Computational Social Choice

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(1)Computational social choice Jérôme Lang LAMSADE CNRS – Université Paris-Dauphine. ICAART-16 Rome, February 26, 2016.

(2) Plan. 1. From social choice to computational social choice 2. Voting 3. Voting rules: easy to compute 4. Voting rules: hard to compute 5. Combinatorial domains 6. Strategic behaviour 7. Communication issues and incomplete preferences 8. Other issues.

(3) Plan. 1. Social choice and computational social choice 2. Voting 3. Voting rules: easy to compute 4. Voting rules: hard to compute 5. Combinatorial domains 6. Strategic behaviour 7. Communication issues and incomplete preferences 8. Other issues.

(4) Social choice theory. I. Social choice: designing and analysing methods for collective decision making. I. Some examples of social choice problems: I I I I. I. elections deciding where and when to have dinner altogether tonight Doodle polls (find a date for a meeting) in a divorce settlement: deciding how to divide the bank account, who will have the chidren’s custody, who keeps the stereo and who keeps the cat. in a jury: agreeing on a verdict..

(5) Social choice theory. I. Social choice: designing and analysing methods for collective decision making. I. Some examples of social choice problems: I I I I. I. elections deciding where and when to have dinner altogether tonight Doodle polls (find a date for a meeting) in a divorce settlement: deciding how to divide the bank account, who will have the chidren’s custody, who keeps the stereo and who keeps the cat. in a jury: agreeing on a verdict aggregating preferences.

(6) Social choice theory. I. Social choice: designing and analysing methods for collective decision making. I. Some examples of social choice problems: I I I I. I. elections deciding where and when to have dinner altogether tonight Doodle polls (find a date for a meeting) in a divorce settlement: deciding how to divide the bank account, who will have the chidren’s custody, who keeps the stereo and who keeps the cat. in a jury: agreeing on a verdict aggregating preferences vs. aggregating opinions/beliefs.

(7) Social choice theory. I. Formally: 1. a set of agents N = {1, ..., n}; 2. a set of alternatives X = {x1 , . . . , xm }; 3. each agent i has some preferences on the alternatives. ⇒ choosing a socially preferred alternative I. Two important subdomains of social choice: I. I. Voting: agents (voters) express their preferences on a set of alternatives (for instance candidates) and must choose one collectively. Fair division: agents express their preferences over combinations of resources they may receive and an allocation must be found..

(8) Social choice theory. I. each agent i has some preferences on the alternatives. Usual models for preferences: I. cardinal preferences: each agent has a utility function u : X → IR. I. ordinal preferences: each agent has a preference relation on X (most common assumption in social choice). I. dichotomous preferences: each agent has a partition {Good , Bad } of X.

(9) A very rough history of social choice. 1. end of 18th century: early stage, with Condorcet and Borda (talk at ICAART-1789) 2. 1951: birth of modern social choice I. results are mainly axiomatic (economics/mathematics) I. impossibility theorems: incompatibility of a small set of seemingly innocuous conditions, such as Arrow’s theorem: With at least 3 alternatives, an aggregation function satisfies unanimity and independence of irrelevant alternatives if and only if it is a dictatorship.. I. computational issues are neglected. 3. early 90’s: computer scientists come into play ⇒ Computational social choice.

(10) What is computational social choice? Tow research streams: from computer science to social choice Using computational notions and techniques (mainly from Artificial Intelligence and Operations Research) for solving complex collective decision problems. from social choice to computer science Using social choice concepts and procedures for solving questions arising in CS / AI application domains, especially in multi-agent systems. I managing societies of autonomous agents I fair division of computational resources I ranking systems for internet search engines I group recommendation I etc..


(12) Voting 1. a finite set of voters N = {1, ..., n}; 2. a finite set of alternatives (candidates) X = {x1 , . . . , xm }; 3. a profile = a collection of n preference relations on X P = (1 , . . . , n ). i = linear order over X = vote expressed by voter i . Here is a 100-voter profile over X = {a, b, c , d , e} 33 votes: 16 votes: 3 votes: 8 votes: 18 votes: 22 votes:. abcd e bd cea cd bae cebd a d ecba ecbd a.

(13) Aggregation functions, rules, correspondences 1. a finite set of voters N = {1, ..., n}; 2. a finite set of alternatives (candidates) X = {x1 , . . . , xm }; 3. a profile = a collection of n preference relations on X P = (1 , . . . , n ). i = linear order over X = vote expressed by voter i . Resolute voting rule P 7→ r (P ) ∈ X : collectively preferred candidate Irresolute voting rule / set of collectively preferred P 7→ C (P ) ∈ 2X \ {0}: candidates, or co-winners. (Only one of them will be the final winner but the rule does not specify it.).

(14) Resolute vs. irresolute rules I. Suppose we have only two candidates a, b. I. Obvious voting rule: majority. I. Suppose that P consists of two votes: P = (a b , b a ). I. With irresolute rules: this is not a problem: a, b cowinners C (P ) = {a, b}. I. With voting rules: we need a tie-breaking mechanism I. I. give up neutrality: use a predefined priority relation on candidates (e.g. preference for the oldest candidate) give up anonymity: use a predefined strict relation on voters (e.g. priority given to the chair).

(15) Resolute vs. irresolute rules The usual way of defining voting rules: I. we first define an irresolute rule F. I. a resolute rule is defined from F by using a tie-breaking priority T. I. usual assumption: T = linear order on X. I. FT (P ) = max(T , F (P )): FT resolute rule. Example: I. P = ha b, b ai. I. Maj irresolute voting rule: Maj (P ) = {a, b}. I. Maja>b and Majb>a resolute voting rules. I. Maja>b (P ) = a. In the rest of the talk, we usually define irresolute rules, from which resolute rules are induced by a tie-breaking priority..

(16) Voting with more than three candidates. X = {a, b, c , d , e}. 33 votes: 16 votes: 3 votes: 8 votes: 18 votes: 22 votes: Who should be elected?. abcd e bd cea cd bae cebd a d ecba ecbd a.

(17) Voting with more than three candidates Generalizing simple majority: pairwise majority given any two alternatives x , y ∈ X , use simple majority to determine whether the group prefers x to y or vice versa. Does this work? Sometimes yes: 33 votes: 16 votes: 3 votes: 8 votes: 18 votes: 22 votes:. abcd e bd cea cd bae cebd a d ecba ecbd a. associated majority graph a. d. b. e. c. Collective preference relation: c b d e a Winner: c.

(18) Voting with more than three candidates Generalizing simple majority: pairwise majority given any two alternatives x , y ∈ X , use simple majority to determine whether the group prefers x to y or vice versa. Does this work? Sometimes no: 33 votes: 16 votes: 3 votes: 8 votes: 18 votes: 22 votes:. abdce bd cea cd bae cebd a d ecba ecbd a. associated majority graph a. d. b. e. c. Collective preference relation: {b d c b ...} e a; Winner: ?.

(19) Condorcet winner I. N (x , y ) = #{i , x i y } number of voters who prefer x to y .. I. x Condorcet winner if for all y 6= x , N (x , y ) > a. d. c. n 2. a. d. b. e. c b. e. c Condorcet winner. no Condorcet winner. I. sometimes there is no Condorcet winner. I. when there is a Condorcet winner, it is unique. I. a rule is Condorcet-consistent if it outputs the Condorcet winner whenever there is one..

(20) Approval voting. I. not all rules take rankings as input.. I. approval vote = a subset of (approved) candidates A ⊆ X. I. approval profile = a collection of approval votes P = hA1 , . . . , An i. I. Winner(s): candidate(s) approved most often. I I I I I. X = {a, b, c , d , e} n=5 / {a, b, c , d , e}, {d }i P = h{a, c }, {b, c , d }, 0, c and d approved by 3 voters: a and b by 2 voters; e by 1. cowinners: {c , d }.


(22) Positional scoring rules I. n voters, m candidates. I. fixed list of m integers s1 ≥ . . . ≥ sm , with s1 > sm. I. if voter i ranks candidate x in position j then scorei (x ) = sj. I. winner(s): candidate(s) maximizing n. s (x ) =. ∑ scorei (x ). i =1. Three important examples: plurality s1 = 1, s2 = . . . = sm = 0 veto s1 = s2 = . . . = sm−1 = 1, sm = 0 Borda s1 = m − 1, s2 = m − 2, . . . sm = 0.

(23) Positional scoring rules 33 16 3 8 18 22 I. I. abcd e bd cea cd bae cebd a d ecba ecbd a. plurality: a 7→ 33, b 7→ 16, c 7→ 11, d 7→ 18, e 7→ 22 winner: a Borda: a 7→ (33 × 4) + (3 × 1) = 135 b 7→ 247; c 7→ 244; d 7→ 192; e 7→ 182. winner: b I. veto: a 7→ 36, b 7→ 100, c 7→ 100, d 7→ 100, e 7→ 64 cowinners: b, c , d.

(24) Positional scoring rules Theorem (Fishburn, 73) No positional scoring rule is Condorcet-consistent 6 3 4 4. abc cab bac bca. Without loss of generality, let s3 = 0. I. S (a) = 6s1 + 7s2. I. S (b) = 8s1 + 6s2. I. S (b) − S (a) = 2s1 − s2 = s1 + (s1 − s2 ) > 0. I. S (b) > S (a) whatever the value of s1 and s2. I. but a is a Condorcet winner!.

(25) Condorcet-consistent rules I. P profile 7→ M (P ) directed graph associated with P. I. A voting rule r is based on the majority graph if r (P ) = f (M (P )) for some function f . For the sake of simplicity,we assume an odd number of voters; in this case the majority graph is a complete asymmetric graph: a tournament.. I. Copeland C (x ) = number of candidates y such that M (P ) contains x −→ y . I Copeland winner(s): maximize(s) C . I. a. d. b. e. c. C (a) = 0 C (b) = 3 C (c ) = 3 C (d ) = 3 C (e) = 1. cowinners: b, c , d.

(26) Condorcet-consistent rules I. NP (x , y ) = #{i , x i y } number of voters who prefer x to y .. I. A voting rule r is based on the weighted majority graph if r (P ) = g (NP ) for some function g . maximin winner(s): maximize Sm (x ) = miny 6=x NP (x , y ) NP a b c d e. a. − 67 67 67 66. b 33. − 51 21 48. c 33 49. − 67 40. d 33 79 33. −. e 36 52 60 70. 30. −. Sm (.) 33 49 33 21 30. winner: b.

(27) Condorcet-consistent rules Participation if the winner for profile P is x and P 0 = P ∪ {n+1 } then the winner for P 0 is either x , or a candidate y such that y n+1 x . I I. for m ≥ 4, no Condorcet-consistent rule satisfies participation Proof for maximin: 3 3 5 4. ad cb ad bc d cba bcad. Sm (a) = N (a, c ) = 6; b , d : 5; c : 4; maximin winner: a I I. 3 3 5 4 4. ad cb ad bc d cba bcad cabd. Sm (a) = 6; Sm (b) = 7; etc. maximin winner: b. The four new voters had rather stayed home! Similar paradox for reinforcement (if two disjoint electorates elect the same candidate x then their union still elects x ).

(28) Multiple round rules plurality with runoff I let x , y the two candidates with the highest plurality score (use tie-breaking rule if necessary) I winner: majority winner between x and y 33 16 3 8 18 22. a b c b d c c d b c e b d e c e c b. I. first step: keep a and e. I. winner: e. d e a d b d. e a e a a a.

(29) Multiple round rules Single transferable vote (STV) Repeat x := candidate ranked first by the fewest voters; eliminate x from all ballots {votes for x transferred to the next best remaining candidate} Until some candidate y is ranked first by more than half of the votes; Winner: y I. When there are only 3 candidates, STV coincides with plurality with runoff.. I. STV is used for political elections in several countries (at least Australia and Ireland).

(30) Single transferable vote (STV) 33 16 3 8 18 22. a b c b d c c d b c e b d e c e c b 33 16 3 8 18 22. a d d e d e. d e a d b d. e a e a a a. 33 16 3 8 18 22. a b d e d e. b d b b e b. d e a d e d. e a e a a a. 33 16 3 8 18 22. a d d d d d. d a a a a a. winner: d. d e a d b d. e a e a a a.

(31) Single transferable vote (STV) I. (previous example) winner: d. I. recall that c is the Condorcet winner. I. therefore: STV is not Condorcet-consistent [same thing for majority with runoff]. I. plurality with runoff and STV also fail monotonicity: abc cba bcc cab. 6 4 5 2 I. b eliminated. I. winner: a.

(32) Single transferable vote (STV) I. (previous example) winner: d. I. recall that c is the Condorcet winner. I. therefore: STV is not Condorcet-consistent [same thing for majority with runoff]. I. plurality with runoff and STV also fail monotonicity: abc cba bcc cab. 6 4 5 2. I. abc cba bcc acb. 6 4 5 2. I. b eliminated. I. c eliminated. I. winner: a. I. winner: b. they also fail participation and reinforcement.


(34) Computing voting rules. The voting rules you have seen so far can be computed in polynomial time: I. scoring rules, plurality with runoff, approval: O (nm). I. Copeland, maximin, STV: O (nm2 ).. But some voting rules are NP-hard..

(35) Hard rules: Kemeny. I. Kemeny distance: dK (, 0 ) = number of (x , y ) ∈ X 2 on which and 0 disagree dK (, h1 , . . . , n i) =. ∑. dK (, i ). i =1,...,n. I. Kemeny consensus = order minimizing dK (., h1 , . . . , n i). I. Kemeny winner = candidate ranked first in a Kemeny consensus.

(36) Hard rules: Kemeny Observation: Kemeny is based on the weighted majority graph. 4 3 2. abc bca cab. Computing d (a b c , h1 , . . . , 9 i): N a b c. a. − 3 5. b 6. −. c 4 7. 2. −. I. 3 voters disagree with a b. I. 5 voters disagree with a c. I. 2 voters disagree with b c. I. hence d (a b c , h1 , . . . , 9 i) = 10.. Kemeny scores: abc 10. acb 15. bac 13. bca 12. cab 14. Kemeny consensus: abc ; Kemeny winner: a. cba 17.

(37) Hard rules: Kemeny. I. NP-hard (Bartholdi et al., 89; Hudry, 89). I. exact complexity: deciding whether a candidate is a Kemeny winner is ΘP2 -complete (Hemaspaandra et al., 04). I. existence of polynomial approximation algorithms. I. good heuristics. I. otherwise, satisfy many good properties. I. several other important rules are also hard to compute.

(38) Is there a life after NP-hardness?. I. efficient computation: design algorithms that do as well as possible, possibly using heuristics, or translations into well-known frameworks (such as integer linear programming).. I. fixed-parameter complexity: isolate the components of the problem and find the main cause(s) of hardness approximation: design algorithms that produce a (generally suboptimal) result, with some performance guarantee.. I. I. The approximation of a voting rule is a new voting rule that may be interesting per se..


(40) Voting in combinatorial domains Key question: structure of the set X of candidates. Example 1 choosing a common menu: X= {asparagus risotto, foie gras} × {roasted chicken, vegetable curry} × {white wine, red wine} Example 2 multiple referendum: a local community has to decide on several interrelated issues (should we build a swimming pool or not? should we build a tennis court or not?) Example 3 recruiting committee (3 positions, 6 candidates): X = {A | A ⊆ {a, b, c , d , e, f }, |A| ≤ 3}. Combinatorial domains: I. V = {X1 , . . . , Xp } set of variables, or issues; X = D1 × ... × Dp. I. for each i , Di is a finite value domain for variable Xi. I.

(41) Voting in combinatorial domains I. Two binary variables: I I. I. S (build a new swimming pool or npt) T (build a new tennis court or not). 5 voters with their preferences: voters 1 and 2 voters 3 and 4 voter 5. ST ST ST ST ST ST ST ST ST ST ST ST.

(42) Voting in combinatorial domains I. Two binary variables: I I. I. I. S (build a new swimming pool or npt) T (build a new tennis court or not). 5 voters with their preferences: voters 1 and 2 ST ST ST ST voters 3 and 4 ST ST ST ST voter 5 ST ST ST ST Problem 1: voters 1-4 feel ill at ease reporting a preference on {S , S } and {T , T }.

(43) Voting in combinatorial domains I. Two binary variables: I I. I. I. I. S (build a new swimming pool or npt) T (build a new tennis court or not). 5 voters with their preferences: voters 1 and 2 ST ST ST ST voters 3 and 4 ST ST ST ST voter 5 ST ST ST ST Problem 1: voters 1-4 feel ill at ease reporting a preference on {S , S } and {T , T } Problem 2: suppose they do so by an “optimistic” projection I I. voters 1, 2 and 5: S ; voters 3 and 4: S ⇒ decision = S ; voters 3,4 and 5: T ; voters 1 and 2: T ⇒ decision = T .. Alternative ST is chosen although it is the worst alternative for all but one voter! I. Several classes of methods for solving this problem..


(45) Manipulation and strategyproofness I. I. Manipulation: a coalition of voters report insincere preferences so as to elect a better candidate. Example: r = plurality with runoff 2+6 4 5. abc cba bac. 1st round: c eliminated 2nd round: b elected.

(46) Manipulation and strategyproofness I. I. Manipulation: a coalition of voters report insincere preferences so as to elect a better candidate. Example: r = plurality with runoff 2+6 4 5. abc cba bac. 1st round: c eliminated 2nd round: b elected I. 2 6 4 5. cab abc cba bac. 1st round: b eliminated 2nd round: a elected. Is this a specific flaw of plurality with runoff?.

(47) Manipulation and strategyproofness I. I. Manipulation: a coalition of voters report insincere preferences so as to elect a better candidate. Example: r = plurality with runoff 2+6 4 5. abc cba bac. 1st round: c eliminated 2nd round: b elected I I. 2 6 4 5. cab abc cba bac. 1st round: b eliminated 2nd round: a elected. Is this a specific flaw of plurality with runoff? Unfortunately no.... Gibbard and Satterthwaite’s theorem (73/75) If |X | ≥ 3, any nondictatorial, surjective voting rule is manipulable for some profiles..

(48) Escaping Gibbard and Satterthwaite One solution for nearly escaping Gibbard and Satterthwaite: Computational barrier I make manipulation hard to compute. I the harder it is to find a manipulation, the better the voting rule I (similar approach in cryptography) Constructive manipulation for voting rule r : Input A set of m candidates X , a candidate x ∈ X , votes of voters 1, . . . , k < n Question Is it possible for voters k + 1, . . . , n to cast their votes so that the winner is x ? First papers on the topic: Bartholdi, Tovey & Trick (89); and lots of papers since then..

(49) Complexity of manipulation. I. Manipulating the Borda rule by a single voter I. I. 4 voters:. abd ce baed c ceabd d cbae. Current Borda scores a 7→ 10. b 7→ 10. c 7→ 8. d 7→ 7. e 7→ 5. Can the last voter find a vote so that the winner is ... a?.

(50) Complexity of manipulation. I. Manipulating the Borda rule by a single voter I. I. 4 voters:. abd ce baed c ceabd d cbae. Current Borda scores a 7→ 10. b 7→ 10. c 7→ 8. d 7→ 7. e 7→ 5. Can the last voter find a vote so that the winner is ... a?. b?.

(51) Complexity of manipulation. I. Manipulating the Borda rule by a single voter I. I. 4 voters:. abd ce baed c ceabd d cbae. Current Borda scores a 7→ 10. b 7→ 10. c 7→ 8. d 7→ 7. e 7→ 5. Can the last voter find a vote so that the winner is ... a?. b?. c?.

(52) Complexity of manipulation. I. Manipulating the Borda rule by a single voter I. I. 4 voters:. abd ce baed c ceabd d cbae. Current Borda scores a 7→ 10. b 7→ 10. c 7→ 8. d 7→ 7. e 7→ 5. Can the last voter find a vote so that the winner is ... a?. b?. c?. d?.

(53) Complexity of manipulation. I. Manipulating the Borda rule by a single voter I. I. 4 voters:. abd ce baed c ceabd d cbae. Current Borda scores a 7→ 10. b 7→ 10. c 7→ 8. d 7→ 7. e 7→ 5. Can the last voter find a vote so that the winner is ... a?. b?. c?. d?. e?.

(54) Complexity of manipulation. I. Manipulating the Borda rule by two voters I I. Borda + tie-breaking priority a > b > c > d > e. Current Borda scores: a 7→ 12. I. b 7→ 10. c 7→ 9. d 7→ 9. e 7→ 4. f 7→ 1. Is there a constructive manipulation by two voters for e?.

(55) Complexity of manipulation. Existence of a manipulation for the Borda rule: I. for a single voter : in P. I. for a coalition of at least two voters : NP-complete.

(56) Complexity of manipulation. Number of manipulators Copeland STV veto cup maximin ranked pairs Bucklin Borda. 1 P NP-complete P P P NP-complete P P. at least 2 NP-complete NP-complete P P NP-complete NP-complete P NP-complete.

(57) Complexity of manipulation. An important concern: I. a worst-case NP-hardness results only says that sometimes (maybe rarely), computing a manipulation will be hard. I. negative results about the average hardness of manipulation: under reasonable assumptions, with a high probability, if there is a manipulation then it can be found by a greedy algorithm a manipulation with high probability..


(59) Incomplete knowledge and communication Given some incomplete description of the voters’ preferences, I. is the outcome of the voting rule determined?. I. if not, whose information about which candidates is needed?. 4 2 2 1. voters: voters: voters: voter:. cd ab abd c bacd. ? ? ? ?. plurality winner already known (c ).

(60) Incomplete knowledge and communication Given some incomplete description of the voters’ preferences, I. is the outcome of the voting rule determined?. I. if not, whose information about which candidates is needed?. 4 2 2 1. voters: voters: voters: voter:. cd ab abd c bacd. ? ? ? ?. plurality winner already known (c ) Borda. I. I. partial scores for 8 voters: a: 14 ; b: 10 ; c : 14; d : 10 only need to know the last voters’s preference between a and c.

(61) Possible and necessary winners. I. incomplete knowledge of the voters’ preferences.. I. For each voter: a partial order on the set of candidates: P = hP1 , . . . , Pn i incomplete profile. I. Completion of P : full profile T = hT1 , . . . , Tn i of P , where each Ti is a linear ranking extending Pi .. Given a voting rule r , an incomplete profile P , and a candidate c : I. c is a possible winner if there exists a completion of P in which c is elected.. I. c is a necessary winner if c is elected in every completion of P..

(62) Possible and necessary winners. a b, a c abc abc abc acb acb acb. ba cba bca bac cba bca bac. cab cab cab cab cab cab cab. plurality with tie-breaking b > a > c c b b c b c. I. Possible Condorcet winners: {a, c }. I. possible pluralityb>a>c -winners: {b, c }.. Condorcet c a c c a.

(63) Possible and necessary winners Two particular cases: possible/necessary winners with respect to addition of voters I A subset of voters A have reported a full ranking; the other ones have not reported anything. I x is a possible winner if the coalition N \ A has a constructive manipulation for x . possible/necessary winners with respect to addition of candidates I. The voters have reported a full ranking on a subset of candidates X (and haven’t said anything about the remaining candidates)..

(64) Possible and necessary winners with respect to addition of candidates I. I I. New candidates sometimes come while the voting process is going on: I. Doodle: new dates become possible. I. recruiting committee: a preliminary vote can be done before the last applicants are inrerviewed. Question: who among the initial candidates can win? Example : I I I I I. I. 12 voters; initial candidates {a, b, c }; one new candidate y voting rule = plurality with tie-breaking priority a > b > c > y plurality scores before y is a candidate: a 7→ 5, b 7→ 4, c 7→ 3 y and a possible winners (obvious) b possible winner: if three of the voters who voted for a now vote for y then the new scores are a 7→ 2, b 7→ 4, c 7→ 3, y 7→ 3 c not a possible winner; we’d need 4 of the voters who voted for a and 2 of the voters who voted for b to vote for y ; but then y would have a score ≥ 5 and c has a score ≤ 3..


(66) Computational social choice: beyond voting. I. Iterated voting. I. Fair division of indivisible resources. I. Fair division of divisible resources (cake-cutting protocols). I. Coalition formation, stable matching. I. Judgment aggregation.

(67) Bibliography. I. Handbook of Computational Social Choice (F. Brandt, V. Conitzer, U. Endriss, J. Lang, A. Procaccia, eds.). Cambridge University Press, 2016 (to appear extremely soon).. An experimental voting platform: Whale (developed by Sylvain Bouveret, University of Grenoble) http://whale3.noiraudes.net/ Contacting me: lang@lamsade.dauphine.fr.

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