Equilibrium Finding and
Equilibrium Index in
Two-Player Games
Bernhard von Stengel
Department of Mathematics
London School of Economics
Overview
•
Nash Equilibrium of a bimatrix game: elementary
existence proof with the Lemke-Howson method
•
Geometric view of equilibrium [Shapley 1974]
•
Definition of the Index of an Equilibrium via an
orientation of the Lemke-Howson path
Nash equilibria of bimatrix games
0 6
A= 2 5
3 3
B=
2
1
4
Nash equilibrium =
pair of strategies x , y with
x best response to y and
y best response to x.
1
3
3
Mixed equilibria
0 6
A= 2 5
3 3
2/3
x = 1/3
0
4
Ay = 4
3
B=
2
1
4
1
3
3
xTB = 5/3 5/3
yT = 1/3 2/3
only pure best responses can have probability > 0
Best response condition
Let x and y be mixed strategies of player I and II, respectively.
Then x is a best response to y
⇐⇒ for all pure strategies i of player I:
xi > 0 =⇒ (Ay)i = u = max{(Ay)k | 1 ≤ k ≤ m}.
Here, (Ay)i is the i th component of Ay , which is the expected
payoff to player I when playing row i .
Proof.
m
m
i=1
m
i=1
m
m
i=1
i=1
i=1
xAy = ∑ xi (Ay)i = ∑ xi (u − (u − (Ay)i)
= ∑ xi u − ∑ xi (u − (Ay)i) = u − ∑ xi (u − (Ay)i) ≤ u,
because xi ≥ 0 and u − (Ay)i ≥ 0 for all i . Furthermore,
xAy = u ⇐⇒ xi > 0 implies (Ay)i = u , as claimed.
Best responses to mixed strategy of player 2
4
5
1
0
6
2
2
5 =A
3
3
3
payoffs to
player I
1,0
0,1
Best responses to mixed strategy of player 2
4
5
1
0
6
2
2
5 =A
3
3
3
payoffs to
player I
1,0
0,1
Best responses to mixed strategy of player 2
6
1
4
5
1
0
6
2
2
5 =A
3
3
3
payoffs to
player I
0
1,0
0,1
Best responses to mixed strategy of player 2
5
2
2
1,0
0,1
4
5
1
0
6
2
2
5 =A
3
3
3
payoffs to
player I
Best responses to mixed strategy of player 2
3
3
1,0
3
0,1
4
5
1
0
6
2
2
5 =A
3
3
3
payoffs to
player I
Best responses to mixed strategy of player 2
6
1
3
3
5
2
3
2
0
1,0
0,1
4
5
1
0
6
2
2
5 =A
3
3
3
payoffs to
player I
Best responses to mixed strategy of player 2
1
3
2
4
5
1
0
6
2
2
5 =A
3
3
3
payoffs to
player I
best response polyhedron
1,0
0,1
Best responses to mixed strategy of player 2
5
4
1
3
2
4
5
1
0
6
2
2
5 =A
3
3
3
payoffs to
player I
best response polyhedron
with facet labels
1,0
0,1
Best responses to mixed strategy of player 2
5
4
1
3
5
3
2
2
4
5
1
0
6
2
2
5 =A
3
3
3
payoffs to
player I
1
4
Best responses to mixed strategy of player 2
4
5
1
0
6
2
2
5 =A
3
3
3
payoffs to
player I
5
3
2
1
4
Best responses to mixed strategy of player 1
4
5
1
2
1
2
1
3
3
4
3
=B
payoffs to
player II
0
0
1
1
0
0
0
1
0
Best responses to mixed strategy of player 1
4
5
1
2
1
2
1
3
3
4
3
=B
payoffs to
player II
0
0
1
1
0
0
0
1
0
Best responses to mixed strategy of player 1
4
4
5
1
2
1
2
1
3
3
4
3
=B
payoffs to
player II
2
1
0
0
0
0
1
1
0
1
0
Best responses to mixed strategy of player 1
4
4
5
1
2
1
2
1
3
3
4
3
=B
payoffs to
player II
2
1
0
0
0
0
1
1
0
1
0
Best responses to mixed strategy of player 1
3
3
1
1
0
0
0
0
1
0
1
0
4
5
1
2
1
2
1
3
3
4
3
=B
payoffs to
player II
Best responses to mixed strategy of player 1
3
3
1
1
0
0
0
0
1
0
1
0
4
5
1
2
1
2
1
3
3
4
3
=B
payoffs to
player II
Best responses to mixed strategy of player 1
4
5
1
2
1
2
1
3
3
4
3
=B
payoffs to
player II
0
0
1
1
0
0
0
1
0
Best responses to mixed strategy of player 1
4
5
1
2
1
2
1
3
3
4
3
=B
payoffs to
player II
0
0
1
1
0
0
0
1
0
Best responses to mixed strategy of player 1
1
2
4
5
0
0
1
1
0
0
3
4
5
1
2
1
2
1
3
3
4
3
=B
payoffs to
player II
best response
polyhedron
with facet labels
(front)
0
1
0
Alternative view
Chop off Toblerone prism
Chop off Toblerone prism
Chop off Toblerone prism
Chop off Toblerone prism
Chop off Toblerone prism
Best responses to mixed strategy of player 1
4
5
1
2
1
2
1
3
3
4
3
=B
payoffs to
player II
0
0
1
1
0
0
0
1
0
Best responses to mixed strategy of player 1
4
5
1
2
1
2
1
3
3
4
3
=B
payoffs to
player II
2
4
5
3
1
Equilibrium = completely labeled strategy pair
5
2
1
4
5
3
3
2
1
4
Equilibrium = completely labeled strategy pair
5
2
1
4
5
3
3
2
1
4
Equilibrium = completely labeled strategy pair
5
2
1
4
5
3
3
2
1
4
Constructing games using geometry
low dimension: 2, 3, (4) pure strategies:
subdivide mixed strategy simplex into
response regions, label suitably
high dimension:
use polytopes with known combinatorial structure
e.g. for constructing games with many equilibria,
or long Lemke-Howson computations
[Savani & von Stengel, FOCS 2004,
Econometrica 2006]
Construct isolated non−quasi−strict equilibrium
5
5
4
3
2
1
2
4
6
1
6
3
Construct isolated non−quasi−strict equilibrium
5
5
4
3
2
1
2
4
6
1
3
6
4
0 2 0
A= 2 0 0
1 1 1
1
2
3
5
6
1 0 0
B= 1 0 2
1 1 0
Best response polyhedron H2 for player 2
y4
°
1 3
°
2 2
°
3 0
y5
3
5 =A
6
u
4
5
H2
H2 = { (y4, y5, u) |
°
1 : 3y4 + 3y5 ≤ u
°
2 : 2y4 + 5y5 ≤ u
°
3 :
6y5 ≤ u
6
y4 + y5 = 1
3
3
5
2
1
2
°
4 : y4
°
5 :
≥ 0
y5 ≥ 0 }
y4
0
0
1
Best response polytope Q for player 2
y4
°
1 3
°
2 2
°
3 0
y5
3
5 =A
6
Q = { y | Ay≤1, y≥0 }
y5
Q = { (y4, y5) |
°
1 : 3y4 + 3y5 ≤ 1
°
2 : 2y4 + 5y5 ≤ 1
°
3 :
6y5 ≤ 1
°
4 : y4
°
5 :
≥ 0
y5 ≥ 0 }
3
2
4
1
5
y4
Projective transformation
°
4
H2, Q same face incidences
u
u
1
0
6
6
s
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-
0
yj
yj
£
£
£
£
£
£
£
£
£
£
H2
s
£B
£ B
B
B
B
B
B
B
°
5
°
3 BB
B
Bs
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°
2
y
1
0
°
1
Q
y4
The Lemke−Howson algorithm
5
4
3
2
1
4
5
3
2
1
The Lemke−Howson algorithm
5
4
3
2
1
4
5
3
2
1
The Lemke−Howson algorithm
0
0
0
artificial equilibrium
0,0
5
4
3
2
1
4
5
3
2
1
The Lemke−Howson algorithm
0
0
0
artificial equilibrium
0,0
5
4
3
2
2
1
1
missing label
4
5
3
2
The Lemke−Howson algorithm
0
0
0
0,0
5
4
3
2
2
1
1
missing label
4
5
3
2
The Lemke−Howson algorithm
0
0
0
0,0
5
4
3
2
2
1
1
missing label
4
5
3
2
The Lemke−Howson algorithm
0
0
0
0,0
5
4
3
2
2
1
1
missing label
4
5
3
2
The Lemke−Howson algorithm
0
0
0
0,0
5
4
3
2
2
1
1
missing label
4
5
3
2
The Lemke−Howson algorithm
0
0
0
0,0
5
4
3
2
2
1
1
found label
4
5
3
2
Why Lemke-Howson works
LH finds at least one Nash equilibrium because
•
finitely many "vertices"
for nondegenerate (generic) games:
•
unique starting edge given missing label
•
unique continuation
precludes "coming back" like here:
The Lemke−Howson algorithm
start at Nash equilibrium
5
4
3
2
2
1
1
missing label
4
5
3
2
The Lemke−Howson algorithm
start at Nash equilibrium
5
4
3
2
2
1
1
missing label
4
5
3
2
The Lemke−Howson algorithm
start at Nash equilibrium
5
4
3
2
2
1
1
missing label
4
5
3
2
Odd number of Nash equilibria!
start at Nash equilibrium
5
4
3
2
2
1
1
found label
4
5
3
2
Nondegenerate bimatrix games
Given:
m × n bimatrix game (A,B)
X = { x ∈ Rm | x ≥ 0, x1 + . . . + xm = 1 }
Y = { y ∈ Rn | y ≥ 0, y1 + . . . + yn = 1 }
supp(x) = { i | xi > 0 }
supp(y) = { j | yj > 0 }
(A,B) nondegenerate
⇔ ∀ x ∈X,
y ∈Y:
| { j | j best response to x } | ≤ | supp(x) |,
| { i | i best response to y } | ≤ | supp(y) |.
Nondegeneracy via labels
m × n bimatrix game (A,B) nondegenerate
⇔
no x  X has more than m labels,
no y  Y has more than n labels.
E.g. x with > m labels,
s labels from { 1 , . . . , m } ,
⇒ > m−s labels from { m+1 , . . . , m+n }
⇔ > |supp(x)| best responses to x.
⇒ degenerate.
Example of a degenerate game
4
5
1
2
1
2
1
3
3
2
1
4
5
3
4 4
=B
payoffs to
player II
Handling degenerate games
Lemke–Howson implemented by pivoting, i.e., changing from
one basic feasible solution of a linear system to another by choosing an entering and a leaving variable.
Choice of entering variable via complementarity (only difference
to simplex algorithm for linear programming).
Leaving variable is unique in nondegenerate games.
In degenerate games: perturb system by adding (ε, . . . , εn)> ,
creates nondegenerate system.
Implemented symbolically by lexicographic rule.
2-player game: find one Nash equilibrium
2-NASH ∈ PPAD (Polynomial Parity Argument with Direction)
Implicit digraph with indegrees and outdegrees ≤ 1 is a set of
[nodes], paths and cycles:
0
Parity argument: number of sources of paths = number of sinks
Comput. problem: given one source 0, find another source or sink
[Chen/Deng 2006] 2-NASH is PPAD-complete.
2
Symmetric Nash equilibria of symmetric games
square game matrix A = payoffs to row player
0 3 0
A= 2 2 2
3 0 0
3
Symmetric Nash equilibria of symmetric games
equilibrium: only optimal strategies are played
1/3 2/3 0
0 3 0 2
A= 2 2 2 2
3 0 0 1
3
Symmetric Nash equilibria of symmetric games
plot polytope with strategy weights z1 , z2 , z3
z3
z ≥0,
z1 z2 z3
0 3 0
A= 2 2 2
3 0 0
z2
z1
3
Symmetric Nash equilibria of symmetric games
with payoffs (scaled to 1) and labels for binding inequalities
z3
{ z |z ≥0, A z≤ 1 }
1 (back)
2
z1 z2 z3
0 3 0
A= 2 2 2
3 0 0
2
1
3
z2
3 (bottom)
z1
3
Symmetric Nash equilibria of symmetric games
equilibrium = completely labeled point
z3
{ z |z ≥0, A z≤ 1 }
1 (back)
1/6 1/3 0
2
0 3 0 1
A= 2 2 2 1
3 0 0 1/2
2
1
3
z2
3 (bottom)
z1
3
Symmetric Nash equilibria of symmetric games
start path with artificial equilibrium z=0
z3
{ z |z ≥0, A z≤ 1 }
1 (back)
2
0 3 0
A= 2 2 2
3 0 0
2
1
3
z2
3 (bottom)
z1
3
Symmetric Nash equilibria of symmetric games
start path with artificial equilibrium z=0, choose e.g.
z3
missing label 1
{ z |z ≥0, A z≤ 1 }
1 (back)
2
0 3 0
A= 2 2 2
3 0 0
2
1
3
z2
3 (bottom)
z1
3
Symmetric Nash equilibria of symmetric games
leave facet with label 1, find duplicate label 3
z3
missing label 1
{ z |z ≥0, A z≤ 1 }
1 (back)
2
0 3 0
A= 2 2 2
3 0 0
2
1
3
z2
3 (bottom)
z1
3
Symmetric Nash equilibria of symmetric games
leave facet with old label 3, find duplicate label 2
z3
missing label 1
{ z |z ≥0, A z≤ 1 }
1 (back)
2
0 3 0
A= 2 2 2
3 0 0
2
1
3
z2
3 (bottom)
z1
3
Symmetric Nash equilibria of symmetric games
leave facet with old label 2, find duplicate label 3
z3
missing label 1
{ z |z ≥0, A z≤ 1 }
1 (back)
2
0 3 0
A= 2 2 2
3 0 0
2
1
3
z2
3 (bottom)
z1
3
Symmetric Nash equilibria of symmetric games
leave facet with old label 3, find missing label 1
z3
found label 1
{ z |z ≥0, A z≤ 1 }
1 (back)
2
0 3 0
A= 2 2 2
3 0 0
2
1
3
z2
3 (bottom)
z1
3
Symmetric Nash equilibria of symmetric games
equilibria (including artificial equilibrium) = endpoints of paths
z3
missing label 1
{ z |z ≥0, A z≤ 1 }
1 (back)
1/6 1/3 0
2
0 3 0 1
A= 2 2 2 1
3 0 0 1/2
2
1
3
z2
3 (bottom)
z1
3
The castle where each room has at most two doors
4
The castle where each room has at most two doors
4
The castle where each room has at most two doors
4
The castle where each room has at most two doors
4
Path of “almost completely labeled” edges
two completely labeled vertices
2
2
1
3
1
3
5
Path of “almost completely labeled” edges
path because at most two neighbours (“doors” in castle)
2
2
1
3
1
3
5
Path of “almost completely labeled” edges
orientation of edges: 2 on left, 3 on right
2
2
1
3
1
3
5
Path of “almost completely labeled” edges
opposite orientation (“sign”) of endpoints
2
2
1
3
1
3
5
Path of “almost completely labeled” edges
equilibrium sign
−
or
+
does not depend on path
2
2
3
1
1
3
+
−
5
Path of “almost completely labeled” edges
equilibrium sign
−
or
+
does not depend on path
2
2
3
1
1
3
+
−
5
Path of “almost completely labeled” edges
equilibrium sign
−
or
+
does not depend on path
2
2
3
1
1
3
+
−
5
Path of “almost completely labeled” edges
equilibrium sign
−
or
+
does not depend on path
2
2
3
1
1
3
+
−
5
Path of “almost completely labeled” edges
equilibrium sign
−
or
+
does not depend on path
2
2
3
1
1
3
+
−
5
Labeled polytope P
Let aj ∈
R m , βj
∈ R,
P = {x ∈
Rm | a j x
≤ βj , 1 ≤ j ≤ n },
let facet
Fj = { x ∈ P | aj x = βj } have
label
l(j) ∈ {1, . . . , m}.
Assume P is a simple polytope (no x ∈ P on > m facets)
⇒ each vertex x on m facets = m linearly independent equations.
x completely labeled ⇔ {l(j) | x ∈ Fj } = {1, . . . , m}.
6
Completely labeled points come in pairs
Theorem [ Parity Argument ]
Let P be a labeled polytope.
Then P has an even number of completely labeled vertices.
7
Completely labeled points come in pairs
of opposite sign
Theorem [ Parity Argument with Direction ]
Let P be a labeled polytope.
Then P has an even number of completely labeled vertices.
Half of these have sign − , half have sign + .
7
Completely labeled points come in pairs
of opposite sign
Theorem [ Parity Argument with Direction ]
Let P be a labeled polytope.
Then P has an even number of completely labeled vertices.
Half of these have sign − , half have sign + .
sign of completely labeled x is sign of determinant of facet
normal vectors in order of their labels: if (e.g.) facet ai x = βi has
label i = 1, 2, ..., m, then
sign(x) = sign |a1 a2 · · · am |
7
Pivoting changes signs
Lemma
Let x, y ∈
Rm be adjacent vertices of a simple polytope P
x
y
8
Pivoting changes signs
Lemma
Let x, y ∈ Rm be adjacent vertices of a simple polytope P
with facet normals c, a2 , . . . , am for x and d, a2 , . . . , am for y.
a2
c
y
x
d
a3
8
Pivoting changes signs
Lemma
Let x, y ∈ Rm be adjacent vertices of a simple polytope P
with facet normals c, a2 , . . . , am for x and d, a2 , . . . , am for y.
Then |c a2 · · · am | and |d a2 · · · am | have opposite sign.
a2
c
y
x
d
a3
8
Pivoting changes signs
Proof :
cx = β0
a2 x = β2
..
.
dy = β1
a2 y = β2
..
.
am x = βm
a m y = βm
9
Pivoting changes signs
Proof :
cx = β0
a2 x = β2
..
.
dy = β1
a2 y = β2
..
.
am x = βm
a m y = βm
Let (γ, δ, α2 , . . . , αm ) 6= (0, 0, 0, . . . , 0) with
γc + δd + α2 a2 + · · · + αm am = 0
9
Pivoting changes signs
Proof :
cx = β0
a2 x = β2
..
.
dy = β1
a2 y = β2
..
.
am x = βm
a m y = βm
Let (γ, δ, α2 , . . . , αm ) 6= (0, 0, 0, . . . , 0) with
γc + δd + α2 a2 + · · · + αm am = 0
⇒ γ 6= 0, δ 6= 0,
(γc + δd)x = (γc + δd)y
9
Pivoting changes signs
Proof :
cx = β0
cy < β0
dx < β1
a2 x = β2
..
.
dy = β1
a2 y = β2
..
.
am x = βm
a m y = βm
Let (γ, δ, α2 , . . . , αm ) 6= (0, 0, 0, . . . , 0) with
γc + δd + α2 a2 + · · · + αm am = 0
⇒ γ 6= 0, δ 6= 0,
(γc + δd)x = (γc + δd)y,
γ(cx − cy) = δ(dy − dx)
9
Pivoting changes signs
Proof :
cx = β0
cy < β0
dx < β1
a2 x = β2
..
.
dy = β1
a2 y = β2
..
.
am x = βm
a m y = βm
Let (γ, δ, α2 , . . . , αm ) 6= (0, 0, 0, . . . , 0) with
γc + δd + α2 a2 + · · · + αm am = 0
⇒ γ 6= 0, δ 6= 0,
(γc + δd)x = (γc + δd)y,
γ(cx − cy) = δ(dy − dx)
⇒ γ and δ have same sign
9
Pivoting changes signs
Proof :
cx = β0
cy < β0
dx < β1
a2 x = β2
..
.
dy = β1
a2 y = β2
..
.
am x = βm
a m y = βm
Let (γ, δ, α2 , . . . , αm ) 6= (0, 0, 0, . . . , 0) with
γc + δd + α2 a2 + · · · + αm am = 0
⇒ γ 6= 0, δ 6= 0,
(γc + δd)x = (γc + δd)y,
γ(cx − cy) = δ(dy − dx)
⇒ γ and δ have same sign,
|(γc + δd) a2 · · · am | = γ |c a2 · · · am | + δ |d a2 · · · am | = 0
9
Pivoting changes signs
Proof :
cx = β0
cy < β0
dx < β1
a2 x = β2
..
.
dy = β1
a2 y = β2
..
.
am x = βm
a m y = βm
Let (γ, δ, α2 , . . . , αm ) 6= (0, 0, 0, . . . , 0) with
γc + δd + α2 a2 + · · · + αm am = 0
⇒ γ 6= 0, δ 6= 0,
(γc + δd)x = (γc + δd)y,
γ(cx − cy) = δ(dy − dx)
⇒ γ and δ have same sign,
|(γc + δd) a2 · · · am | = γ |c a2 · · · am | + δ |d a2 · · · am | = 0
⇒ |c a2 · · · am | and |d a2 · · · am | have opposite sign, QED.
9
General Parity Argument with Direction
Facet normal vectors a1 a2 a3 c1 c2 c3 , labels 1 2 3 1 2 3
c2
a1
a2
c3
c1
a3
10
General Parity Argument with Direction
Start with a1 a2 a3 , sign
−
−
a1 a 2 a3
c2
a1
a2
c3
c1
a3
10
General Parity Argument with Direction
Start with a1 a2 a3 , sign
− , label 1 missing, a1 → c3 gives sign +
−
+
a1 a 2 a3
c3 a 2 a3
c2
a1
a2
c3
c1
a3
10
General Parity Argument with Direction
Switch columns c3 and a3 in determinant: back to sign
a2
c3
−
+
a1 a 2 a3
c3 a 2 a3
a3 a 2 c3
c2
a1
−
c1
a3
10
General Parity Argument with Direction
next pivot a3 → c2 gives sign
c2
a1
a2
c3
+
−
+
a1 a 2 a3
c3 a 2 a3
a3 a 2 c3
c2 a 2 c3
c1
a3
10
General Parity Argument with Direction
Switch columns c2 and a2 in determinant: back to sign
c2
a1
a2
c3
c1
−
−
+
a1 a 2 a3
c3 a 2 a3
a3 a 2 c3
c2 a 2 c3
a2 c 2 c3
a3
10
General Parity Argument with Direction
next pivot a2 → a3 gives sign
c2
a1
a2
c3
c1
+
−
+
a1 a 2 a3
c3 a 2 a3
a3 a 2 c3
c2 a 2 c3
a2 c 2 c3
a3 c 2 c3
a3
10
General Parity Argument with Direction
Switch columns a3 and c3 in determinant: back to sign
c2
a1
a2
c3
c1
a3
−
−
+
a1 a 2 a3
c3 a 2 a3
a3 a 2 c3
c2 a 2 c3
a2 c 2 c3
a3 c 2 c3
c3 c 2 a3
10
General Parity Argument with Direction
Last pivot c3 → c1 gives sign
c2
a1
a2
c3
c1
a3
+ , opposite to starting sign − .
−
+
a1 a 2 a3
c3 a 2 a3
a3 a 2 c3
c2 a 2 c3
a2 c 2 c3
a3 c 2 c3
c3 c 2 a3
c1 c 2 a3
10
General Parity Argument with Direction
Only need: sign-switching of pivots and column exchanges
c2
a1
a2
c3
c1
a3
−
+
a1 a 2 a3
c3 a 2 a3
a3 a 2 c3
c2 a 2 c3
a2 c 2 c3
a3 c 2 c3
c3 c 2 a3
c1 c 2 a3
10
A more abstract example
−
a1 a 2 a3 a 4 a5
11
A more abstract example
−
+
a1 a 2 a3 a 4 a5
c3 a 2 a3 a 4 a5
11
A more abstract example
−
+
a1 a 2 a3 a 4 a5
c3 a 2 a3 a 4 a5
a3 a 2 c3 a 4 a5
11
A more abstract example
−
+
a1 a 2 a3 a 4 a5
c3 a 2 a3 a 4 a5
a3 a 2 c3 a 4 a5
c4 a 2 c3 a 4 a5
11
A more abstract example
−
+
a1 a 2 a3 a 4 a5
c3 a 2 a3 a 4 a5
a3 a 2 c3 a 4 a5
c4 a 2 c3 a 4 a5
a4 a 2 c3 c 4 a5
11
A more abstract example
−
+
a1 a 2 a3 a 4 a5
c3 a 2 a3 a 4 a5
a3 a 2 c3 a 4 a5
c4 a 2 c3 a 4 a5
a4 a 2 c3 c 4 a5
c5 a 2 c3 c 4 a5
11
A more abstract example
−
+
a1 a 2 a3 a 4 a5
c3 a 2 a3 a 4 a5
a3 a 2 c3 a 4 a5
c4 a 2 c3 a 4 a5
a4 a 2 c3 c 4 a5
c5 a 2 c3 c 4 a5
a5 a 2 c3 c 4 c5
11
A more abstract example
−
+
a1 a 2 a3 a 4 a5
c3 a 2 a3 a 4 a5
a3 a 2 c3 a 4 a5
c4 a 2 c3 a 4 a5
a4 a 2 c3 c 4 a5
c5 a 2 c3 c 4 a5
a5 a 2 c3 c 4 c5
c1 a 2 c3 c 4 c5
11
Nash equilibria of bimatrix games
Recall: m × m matrix C,
P = {z ∈
Rm | −z
≤ 0, Cz ≤ 1 }
with 2m inequalities labeled 1, . . . , m, 1, . . . , m.
12
Nash equilibria of bimatrix games
Recall: m × m matrix C,
P = {z ∈
Rm | −z
≤ 0, Cz ≤ 1 }
with 2m inequalities labeled 1, . . . , m, 1, . . . , m.
Completely labeled z 6= 0 ⇔
Nash equilibrium (z, z) of game (C, C > )
12
Nash equilibria of bimatrix games
Recall: m × m matrix C,
P = {z ∈
Rm | −z
≤ 0, Cz ≤ 1 }
with 2m inequalities labeled 1, . . . , m, 1, . . . , m.
Completely labeled z 6= 0 ⇔
Nash equilibrium (z, z) of game (C, C > )
Normalize sign of “artificial equilibrium” 0 to
− , in general
index(z) = sign(z) · (−1)m+1
12
Nash equilibria of bimatrix games
Recall: m × m matrix C,
P = {z ∈
Rm | −z
≤ 0, Cz ≤ 1 }
with 2m inequalities labeled 1, . . . , m, 1, . . . , m.
bimatrix game (A, B):
0 A
, z = (x, y) :
B> 0
C=
Completely labeled (x, y) 6= (0, 0) ⇔
Nash equilibrium (x, y) of game (A, B)
12
Index of an equilibrium
Theorem [Shapley 1974]
A nondegenerate bimatrix game (A, B) has an odd number of
equilibria, one more of index + than of index − .
13
Index of an equilibrium
Theorem [Shapley 1974]
A nondegenerate bimatrix game (A, B) has an odd number of
equilibria, one more of index + than of index − .
[Proof: Endpoints of pivoting paths have opposite index
−
and
+ .]
13
Index of an equilibrium
Theorem [Shapley 1974]
A nondegenerate bimatrix game (A, B) has an odd number of
equilibria, one more of index + than of index − .
[Proof: Endpoints of pivoting paths have opposite index
Equilibria of index
+
−
and
+ .]
include every
•
pure-strategy equilibrium
•
unique equilibrium
•
dynamically stable equilibrium [Hofbauer 2003]
13
Dynamically stable equilibrium: only if index
1
2
2
3
1
+
1
2
3
1
3
0
0
2
2
2
2
3
0
3
0
3
14
Dynamically stable equilibrium: only if index
1
2
2
3
1
+
1
2
3
1
3
0
0
2
2
2
2
3
0
3
0
3
14
Dynamically stable equilibrium: only if index
1
2
+
2
+
3
1
−
+
1
2
3
1
3
0
0
2
2
2
2
3
0
3
0
3
14
Dynamically stable equilibrium: only if index
1
2
+
2
+
3
1
−
+
1
2
3
1
3
0
0
2
2
2
2
3
0
3
0
3
−
14
Dynamically stable equilibrium: only if index
1
2
+
2
+
3
1
−
+
1
2
3
1
3
0
0
2
2
2
2
3
0
3
0
3
14
Dynamically stable equilibrium: only if index
1
2
+
2
+
3
1
−
+
1
2
3
1
3
0
0
2
2
2
2
3
0
3
0
3
14
Strategic characterization of the index
Theorem [von Schemde / von Stengel 2004]
An equilibrium of a nondegenerate bimatrix game has index
+
⇔ it is the unique equilibrium in a larger game that has suitable
additional strategies for one player.
15
Strategic characterization of the index
Theorem [von Schemde / von Stengel 2004]
An equilibrium of a nondegenerate bimatrix game has index
+
⇔ it is the unique equilibrium in a larger game that has suitable
additional strategies for one player.
Theorem [Balthasar / von Stengel 2009]
A symmetric equilibrium of a nondegenerate symmetric bimatrix
game has symmetric index +
⇔ it is the unique equilibrium in a larger symmetric game that
has suitable additional strategies for both players.
15