Combined Presentation Files from ISI-CAGT workshop June 2014

1. Nash Equilibrium
vs.
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 TheoryWorkshop on Introduction to Computational Aspects of Game Theory
June 20, 2014June 20, 2014

2. 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.

3. Battle of Sexes
• H denotes Husband and W
denotes wife.
• Nash Equilibrium: (C,C) and
(O,O)
• Pareto Efficient Outcome:
P2
P1 (W)
(H)
Cricket
(C)
Opera
(O)
Cricket 2,1 0,0 • Pareto Efficient Outcome:
(C,C) and (O,O)
• Thus Nash equilibrium
coincides with Pareto
efficient outcome.
Cricket
(C)
2,1 0,0
Opera
(O)
0,0 1,2

4. 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.

5. 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
P2
P1
Confess
(C)
Don’t
confess
(NC)
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
(C)
-3,-3 0, -10
Don’t
confess
(NC)
-10,0 -1,-1

6. 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

7. 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.

8. 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
)(
0if1.0
)(
m
mF
s
sp
i
i
m
mF )(
2
1.01.1)( mmmF −=

9. 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.

10. Braess’s Paradox
D
B
C
B45
x/100
45x/100
D
A
CD
A
x/10045
x/100
45
0

11. 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.

12. 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 toCD, 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.

13. 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 Nashchange 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.

14. 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 Paretosuch 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.

15. 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”.

16. Anything to ask?

17. 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
Workshop on Introduction to Computational Aspects of Game TheoryWorkshop on Introduction to Computational Aspects of Game Theory
June 20, 2014June 20, 2014
17

18. Motivation
• Computing Nash equilibrium is PPAD
complete.
• But Zero-sum games has efficient algorithm.
What, other games has polynomial timeWhat, other games has polynomial time
algorithms ?
• Can Graph Theory help to find the answer !?!
6/20/2014 18

19. Is it possible that. . . . .
6/20/2014 19

20. Yes!
• “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

21. 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

22. 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

23. 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

24. Quest
• 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

25. 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

26. Prove or Disprove . . .
• ( Games x Hereditary Graph properties)
Efficient solutionsEfficient solutions
6/20/2014 26

27. Games on Triangulation
Sujoy Kr. Bhore
M. Sc. Student
Computer Science DepartmentComputer Science Department
R.K.M.V.U., Belur, India
Presentation at
Workshop on Introduction to Computational Aspects of Game TheoryWorkshop on Introduction to Computational Aspects of Game Theory
June 20, 2014June 20, 2014

28. 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.

29. 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).

30. 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.

31. 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.

32. 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.

33. 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.

34. 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.

35. 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…

36. 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

37. 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 andNow, 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…

38. 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 findQ: Why you have not tried to find
NE for your problme…
I went to sleep…

39. I liked Game Theory …
• The essence of all understanding here – is generalization… so I
feel it has connections with other fields…
• QUESTIONs

40. Cake Cutting problem
Sanchayan Santra
Indian Statistical Institute
Presented on
Workshop on Computational Aspects of Game Theory, 2014

41. Scenario

42. Scenario

43. Scenario

44. Scenario

45. You Cut I Choose
One possible way to resolve this dispute is “You cut I choose”
method.
The first player starts by dividing the cake into two pieces.
The second player then chooses the piece he prefers, and the
first player receives the remaining piece.
The first player will try to divide the cake into two pieces
“equally”. Otherwise, he/she will be the loser.

46. Three Players

47. More Players

48. Model
We have a set of agents N = {1, · · · , n} and one divisible good
X, usually represented by the interval [0, 1]. Each agent i has a
valuation function Vi, where Vi : [0, 1] → R.
The following conditions hold ∀i ∈ N
Normalization Vi([0, 1]) = 1.
Divisibility For every sub-interval [x, y] and 0 ≤ λ ≤ 1 there
exits 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 should
feel being deceived. → Fairness

49. Fairness
Subjective fairness can be defined as follows, where the Xi denote
the 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.

50. Algorithm for N - players
Dubins-Spanier “moving-knife” protocol
Selfridge-Conway (N=3) (not discussed)

51. Moving knife protocol
Assumption: Cake is a long and rectangular and there is one
referee whi will cut the cake.
In each stage -
Referee moves the knife over the cake in, say, left to right
direction.
One of the player shouts “stop”.
The cake is cut at that portion and the piece to the left is
given to that player.
The player and the piece is removed and the process is
repeated with remaing players and the remaining cake, till one
player remains.
The last player receives the last piece.

52. Moving knife protocol
This produces proportional allocation -
Each player try to take what he/she thinks frac1n piece of
the cake.
Otherwise he/she will be the loser.
One problem
This does not gurantee Envy-free solution.
One player may think some other player got a bigger piece.

53. Other problems
Fair Resource Allocation problem
Resources - Homogeneous, Heterogeneous

54. Acknowledgement
I would like to thank
Prithviraj Dasgupta - for
a wonderful introductory talk
a hands-on approach of solving problems
also letting us present a topic
ECSU
Organizing the Workshop on Computational Aspects of Game
Theory

55. Thank you.