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Nash Equilibrium

Pareto Optimality

Iman PalIman Pal

PG-1 Student

M.Sc Applied Economics

Presidency University, Kolkata, India

Presentation at

Workshop on Introduction to Computational Aspects of Game Theory Workshop on Introduction to Computational Aspects of Game Theory June 20, 2014June 20, 2014

Introduction

• A pair of strategies is an Nash equilibrium (NE) for a two

player game if no player can improve his payoff by changing

his strategy from his equilibrium strategy to another strategy

provided his opponent plays his equilibrium strategy.

• Unilateral deviations are unprofitable.• Unilateral deviations are unprofitable.

• A pair of strategies in a two-player game, is not Pareto

Optimal (PO) if there exists another choice of strategies such

that both players are no worse off switching from the initial

choice to the final and at least one of the player is strictly

better off.

Battle of Sexes

• H denotes Husband and W

denotes wife.

• Nash Equilibrium: (C,C) and

• Pareto Efficient Outcome:

P1 (W)

Cricket

Cricket 2,1 0,0 • Pareto Efficient Outcome:

(C,C) and (O,O)

• Thus Nash equilibrium

coincides with Pareto

efficient outcome.

Cricket

2,1 0,0

0,0 1,2

Does NE imply a socially optimum

outcome?

• Battle of Sexes is a standard game where NE is

the socially desirable outcome. The players

maximize their joint payoff at NE which in turn

is the PO outcome.is the PO outcome.

• In many games PO outcome differs from NE.

• Examples include Prisoner’s Dilemma and

Tragedy of Commons.

Example 1: Prisoner’s Dilemma

� The police have arrested two suspects for a crime.

� They tell each prisoner they’ll reduce his/her prison sentence if he/she betrays the other prisoner.

� Each prisoner must choose

Confess

Don’t

confess

Confess -3,-3 0, -10� Each prisoner must choose

between two actions:

• Confess

• Don’t confess

� In this game, pure Nash equilibrium is at (confess, confess).

Confess

-3,-3 0, -10

Don’t

confess

-10,0 -1,-1

Is Nash Equilibrium Pareto Optimal?

• Strategy profile S pareto dominates a strategy profile S′ if no agent gets a

worse payoff with S than with S′, i.e., Ui(S) ≥ Ui(S′) for all i , and at least

one agent gets a better payoff with S than with S′, i.e., Ui(S) > Ui(S′) for at

least one i.

• In Prisoner’s Dilemma,

� (NC,NC) is Pareto optimal as no profile gives both players a higher payoff.� (NC,NC) is Pareto optimal as no profile gives both players a higher payoff.

� (C,NC) is Pareto optimal as no profile gives player 1 a higher payoff.

� (NC,C) is Pareto optimal by the same argument.

� (NC,NC) is Pareto dominated by (C,C).

But ironically, (NC,NC) is the dominant strategy Nash equilibrium

Example 2: Tragedy of Commons

• Common resources: goods that are not excludable (people cannot be prevented from using them) but are rival (one person’s use of them diminishes another person’s enjoyment of it).

• Examples include congested toll-free roads, fish in the ocean, the environment, . . .,Examples include congested toll-free roads, fish in the ocean, the environment, . . .,

• Problem: Overuse of such common resources leads to their destruction.

• This phenomenon is called the tragedy of the commons.

Looking into this game…

• In this game there are N players.

• two strategies:1 (use the resource), 0 (don’t use),

• payoff function is defined as follows:

where m = Sum over all sj and

• is strictly decreasing function.

otherwise)(

0 if1.0

21.01.1)( mmmF −=

Playing the game

• Using numerical values for 9, 10 and 11 in the

given functional form, we get,

F(9)/9 = 0.2, F(10)/10 = 0.1, F(11)/11 = 0.

• Nash equilibria:• Nash equilibria:

� n < 10: all players use the resource,

� n ≥10: 9 or 10 players use the resource,

• Social optimum: 5 players use the resource.

Braess’s Paradox

B45x/100

45x/100

x/10045

At Equilibrium

For the first network diagram

• suppose, we have 4000 cars, if each car takes

upper route, then it takes 85 mins; if they

divide evenly, then 65 mins.divide evenly, then 65 mins.

• Nash equilibrium - is when the cars divide up

evenly: with even balance between two

routes, no driver has an incentive to switch

over to the other route.

Paradox

• A small change to the network can lead to a

counterintuitive situation

• Adding CD - a fast highway (0 mins to drive). With

CD, there is a unique Nash equilibrium and it leads to CD, there is a unique Nash equilibrium and it leads to

a worse travel time to everyone - 80 mins

• This phenomenon is Braess’s Paradox: adding

resources to a transportation network can

sometimes hurt performance at equilibrium.

Why a Paradox?

• This is because the Nash equilibrium of such a

system is not necessarily Pareto optimal.

• At Nash equilibrium, drivers have no incentive to

change their routes. If the system is not in Nash change their routes. If the system is not in Nash

equilibrium, selfish drivers must be able to improve

their respective travel times by changing the routes

they take. In the case of Braess's paradox, drivers will

continue to switch until they reach Nash equilibrium,

despite the reduction in overall performance.

Conclusion

• Nash equilibrium does not always converge to

socially desirable outcome.

• There is a chance of Pareto improvement in

such cases. Players can move to Pareto such cases. Players can move to Pareto

optimality by playing a different strategy.

• Braess’s Paradox arise due to this difference in

Pareto optimal outcome and NE.

Acknowledgements

• I would like to thank our instructor Dr. Prithviraj Dasgupta for

sharing with us the computational aspects of Game Theory.

• I would like to thank the following:

� Emil Simion for the paper on Braess Paradox, Operational

Research and Optimization (Master EEJSI).Research and Optimization (Master EEJSI).

� Maria Grineva for her lecture notes on Modeling Network

Traffic Using Game Theory of march, 2011.

� Krzysztof R. Apt of University of Amsterdam for the numerical

example in “Nash Equilibria and Pareto Efficient Outcomes”.

Anything to ask?

Games to Graph: Transition in

Quest of Efficiency

Dibyayan ChakrabortyDibyayan Chakraborty

M. Sc. Student

Computer Science Department

RKMVU, Belur, India

6/20/2014

Presentation at

Motivation

• Computing Nash equilibrium is PPAD

complete.

• But Zero-sum games has efficient algorithm.

What, other games has polynomial time What, other games has polynomial time

algorithms ?

• Can Graph Theory help to find the answer !?!

6/20/2014 18

Is it possible that. . . . .

6/20/2014 19

• “A Polynomial Time Algorithm for Finding

Nash Equilibria in Planar Win-Lose Games”:

Louigi Addario-Berry, Neil Olver, and Adrian

Vetta .Vetta .

6/20/2014 20

Idea in brief

• A win-lose game is a game in which the pay-

off to every player is either zero or one. In the

paper they consider two-player win-lose

games. games.

• Here pay-offs are given by two m × n zero-one

matrices A and B for players I and II,

respectively.

6/20/2014 21

Idea in brief

• We have one vertex for each pure strategy;

that is, our digraph G has one vertex for each

row ri and one vertex for each column cj. We

have an arc (ri, cj) if the entry aij = 1; have an arc (ri, cj) if the entry aij = 1;

• Assumption: The resulting graph is planar.

6/20/2014 22

Interesting observation

• Their result: There is a polynomial time algorithm for finding a Nash equilibrium in a two-player planar win-lose game.

• Interesting part is that all they had to do is to • Interesting part is that all they had to do is to detect induced cycle which had no two incoming arcs.

• Such induced cycles always exists in the graph, so NE can be always found in polynomial time.

6/20/2014 23

• Can we say that the hereditary graph

properties has strong relation with NE

solutions ??

• Hereditary properties are those which remains

invariant even if vertices are deleted or edges

are contracted from the graph.

6/20/2014 24

It might be possible because . . . . .

1. Such properties often uniquely characterizes the graph family. So, it always exists . . . . Just like NE always exists for finite actions and players .

2. Deletion of vertices does not remove the property . This might correspond to the reduction of actions without changing game structure .

6/20/2014 25

Prove or Disprove . . .

• ( Games x Hereditary Graph properties)�

Efficient solutionsEfficient solutions

6/20/2014 26

Games on Triangulation

Sujoy Kr. Bhore

M. Sc. Student

Computer Science DepartmentComputer Science Department

R.K.M.V.U., Belur, India

Presentation at

Triangle

Q:Why me?

• Elementary geometric shape – So, complex

geometric structure can be decomposed into

Collection of Triangles.Collection of Triangles.

• Triangulation (Definition): A triangulation of a finite

planar point set S is a simplicial decomposition of its

convex hull whose vertices are precisely the points in

S. i.e., no three points in S are collinear.

Let’s play with triangles

Games on triangulations come in three main

favors

� Constructing (a triangulation).� Constructing (a triangulation).

� Transforming (a triangulation).

�Marking (a triangulation).

Complexity

Q: What about me?

• Optimal strategies for Classical sequential games (ex- Chess) –

We construct a game tree. Even allowing repeatation – depth

O(32^65).O(32^65).

• Finding winning strategy in Combination game theory

(NP,PSPACE,EXPSPACE,EXPTIME).

• Polynomial algorithm can be given by using a deceptive

property.

Impartial Games …Democracy

• the allowable moves depend only on the position and not on which of the two players is currently moving, and where the payoffs are symmetric.

• Sprague-Grundy Theorem.Sprague-Grundy Theorem.

• Chess is impartial ? No, as each player can only move pieces of their own color.

• Triangulation? Yes… One player can select any from the remaining …Utility also same.

Triangulation Coloring Game and

Kayles

• Kayles :

• Triangulation Coloring Game : Two players • Triangulation Coloring Game : Two players

move inturn by coloring an edge of T(S) green,

and the first player who completes an empty

green triangle wins.

That’s fine…But… How to find an optimal

strategy ?

• Obviously, this game terminates after a linear number of • Obviously, this game terminates after a linear number of

moves and there are no ties.

• Consider the dual of the triangulation T(S). An inner triangle

consists entirely the diagonal of T(S) and therefore it does not

use an edge of the convex hull of S.

Winning Strategy (Cont.)

• Why Dual graph ? A player can’t choose an edge from a triangle where his

opponent has picked any edge … So, whoever markes the last vertex

wins…

• If the triangulation is serpentine (equivalently, a single array of boxes • If the triangulation is serpentine (equivalently, a single array of boxes

without branches). From Tweedledum-Tweedledee Argument –

an odd number of triangles, first takes the central triangle by coloring the

edge of this triangle that belongs to the convex hull.

For an even the number of triangles, first takes both triangles adjacent to the

central diagonal by coloring this diagonal.

Winning strategy (cont.)

• That will separate it into identical two partition. Now, player one just can

mimic the opponent’s move by simply coloring the corresponding edge in

the other triangulation.

• Life is Complex

• For branching … let’s take no two inner triangle shares common diagonal. -

--remove the enemy (inner triangle)…

Now it’s the Kayle problem…

Acceptor

LOVES THEOREM Theorem :

Deciding whether the Triangulation Coloring Game on a simple-

branching triangulation on n points in convex position is a first-

player win or second-player win, as well as finding moves

leading to an optimal strategy, can be solved in time linear in

the size of the triangulation.the size of the triangulation.

Note :

it is shown that there are polynomial-time

algorithms to determine the winner in Kayles

on graphs with bounded asteroidal number, on

cocomparability graphs.

For outer planer graph th. Can be rephrashed

I have a problem…

• Given a convex point set we want to draw eularian

traingulation …

Now, there is player 1 and 2 will take sequential action and Now, there is player 1 and 2 will take sequential action and

who will have no edge to draw will loose…

We love to draw pic. In a single shot… but here it’s not trivial…

Why I believe it has connection with NE …

• Theorem 1 : A connected topological spaceis homeomorphic to a Nash

equilibrium component if and only if it is has a triangulation.

• John nash

Q: Why you have not tried to find Q: Why you have not tried to find

NE for your problme…

I went to sleep…

I liked Game Theory …

• The essence of all understanding here – is generalization… so I

feel it has connections with other fields…

• QUESTIONs

Cake Cutting problem

Sanchayan Santra

Indian Statistical Institute

Presented onWorkshop on Computational Aspects of Game Theory, 2014

Scenario

You Cut I Choose

One possible way to resolve this dispute is “You cut I choose”method.

I The first player starts by dividing the cake into two pieces.

I The second player then chooses the piece he prefers, and thefirst player receives the remaining piece.

I The first player will try to divide the cake into two pieces“equally”. Otherwise, he/she will be the loser.

Three Players

More Players

We have a set of agents N = {1, · · · , n} and one divisible goodX, usually represented by the interval [0, 1]. Each agent i has avaluation function Vi, where Vi : [0, 1]→ R.The following conditions hold ∀i ∈ NNormalization Vi([0, 1]) = 1.

Divisibility For every sub-interval [x, y] and 0 ≤ λ ≤ 1 thereexits z ∈ [x, y] such that Vi([x, z]) = λVi([x, y]).

Non-negativity For every sub-interval I, Vi(I) ≥ 0.

GOAL: We want to achieve a divisions such that no one shouldfeel being deceived. → Fairness

Fairness

Subjective fairness can be defined as follows, where the Xi denotethe portion allocated to player i

proportional Vi(Xi) ≥ 1/n ∀i.envy-free Vi(Xi) ≥ Vi(Xj) ∀i, j.equitable Vi(Xi) = Vj(Xj) ∀i, j.

Algorithm for N - players

I Dubins-Spanier “moving-knife” protocol

I Selfridge-Conway (N=3) (not discussed)

Moving knife protocol

Assumption: Cake is a long and rectangular and there is onereferee whi will cut the cake.In each stage -

I Referee moves the knife over the cake in, say, left to rightdirection.

I One of the player shouts “stop”.

I The cake is cut at that portion and the piece to the left isgiven to that player.

I The player and the piece is removed and the process isrepeated with remaing players and the remaining cake, till oneplayer remains.

I The last player receives the last piece.

Moving knife protocol

This produces proportional allocation -

I Each player try to take what he/she thinks frac1n piece ofthe cake.

I Otherwise he/she will be the loser.

One problem

I This does not gurantee Envy-free solution.

I One player may think some other player got a bigger piece.

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