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Comp/Math 553: Algorithmic Game Theory Comp/Math 553: Algorithmic Game Theory

Comp/Math 553: Algorithmic Game Theory - PowerPoint Presentation

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Comp/Math 553: Algorithmic Game Theory - PPT Presentation

Fall 2016 Yang Cai Lecture 05 Overview so far Recap Games rationality solution concepts Existence Theorems for Nash equilibrium Nashs theorem for general games via Brouwer ID: 651514

player equilibrium games nash equilibrium player nash games polytope game support algorithms lemke symmetric howson algorithm democracy represented strategy

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Slide1

Comp/Math 553: Algorithmic Game Theory

Fall 2016

Yang Cai

Lecture

05Slide2

Overview so far

Recap:

Games, rationality, solution concepts

Existence Theorems for Nash equilibrium:

Nash’s theorem for general games (via

Brouwer

)

von Neumann’s theorem for 2-player ZS games (via LP duality)

This lecture:

algorithms for Nash equilibrium beyond two player zero-sum games (which can be solved using linear programming)

Overarching goal:

How complex computationally is it to find the Nash equilibrium of the game, for the game theorist and the players themselves? Slide3

Algorithms for Nash

Equilibria

Support Enumeration Algorithms

Algorithms for Symmetric Games

The Lemke-

Howson

AlgorithmSlide4

Algorithms for Nash

Equilibria

Algorithms for Symmetric Games

The Lemke-

Howson

Algorithm

Support Enumeration AlgorithmsSlide5

Support Enumeration Algorithms

How better would my life be if I knew the support of the Nash equilibrium?

… and the game is 2-player?

Claim:

Can find Nash equilibrium (

x

,

y

) using linear programming.

Setting:

Let (

R

,

C

) be an

m

by

n

game, and suppose a friend revealed to us the supports and respectively of the Row and Column players’ mixed strategies at some equilibrium of the game.

s.t

.

andSlide6

Support Enumeration Algorithms

How better would my life be if I knew the support of the Nash equilibrium?

… and the game is 2-player?

for guessing the support of each player’s strategy at a Nash

eq

for solving the LP

Runtime:

Corollary:

Existence of rational

equilibria

in 2-player games.

Proof:

Follows from the correctness of the support enumeration algorithm. If there is a Nash equilibrium with supports

S

R

and

S

C

for the two players, then the

polytope

of the corresponding LP is non-empty. Any vertex of that

polytope

is a Nash equilibrium, whose coordinates are rational and whose bit complexity is polynomial in the description of the game.Slide7

Support Enumeration for

n-player games

How better would my life be if I knew the support of the Nash equilibrium?

… and the game is

n

-player?

Challenge:

Even if the support

Σ

p

of each player at a Nash equilibrium is given to us, the problem we need to solve is still not an LP.

Polynomial of degree (

n

-1) in variables

x

11

, x

12

,…,

x

nk

(suppose

k

strategies per player)

So need to solve a system of polynomial equations and inequalities on

n

k

variables, where the involved polynomials have degree

n

-1.

Can be done to accuracy

ε

in time polynomial in

n

nk

log (1/ε). (using tools from the existential theory of the

reals

)

Overall time: polynomial in

nnk

log (1/ε) (c.f. input size: n

kn

= n

2n log k )Slide8

Algorithms for Nash

Equilibria

The Lemke-

Howson

Algorithm

Support Enumeration Algorithms

Algorithms for Symmetric GamesSlide9

Symmetries in Games

Def:

An n-player game is

symmetric

iff

:

- all players have the same strategy set:

S

= {1

,…,

k

}

- there exists a function

f

such that every player’s utility can be written

as:

u

p

(s

) =

f

(

s

p

;

n

1

(

s

-p

)

,…,

n

k

(

s

-p

) )

number of the other players choosing each strategy in

S

E.g. :

- congestion games, with same source destination pairs for each player

Thm

[Nash ’51]:

Always exists a Nash equilibrium in which every player uses the same mixed strategy.

- Rock-Paper-Scissors, guess 2/3 of the average

Description Size:

O(min

{

k

n

k-1

,

k

n

})Slide10

Existence of a Symmetric Equilibrium

Recall Nash’s function:

restrict Nash’s function on the set:

Thought

Experiment:

crucial observation:

Nash’s function maps points of the above set to itself as every player will perform the same updateSlide11

Algorithms for

n-player symmetric games

A symmetric equilibrium

x

= (

x

1

,

x

2

, …,

x

k

)

can be found as follows [Papadimitriou-

Roughgarden

’04]:

- guess the support of

x

: 2

k

possibilities

- write down a set of polynomial equations and inequalities corresponding to the equilibrium conditions, for the guessed support

- polynomial equations and inequalities of degree

n

in

k

variables (

cf

kn

for general games)

can be solved approximately in time polynomial in

n

k

log(1/ε)

using tools from the existential theory of the

reals

polynomial in the size of the input for

k

up to about

n

(

n

1-

δ

for all

δ

)Slide12

Symmetrization

R

,

C

C

T

,

R

T

R

,

C

x

/ |x|

1

y

/ |y|

1

x

y

x

y

Symmetric Equilibrium

Equilibrium

0, 0

0, 0

Any Equilibrium

Equilibrium

In fact we show that

[

Gale-Kuhn-Tucker 1950

]

w.l.o.g

. suppose that R, C have positive entries

Proof: On the board.Slide13

Symmetrization

R

,

C

R

T

,C

T

C, R

x

y

x

y

x

y

Symmetric Equilibrium

Equilibrium

0,0

0,0

Any Equilibrium

Equilibrium

In fact

[…]

Hence, essentially as hard to solve symmetric 2-player games as it is to solve general 2-player games

Open:

- Reduction from 3-player games to symmetric 3-player games?Slide14

Algorithms for Nash

Equilibria

The Lemke-

Howson

Algorithm

Support Enumeration Algorithms

Parenthesis: Symmetric GamesSlide15

Polytopes

101

Convex Polytope in

R

n

:

The intersection of

n

-dimensional

hyperplanes

(or half-spaces) of the form

a

T

x ≥ c, where a is a vector in

R

n

and

c

a scalar.

Often described compactly in matrix form:

A

x

b

, where

A

is an

m

n

matrix and

b

is a vector

in

R

n

.

This inequality specifies the

polytope

that is the intersection of half-spaces:

A

iT

 x ≥ b

i ,

i=1,…,m

(*)

Extreme-point of a Convex

Polytope

:

A point

x

such that

(*) is satisfied with at least

n

(linearly-independent

1

) inequalities tight.

Edge of a Convex

Polytope

:

Subset of the

polytope

where

n

-1 (linearly independent) inequalities are tight.

1

a collection of inequalities

B

i

T

x

d

i

,

i

= 1, …,

k

are called

linearly independent

if the vectors

B

1

,…,

B

k

are linearly independent vectors of

R

nSlide16

The Lemke-

Howson Algorithm (1964)

Problem: Find an exact equilibrium of a 2-player game.

Since there exists a rational equilibrium this task is feasible.

Idea of LH:

Perform pivoting steps between the corners of a

polytope

related to the game until a Nash equilibrium is found.

Assumption (

w.l.o.g

.):

The game given in the input is a

symmetric game

, i.e.

Polytope

of Interest:

Assumption 2 (

w.l.o.g

):

At every corner of the

polytope

exactly

n

out of the 2

n

inequalities are tight.

(perturb original payoff entries with exponentially small noise to achieve this;

equilibria

of the new game are approximate eq. of original game of very high accuracy, and these can be converted to exact

equilibria

(exercise) )Slide17

The Lemke-

Howson Algorithm

At corner (0,0,…,0) all pure strategies are present. Call any corner of the

polytope

where this happens a

democracy.

Lemma:

If a vertex

z

≠0

of the

polytope

is a democracy, then is a Nash eq.

Proof:

Hence:

Def: Pure strategy

i

is

represented

at a corner

z

of the

polytope

if at least one of the following is tight:

At a democracy we have the following implication:Slide18

The Lemke-

Howson Algorithm

Start at the corner (0,0,…,0).

By non-degeneracy there are exactly

n

edges of the

polytope

adjacent to the (0,0,…,0) corner. Each of these edges corresponds to un-tightening one of the inequalities.

Select an arbitrary pure strategy, say pure strategy

n

, and un-tighten . This corresponds to an edge of the

polytope

adjacent to 0. Jump to the other endpoint of this edge.

If the obtained vertex

z

is a democracy, then a Nash equilibrium has been found because

z

≠0.

Otherwise, one of the strategies 1,…,

n

-1, say strategy

j

, is represented twice, by both

was already tight

just became tight

Question:

I will

untighten

one of the above. What happens if I require ?

A: I am going to return to (0,0,…,0), since I would be walking on the edge of the

polytope

that brought me here.

So let me

untighten

the other one, requiring .Slide19

The Lemke-

Howson Algorithm

If the obtained vertex is a democracy, then a Nash equilibrium has been found.

Otherwise, one of the strategies 1,…,

n

-1, is represented twice. This strategy is doubly represented because one of its inequalities was tight before the step, and the other one became tight after the step was taken. To proceed, un-tighten the former.

This defines a directed walk on the vertices of the

polytope

, starting at the democracy (0,0,…,0), and with every intermediate vertex having all of 1,…,

n

-1 represented, and exactly one of them represented twice.

The walk proceeds by un-tightening one of the two inequalities of the doubly represented strategy, namely the one that does not bring it back to where it came from.

The walk can keep going unless a democracy is encountered.

Claim:

The walk will settle on a democracy ≠ 0.

Proof:

Next slide. Slide20

Proof that Lemke-

Howson Terminates

Claim: The walk will settle on a democracy ≠ 0.

Proof:

Consider the vertices of the

polytope

that are either democracies or have only 1,…,

n

-1 represented (and exactly one of them represented twice).

Define a graph whose nodes are the aforementioned vertices, and whose edges are defined as follows:

a vertex where

n

is not represented and where

j

is represented twice has two neighbors corresponding to the vertices reached when un-tightening either

zj

≥ 0 or

R

j

z

≤1

a vertex that is a democracy has one neighbor corresponding to un-tightening whichever of

z

n

≥ 0 or

R

n

z

≤1 is tight

Clearly every node in this graph has degree ≤ 2, hence it comprises paths and cycles. In fact, nodes with degree 1 are democracies and only democracies have degree 1.

(0,0,…,0) has degree 1 so it is sitting on a path of this graph, whose other endpoint is another democracy ≠0

Lemke-

Howson

visits the vertices on the path where (0,0,…,0) sits, by the definition of the graph and the definition of the algorithm.Slide21

Lemke-

Howson Example

3 0 0

2 2 2

0 3 0Slide22

Post Mortem

The Lemke-

Howson algorithm:

- provides an alternative proof that a Nash equilibrium exists in 2-player games;

- moreover, it shows that there always exists a rational equilibrium in 2-player games;

it works by virtue of a

parity argument

: it identifies a directed path on the vertices of the

polytope

; since that path has a source, it must also have a sink.

worst-case running time: exponential in the number of strategies

[

Savani

-von Stengel’04]

.

there are analogs of the Lemke-

Howson

algorithm for multi-player games working with manifolds instead of

polytopes

( [

Rosenmuller

’71] and [Wilson ’71])