Weighted Voting Games

Transcription

1 Weighted Voting Games R.Chandrasekaran, MostofthisfollowsCEWandfewpapers Assume that we are given a game in characteristic function form with v:2 N R. Definition1 Agamev isasimplegameifitsatisfiestherelation v(s) {0,1} foralls N Coalitions S such that v(s) = 1(0) are called winning(losing) coalitions. Definition2 Agameismonotonicifitsatisfiestherelation: [S T] v(s) v(t) Definition3 Letv beagameonasetofplayersn ={1,2,...,n}whichis definedbyaninput{w R n +,q R + }asfollows: { 1 v(s)= 0 if j S w j q otherwise Thisgamerepresentedby{N,w,q}isknownasaweightedvotinggame. Such games are clearly monotonic games. Example 1 The following monotone simple game is NOT a weighted voting game(from CEW) v(s)= 1 0 if S 3;orS {(1,2),(2,3),(1,4),(3,4)} otherwise UniversityofTexasatDallas,Richardson,Texas,U.S.A.. 1

2 Itispossiblethat(N,w,q)and(N,ŵ,ˆq)representthesamegameeven if(w,q) (ŵ,ˆq). Theorem4 Givenaweightedvotinggame[N,w,q ],thereexistsanidenticalweightedvotinggame[n,w,q]inwhichw i =O(2 nlgn )andq=o(2 nlgn ). Hencethenumberofbitsneededfortheinputispolynomialinn. Definition 5 For a monotone simple game[n, v], the following partial order on v onn iscalledadesirabilityrelationbetweenplayers: [i v j] v(s {i}) v(s {j})foralls suchthati / S,j / S [i v j] v(s {i})=v(s {j})foralls suchthati / S,j / S [i v j] v(s {i}) v(s {j})foralls suchthati / S,j / S [i v j] [i v j]and [i v j] [i v j] [i v j]and [i v j] Ifnoneoftheseholdforapairi,j,wesaythatthesenotcomparable. Definition6 A monotone simple game in which no pair of players is not comparable under is called a linear game. All weighted voting games are linear(under that order that corresponds to increasing order of weights.); but the converse is not true. Definition7 A linear game is canonical if n. A weighted votinggameiscanonicalifw 1 w 2... w n. Recallthatinasimplegameplayeriiscalledavetoplayer if andplayeriisadummyplayerif i / S v(s)=0 i / S v(s {i})=v(s) Supposewewanttoknowifplayeriisaveto(dummy)playerinaweighted votinggame[n,w,q]. Clearly,checkingifplayeriisavetoplayerrequires us to check only that v(n {i}) = 0 for a simple monotone game. For aweightedvoting game this requires thatwe checkif j i w j < q or not. Hence this is polynomial-time solvable. On the other hand checking if player iisadummyplayerinaweightedvotinggameisnp-complete. 2

3 Problem 8 DUMMY Input: Aweightedvotinggame[N,w,q];andaplayeri N Question: Isiadummyin[N,w,q]? (i) DUMMY co-np: There is a polynomial length certificate and a verificationalgorithmfornoinstances. ToshowthatthegiveninstanceisaNO instance,weneedtoshowthat S N;i / Sthatsatisfiesv(S {i}) v(s). Inthiscase,sincethegameissimple,weneedS:i / S;v(S)=0;v(S {i})= 1. OurcertificateisthesetSitself. Tocheck(byourverificationalgorithm) wechecktoseethattherelations i / S w j <q j S j S {i} w j q holdandthiscanclearlybedoneinpolynomialtime. (ii) To show that DUMMY co-np-complete: we take a problem known to be NP-complete and reduce it(by a polynomial transformation) to DUMMY. We use PARTITION for this purpose Problem 9 PARTITION Input:Positiveintegersx 1,x 2,...,x n,kthatsatisfytherelation n j=1 x j= 2K Question: IsthereasubsetS {1,2,...,n}suchthat j S x j=k? Now we must show the transformation that converts an arbitrary instance of PARTITION to a corresponding instance of DUMMY. Here"corresponding"meansthefollowing: iftheinstanceofpartitionis"yes"thenthe corresponding instance of DUMMY is"no". Corresponding instance of DUMMY: Input: G=[N,w,q]: where N ={1,2,...,n,n+1};w j =2x j ;1 j n;w n+1 =1;q=2K+1. Itiseasytoseethat{x 1,x 2,...,x n,k}isa"yes"instanceofpartition player n+1isnotadummyinthecorrespondinginstanceofdummy. Sinceweightedvotinggamesaremonotone,iisadummyiffhisShapley valueis0. NextweshowhowtocomputeShapleyvalueintimepoly(n,w max ) making such an algorithm pseudo-polynomial. There is a similar result for PARTITION as well based on dynamic programming technique. 3

4 Power Indices for Weighted Voting Games: For a weighted voting game [N,w,q] it is easy to show that [w i w j ] [ϕ i ϕ j ]; but it is possiblethatw i <w j andϕ i =ϕ j. Proposition10 Given two games G=[N,w,q] and G =[N,w,q ] where q = n j=1 w j q+1 (here we assume that all input is integral), we have ϕ i (G)=ϕ i (G )foralli N. Since checking whether a player is a dummy is co-np-hard, checking whether a player s Shapley Value is 0 is also equally difficult. Similar results also hold for Banzhaf index. However, the following result is somewhat useful in practice. Theorem11 Givenann-playerweightedvotinggameG=[N,w,q] anda playeri N,wecancomputeBanzhafindexβ i (G)andShapleyValueϕ i (G) intimeo(n 2 w max )ando(n 3 w max )respectively. Proof. Wedescribetheprocessforcomputingϕ i (G)firstandindicatemodificationsthatreducetheorderforcomputingβ i (G). Withoutlossassume i = n. [This is done for convenience only.] If w n = 0, then β n = ϕ n = 0. So we will assume from now on that w n > 0. For s = 0,1,2,...,n 1, let N s denote the number of subsets of N {n} with size s that have weight W {q w n,q w n 1,...,q 1}. If S is any coalition such that n / S, and W = j S w j < q;w +w n q, it is clear that W {q w n,q w n 1,...,q 1} and conversely. So coalitions for which n is a swing player, are all of this type. Of course the members of such coalitionmay"arrive"inanyorderfollowedbynandthentherestincomputing Shapleyvalue. Thereares!(n s 1)!suchpermutationsofthemembersof suchaset. Hencewehave: ϕ n (G)= 1 n 1 s!(n s 1)!N s n! s=0 Hence if we can compute the values of N s for s = 0,1,2,...,n 1, we can compute Shapley value. We do this by dynamic programming. Let M[j,W,s] = number of s-element subsets of {1,2,...,j} that have weight equal to W. Here j {1,2,...,n 1};s {0,1,...,n 1};W {0,1,..., n j=1 w j}. Thenumberofthevaluescomputedequalsn 2 n j=1 w j= O(n 3 w max ). Wedothisincreasingorderofvaluesofj+W+s. 4

5 Basis Cases: { 1 M[j,W,0]= 0 { 1 M[1,W,s]= 0 General Case: ifw =0 otherwise ifw =w 1 ands=1 otherwise Finally, M[j,W,s]=M[j 1,W,s]+M[j 1,W w j,s 1] N s = q 1 W=q w n M[n 1,W,s] ForcomputingBanzhaf indexweomitthe dependenceonsandhencethe algorithmhasoneorderofnless. Stability: Definition12 Given a game with a player set N, a coalition structure CS is a partition of the players into nonempty subsets. CS={S 1,S 2,...S k };S i ϕforalli; [i j] S i S j =ϕ k i=1 S i=n Definition13 Given a set N of players we denote by CS N the set of all coalition structures on N. Recall that in a superadditive game (which is necessarily monotone), the only stable coalition structure is {(N)}. If a simple game is superadditive,thentherecannotexisttwodisjointwinningcoalitionss andt. In nj=1 w j 2, the game will be superadditive. aweightedvotinggame,ifq> Question#1: Is the above condition necessary? Question#2: Is checking whether a weighted voting game superadditive poly-time solvable? Question#3: Is checking whether a monotone simple game superadditive poly-time solvable? 5

6 Wehavealreadyshownthatifasimplegameissuperadditive, ithasa nonemptycoreiffthereisatleastonevetoplayer. Checkingifasimplegame has a veto player is poly-time solvable as mentioned before. Hence, checking if a weighted voting game has a nonempty core is poly-time solvable. Checking whetherapayoffvectorxisinthecoreisalsopoly-timesolvable. However, a weighted voting game may not be superadditive. This kind of situation arises in multi-agent systems. Now we allow for payoff configurations[cs,x]wherecs {(N)}. Definition14 LetacoalitionstructurebegivenbyCS={S 1,S 2,...,S k }. A payoffconfiguration[cs,x]issaidtobeefficientif x j =v(s i ) fori=1,2,...,k j S i Definition15 A payoff vector x for a coalition structure CS CS N is saidtobeanimputationifitisefficientforcs andsatisfiestheindividual rationality condition: x j v({j})forallj N WedenotebyE CS (v)thesetofallimputationsforthegamevwithcoalition structure CS. Definition16 GivenacoalitionstructureCS foragameonn,thesetof imputations that satisfy the relations: x j v(t)forallt N j T iscalledthecs-coreofthegamev. [Pleasenote: Whenwetalkofcorewe wantittobeanimputationwithrespecttothecoalitionstructurecs={(n)} andhencerequireonlythat n j=1 x j=v(n).] Hence it is possible that for games that are not superadditive, there may bestablepayoffconfigurationsoftheform[cs,x]wherecs {(N)}. These are the subject of the CS core. 6

7 Lemma17 Let G=[N,w,q] be weighted voting game. Suppose there is a coalitionstructurecs={s 1,S 2,...,S k }suchthat j S i w j =q fori=1,2,...,k ThenCS-coreofGisnonempty. Proof. Letx j = w j ;j=1,2,...,n. Considerthepayoffconfiguration[CS,x]. q w j q =1fori=1,2,...,k j S i [v(t)=1] [ j T W j q] j T x j 1=v(T) Hence the result follows. Problem18(-CS-CORE) Input: AweightedvotinggameG=[N,w,q] Question: Is the CS-core of G nonempty? Theorem 19 CS-CORE is NP-hard Proof. WewillshowthatPARTITION p CS-CORE GivenaninstanceofPARTITION:Positiveintegersx 1,x 2,...,x n,k that satisfytherelation n j=1 x j=2k Weassumethatx j K forallj [otherwise,theanswertopartition instance is NO]. CorrespondinginstanceofweightedvotinggameisG=[N,w,q]where N ={1,2,...,n};w j =x j forj=1,2,...,n;andq=k. Now if the PARTITION instance is a "YES" instance, then there is a subsets N suchthat j S x j=k andhencethereisnonemptycs-core forcs={s,n S}. If the PARTITION instance is a "NO" instance, then in any coalition structure CS of N, there is at most one wining coalition. Hence the total payoff for all players is at most 1 for any payoff configuration. If CS has no winning coalition, then the coalition structure is not stable since grand coalitionwillhaveavalueequalto1. Considerapayoffconfiguration[CS,x] withcs={s 1,S 2,...,S k }where v(s i )=0fori 1 7

8 Hence if [CS,x] is in CS-Core, then i S 1 such that x i > 0. Hence j i x j < 1. But since w i K, it follows that j i w j > q, and hence v({n {i})=1. Hence[CS,x]isnotinCS-Core. HenceCS-Coreisempty in this case. The reduction is complete. Checking whether [CS, x] is in CS-Core is also co-np-complete, but solvable by a pseudo-polynomial time algorithm. 8