Computational aspects of stability in

weighted voting games

Edith Elkind (NTU, Singapore)

Based on joint work with Leslie Ann Goldberg, Paul W. Goldberg,

Michael Wooldridge, and Dmitrii Pasechnik

Cooperative vs. non-cooperative games

• Non-cooperative games:– each player can choose an action– payoffs are determined by the action profile

• Cooperative (coalitional) games:– players can form groups (coalitions)– payoff to a group determined by its composition– players have to share group payoff

Coalitional games: formal model

• G = (N, v)• N={1, ..., n}: set of players• v: 2N→R: characteristic function

– v(S): payoff available to S, has to be shared among members of S

• p=(p1, ..., pn) is an imputation if– pi ≥ 0 for i =1, ..., n

– p(N) := S i in N pi = v(N)

Coalitional games with compact representations

• Weighted voting games (subject of this talk):– G = (w1, ..., wn; T)

– each player i has a weight wi

– threshold T– v(S)=1 if w(S) ≥ T, v(S)=0 otherwise

• Network flow games:– players are edges of a network– value of a coalition = size of the flow it can carry

• Minimum spanning tree games, matching games, etc.

Coalitional games: stability

• Which imputations are stable?– no subset of players should want to deviate

• Core: set of all stable imputations(p1, ..., pn) is in the core if p(S) ≥ v(S) for all S N

• Problem: core may be emptyweighted voting game G=(1, 1, 1; 2) suppose wlog p1 > 0

then p({2, 3}) < 1 v({2, 3}) = 1

recall: p(S) = S iS pi

When is the core non-empty?

• Def: G=(N, v) is simple if v(S){0, 1} for all S– WVGs are simple games

• Def: in a simple game, i is a veto player if v(S) = 0 for any S N \ { i }

• Claim: a simple game has a non-empty core iffthere is a veto player.

Also, (p1, ..., pn) is in the core iffpi = 0 for all non-veto players

pi > 0

N

S

e-core and least core

Need to relax the notion of the core: core: p(S) ≥ v(S) for all S N e-core: p(S) ≥ v(S) - e for all S Nleast core: smallest non-empty e-core

– minimizing the worst deficit v(S) - p(S)

G=(1, 1, 1; 2):– 1/3-core is non-empty: (1/3, 1/3, 1/3) 1/3-core– e-core is empty for any e < 1/3– least core = 1/3-core

Can we compute the core,

e-core and the least core of weighted voting games?

Computational issuesOur Results (E., Goldberg, Goldberg, Wooldridge, AAAI’07)

• Is the core non-empty? – poly-time: use the lemma

• For a given e, is the e-core non-empty?

• For a given e, is a given imputation p in the e-core?

• Is a given imputation p in the least core?

• Construct an imputation in the least core. – p

Given a WVG G = (w1, ..., wn; T)

reductions from Partition

Computational issues

• Is the core non-empty? – poly-time: use the lemma

• For a given e, is the e-core non-empty? – coNP-hard

• For a given e, is a given imputation p in the e-core? – coNP-hard

• Is a given imputation p in the least core? – NP-hard

• Construct an imputation in the least core. – NP-hard

Given a WVG G = (w1, ..., wn; T)

reductions from Partition

A pseudopolynomial algorithm?

• Hardness reduction from Partition assumes large weights– recall: wi are given in binary, – poly-time algorithm <=>

runs in time poly (n, log wmax)

• What if weights are small?– e.g., at most poly(n)?– we are happy with algorithms

that run in time poly (n, wmax)

min cp1+…+ pn = 1

pi ≥ 0 for all i = 1, …, n

S iJ pi ≥ 1 - c for any J s.t. w(J) ≥ T

linear program exponentially many ineqs

Claim: least core = c-core

LP for the least core

LPs and separation oracles

• Separation oracle: – input: (p, c)– output: “yes” if (p, c) satisfies the LP,

violated constraint otherwise

• Fact: LPs with poly-time separation oraclescan be solved in poly-time.

• Our case: given (p, c), is there a J with w(J) ≥ T, p(J) < 1-c?– reduces to Knapsack => solvable in time poly (n, wmax)

• Works for other problems listed above

An approximation algorithm

• Back to large weights…• Theorem: suppose least core value = e

Then for any d we can compute e’ s.t. e ≤ e’ ≤ (1+ )e d and e’-core is non-empty in time poly (n, log wmax, 1/d)

(FPTAS) • Proof idea: use FPTAS for Knapsack inside the

separation oracle

min cp1+…+ pn = 1

pi ≥ 0 for all i = 1, …, nS iJ pi ≥ 1-c for any J s.t. w(J) ≥ T

p1+…+ pn = 1

pi ≥ 0 for all i = 1, …, nS iJ pi ≥ kd for any J s.t. w(J) ≥ T

Approximating the least core

LPk, k=1, ..., 1/d

LPk and the least core value

p1+…+ pn = 1

pi ≥ 0 for all i = 1, …, nS iJ pi ≥ kd for any J s.t. w(J) ≥ T

Claim: let k* be the largest value of k such that LPk has a feasible solution. Then the value e of the least core satisfies 1 – e - d ≤ k*d ≤ 1- e

Can we find k*?– maybe not, but we can find k’ {k*, k*-1}

LPk

An “almost” separation oracle• Claim: for each k =1, …, 1/d, there is a procedure SOk

that runs in time poly (n, 1/d) and either correctly implements a separation oracle for LPk or stops and produces a feasible solution to LPk-1

• Try to solve LP1, LP2, … , LP1/d using SO1, SO2, … , SO1/d – k’: the largest value of k for which we find a feasible

solution (reported by SOk’ or SOk’+1)• Claim: k’ {k*, k*-1}

e ≤ 1 – k*d ≤ e+ d implies e ≤ 1 – k’d ≤ e+2 d

Implementing SOk

Input to SOk: p1, …, pn

Need to check: S iJ pi ≥ kd for any J s.t. w(J) ≥ T– p’i = max { jd/n : jd/n ≤ pi }

– |p’i - pi| < d/n– check if there is a J with p’(J) < (k-1)d, w(J) ≥ T

(DP for Knapsack)• if not, p1, …, pn is a feasible solution for LPk-1

• if yes, p(J) ≤ p’(J)+d, so p(J) < kd, and hence J is a violated constraint for LPk

Additive => multiplicative

• We have shown: can compute e’ s.t. e ≤ e’ ≤ + e d

• Need: e’ s.t. e ≤ e’ ≤ (1+ ) e d• Claim: if e > 0 then e ≥ 1/n

– proof: • some player i is paid at least 1/n• N \ { i } is a winning coalition

• Given d, run our algorithm with d’ = d/n: e ≤ e’ ≤ + e d/n ≤ + e de= (1+ )e d

Most stable point?

• Least core may contain more than one point, not all of them equally good

• G = (3, 3, 2, 2, 2; 6)p = ( 1/4, 1/4, 1/6, 1/6, 1/6);q = (1/3, 1/6, 1/6, 1/6, 1/6);– p and q are both in the least core = 1/2-core– under p, only 2 coalitions have deficit 1/2– under q, 5 coalitions have deficit 1/2

Nucleolus: definition

• deficit: d(p, S) = v(S) - p(S)• least core: minimizes worst deficit• nucleolus:

– minimize worst deficit;– given this, minimize 2nd worst deficit, etc.

• deficit vector: d(p) = (d(p, S1), ..., d(p, S2n)), ordered from largest to smallest

• Def: nucleolus is an imputation p withlex-minimal deficit vector d(p)

Nucleolus: properties

• Introduced by Schmeidler (1969)• Nucleolus is unique• Always in the least core• “Most stable” imputation

Can we compute the nucleolus of weighted voting games?

Small vs. large weights

• Binary weights: NP-hard to compute– reduction from Partition (E., Goldberg, Goldberg, Wooldridge, AAAI’07)

• Unary weights?– pseudopolynomial algorithm (E., Pasechnik, SODA’09)

Computing nucleolus: general scheme

Sequence of linear programs:• LP1: min (p, e) e

p(S) ≥ v(S) – e for all S 2N

(p1, e1): interior optimizer for LP1

S1: set of tight constraints for (p1, e1)

• LP2: min (p, e) e

p(S) = v(S) – e1 for all S S1

p(S) ≥ v(S) – e for all S 2N \ S1

• LP3, LP4, etc. – till there is a unique solution

Solving LP1: weighted voting games

Solving LP1 = finding the least core LP1: min (p, e) e p1+…+ pn = 1

pi ≥ 0 for all i = 1, …, n

p(S) ≥ v(S) – e for all S 2N

• Can find (p, e) in poly-time. • S1? cannot list explicitly...

How to solve LP2?

LP 2: min (p, e) e p(S) = v(S) – e1 for all S S1

p(S) ≥ v(S) – e for all S 2N \ S1

Is there a poly-time separation oracle for LP2?– input: p1, ..., pn, e; can assume e < e1 – suppose we have found S with w(S) ≥ T, p(S) < 1-e– this is only useful if S is not in S1

– difficulty: S1 can be exponentially large

How to solve LPj?

LP j: min (p, e) e p(S) = v(S) – e1 for all S S1

... p(S) = v(S) – e j-1 for all S S j-1

p(S) ≥ v(S) – e for all S 2N \(S1 U ... U S j-1) S1 ,..., S j-1 can be exponentially large

Idea: listing → counting

• We know e1, ..., e j-1

• Can assume that we know s1=|S1|, ..., s j-1=|S j-1|• Given a candidate solution (p, e),

suppose we can compute in poly-time– top j distinct deficits m1, ..., mj

– nj = number of coalitions with deficit mj

• Check if – mt = et, nt = st for t=1, ..., j-1 – mj ≤ e

• Thm: answer is “yes” iff (p, e) is feasible for LPj

Missing pieces

• How to implement the counting?– WVGs with unary weights:

dynamic programming• If the answer is “no”, need to identify a

violated constraint– WVGs with unary weights:

more dynamic programming

General recipe?

Meta-theorem: given a coalitional game G, suppose that we can, for any p and j=1, ..., n, identify top j deficits under p andcount how many coalitions have those deficitsin poly-time.Then we can compute the nucleolus of G in poly-time.

Question: for which classes of games can we do this?

Conclusions

• Stability in weighted voting games– core: poly-time computable– e-core, least-core

• weakly NP-hard • pseudopolynomial algorithm• FPTAS

– nucleolus• weakly NP-hard• pseudopolynomial algorithm• approximation???

An “almost” separation oracle• k’: max {k : SOk has a feasible solution}• k’’: max {k : our procedure finds a feasible solution}• Claim: k’’ {k’, k’-1}

– for k = 1, …, k’:• if SOk works, it produces a feasible solution for k

• if SOk fails, it produces a feasible solution for k-1– for k = k’+1

• if SOk works, it tells us there is no feasible solution for k

• if SOk fails, it produces a feasible solution for k-1– for k > k’+1

• SOk works and tells us there is no feasible solution for k