2. B.5 Proof of Theorem 4.12.1
Theorem 4.12.
The problem f -CCAV is NP-complete if f is one of the following puregenerators:1. f = ( α , α , α , α , . . . , α , α , with α > α > .2. f = ( α , α , . . . , α , with α / ∈ { α , α } , α > .3. f = ( α , α , . . . , α , α , with α > α > α .4. f = ( α , . . . , α , α , with α > α > .Proof. Again, we reduce from 3DM. Let n = k M k , and let k = k X k = k Y k = k Z k . Weintroduce the following candidates: • The preferred candidate p , • for each c ∈ X ∪ Y ∪ Z a candidate c , • two dummy candidates d and d .Using Lemma 4.7, we construct the registered voters such that the relative score of thecandidates before adding voters is as follows: • score ( p ) = − kα , • score ( d ) , score ( d ) < − k ( α + α ), 32 score ( x ) = − α − ( k − α for each x ∈ X , • score ( y ) = − α − ( k − α for each y ∈ Y , • score ( z ) = − ( k − α for each z ∈ Z .For each ( x, y, z ) ∈ M , we add one available voter voting x |{z} α > y |{z} α > d |{z} α > p > . . . | {z } α > d |{z} α > z |{z} . We claim that p can be made a winner by adding at most k voters if and only if thereis a cover C ⊆ M with k I k = k .First, assume that there is such a cover C . Then, adding the voters corresponding tothe elements of C changes the scores as follows: • p gains kα points and thus ends up with 0 points, • each x ( y ) gains α + ( k − α points ( α + ( k − α points), which leads to 0 points, • each z gains ( k − α points, also finishing with 0 points, • by construction, d and d cannot win the election.For the other direction, assume that C is a set of added voters that makes p win theelection with k C k ≤ k . Since α >
0, we know that p must gain one point against everycandidate z ∈ Z , and can gain against one of these candidates for each added vote, at least k Z k = k votes must be added, so k C k = k . We prove that C is a cover. By the above, C covers Z . Indirectly assume that there is some x ( y ) that is voted in the first (second) placein more than one of the voters from C . In this case, x ( y ) gains at least 2 α + ( k − α (2 α + ( k − α ) points, and so ends up with − α − ( k − α + 2 α + ( k − α = α − α points ( α − α points). Since α ≥ α > α , p does not win the election in this case, acontradiction. This completes the proof. B.6 Proof of Theorem 4.12.2
Theorem 4.12.
The problem f -CCAV is NP-complete if f is one of the following puregenerators:1. f = ( α , α , α , α , . . . , α , α , with α > α > .2. f = ( α , α , . . . , α , with α / ∈ { α , α } , α > .3. f = ( α , α , . . . , α , α , with α > α > α .4. f = ( α , . . . , α , α , with α > α > . α > α or α < α applies. We start with the case α > α : Theorem B.3. f -CCAV is NP-complete for the generator f = ( α , α , . . . , α , with α > α > .Proof. We again reduce from 3DM. So let k = k X k = k Y k = k Z k . We introduce thefollowing candidates: • the preferred candidate p , • a candidate c for each c ∈ X ∪ Y ∪ Z , • for each S i ∈ M , a candidate S i , • a dummy candidate d (used for the application of Lemma 4.7).Without loss of generality, due to Proposition 2.1, we can assume that for the relevantlength, gcd α , α = 1, and α , α ∈ N . In particular, the number 1 can be obtained as alinear combination of α and α . We thus can apply Lemma 4.7 to construct a set of voters V such that the relative scores are as follows: • score ( p ) = 0, • score ( y ) = α for each y ∈ Y , • score ( x ) = score ( z ) = α − α for each x ∈ X and z ∈ Z , • score ( S i ) = 2 α − α + 1 for each S i ∈ M , • score ( d ) < − kα .For each S i = ( x i , y i , z i ) ∈ M , we add the following available voters: • S i > · · · > y i , • z i > · · · > S i , • x i > · · · > S i .We say that a vote vetoes the candidate it places in the last position. We claim that p can be made a winner of the election with adding at most 3 k votes if and only if the3DM-instance is positive. By construction, the dummy candidate d cannot win the electionby adding at most 3 k votes, so we ignore d in the sequel.Note that since p is voted in one of the middle positions in all of the potentially addedvoters, the score of p only depends on the number of added voters, and not on the concretechoice of added votes. We thus consider only the change of relative scores between thecandidates and p when studying the effect of added voters.First assume that there is a cover C with k C k = k . For all S i ∈ C , we add all votesvoting S i in the first or in the last position. Then the scores (relatively to p ) change asfollows: 34 each y i is vetoed once, and so loses α points relative to p , so has final score 0, • each S i with i ∈ I appears once in the first position and twice in the last position, so S i gains α − α − α = α − α points. So the final score of such an S i is 2 α − α + 1 + α − α = − α + 1 ≤
0, since α ≥ • for each S i with i / ∈ I , the relative points of S i and p do not change. Since α > α and α , α ∈ N , it follows that 2 α − α + 1 ≤ S i does not beat p . • each z i and each x i gain exactly α − α points against p , and so tie with p .Thus p wins the election after adding these voters.For the converse, assume that there is a set V of available voters that we can add with k V k ≤ k such that p wins the resulting election. Since each y ∈ Y is currently winningagainst p , and the only way for y to lose points relatively to p is adding the vote voting y last, we know that for each y ∈ Y , there is one vote in V that votes y last. In particular,we have k V k ≥ k . So there are votes v , . . . , v k ∈ V , each voting a different y in the lastposition, and thus each voting a different S i in the first position.For each such vote, one of the S i gains ( α − α ) points against p . If such an S i is vetoedonly once, then its final score is 2 α − α + 1 + α − α − α = 1, and so p does not win theelection. Thus each S i gaining points in one of the votes v , . . . , v k must be vetoed at leasttwice. Since k V k ≤ k , we know that k V k = 3 k . Let I = { i | S i > · · · > y i ∈ V } . We claimthat I is a cover. By the above, I covers each y ∈ Y . Now assume that for i = j ∈ I , wehave x i = x j . Then x i gains 2( α − α ) points against p , and so p does not win the election,since α > α . The same argument holds for y . Thus, due to cardinality, we have a coverof X , Y , and Z .The case α < α is similar: Theorem B.4. f -CCAV is NP-complete for the generator f = ( α , α , . . . , α , with α < α .Proof. Very similar to the proof of Theorem B.3. We again reduce from 3DM. We use thesame candidate set as in the proof of Theorem B.3 (including the dummy candidate for theapplication of Lemma 4.7), and set up the scores of the nondummy candidates as follows: • score ( p ) = 0, • score ( y ) = α − α for each y ∈ Y , • score ( x ) = score ( z ) = α for each x ∈ X and each z ∈ Z , • score ( S i ) = min (0 , α − α ).For each tuple S i = ( x i , y i , z i ), we add three available voters • y i > · · · > S i , 35 S i > · · · > x i , • S i > · · · > z i .We say that a vote as above approves ( vetoes ) the candidate put in its first (last)position.If there is a cover with size at most k , we choose the 3 voters associated with eachelement of the cover. This lets each x and each z lose α points against p , each y i gains( α − α ) points against p , and each S i in the cover ends up with score ( S i )+2( α − α ) − α = score ( S i ) + 2 α − α ≤ α − α + 2 α − α = 0 points (all points counted relative to p ). Thus all candidates tie and p is a winner of the resulting election.For the converse, assume that p can be made a winner by adding at most 3 k voters,let V be the corresponding set. Clearly, each x ∈ X and each z ∈ Z need to be vetoedby some vote in V . Thus there are at least 2 k votes in V that vote some S i in the firstposition. Let C contain all indices S i such that there is a vote in V approving of S i . Clearly, k C k ≥ k . Then, each S i ∈ C gains at least ( α − α ) points from these votes, and thus has min (0 , α − α ) + α − α points. If the minimum is 0, then this clearly is more than 0,if the minimum is 3 α − α , then the score adds up to 3 α − α + α − α = 2 α − α ,which also exceeds 0 since α < α . Thus each S i ∈ C must be vetoed at least once.In particular, there must be at least k votes vetoing some S i , and since due to cardinalityreasons, there can be only k votes vetoing some S i , we know that k C k ≤ k . Due to theabove, we also know k C k ≥ k , so k C k = k .It remains to show that C covers X , Y , and Z . As argued above, each x and each z need to be vetoed once, so C covers X and Z . To show that C also covers Y , it suffices toshow that no y can be approved twice. This trivially follows since score ( y ) = α − α , andeach approval lets y gain α − α > p .The proof of Theorem 4.12.2 now directly follows from the above results: Proof. If α ≥ α > α / ∈ { α , α } , then one of the following cases applies: • α > α >
0, in this case the result follows from Theorem B.3, • α > α > α >
0, in this case the result follows from Theorem B.4.
B.7 Proof of Theorem 4.12.3
Theorem 4.12.
The problem f -CCAV is NP-complete if f is one of the following puregenerators:1. f = ( α , α , α , α , . . . , α , α , with α > α > .2. f = ( α , α , . . . , α , with α / ∈ { α , α } , α > .3. f = ( α , α , . . . , α , α , with α > α > α . . f = ( α , . . . , α , α , with α > α > .Proof. We again reduce from 3DM, let k = k X k = k Y k = k Z k . Using Lemma 4.7, we setup the the relative scores as follows: • score ( p ) = 0, • score ( x ) = − ( α − α ) for each x ∈ X , • score ( y ) = α − α for each y ∈ Y , • score ( z ) = α for each z ∈ Z , • score ( d ) < − kα , here d again is a dummy candidate required for the application ofLemma 4.7, who cannot win the election and whom we ignore in the sequel.The available voters are as follows: For each ( x, y, z ) in M , there is an available votervoting x > · · · > y > z. We claim that p can be made a winner of the election if and only if the 3DM-instance ispositive. First assume that the instance is positive, we then add the k votes correspondingto the cover. Then: • each x ∈ X gains exactly α − α points relatively to p , • each y ∈ Y loses exactly α − α points relatively to p , • each z ∈ Z loses exactly α points relatively to p .Thus all candidates tie and p is a winner of the election. For the converse, assume that p can be made a winner by adding at most k voters, let C be the corresponding subset of M . Since each y ∈ Y and each z ∈ Z need to lose points relatively to p , C covers each Y and Z . C cannot cover any x ∈ X twice, since otherwise x would gain 2( α − α ) pointsand thus beat p in the election. Thus C covers each x at most once, and so covers each x exactly once, thus C is a cover as required. B.8 Proof of Theorem 4.12.4
Theorem 4.12.
The problem f -CCAV is NP-complete if f is one of the following puregenerators:1. f = ( α , α , α , α , . . . , α , α , with α > α > .2. f = ( α , α , . . . , α , with α / ∈ { α , α } , α > .3. f = ( α , α , . . . , α , α , with α > α > α . . f = ( α , . . . , α , α , with α > α > .Proof. We again reduce from 3DM. Let M be a 3DM-instance with k M k = n , and let k = k X k = k Y k = k Z k . In addition to the preferred candidate p , we introduce a candidate c for each c ∈ X ∪ Y ∪ Z , and for each S i ∈ M , we introduce a candidate S i and a candidate S ′ i . We use Lemma 4.7 to set up the registered voters such that, relatively to p , we havethe following scores: • score ( p ) = 0, • score ( x ) = score ( y ) = score ( z ) = α for each x ∈ X , y ∈ Y , and z ∈ Z , • score ( S i ) = min ( α , α − α )) for each S i ∈ M , • score ( S ′ i ) = α − α for each S i ∈ M , • score ( d ) < − ( n +2 k ) α , where d again is a dummy candidate needed to apply Lemma 4.7,whom we ignore from now on.Note that since α > α >