1. Bounding the Cost of Stability in Games with Restricted Interaction Reshef Meir, Yair Zick, Edith Elkind and Jeffrey S. Rosenschein COMSOC 2012 (to appear)

2. Cooperative TU Games Agents divide into coalitions; generate profit. 1 6 4 5 3 2 Coalition members can freely divide profits. How should profits be divided? $5 $3 $2

3. TU Games - Notations Agents: N = {1,…, n } Coalition: S µ N Characteristic function: v : 2 N → R A TU game is simple, if every coalition either wins or loses, i.e. v : 2 N → {0,1} A TU game is monotone, if the value of a coalition can only increase by adding more agents to it.

4. Payoffs Agents may freely distribute profits. An outcome is a coalition structure CS and a vector x = ( x 1,…, x n ) such that Σ i 2 S x i = v ( S ) for all S in CS Individual rationality: each agent gets at least what she can make on her own: x i ≥ v ({ i })

5. The Core The core is 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.

6. 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 3 4 5 6 7 8 9 10 11 12 Restricted Cooperation

7. Restricted cooperation - example The coalition {2,9,10,12} is allowed The coalition {3,6,7,8} is not allowed 1 2 3 4 5 6 7 8 9 10 11 12

8. 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 3 4 5 6 7 8 9 10 11 12

9. 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 )

10. The Cost Of Stability (CoS) Observation: With a big enough payment, any game can be stabilized α ≤ n The Cost of Stability (CoS) is the minimal subsidy α that stabilizes the game. i.e. allows a non-empty core in G (α) (Bachrach et al., SAGT’09)

11. 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

12. 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 H contains a single cycle, then CoS ( G | H ) ≤ 2, and this is tight

13. 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 3 4 5 6 7 8 9 10 11 1,2,3 2,4 2,5,9 5,9,10 5,8,10 5,6,8 6,7,8 9,11

14. Conjectured Connections Conjecture [MRM’11]: Let d be the maximal degree in H, then CoS ( G | H ) ≤ d Conjecture: Let k be the tree-width of H, then CoS ( G | H ) ≤ k There are games on a 3-dimensional grid ( d = 6 ) with unbounded CoS This is “almost” true.

15. Our Main Result Theorem: For any G with an interaction graph H CoS ( G | H ) ≤ tw ( H ) + 1 and this bound is tight. 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.

16. Step 1 – Simple Games 1,2,32,4 2,5,9 5,9,10 5,8,10 5,6,8 6,7,8 9,11 {5,6,8,10} 1.Traverse the nodes from the leaves up. 2.Once the subtree contains a winning coalition, pay 1 to all agents in its root. 3.Delete agents. (x1(x1 x2x2 x3x3 x4x4 x5x5 x6x6 x7x7 x8x8 x9x9 x 10 ) ( 0000000000 ) 1 11 9 2,9

17. Stability: every winning coalition intersects a node in the tree decomposition that was paid by the algorithm; thus gets at least 1. Lemma: For any simple G with an interaction graph H, the algorithm produces a stable imputation x such that x ( N ) ≤ ( tw ( H ) + 1) OPT ( G | H ) Lemma: For any simple G with an interaction graph H, the algorithm produces a stable imputation x such that x ( N ) ≤ ( tw ( H ) + 1) OPT ( G | H )

18. Bounded payoff: let S t be the set of agents that were removed at time t.  S t contains a winning coalition W t  We can partition the agents into a coalition structure CS = {{ W t } t 2 T*, 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 )

19. Step 2 – The General Case 1.Given a general (integer) game, split it into simple games and stabilize each individually. 2.Sum the resulting stable imputations. v ({1}) v ({2}) v ({3}) v ({1,2}) v ({1,3}) v ({2,3}) v(N)v(N)

20. Tightness a1a1 a2a2 a4a4 a3a3 c1c1 c2c2 c4c4 c3c3 z1z1 z3z3 z2z2 b1b1 b2b2 b4b4 b3b3 W 1,1 = { z 1 ; a 1 ; a 4 ; b 3 ; b 1 } W 1,2 = { z 1 ; a 2 ; a 3 ; b 2 ; b 4 } W 2,1 = { z 2 ; b 1 ; b 4 ; c 3 ; c 1 } W 2,2 = { z 2 ; b 2 ; b 3 ; c 2 ; c 4 } W 3,1 = { z 3 ; c 1 ; c 4 ; a 3 ; a 1 } W 3,2 = { z 3 ; c 2 ; c 3 ; a 2 ; a 4 } Any two winning coalitions intersect: optimal value is 1.

21. Tightness a1a1 a2a2 a4a4 a3a3 c1c1 c2c2 c4c4 c3c3 z1z1 z3z3 z2z2 b1b1 b2b2 b4b4 b3b3 W 1,1 = { z 1 ; a 1 ; a 4 ; b 3 ; b 1 } W 1,2 = { z 1 ; a 2 ; a 3 ; b 2 ; b 4 } W 2,1 = { z 2 ; b 1 ; b 4 ; c 3 ; c 1 } W 2,2 = { z 2 ; b 2 ; b 3 ; c 2 ; c 4 } W 3,1 = { z 3 ; c 1 ; c 4 ; a 3 ; a 1 } W 3,2 = { z 3 ; c 2 ; c 3 ; a 2 ; a 4 } x(W1,1) ¸ 1x(W1,1) ¸ 1 x ( W 1,2 ) ¸ 1 x z 1 ¸ 1 - ½ ( x ( A ) + x ( B )) A B C Z x ( Z ) ¸ 3 - ( x ( A ) + x ( B ) + x ( C )) x ( N ) ¸ 3

22. 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?

23. Thank you! Questions?