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Bilinear Games: Polynomial Time Algorithms for Rank Based S
Bilinear Games: Polynomial Time Algorithms for Rank Based S

Bilinear Games: Polynomial Time Algorithms for Rank Based S - PowerPoint Presentation

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Bilinear Games: Polynomial Time Algorithms for Rank Based S - Description

Ruta Mehta Indian Institute of Technology Bombay Joint work with Jugal Garg and Albert X Jiang A Game RockPaperScissor RockPaperScissor A Play Winner 1 RockPaperScissor ID: 540490 Download Presentation

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Presentation on theme: "Bilinear Games: Polynomial Time Algorithms for Rank Based S"— Presentation transcript

Slide1

Bilinear Games: Polynomial Time Algorithms for Rank Based Subclasses

Ruta

Mehta

Indian Institute of Technology, Bombay

Joint work with

Jugal

Garg

and Albert X. JiangSlide2

A Game: Rock-Paper-ScissorSlide3

Rock-Paper-Scissor: A Play

Winner

$

1Slide4

Rock-Paper-Scissor:

A Play

Winner

$

1Slide5

Rock-Paper-Scissor:

A Play

Winner

$

1Slide6

0,0

-1,1

1,-1

1,-1

0,0

-1,1

-1,1

1,-1

0,0

Rock-Paper-Scissor PayoffsSlide7

R

P

C

R

0

-1

1

P

1

0

-1

C

-1

1

0

Bimatrix

Game

Steady State:

No

p

layer gains by unilateral deviation

R

P

C

R

0

1

-1

P

-1

0

1

C

1

-1

0

S

1

= { R, P, C }

S

2

= { R, P, C }

A

BSlide8

R

P

C

R

0

-1

1

P

1

0

-1

C

-1

1

0

Bimatrix

Game

No Steady State

R

P

C

R

0

1

-1

P

-1

0

1

C

1

-1

0

S

1

= { R, P, C }

S

2

= { R, P, C }

A

BSlide9

R

1/3

P

1/3

C

1/3

R

0

-1

1

P

1

0

-1

C

-1

1

0

Mixed Play

Steady State

R

P

C

R 1/3

0

1

-1

P 1/3

-1

0

1

C 1/3

1

-1

0

S

1

= { R, P, C }

A

B

1

={r

1

, p

1

, c

1

≥0;

r

1

+p

1

+c

1

=1}

S

1

= { R, P, C }

2

={r

2

, p

2

, c

2

≥0;

r

2

+p

2

+c

2

=1}Slide10

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.Slide11

Nash Equilibrium ComputationPapadimitriou (JCSS’94)

: PPAD-class P

roblems

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.Slide12

Rank and ComputationKannan

and 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.Slide13

Bilinear Games

Bimatrix Game with polyhedral strategy sets.Two players: 1

and 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

xTAy

to

player

1

, and

x

T

By

to

player

2

.

Motivation:

Koller

et al. (STOC’94) for two-player extensive form game with perfect recall.Slide14

Nash Equilibrium in Bilinear

NE: No player gains by unilateral deviation.Existence: Corollary of

Glicksberg’s

result.

Symmetric Game:

B=A

T

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.Slide15

Bilinear Contains:Bimatrix

, Polymatrix, Bayesian, etc.

Bimatrix

:

X =

1, Y =

∆2

Polymatrix:

N players.

Each pair plays a

bimatrix

game.

Player

i

:

S

i

finite strategy

set,

i

Mixed strategy set.

Goal of

i

: Choose x

i

from

i

to maximize total payoff.

A

ij

i

jSlide16

Polymatrix to Bilinear

M= |S1|+ … + |S

n

|. X = {(x

1

,…,

xn) | xi in ∆

i}, Y=X.A , B=AT

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

polymatrix

game

0

0

A

ij

0

0

i

j

A =

Slide17

Best Response (Koller et al.)

Fix a strategy y of player 2.Player 1 solves

max:

x

T

(Ay) min:

eTp

Ex = e pTE

≥ (Ay)T

x ≥ 0

At optimal: p

s.t

.

A

i

y

pTE

i

&

x

i

> 0 =>

A

i

y

=

p

T

E

i

Given x

X

, for player 2 we get

At optimal: q

s.t

.

B

j

x ≤ qT

Fj &

yj > 0 =>

qT

F

j

=

B

jxSlide18

Best Response Polytopes (BRPs)

(x,y) is a NE

iff

p

: Ay ≤

E

Tp

; xi

> 0 =>

A

i

y

=

p

T

E

i

q:

x

T

B

q

T

F

;

y

j

> 0 =>

q

T

F

j

=

B

j

x

x

T

(Ay

-

E

T

p

)

≤ 0 and (xTB - qTF)y ≤ 0xT(A+B)y – eTp – fTy ≤ 0Slide19

Nash Equilibrium in BRPs

NE

iff

x

T

(Ay -

ETp

)=0 and (xT

B

-

q

T

F

)y=0

x

T

(A+B)y –

e

T

p

f

T

y

=0

Assumption: P and Q are non-

degnerate

.

(u, v) of P x Q gives a NE => (u, v) is a vertex.Slide20

QP Formulation

max: x

T

(A+B)y –

e

T

p

– fT

y

s.t

.

(y, p) P

(x, q) Q

Optimal value 0.

Only vertex solutions.

Slide21

Our ResultsRank-1 games: rank(A+B)=1Extend

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.Slide22

Rank-1 CaseZero-sum ~ rank(A+B)=0: LP formulation (Charnes’53)rank(A+B)=1 then A+B =

a.

b

T

The QP formulation:

max: (x

Ta)(b

Ty) – e

T

p

f

T

y

s.t. (y, p) P

(x, q) QSlide23

Rank-1 CaseReplace (

xTa) by z. Recall B

= -A +

a

.

b

T

xT(A+B)y

e

T

p

f

T

y

=0 z(

b

T

y

)

e

T

p

f

T

y

=0

N

= Points of P x Q’ with

z(

b

T

y

) –

e

T

p

f

T

y=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. Slide24

Parameterized LP

LP(z) = max: z(

b

T

y

) –

eTp – f

Ty

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. Slide25

Rank-1: The AlgorithmNE:

Intersection of N

and H: z –

x

T

a

=0.

. c1=amin

, c2=amax

H

N

H

H

+

NE

N

(c

1

)

N

(c

2

)Slide26

Rank-1: Binary Search Algorithm

NE of (A,B): Points in intersection of N and H.

c=c

1

+c

2

/2.

H

NE

N

(c

1

)

N

(c

2

)

N

N

(c)

H

+

H

–Slide27

Rank-1: Binary Search AlgorithmNE of (A,B): Points in intersection of

N and H.c=c

1

+c

2

/2.

If N(c) in H

–,then c1

=c else c2=c.

H

NE

N

(c

2

)

N

N

(c

1

)

H

+

H

–Slide28

AnalysisTerminates 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.Slide29

ConclusionsBilinear 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.Slide30

Thank You