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Jeffrey S. Rosenschein The Hebrew University of Jerusalem Joint work with: Ariel Procaccia, Aviv Zohar, Michael Zuckerman, Yoni Peleg, Gal Kaminka*, Yoram Bachrach, and Reshef Meir Hebrew University of Jerusalem *Bar Ilan University
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Introduction to voting and power indices A Few Examples of Voting Protocols A Few Criteria for Evaluating Voting Protocols Coalitional manipulation Robustness of Voting Rules Automated Design of Scoring Rules and Voting Trees Conclusions
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Agents have to reach a consensus regarding a preferred alternative in a shared environment Examples: Collaborative filtering (votes over movies, books…) Trust and Reputation Systems Google as a platform for (fast) preference aggregation Improving similarity search Creation of Multiagent Plans Social choice theory gives a well-studied framework for preference aggregation 3
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Agents need to reach consensus about what to do next in shared environment 4 56 g4g4 g3g3 g2g2 g1g1 g5g5 g6g6 2 3 1 4
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Search for a joint plan that brings agents to the consensus state that optimizes global utility 4 56 g4g4 g3g3 g2g2 g1g1 g5g5 g6g6 2 3 1 ? 5
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Centralized planning (Generalized) Partial Global Plans Negotiation (in many forms) Market mechanisms Synchronization of pre-existing plans Voting, a means of preference aggregation Agents reveal preferences by ranking candidates Winner determined by a voting protocol 6
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Choosing a consensus state 4 56 g4g4 g3g3 g2g2 g1g1 g5g5 g6g6 2 3 1 a b c d 7
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Studies how decisions or rankings are made among a collection of alternatives, when there are voters with separate opinions Group choice should reflect the individual voters’ desires (by some definition, as much as possible…) 8
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“We reject: kings, presidents and voting. We believe in: rough consensus and running code.” – David Clark, 1992, speaking of the growth of the internet; a comment on people “We reject: kings and presidents. We believe in: voting, weighted voters, coalitions, and computational barriers to manipulation.” – me, now, speaking of the operation of the internet; a comment on agents 9
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Group of voters with ranked ordinal preference over more than two alternatives, decide on an ordering, or on a choice Example (order of preferences over candidates a, b, c, d): 1 voter acb bad dbc cda 10
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A voting protocol is a function from all the voters’ (linearly ordered) preferences to an outcome ordered list, or to a winning candidate Not always just an ordering (can be cardinal preferences), not always linearly ordered… To understand social choice, think of all the functions… 11 1 voter acb bad dbc cda
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There are many possible functions… Voting can take many more varied forms than most people think of There is no “best” function that chooses the “right” answer, but different functions can satisfy different criteria that we might find desirable What might those criteria look like? 12
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Universality: should create a deterministic, complete social preference order from every possible set of individual preference orders Citizen sovereignty: every possible order should be achievable by some set of individual preference orders Non-dictatorship: the social welfare function should be sensitive to more than the wishes of a single voter Monotonicity: change favorable to candidate x does not hurt x Independence of irrelevant alternatives: if we restrict attention to a subset of options and apply the social welfare function only to those, then the result should be compatible with the outcome for the whole set of options No system meets all these criteria when there are two or more voters, and three or more choices 13
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(Regarding systems that choose a single winner) For three or more candidates, one of the following three things must hold for every voting rule: The rule is dictatorial; or There is some candidate who cannot win, under the rule, in any circumstances; or The rule is manipulable 14
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Voters may prefer to reveal their intentions untruthfully Can happen in the full knowledge case (where a manipulating voter knows others’ votes), or strategically (heuristically) without full knowledge of others’ votes This is presumably undesirable, since the outcome may be one that does not maximize social welfare 10 3 2 1 3 11 4 15
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Manipulation (strategic voting) Coalitional manipulation Bribery Micro-bribery Adding candidates Removing candidates Partition of candidates (cascading elections) Run-off partition of candidates Adding voters Removing voters Partition of voters (hierarchical 2-stage election) 16
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Sequential Pairwise Voting Pareto Criterion Plurality Voting Condorcet Winner and Condorcet Loser Criteria Plurality with Run-off Copeland Monotonicity Criterion The Borda Count Scoring Protocols Black’s Procedure Smith’s Generalized Condorcet Criterion Single Transferable Vote 17
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Group of voters with ranked ordinal preference over more than two alternatives, decide on an ordering, or on a choice Example (order of preferences over candidates a, b, c, d): 1 voter acb bad dbc cda 18
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Different possible agendas a ba c c d d Agenda i b cb a a d a Agenda ii a cc b b d b Agenda iii a ba d a c c Agenda iv In this example, anyone can be a winner! Rule of thumb: bring up your favorite as late as possible 19 111 acb bad dbc cda 111 acb bad dbc cda 111 acb bad dbc cda 111 acb bad dbc cda 111 acb bad dbc cda
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The vote, of course, is also susceptible to insincere, manipulative voting Example: third voter votes insincerely for c instead of for b (just in first election): b cc a c d d Agenda ii (insincere third voter) Third voter gets second choice (d) instead of last choice (a), by lying 20 111 acb bad dbc cda
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21 a a c c ? ? ? ? ? ? ? ? b b a a c c c c b b a a a a c c b b b b a a c c c c c c c c b b b b b b a a a a a a
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“If every voter prefers an alternative x to an alternative y, a voting rule should not produce y as a winner” Sequential pairwise voting violates this criterion; for example, in Agenda i, d wins, even though everyone prefers b to d a ba c c d d Agenda i 22 111 acb bad dbc cda
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Each voter votes for one alternative; the one with the most votes wins Example (9 voters): Plurality voting has c winning, even though 5-4 majority rate c last 3 voters2 voters4 voters abc bab cca 23
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In pairwise decisions, b, which came in last in the plurality vote, would beat both c and a; and c, which won the plurality vote, would have lost all pairwise contests: a c b 6-to-3 5-to-4 24 324 abc bab cca
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a beats b in a pairwise election if the majority of voters prefers a to b a is a Condorcet winner if a beats any other candidate in a pairwise election Often there is no Condorcet winner, but if there is, it is obviously unique 25
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Condorcet-consistency: if a Condorcet winner exists, it must be elected Plurality voting (as we have seen) is not Condorcet-consistent In fact, it doesn’t even satisfy the Condorcet- loser criterion (it might choose a candidate that loses to every other candidate in pairwise elections) Sequential pairwise voting, despite its other faults, is Condorcet-consistent 26
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Voter 1Voter 2 Voter 3 c b a a c b b a c 27 a a a b b b c c c a a c c b b
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Copeland: a’s score is the number of other candidates that a beats in pairwise elections The winner is the candidate with the highest score If a is a Condorcet winner, of course, its score = m-1 (where m is the number of candidates), and for any b≠a, score < m-1 28
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Example (17 voters): Plurality voting, a and b are top two, and a beats b in run-off by 11-to-6 But, if last two voters changed their minds in favor of a, i.e., a b c instead of b a c, then a and c are top two, and c beats a by 9-to-8 a gets more first place votes, and loses an election it would have won! 6 voters5 voters4 voters2 voters (last 2 voters, strategic) acbba bacab cbacc 29
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“If x is a winner under a voting rule, and one or more voters change their preferences in a way favorable to x (without the changing the order in which they prefer any other alternatives), then x should still be the winner” Straight plurality voting satisfies monotonicity Plurality with Run-off violates it 30
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Each voter submits preferences over the m alternatives Each alternative receives no points for being ranked last 1 point for being ranked second-to-last … up to m-1 points for being ranked first Points for each alternative are summed across all voters, and the alternative with the highest total is the winner 31
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With Borda count, a gets 3 points from first voter, 2 points from the second, and 0 from the third Final Borda count totals: a:5, b:6, c:4, d:3 b is the Borda winner 3 points 2 points 1 point 0 points 1 voter acb bad dbc cda 32
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Borda counts, a:6, b:7, c:2 b wins, but a is the Condorcet winner Even worse: a has an absolute majority of first place votes! The existence of c allows the 2 voters to weight b over a more heavily than the 3 voters choose to weight a over b, enabling b to win the Borda count 2 points 1 point 0 points 3 voters2 voters ab bc ca 33
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In general, a good heuristic for manipulating the Borda count is: if a voter favors x and believes y is the most dangerous competitor, he can minimize the risk that y will beat x by putting y at the bottom of his preference list (we’ll look at this idea more in a moment…) 34
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= where i ≥ i+1. Candidate receives i points for each voter that ranks it in i’th place Examples: Plurality: Veto: Borda: Sensitive scoring rules have m = 0 Borda is thus one example of the general class of (sensitive) scoring rules 35
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If there’s a Condorcet winner, choose it. Otherwise, use the Borda count. Satisfies Pareto, Condorcet winner, and Monotonicity Criteria. A combined voting method that satisfies the Condorcet Winner criterion: Black’s Procedure 36
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If the alternatives can be partitioned into two sets A and B such that every alternative in A beats every alternative in B in pairwise contests, then a voting rule should not select an alternative in B. Smith’s Criterion implies both the Condorcet Winner and Condorcet Loser Criteria, but not vice versa. Black’s voting procedure does not satisfy Smith’s generalized Condorcet Criterion. 37
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Example (3 voters): 1 voter abc bca xxx yyy zzz www cab If we partition so that A=[a, b, c] and B=[w, x, y, z], then every alternative in A beats every one in B by a 2-to-1 vote. There is no Condorcet winner (a, b, and c beat each other cyclically). Borda counts are a:11, b:11, c:11, w:3, x:12, y:9, z:6, so Black’s method chooses x. 38
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Some voting procedures try to eliminate ‘undesirable’ alternatives, one after another, until a winner is chosen STV is one such example Election proceeds in rounds In each round, each voter awards one point to candidate he ranks highest out of surviving candidates – candidate with least points is eliminated 39
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2 voters 1 voter a a b b a d 40 b b c c d d b b a a d d c c c c b d d b b a a
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criteriavoting rules SequentialPluralityPlurality w/Runoff STVBordaCopeland pareto NYYYYY monotonicity YYNNYY Condorcet Loser YNYYYY Majority YYYYNY Condorcet Winner YNNNNY Smith YNNNNY 41
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33 voters16 voters3 voters 8 voters18 voters22 voters a b c d e b d c e a c d b a e c e b d a d e c b a e c b d a
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Dodgson score of x = minimum number of switches between adjacent candidates needed to make x a Condorcet winner [Bartholdi et al., SC&W 89]: Dodgson score is NP-hard [Caragiannis et al., SODA 09]: O(logm) LP-based randomized rounding algorithm (matches lower bound) 43 Voter 1Voter 2 Voter 3 a b c a c b b a c
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English author and mathematician, better known as Lewis Carroll Suggested choosing a candidate “as close as possible” to a Condorcet winner 44
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Distance of x = minimum # of exchanges between adjacent candidates needed to make x a Condorcet winner 45
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Voter 1Voter 2Voter 3 Voter 4 Voter 5 b d e a c b c a e d d e c a b e d b a c a e c b d P(a,b) P(a,c) P(a,d) P(a,e) 23222322 33223322 33323332 33333333
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Voter 1Voter 2Voter 3 Voter 4 Voter 5 b d e a c b c a e d d e c a b e d b a c a e c b d def(b,a) def(b,c) def(b,d) def(b,e) 00010001 00000000
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Distance of x = minimum # of exchanges between adjacent candidates needed to make x a Condorcet winner Alternatively: total number of positions that the voters push x Elect candidate with minimum dist=score 48
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D ODGSON -S CORE : given candidate x, a preference profile, and a threshold k, is the Dodgson score of x at most k ? [BTT 89] D ODGSON -S CORE is NP-complete, D ODGSON -W INNER is NP-hard [HHR 97] D ODGSON -W INNER is complete for Parallel access to NP [Procaccia, Feldman, Rosenschein]: O(logm) LP-based randomized rounding algorithm (matches lower bound) 49
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Assume that there is a body voting “yes” or “no” on a law. How much voting power does each member have? Example: Four member body, A, B, C, D. If there is a tie, A (the chairman) gets to break the tie. How much extra voting power does A have? 50
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Shapley-Shubik index: Assume all orderings are possible, see when each member is pivotal (i.e., the losing coalition becomes the winning coalition when that member joins it). The pivotal member holds the power. Arrange the members in order of their enthusiasm for the bill. We don’t know what members’ preferences are, so assume a priori that any order is equally likely (big assumption in the real world). 51
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ABCD ADBC BCAD CABD CDAB DBAC ABDC ADCB BCDA CADB CDBA DBCA ACBD BACD BDAC CBAD DABC DCAB ACDB BADC BDCA CBDA DACB DCBA 4 members, 4!=24 orders, A breaks ties: A is pivotal 12/24 = 1/2 of the time, B, C, D each 1/6 of the time. A’s voting power is three times greater than other members. 52
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Sometimes there are bodies where different members’ votes are weighted: [q ; w 1, w 2, w 3, …, w n ] – q is required to pass a law, each member i gets wi weight for his vote. Example: Council of Ministers of the European Community in 1958 (France, Great Britain, Italy, Belgium, Netherlands, Luxembourg): [12 ; 4, 4, 4, 2, 2, 1] – 12 to pass resolution F G I B N L 53
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60 different ways to arrange 4, 4, 4, 2, 2, 1 [6!/(3!2!1!)]. A “4” occupies the pivotal position in 42 orders, a “2” occupies the pivotal position in 18 orders, and a “1” never occupies the pivotal position. Shapley-Shubik index: ( 14/60, 14/60, 14/60, 9/60, 9/60, 0) F G I B N L Luxembourg is a “dummy.” % votes% power France, Germany, Italy23.5%23.3% Belgium, Netherlands11.8%15.0% Luxembourg5.9%0.0.% 54
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[5 ; 2, 2, 1, 1] (A, B, C, D) Six possible orders: 221121212112122112121122 A “2” pivots in five of them, so power indices are: (5/12, 5/12, 1/12, 1/12) [Voting bodies with same winning coalitions are “structurally equivalent”; e.g., body with dummy equivalent to same body without the dummy.] 55
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Minimal WinningSample WeightedShapley-Shubik CoalitionsVoting BodyIndices A[1; 1](1) AB[2; 1, 1](1/2, 1/2) AB, AC, BC[2; 1, 1, 1](1/3, 1/3, 1/3) ABC[3; 1, 1, 1](1/3, 1/3, 1/3) AB, AC[3; 2, 1, 1](4/6, 1/6, 1/6) ABC, ABD, ACD, BCD[3; 1, 1, 1, 1](1/4, 1/4, 1/4, 1/4) ABCD[4; 1, 1, 1, 1](1/4, 1/4, 1/4, 1/4) AB, AC, AD, BCD[3; 2, 1, 1, 1](3/6, 1/6, 1/6, 1/6) ABC, ABD, ACD[4; 2, 1, 1, 1](3/6, 1/6, 1/6, 1/6) AB, AC, AD[4; 3, 1, 1, 1](9/12, 1/12, 1/12, 1/12) AB, ACD, BCD[4; 2, 2, 1, 1](2/6, 2/6, 1/6, 1/6) ABC, ABD[5; 2, 1, 1, 1](5/12, 5/12, 1/12, 1/12) AB, ACD[5; 3, 2, 1, 1](7/12, 3/12, 1/12, 1/12) AB, AC, BCD[5; 3, 2, 2, 1](5/12, 3/12, 3/12, 1/12) 56
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Voting blocs can be thought of as a single member with a weighted vote: [4; 3, 1, 1, 1, 1] has power indices (6/10, 1/10, 1/10, 1/10, 1/10) Single bloc among heterogeneous voters has disproportionate power (another example): [19; 13, 1, 1, …, 1] (23 “1”’s) – 24 distinct orders (“13” listed 1st, …, 24th), “13” is pivotal from 7th through 19th, i.e., 13 out of 24, so has 13/24 = 54% of the voting power. 57
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Things change, however, when there are two blocs in the voting body (bloc E, bloc I): Power of Two Voting Blocs in 11-member Voting Body: Two small blocs gain power at the expense of others. One large bloc gains power at the expense of small bloc and others. Two large blocs lose power to other voters (since when they are opposed, non-bloc members hold balance). Number of Bloc members % of vote of E % of power of E % of vote of I % of power of I |E|=2, |I|=218.219.418.219.4 |E|=4, |I|=236.447.618.214.3 |E|=4, |I|=436.430.036.430.0 58
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Example: 3-member committee (A’s) in 9-member body (6 other B’s). Laws must pass committee, then main body. To be approved, law must have support of 5 members, including at least 2 A’s. Orders: 59
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For a B to be pivotal, it must be preceded by exactly 4 others (including either 2 or 3 A’s): (AABB) B (ABBB) (AAAB) B (BBBB) B’s pivot 28 ways, so A’s pivot 56 ways. Power index of each A is (1/3 * 56/84) = 2/9, each B is (1/6 * 28/84) = 1/18. A’s are four times more powerful than B’s. 60
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Two voting bodies will share power equally: Example: 3-person body (A’s) and 5-person body (B’s), 8 choose 3 = 56 orders. For an A to pivot, must be second A preceded by 3 or 4 or 5 B’s: (ABBB) A (ABB) (ABBBB) A (AB) (ABBBBB) A A = 28 ways Individual A’s more powerful than B’s (A’s 1/6 each of power, B’s 1/10 each of power) 61
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If the voting rules change, however, and A’s must agree unanimously, their power increases dramatically: (AABBB) A BB (AABBBB) A B (AABBBBB) A = 46 ways The A’s together now have 46/56 = 82% of the power; each A has one third of that, i.e., 0.274; each B has 1/5 * 10/56 = 0.036. An A’s power goes from 1.67 to 7.67 of a B. 62
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As we saw, it is often in the voters’ interest to reveal false preferences May lead to the election of a socially bad candidate 63 pab 01 2 3 4 01 2 3 4 01 2 3 4
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Center (running the election) removes candidates Center adds candidates Center adds voters Center bribes voters … 64
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If m=2, Plurality is nonmanipulable Let m 3. The following properties are incompatible: 1. Onto: any candidate can be elected 2. Nondictatorship: there is no single voter who dictates the outcome of the election 3. Nonmanipulability 65
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A grocery store is being built. Each voter (resident of the street) wants it as close as possible to his own house. Need to choose a spot. Nonmanipulable solution: choose the median voter 66
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Circumvent Gibbard-Satterthwaite by: Single-peaked preferences Mechanism design [Bartholdi et al. SC&W 89]: Computational hardness to the rescue! [Bartholdi and Orlin SC&W 91]: STV is NP- hard to manipulate A lot of recent work
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Introduction to voting and power indices A Few Examples of Voting Protocols A Few Criteria for Evaluating Voting Protocols Coalitional manipulation Robustness of Voting Rules Automated Design of Scoring Rules and Voting Trees Conclusions
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A coalition of manipulators cooperates in order to make p C win the election Votes are weighted Formulation as decision problem (CCWM): Instance: a set of weighted votes which have been cast, the weights of the manipulators, p C Question: Can p win the election? Manipulation is (presumably) undesirable Bounded rationality comes to the rescue! Conitzer et al. [JACM 07]: NP-hard for a variety of voting rules, even when m is constant
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Computational complexity can be an obstacle to desirable computations Computational complexity can be an obstacle to undesirable computations Example: RSA encryption Manipulation is (basically) always possible, but if it’s too hard to calculate, perhaps a voting process can be manipulation-resistant 71
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[Bartholdi and Orlin 1991] There are voting protocols that are NP-hard for a single voter to manipulate [Conitzer and Sandholm 2002, 2003a] Some manipulations of common voting protocols are NP-hard, even for a small number of candidates [Conitzer and Sandholm 2003b] Adding a pre-round to some voting protocols can make manipulation hard (even PSPACE-hard in some cases) 72
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Individual manipulation of some protocols is NP-hard when the number of candidates m is large Coalitional manipulation of sensitive scoring protocols is NP-hard, even when m=3 But…this may be a weak guarantee of resistance to manipulation Given a reasonable distribution, how hard is it to manipulate? (ongoing research) 73
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Procaccia and Rosenschein [JAIR 07]: Junta distributions are hard. Susceptibility to manipulation if one can manipulate with high prob. w.r.t. a Junta distribution. Scoring rules are susceptible; very loose bound on the error window of a greedy algorithm Only scoring rules and constant m
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Reminder: in Borda, each voter awards m-k points to candidate ranked k Greedy algorithm for coalitional manipulation [Procaccia and Rosenschein, JAIR 07]: each manipulator ranks p first, and the other candidates by inverse score
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Greedy algorithm: each voter in T ranks p first, and the other candidates in an order inversely proportional to their current score. a b p 2.9 2.1 1 1 2.6 3.1 3.9 2 2 4 4 T = 0.50.51 2.1 1.6 1 1 T = 0.51 2.1 3.1 3 3 Case 1Case 2 76
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Greedy algorithm: each voter in T ranks p first, and the other candidates in an order inversely proportional to their current score. a b p 2.9 2.1 1 1 2.6 3.1 3.9 2 2 4 4 T = 2.1 1.6 1 1 T = 0.51 2.6 3 3 Case 1Case 2 77
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pab 55 10 0 2030 40 0102030 40 0102030 40
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pab 5 10 0 2030 40 0102030 40 0102030 40
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Theorem: Given an instance of CWM with weights W for the manipulators 1. If there is no manipulation, the greedy algorithm will return false 2. If there is a manipulation with weights W \ max W, then the algorithm will find the manipulation with W
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pab 5 10 0 2030 40 0102030 40 0102030 40 10
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Algorithm fails Manipulation exists All instances
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Preferences may be faulty: Agents may misunderstand choices Robots operating in an unreliable environment Want theoretical analysis of robustness of voting rules Informal def: given the worst preference profile and a uniform distribution over faults, the robustness of F is the probability of the outcome not changing
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Fault: a “switch” between two adjacent candidates in the preferences of one voter. Depends on representation; consistent, “quite good” representation.
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3 2 1 rank b b c c a a 3 2 1 c c a a b b c c b b
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Fault: a “switch” between two adjacent candidates in the preferences of one voter Depends on representation; consistent, “quite good” representation D k (<) = prob. dist. over profiles; to sample it, start with < and perform k independent uniform switches The k-robustness of voting rule F at < is: k (F,<) = Pr <’ D k (<) [F(<)=F(<’)]
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F = Plurality. 1-Robustness at this profile is 1/3. 2 1 rank a a b b 2 1 a a b b 2 1 b b a a a a b b b b a a
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Fault: a “switch” between two adjacent candidates in the preferences of one voter. Depends on representation; consistent, “quite good” representation. D k (<) = prob. dist. over profiles; sample: start with < and perform k independent uniform faults. The k-robustness of voting rule F at < is: k (F,<) = Pr <’ Dk(<) [F(<)=F(<’)] The k-robustness of voting rule F is: k (F) = min < k (F,<)
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RuleLower BoundUpper Bound Scoring(m-1-a F )/(m-1)(m-a F )/m Copeland01/(m-1) Maximin01/(m-1) Bucklin(m-2)/(m-1)1 Plurality w. Runoff(m-5/2)/(m-1) (m-5/2)/(m-1)+5m/(2m(m-1))
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Consider a class of voting rules Designer/teacher is presented with preference profiles, and designates the winner in each Philosophical justification Practical justification: designer simply wants to find a concise representation Given “target” voting rule, the goal is to find a voting rule from our class that is “close” We’ll look at two instantiations of the class: scoring rules, and voting trees
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1 2 3 a a b b c c a a b b c c a a b b c c a a 1 2 3 c c a a b b c c b b a a b b c c a a c c a a b b c c a a b b c c b b c c a a a a a a b b c c a a b b c c a a b b c c ? a a a a b b c c a a b b c c b b c c a a a a
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Theorem: If there are at least poly(n,m,1/ ,1/ ) examples in the training set, then any “consistent” scoring rule g achieves the goal Such a rule can be efficiently found using LP Example: Scoring rules are efficiently PAC-learnable a a b b c c a a b b c c a a b b c c a a b b c c a a b b c c b b c c a a a a a a find 1, 2, 3 s.t. 3 1 > 3 2 3 1 > 3 3 2 1 + 3 > 1 + 2 2 2 1 + 3 > 2 + 2 3 1 2 3 0
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Scoring rules are efficiently PAC- learnable This means that you only need to ask the designer a polynomial number of queries to get a scoring rule close to what he had in mind
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96 a a c c ? ? ? ? ? ? ? ? b b a a c c c c b b a a a a c c b b b b a a c c c c c c c c b b b b b b a a a a a a
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Voting trees are everywhere!
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Theorem: An exponential training set is needed to learn voting trees. Restrict to the class of voting trees of polynomial size (at most k leaves). Lemma: If the size of this class is (only) exponential, the following alg achieves the goal with polynomial training set: return the tree which minimizes mistakes on the training set. Theorem: |Voting trees with k leaves| exp(m,k)
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In general, many queries to the designer are needed But if restricted to small voting trees (of polynomial size, at most k leaves) then only a polynomial number of queries are needed
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Computational voting theory can be used as a basis for aggregating preferences of automated agents, e.g., Multi-agent Planning Collaborative Filtering Web ranking methods It presents fascinating open challenges on a wide variety of topics, including those we saw today Coalitional Manipulation Robustness Automated Design It can provide domain-specific guidelines for the design of multiagent systems that aggregate preferences
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Algorithms for the Coalitional Manipulation Problem, Michael Zuckerman, Ariel D. Procaccia, and Jeffrey S. Rosenschein. SODA 2008, San Francisco, January 2008, pages 277-286 On the Robustness of Preference Aggregation in Noisy Environments, Ariel D. Procaccia, Jeffrey S. Rosenschein, and Gal A. Kaminka. AAMAS’07, Honolulu, May 2007, pages 416-422 Automated Design of Scoring Rules by Learning from Examples, Ariel D. Procaccia, Aviv Zohar, and Jeffrey S. Rosenschein. AAMAS’08, Estoril, Portugal, May 2008, pages 951-958 Learning Voting Trees, Ariel D. Procaccia, Aviv Zohar, Yoni Peleg, and Jeffrey S. Rosenschein. AAAI’07, Vancouver, British Columbia, July 2007, pages 110-115 101 http://www.cs.huji.ac.il/~jeff/papers.html
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