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Bounding the Cost of Stability in Games with Restricted Interaction. Reshef Meir, Yair Zick , Edith Elkind and Jeffrey S. Rosenschein COMSOC 2012 (to appear). Cooperative TU Games. Agents divide into coalitions; generate profit. 1. $5. 2. Coalition members can freely divide profits. 3.
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Bounding the Cost of Stability in Games with Restricted Interaction Reshef Meir, YairZick, Edith Elkind and Jeffrey S. Rosenschein COMSOC 2012 (to appear)
Cooperative TU Games Agents divide into coalitions; generate profit. 1 $5 2 • Coalition members can freely divide profits. 3 4 $2 5 6 • How should profits be divided? $3
TU Games - Notations • Agents: N= {1,…,n} • Coalition: S µ N • Characteristic function: v: 2N→ R • A TU game is simple, if every coalition either wins or loses, i.e. v: 2N→ {0,1} • A TU game is monotone, if the value of a coalition can only increase by adding more agents to it.
Payoffs • Agents may freely distribute profits. • An outcomeis a coalition structure CS and a vector x = (x1,…,xn) such that Σi2S xi= v(S) for all SinCS • Individual rationality: each agent gets at least what she can make on her own: xi≥v({i})
The Core • The coreis the set of all stable outcomes: for all S µ N we have x(S)¸ v(S) • May be empty in many games. • Example: the 3-majority game. • Three players; any set of size two or more has a value of 1; singletons have a value of 0.
Restricted Cooperation • Some coalitions may be impossible or unlikely due to practical reasons • an underlying communication network (Myerson’77). • agents are nodes. • A coalition can form only if its agents are connected. 1 2 4 9 11 3 5 6 10 8 12 7
Restricted cooperation - example • The coalition {2,9,10,12} is allowed • The coalition {3,6,7,8} is not allowed 1 2 4 9 11 3 5 6 10 8 12 7
Restricted cooperation increases stability Theorem [Demange’04]:If the underlying communication network H is a tree, then the core is non-empty. Moreover, a core outcome can be computed efficiently. 1 2 4 9 11 3 5 6 10 8 12 7
Using Subsidies to Stabilize the game • Originally, we divided OPT(G)between the agents. • We increase the value of OPT(G), creating a “superimputation”. • Division of the incremented value α∙OPT(G)
The Cost Of Stability (CoS) (Bachrach et al., SAGT’09) • Observation: With a big enough payment, any game can be stabilized • α ≤ n • The Cost of Stability (CoS) is the minimal subsidy αthat stabilizes the game. i.e. allows a non-empty core in G(α)
Back to our example • 3-majority game(core is empty) • By distributing a total payoff of 1½ (rather than 1), the core of G(1½) is non-empty. • x = (½, ½, ½) is a stable superimputation. • CoS(G) ≤ 1½ • This bound is tight! No lower subsidy will stabilize the game. • CoS(G) = 1½
CoS with restricted cooperation • Recall that by [Demange’04] : if H is a tree, then the core is non-empty (i.e. CoS= 1). What is the connection between graph complexity and the cost of stability? Theorem [Meir et al., IJCAI’11]: If Hcontains a single cycle, then CoS(G|H) ≤ 2, and this is tight
Graphs and tree-width • Combinatorial measures to the “complexity” of a graph: • Degree • Path-width • Tree-width • Many NP-hard combinatorial problems become easy when the tree-width is bounded. 1 2 4 9 11 3 5 10 6 8 7 1,2,3 2,4 2,5,9 5,9,10 5,6,8 6,7,8 9,11 5,8,10
Conjectured Connections Conjecture [MRM’11]: Let d be the maximal degree in H, then CoS(G|H) ≤ d There are games on a 3-dimensional grid (d = 6) with unbounded CoS Conjecture: Let k be the tree-width of H, then CoS(G|H) ≤ k This is “almost” true.
Our Main Result Theorem: For any G with an interaction graph H CoS(G|H) ≤ tw(H) + 1 and this bound is tight. Also, a stable payoff vector can be found efficiently in the case of simple, monotone games.
Step 1 – Simple Games Traverse the nodes from the leaves up. Once the subtree contains a winning coalition, pay 1 to all agents in its root. Delete agents. 1 1 1 2,5,9 • 2,9 1,2,3 2,4 5,9,10 • 9 5,8,10 9,11 5,6,8 6,7,8 {5,6,8,10}
Lemma: For any simple G with an interaction graph H, the algorithm produces a stable imputation xsuch that x(N) ≤ (tw(H) + 1)OPT(G|H) • Stability: every winning coalition intersects a node in the tree decomposition that was paid by the algorithm; thus gets at least 1.
Bounded payoff: let St be the set of agents that were removed at time t. • St contains a winning coalition Wt • We can partition the agents into a coalition structure CS = {{Wt}t2T*, L}. • T* is the set of all times where sets were pruned by the algorithm. • The value of CS is at most |T*|. • x(N) ≤ (tw(H) + 1) |T*| ≤ (tw(H) + 1)OPT(G|H)
Step 2 – The General Case Given a general (integer) game, split it into simple games and stabilize each individually. Sum the resulting stable imputations.
Tightness a1 a2 b1 b2 a3 a4 b3 b4 z1 z2 z3 W1,1 = {z1; a1; a4; b3; b1} W1,2 = {z1;a2; a3; b2; b4} W2,1 = {z2; b1; b4; c3; c1} W2,2 = {z2; b2; b3; c2; c4} W3,1 = {z3; c1; c4; a3; a1} W3,2 = {z3; c2; c3; a2; a4} Any two winning coalitions intersect: optimal value is 1. c1 c2 c3 c4
Tightness x(W1,1)¸1 a1 a2 b1 b2 xz1 ¸1 - ½(x(A) + x(B)) x(W1,2)¸1 a3 a4 b3 b4 z1 z2 B A Z z3 W1,1 = {z1; a1; a4; b3; b1} W1,2 = {z1;a2; a3; b2; b4} W2,1 = {z2; b1; b4; c3; c1} W2,2 = {z2; b2; b3; c2; c4} W3,1 = {z3; c1; c4; a3; a1} W3,2 = {z3; c2; c3; a2; a4} x(Z) ¸3 - (x(A) + x(B) + x(C)) c1 c2 x(N) ¸3 c3 c4 C
Discussion/Future Work • A slightly better (tight) bound holds for the pathwidth of the interaction graph: we can drop the +1. • Bounded tree-width does not facilitate computations (e.g. Greco et al.’11) • Other graphical models of cooperative games? Non-cooperative games? • Other measures of graph complexity?
Thank you! Questions?