Communication Issues in Collective Decision Making Nicolas Maudet - PowerPoint PPT Presentation

Communication Issues in Collective Decision Making Nicolas Maudet nicolas.maudet@lip6.fr Universit´ e Pierre et Marie Curie EPCL-BTC: 18th of November 2013

Overview of this course EPCL BTC Nicolas Maudet UPMC EPCL-BTC: 18th of November 2013 In this course I discuss communication issues related to different Voting collective decision making problems. Two-sided Matching Resource 1. This morning I introduce three different collective decision making allocation problems, and show in particular why they raise interesting computational problems. (Part I) 2. After coffee break, I investigate in particular communication issues related to these problems. (Part II). 2 / 64

Overview of this course EPCL BTC Collective decision-making: Nicolas Maudet UPMC ◮ a set of agents N , a set of“options” O EPCL-BTC: 18th of November 2013 ◮ agents have (potentially conflicting) preferences about the options Voting Two-sided ◮ have to agree on a decision (choice of an option) Matching Resource allocation 1. voting ( O = set of candidates) 2. two-sided matching ( O = set of matchings, preferences about agents from the other side) 3. resource allocation ( O = set of allocations, preferences about bundle of resources they hold) Note: Where is computational logic here? In this talk I will signal a“Logic Alert”with this symbol: 3 / 64

Outline of the Talk EPCL BTC Nicolas Maudet UPMC EPCL-BTC: 18th of November 2013 Voting Voting 1 Two-sided Matching Resource allocation Two-sided Matching 2 Resource allocation 3 4 / 64

Motivation Quest for the “best” voting system EPCL BTC Nicolas Maudet UPMC EPCL-BTC: 18th of November 2013 Voting Two-sided Matching Resource allocation Figure: Referendum on Alternative Vote (UK, 2011) 5 / 64

Motivation Manipulating the system: Gerrymandering EPCL BTC Nicolas Maudet UPMC EPCL-BTC: 18th of November 2013 Voting Two-sided Matching Resource allocation Figure: The salamander of Elbridge Gerry (1812) 6 / 64

Motivation Online voting systems EPCL BTC Nicolas Maudet UPMC EPCL-BTC: 18th of November 2013 Voting Two-sided Matching Resource allocation Figure: Choice of a restaurant 7 / 64

Motivation Meta-search engines EPCL BTC Nicolas Maudet UPMC EPCL-BTC: 18th of November 2013 Voting Two-sided Matching Resource allocation Figure: Aggregating search results 8 / 64

Voting: Definitions EPCL BTC 1. a finite set of voters A = { 1 , ..., n } ; Nicolas Maudet 2. a finite set of candidates (alternatives) O ; UPMC 3. a profile = a preference relation (= linear order) on O for each EPCL-BTC: 18th of November 2013 voter Voting P = ( V 1 , . . . , V n ) = ( ≻ 1 , . . . , ≻ n ) Two-sided Matching V i (or ≻ i ) = vote expressed by voter i . Resource 4. P n set of all profiles. allocation 9 / 64

Voting: Definitions EPCL BTC 1. a finite set of voters A = { 1 , ..., n } ; Nicolas Maudet 2. a finite set of candidates (alternatives) O ; UPMC 3. a profile = a preference relation (= linear order) on O for each EPCL-BTC: 18th of November 2013 voter Voting P = ( V 1 , . . . , V n ) = ( ≻ 1 , . . . , ≻ n ) Two-sided Matching V i (or ≻ i ) = vote expressed by voter i . Resource 4. P n set of all profiles. allocation ◮ Voting rule F : P n → O F ( V 1 , . . . , V n ) = socially preferred (elected) candidate ◮ Voting correspondence C : P n → 2 O \ {∅} C ( V 1 , . . . , V n ) = set of socially preferred candidates. ◮ Social welfare function H : P n → P H ( V 1 , . . . , V n ) = social preference relation ( ≻ P ) Note: Rules can be obtained from correspondences by tie-breaking 9 / 64 (usually by using a predefined priority order on candidates).

Positional scoring rules EPCL BTC ◮ n voters, p candidates Nicolas Maudet ◮ fixed list of p integers s 1 ≥ . . . ≥ s p UPMC EPCL-BTC: 18th of ◮ voter i ranks candidate x in position j ⇒ score i ( x ) = s j November 2013 ◮ winner: candidate maximizing s ( x ) = � n Voting i =1 score i ( x ) Two-sided Matching Examples: Resource allocation ◮ s 1 = 1 , s 2 = . . . = s m = 0 ⇒ plurality ; ◮ s 1 = s 2 = . . . = s m − 1 = 1 , s m = 0 ⇒ veto ; ◮ s 1 = m − 1 , s 2 = m − 2 , . . . s m = 0 ⇒ Borda . plurality Borda 2 voters 1 voter 1 voter a �→ 1 a �→ 6 c a d b �→ 0 b �→ 7 b b a c �→ 2 c �→ 6 a d b d �→ 1 d �→ 4 d c c 10 / 64 c winner b winner

Condorcet winner N ( x , y ) = { i | x ≻ i y } set of voters who prefer x to y . EPCL BTC # N ( x , y ) number of voters who prefer x to y . Nicolas Maudet UPMC Condorcet winner EPCL-BTC: 18th of November 2013 for P = �≻ 1 , . . . , ≻ n � : a candidate x such that ∀ y � = x , # N ( x , y ) > n 2 Voting (a candidate who beats any other candidate by a majority of votes). Two-sided Matching Resource a d c allocation Majority graph b b a a c d c b c a d 2 voters out of 3: a ≻ b 2 voters out of 3: c ≻ a b d 2 voters out of 3: a ≻ d b ≻ c 2 voters out of 3: 2 voters out of 3: b ≻ d d ≻ c 2 voters out of 3: 11 / 64

Condorcet winner N ( x , y ) = { i | x ≻ i y } set of voters who prefer x to y . EPCL BTC # N ( x , y ) number of voters who prefer x to y . Nicolas Maudet UPMC Condorcet winner EPCL-BTC: 18th of November 2013 for P = �≻ 1 , . . . , ≻ n � : a candidate x such that ∀ y � = x , # N ( x , y ) > n 2 Voting (a candidate who beats any other candidate by a majority of votes). Two-sided Matching Resource a d c allocation Majority graph b b a a c d c b c a d 2 voters out of 3: a ≻ b 2 voters out of 3: c ≻ a b d 2 voters out of 3: a ≻ d b ≻ c 2 voters out of 3: → No Condorcet winner 2 voters out of 3: b ≻ d ֒ d ≻ c 2 voters out of 3: 11 / 64

Condorcet winner N ( x , y ) = { i | x ≻ i y } set of voters who prefer x to y . EPCL BTC # N ( x , y ) number of voters who prefer x to y . Nicolas Maudet UPMC Condorcet winner EPCL-BTC: 18th of November 2013 for P = �≻ 1 , . . . , ≻ n � : a candidate x such that ∀ y � = x , # N ( x , y ) > n 2 Voting (a candidate who beats any other candidate by a majority of votes). Two-sided Matching Resource a d c allocation Majority graph b b a a c d a b c c d 2 voters out of 3: a ≻ b 2 voters out of 3: a ≻ c b d 2 voters out of 3: a ≻ d b ≻ c 2 voters out of 3: 2 voters out of 3: b ≻ d d ≻ c 2 voters out of 3: 12 / 64

Condorcet winner N ( x , y ) = { i | x ≻ i y } set of voters who prefer x to y . EPCL BTC # N ( x , y ) number of voters who prefer x to y . Nicolas Maudet UPMC Condorcet winner EPCL-BTC: 18th of November 2013 for P = �≻ 1 , . . . , ≻ n � : a candidate x such that ∀ y � = x , # N ( x , y ) > n 2 Voting (a candidate who beats any other candidate by a majority of votes). Two-sided Matching Resource a d c allocation Majority graph b b a a c d a b c c d 2 voters out of 3: a ≻ b 2 voters out of 3: a ≻ c b d 2 voters out of 3: a ≻ d b ≻ c 2 voters out of 3: → a is the Condorcet winner 2 voters out of 3: b ≻ d ֒ d ≻ c 2 voters out of 3: 12 / 64

Condorcet-consistent rules The Copeland rule ◮ Consistency with Condorcet: the voting rule should elect the EPCL BTC Condorcet winner whenever there is one. Nicolas Maudet UPMC ◮ Example: Copeland rule EPCL-BTC: 18th of November 2013 get 1 pt for each pairwise win, 1 2 for a tie, 0 otherwise Voting 2 1 2 Two-sided Matching Majority graph a d c Resource allocation b b a a c d c b c a d 4 voters out of 5: a ≻ b 3 voters out of 5: c ≻ a b d 4 voters out of 5: a ≻ d 3 voters out of 5: b ≻ c 4 voters out of 5: b ≻ d 3 voters out of 5: d ≻ c 13 / 64

Condorcet-consistent rules The Copeland rule ◮ Consistency with Condorcet: the voting rule should elect the EPCL BTC Condorcet winner whenever there is one. Nicolas Maudet UPMC ◮ Example: Copeland rule EPCL-BTC: 18th of November 2013 get 1 pt for each pairwise win, 1 2 for a tie, 0 otherwise Voting 2 1 2 Two-sided Matching Majority graph a d c Resource allocation b b a a c d c b c a d 4 voters out of 5: a ≻ b 3 voters out of 5: c ≻ a b d 4 voters out of 5: a ≻ d C(a) = 2 3 voters out of 5: b ≻ c C(b) = 2 4 voters out of 5: b ≻ d C(c) = 1 3 voters out of 5: d ≻ c C(d) = 1 13 / 64

Condorcet-consistent rules The Simpson rule ◮ Consistency with Condorcet: the voting rule should elect the EPCL BTC Condorcet winner whenever there is one. Nicolas Maudet UPMC ◮ Example: Simpson rule EPCL-BTC: 18th of November 2013 pick the candidate who minimizes the max pairwise defeat Voting 2 1 2 Two-sided Matching (Weighted) Majority graph a d c Resource 3 allocation b b a a c d c b 4 c a d 4 3 3 4 voters out of 5: a ≻ b 3 voters out of 5: c ≻ a b d 4 4 voters out of 5: a ≻ d 3 voters out of 5: b ≻ c 4 voters out of 5: b ≻ d 3 voters out of 5: d ≻ c 14 / 64