• Voting Input: Preference of agents over a set of candidates or outcomes Output: one candidate or outcome (or a set) Introduction to Tournaments • Tournament Stéphane Airiau Input: Binary relation between outcomes or candidates Output: One candidate or outcome (or a set) ILLC COMSOC 2009 When no ties are allowed between any two alternatives. Either x beats y or y beats x . which are the best outcomes? Stéphane Airiau (ILLC) COMSOC 2009 1 / 47 Stéphane Airiau (ILLC) COMSOC 2009 2 / 47 Notations Definition (Tournament) The relation T is a tournament iff 1 ∀ x ∈ X ( x, x ) / ∈ T 2 ∀ ( x, y ) ∈ X 2 x � = y ⇒ [(( x, y ) ∈ T ) ∨ (( y, x ) ∈ T )] X is a finite set of alternatives. 3 ∀ ( x, y ) ∈ X 2 ( x, y ) ∈ T ⇒ ( y, x ) / ∈ T . T is a relation on X , i.e, T ⊂ X 2 . A tournament is a complete and asymmetric binary relation notation: ( x, y ) ∈ T ⇔ xTy ⇔ x → y ⇔ x “beats" y T ( X ) is the set of tournaments on X Majority voting and tournament: T + ( x ) = { y ∈ X | xTy } : successors of x . • I finite set of individuals. The preference of an individual i is T − ( x ) = { y ∈ X | yTx } : predessors of x . represented by a complete order P i defined on X . s ( x ) = # T + ( x ) is the Copeland score of x . • The outcome of majority voting is the binary relation M ( P ) on X such that ∀ ( x, y ) ∈ X, xM ( P ) y ⇔ # { i ∈ I | xP i y } > # { i ∈ I | yP i x } If initial preferences are strict and number of individual is odd, M ( P ) is a tournament. Stéphane Airiau (ILLC) COMSOC 2009 3 / 47 Stéphane Airiau (ILLC) COMSOC 2009 4 / 47
Definition (isomorphism) Example (cyclone of order n ) Let X and Y be two sets, T ∈ T ( X ) , U ∈ T ( Y ) two tournaments on X and Y . A mapping φ : X → Y is a tournament isomorphism iff 2 1 φ is a bijection ∀ ( x, y ) ∈ X 2 , xTx ′ ⇔ φ ( x ) Uφ ( x ′ ) Z n set of integers modulo n . 3 1 , . . . , n − 1 � � xC n y ⇔ y − x ∈ 2 7 n · ( n − 1) T + (1) = { 2 , 3 , 4 } On a set X of cardinal n , there are 2 tournaments, but many of 2 T − (1) = { 5 , 6 , 7 } them are isomorphic. 4 6 number of n ( n − 1) 5 n 2 2 non-isomorphic tournaments 8 268,435,456 6,880 10 35,184,372,088,832 9,733,056 Stéphane Airiau (ILLC) COMSOC 2009 5 / 47 Stéphane Airiau (ILLC) COMSOC 2009 6 / 47 Outline Condorcet principle 1 Introduction: Reasoning about pairwise competition 2 Desirable properties of solution concepts Definition (Condorcet winners) 3 Solution based on scoring and Ranking Let T ∈ T ( X ) . The set of Condorcet winners of T is 4 Solutions based on Covering C ondorcet ( T ) = { x ∈ X | ∀ y ∈ X, y � = x ⇒ xTy } 5 Solution based on Game Theory Property 6 Contestation Process Either C ondorcet ( T ) = ∅ or C ondorcet ( T ) is a singleton. 7 Knockout tournaments 8 Notes on the size of the choice set Stéphane Airiau (ILLC) COMSOC 2009 7 / 47 Stéphane Airiau (ILLC) COMSOC 2009 8 / 47
A first solution: the Top Cycle (TC) Definition (Tournament solution) A tournament solution S associates to any tournament T ( X ) a subset S ( T ) ⊂ X and satisfies ∀ T ∈ T ( X ) , S ( T ) � = ∅ Definition (Top Cycle) For any tournament isomorphism φ , φo S = S oφ (anonymity) The top cycle of T ∈ T ( X ) is the set TC defined as ∀ T ∈ T ( X ) , C ondorcet ( T ) � = ∅ ⇒ S ( T ) = C ondorcet ( T ) ∃ ( z 1 , . . . , z k ) ∈ X k , z 1 = x, z k = y , For S , S 1 , S 2 tournament solutions. TC ( T ) = x ∈ X | ∀ y ∈ X, ∃ k > 0 and S 1 o S 2 ( T ) = S 1 ( T/ S 2 ( T )) = S 1 ( S 2 ( T )) 1 ≤ i < j ≤ k ⇒ z i Tz j S 1 = S , S k +1 = S o S k , S ∞ = lim k →∞ S k The top cycle contains outcomes that beat directly or indirectly every solutions may be finer/more selective: other outcomes. S 1 ⊂ S 2 ⇔ ∀ T ∈ T ( X ) S 1 ( T ) ⊂ S 2 ( T ) than S 2 . solutions may be different: z 2 z 3 y x S 1 ∅ S 2 ⇔ ∃ T ∈ T | S 1 ( T ) ∩ S 2 ( T ) = ∅ solution may have common elements: S 1 ∩ S 2 ⇔ ∀ T ∈ T | S 1 ( T ) ∩ S 2 ( T ) � = ∅ Stéphane Airiau (ILLC) COMSOC 2009 9 / 47 Stéphane Airiau (ILLC) COMSOC 2009 10 / 47 Properties of Solutions Definition (Regular tournament) A tournament is regular iff all the points have the same Copeland score. Regular Definition (Monotonous) Monotonous A solution S is monotonous iff ∀ T ∈ T ( X ) , ∀ x ∈ S ( T ) , ∀ T ′ ∈ T ( X ) Independent of the losers � T ′ /X \ { x } = T/X \ { x } Strong Superset Property such that ∀ y ∈ X , xTY ⇒ xT ′ y Idempotent one has x ∈ S ( T ′ ) Aïzerman property “Whenever a winner is reinforced, it does not become a loser.” Composition-consistent and weak composition-consistent Stéphane Airiau (ILLC) COMSOC 2009 11 / 47 Stéphane Airiau (ILLC) COMSOC 2009 12 / 47
Definition (Idempotent) Definition (Independence of the losers) A solution S is idempotent iff S o S = S . A solution S is independent of the losers iff ∀ T ∈ T ( X ) , ∀ T ′ ∈ T ( X ) such that ∀ x ∈ S ( T ) , ∀ y ∈ X , xTy ⇔ xT ′ y S ( T ) one has S ( T ) = S ( T ′ ) . X � winners to winners “the only important relations are ” winners to losers “What happens between losers do not matter.” Definition (Aïzerman property) A solution S satisfies the Aïzerman property iff ∀ T ∈ T ( X ) , ∀ Y ⊂ X Definition (Strong Superset Property (SSP)) S ( T ) ⊂ Y ⊂ X ⇒ S ( T/Y ) ⊂ S ( T ) A solution S satisfies the Strong Superset Property (SSP) iff ∀ T ∈ T ( X ) , ∀ Y | S ( T ) ⊂ Y ⊂ X S ( T ) one has S ( T ) = S ( T/Y ) Y X “We can delete some or all losers, and the set of winners does not change” Stéphane Airiau (ILLC) COMSOC 2009 13 / 47 Stéphane Airiau (ILLC) COMSOC 2009 14 / 47 Solution Concepts Copeland solution (C) UC ∞ TC UC MC BP B TEQ SL C the Long Path (LP) Monotonicity ? method for ranking Independence of the losers ? Markov solution (MA) Idempotency ? Aïzerman property ? Slater solution (SL) Strong superset property ? Uncovered set (UC) Composition-consistency based on the notion of Iterations of the Uncovered set ( UC ∞ ) Weak Comp.-consist. covering Regularity Dutta’s minimal covering set (MC) Copeland value 1 1 1/2 1/2 1/2 ≤ 1/3 ≤ 1/3 1/2 1 O ( n 2 ) O ( n 2 . 38 ) O ( n 2 ) Complexity P NP -hard NP -hard NP -hard Bipartisan set (BP) Game theory based Bank’s solution (B) Based on Contestation Tournament equilibrium set (TEQ) Stéphane Airiau (ILLC) COMSOC 2009 15 / 47 Stéphane Airiau (ILLC) COMSOC 2009 16 / 47
Outline UC ∞ TC UC MC BP B TEQ C 1 Introduction: Reasoning about pairwise competition ⊂ UC UC ∞ ⊂ ⊂ 2 Desirable properties of solution concepts MC ⊂ ⊂ ⊂ BP ⊂ ⊂ ⊂ ⊂ 3 Solution based on scoring and Ranking B ⊂ ⊂ ∩ ∩ a ⊂ ⊂ ⊂ ⊂ TEQ b a 4 Solutions based on Covering ∅ ∅ ∅ ∅ ∅ C ⊂ ⊂ ∅ ∅ ∅ ∅ ∅ ∅ SL ⊂ ⊂ 5 Solution based on Game Theory 6 Contestation Process a ∃ T ∈ T 29 | B ( T ) ⊂ BP ( T ) and B ( T ) � = BP ( T ) ∃ T ′ ∈ T 6 | BP ( T ′ ) ⊂ B ( T ′ ) and B ( T ′ ) � = BP ( T ′ ) . It is unknown if B ∩ BP can be empty. 7 Knockout tournaments Same for TEQ and BP. b TEQ ⊂ MC is a conjecture 8 Notes on the size of the choice set Stéphane Airiau (ILLC) COMSOC 2009 17 / 47 Stéphane Airiau (ILLC) COMSOC 2009 18 / 47 Recall: Copeland score s ( x ) = | T + ( x ) | = |{ y ∈ X | xTy }| Definition (Slater, Kandall, or Hamming distance) s ( x ) is the number of alternatives that x beats. Let ( T, T ′ ) ∈ T ( X ) ∆( T, T ′ ) = 1 ( x, y ) ∈ X 2 | xTy ∧ yT ′ x Definition (Copeland solution (C)) � � 2# Copeland winners of T ∈ T ( X ) is How many arrows are flipped in the tournament graph? C ( T ) = { x ∈ X | ∀ y ∈ X, s ( y ) = s ( x ) } Definition (Slater order) 3 Let T ∈ T ( X ) . a A Slater order for T is a linear order U ∈ L ( X ) such that 2 2 c b V ∈ L ( X ) { ∆( T, V ) } ∆( T, U ) = min where L ( X ) is the set of linear order over X . The set of Slater winners of T , noted SL ( T ) , is the set of alternatives e d 1 2 in X that are Condorcet winner of a Slater order for T . idea: approximate the tournament by a linear order. Stéphane Airiau (ILLC) COMSOC 2009 19 / 47 Stéphane Airiau (ILLC) COMSOC 2009 20 / 47
Recommend
More recommend
