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

6

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5

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2Coalition members can freely divide profits.

How should profits

be divided?

$5

$3

$2

TU Games - NotationsAgents: N = {1,…,n}Coalition: S µ NCharacteristic function: v: 2N → RA 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 outcome is a coalition structure CS

and a vector x = (x1,…,xn) such that

Σi2S xi= v(S) for all S in CS

Individual rationality: each agent gets at least what she can make on her own: xi ≥ v({i})

The CoreThe 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.

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.

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Restricted Cooperation

Restricted cooperation - example

The coalition {2,9,10,12} is allowedThe coalition {3,6,7,8} is not allowed

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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.

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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)

Observation: With a big enough payment, any game can be stabilizedα ≤ nThe 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)

Back to our example3-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 H contains a single cycle, then CoS(G|H) ≤ 2, and this is tight

Graphs and tree-widthCombinatorial measures to the

“complexity” of a graph:DegreePath-widthTree-width

Many NP-hard combinatorial problems become easy when the tree-width is bounded.

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1,2,3 2,4

2,5,9

5,9,10

5,8,10

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6,7,8 9,11

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.

Our Main Result

Theorem: For any G with an interaction graph H

CoS(G|H) ≤ tw(H) + 1and this bound is tight.

Also, a stable payoff vector can be found efficiently in the case of simple,

monotone games.

Step 1 – Simple Games

1,2,3 2,4

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 x2 x3 x4 x5 x6 x7 x8 x9 x10)

( 0 0 0 0 0 0 0 0 0 0 )1 1 1

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2,9

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)

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 Case1.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)

Tightness

a1 a2

a4a3

c1 c2

c4c3

z1

z3

z2

b1 b2

b4b3

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.

Tightness

a1 a2

a4a3

c1 c2

c4c3

z1

z3

z2

b1 b2

b4b3

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(W1,1) ¸ 1

x(W1,2) ¸ 1xz1

¸ 1 - ½ (x(A) + x(B))

A B

C

Z

x(Z) ¸ 3 - (x(A) + x(B) + x(C))

x(N) ¸ 3

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?