Bilinear Games: Polynomial Time Algorithms for Rank Based Subclasses

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Bilinear Games: Polynomial Time Algorithms for Rank Based Subclasses. Ruta Mehta Indian Institute of Technology, Bombay Joint work with Jugal Garg and Albert X. Jiang. A Game: Rock-Paper-Scissor. Rock-Paper-Scissor: A Play. Winner. \$ 1. Rock-Paper-Scissor: A Play. Winner. \$ 1.

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Ruta Mehta

Indian Institute of Technology, Bombay

Joint work with JugalGarg and Albert X. Jiang

Bimatrix Game

S1 = { R, P, C }

S2 = { R, P, C }

A

B

Steady State: No player gains by unilateral deviation

Bimatrix Game

S1 = { R, P, C }

S2 = { R, P, C }

A

B

Mixed Play

S1 = { R, P, C }

S1 = { R, P, C }

∆1={r1, p1, c1≥0;

r1+p1+c1=1}

∆2={r2, p2, c2≥0;

r2+p2+c2=1}

B

A

John Nash (1951)
• Finite Game: Finitely many players, each with finitely many strategies.
• Nash: Every finite game has a steady state in mixed strategy.

Hence forth called Nash equilibrium (NE)

• Proved using Kakutani fixed point theorem: Highly non-constructive.
Nash Equilibrium Computation
• Problems where existence is guaranteed like fixed point, Sperner’s Lemma, Nash equilibrium.
• Chen and Deng (FOCS’06): It is PPAD-hard.
• CDT (FOCS’06): Even approximation is PPAD-hard.
Rank and Computation
• Kannanand Theobald(SODA’07):
• Define rank of (A,B) as rank(A+B).
• FPTAS for fixed rank games.
• Polynomial time algorithms for exact Nash.
• Dantzig(1963): Zero-sum (rank-0) is equiv. to LP.
• AGMS (STOC’11): Rank-1 games.
Bilinear Games

Bimatrix Game with polyhedral strategy sets.

• Two players: 1and 2
• Polyhedral strategy sets:
• X={x | Ex = e; x ≥ 0}, Y={y | Fy=f; y ≥ 0}
• Payoff matrices: A, B
• Bilinear Payoff: (x, y) fetches xTAyto player 1, and xTBy to player 2.

Motivation: Koller et al. (STOC’94) for two-player extensive form game with perfect recall.

Nash Equilibrium in Bilinear
• NE: No player gains by unilateral deviation.
• Existence: Corollary of Glicksberg’s result.
• Symmetric Game:B=AT and Y=X.
• (x, y) is a symmetric profile if y=x.
• Existence of symmetric NE: An adaptation of Nash’s proof for symmetric bimatrix games.
Bilinear Contains:
• Bimatrix, Polymatrix, Bayesian, etc.
• Bimatrix: X = ∆1, Y = ∆2
• Polymatrix:
• N players. Each pair plays a bimatrix game.
• Player i: Si finite strategy set, ∆i Mixed strategy set.
• Goal of i: Choose xi from ∆i to maximize total payoff.

i

Aij

j

Polymatrix to Bilinear
• M= |S1|+ … + |Sn|. X = {(x1,…,xn) | xi in ∆i}, Y=X.
• A , B=AT

Symmetric NE of (A,B) maps to a NE of the polymatrix game

j

i

A =

Best Response (Koller et al.)
• Fix a strategy y of player 2.
• Player 1 solves

max: xT(Ay) min: eTp

Ex = e pTE≥ (Ay)T

x ≥ 0

At optimal: p s.t. Aiy ≤ pTEi& xi > 0 => Aiy = pTEi

• Given x X, for player 2 we get

At optimal: q s.t. Bjx ≤ qTFj& yj> 0 => qTFj = Bjx

Best Response Polytopes (BRPs)
• (x,y) is a NE iff

p: Ay ≤ETp; xi > 0 => Aiy = pTEi

q: xTB≤qTF; yj> 0 => qTFj = Bjx

xT(Ay - ETp) ≤ 0 and (xTB - qTF)y ≤ 0

xT(A+B)y – eTp – fTy ≤ 0

Nash Equilibrium in BRPs

NE iffxT(Ay - ETp)=0 and (xTB - qTF)y=0

xT(A+B)y – eTp – fTy=0

Assumption: P and Q are non-degnerate.

(u, v) of P x Q gives a NE => (u, v) is a vertex.

QP Formulation

max: xT(A+B)y – eTp – fTy

s.t. (y, p) P

(x, q) Q

• Optimal value 0.
• Only vertex solutions.
Our Results
• Rank-1 games: rank(A+B)=1
• Extend Adsul et al. algorithm for exact NE.
• Fixed rank games: rank(A+B)=k
• Extend FPTAS of Kannan et al.
• Rank of A or B is constant
• Enumerate all NE in polynomial time.
Rank-1 Case
• Zero-sum ~ rank(A+B)=0: LP formulation (Charnes’53)
• rank(A+B)=1 then A+B = a.bT
• The QP formulation:

max: (xTa)(bTy) – eTp – fTy

s.t. (y, p) P

(x, q) Q

Rank-1 Case
• Replace (xTa) by z. Recall B = -A + a.bT

xT(A+B)y – eTp – fTy=0 z(bTy) – eTp – fTy=0

• N = Points of P x Q’ with z(bTy) – eTp – fTy=0
• Forms paths and cycles, since z gives one degree of freedom.

NE of (A,B): Points in intersection of N and z – xTa =0.

Parameterized LP

LP(z) = max: z(bTy) – eTp – fTy

s.t. (y, p) P

(x, z, q) Q’

• Given any c, Optimal value of LP(c) is 0.
• OPT(c) lies on N, and
• Let N(c)={Points of N with z=c}, then OPT(c)=N(c).
• N is a single path on which z is monotonic.
Rank-1: The Algorithm
• NE: Intersection of N and H: z – xTa =0.
• . c1=amin, c2=amax

H

N(c1)

N

H+

H–

NE

N(c2)

Rank-1: Binary Search Algorithm
• NE of (A,B): Points in intersection of N and H.
• c=c1+c2/2.

H

N(c1)

N

H+

H–

NE

N(c)

N(c2)

Rank-1: Binary Search Algorithm
• NE of (A,B): Points in intersection of N and H.
• c=c1+c2/2. If N(c) in H–,then c1=c else c2=c.

H

N

H+

H–

NE

N(c1)

N(c2)

Analysis
• Terminates because,
• z is monotonic on N.
• Increase in z on each edge is lower bounded by 1/d where d is polynomial sized in the input.
• Time complexity:
• Solve LP(c) to get N(c) in each pivot.
• log(d) * log(amax – amin) pivots.
Conclusions
• Bilinear games:
• Bimatrix with polytopal strategy sets.
• Fairly general. Contains polymatrix, bayesian, etc.
• Polynomial time algorithm for rank based subclasses.
• Open problems:
• Designing a Lemke-Howson type algorithm.
• Degree, index, stability concepts.
• Computation of approximate equilibrium.