Exponentially many steps for finding a NE in a bimatrix game

Rahul Savani, Bernhard von Stengel (2004)

Presentation: Angelina Vidali

Previous Work• Morris, W. D., Jr. (1994), Lemke paths on simple

polytopes. Math. of Oper. Res. 19, 780–789.

uses duals of cyclic polytopes

but with different labeling• The LH method for finding a symmetric

equilibrium of a bimatrix game can take exponentially long.

• But these Games have non-symmetric equilibria that can be found very quickly.

Our Result

• A family of dxd games with a unique equilibrium for which:

the LH algorithm takes an exponential number of steps, for any dropped label.

Why Duals of cyclic polytopes?

• Idea: Look for examples for long Lemke Paths in polytopes with many vertices:

• Duals of cyclic polytopes have the maximum number of vertices for d-polytopes with a fixed number of facets.

• & a convinient combinatorial description!

Gale Evenness

• A bitstring represents a vertex iff any substring of the form 01…10 has an even length.

• Not allowed: 010, 001, 0101, 01110, …

• Allowed: 0110, 0011, 1110001, 011110,…

• Note we use cyclic symmetry.

Dual cyclic polytopesspace dimension: d d even

The vertices of the dual cyclic polytopes are bitstrings (u1,…,ud, ud+1,…,u2d) that:

1. Fullfill the Gale evenness condition2. Have exactly d ones and d zeros.

i.e.: 00001111, 11000011, 11001100,…Each vertex is on exactly d facets.

(simple polytope)

Facet labels=Pemutations l (for P) and l’ (for Q), of 1,…,2d

l : the identity permutation (l(k)=k)

Fixed points: l’ (1)=1, l’ (d)=dOtherwise: exchanges adjacent numbers:

l’•1,2,3,4,5,6,7,8 1,3,2,4,6,5,8,7

Artificial equilibrium: e0=(1d0d ,0d1d)

• The artificial equilibrium e0 is a vertex pair (u,v) such that:

• u has labels 1,…,d

• v has labels d+1,…,2d

(u,v)Completely labeled

(u1,…,ud, ud+1,…,u2d)1 1 0 0

Labels 1 … d d+1 … 2d 1 … d d+2 … 2d-1

(v1,…,vd, vd+1,…,v2d)0 0 1 1

Lemke-Howson on dual cyclic polytopesdropping label 1

Unique NE: (0d1d,1d0d) Lemma: Let e1=(0d1d,1d0d) This is the only NE of the Game.

Proof outline:

let (u,v) be a completely labeled vertex pair

(u0,…,ud,…,v0,…) (u,v)=e1=(0d1d,1d0d)

(u0,…,ud,…,v0,…) (u,v)= e0=(1d0d ,0d1d)

using: Definition of l’

0

1

01?0 0110& Gale Evenness:

Names for LH paths

• π(d,l)=LH path dropping label l in dim d

• π(d,1)=LH path dropping label 1 in dim d=A(d)

• π(d,2d)=LH path dropping label 2d in dim d=B(d)

Symmetries of π(d,1)Just mirror the image down!

Symmetries of π(d,2d)

Remaining path: B(d),

excluding the zero columns

is point symmetric around

A(4)=π(4,1)Label 1 is dropped

B(6)=π(d=6,12) drop label 12

A(4) is preffix of B(6)A(d) is preffix of B(d+2)? -Yes!

A(6)

B(6) is preffix of A(6)B(d) is preffix of A(d+2)? -Yes!

B(6)

C(6)

A(6)=B(6)+C(6)

A(6)

A(4)

B(6)

C(6)=A(4)+

Solution to the Recurrences: Fibonacci numbers

• Length(A(d))=Length(B(d))+Length(C(d))

• Length(C(d))=Length(A(d-2))+Length(C(d-2))

• Length(B(d))=Length(A(d-2))+Length(C(d-2))

d evenlengths ofB(2) C(2) A(2) B(4) C(4) A(4) B(6) C(6) A(6) . . .are the Fibonacci numbers

2 3 5 8 13 21 34 55 89 . . .We know their order of growth is exponential.

Longest paths: drop label 1 or 2d, paths A(d), B(d)

path length Ω( 3d/2 )

Shortest path: drop label 3d/2We can write it as: B(d/2)+B(d/2+1)path length: Ω( 3d/4 )= (1.434...d )

π(4,4) is transformed to π(4,1) using a shift and a reversalπ(4,1)

π(4,4)

The transformation shows that the paths have equal length.

dropping label 1 we don’t use cyclic symmetry

Here we do

Endnote

• a construction of dxd games with a unique equilibrium which is found by the LH algorithm using an exponential number of steps, for any dropped label.

• But: it easily guessed since it has full support.

It is the end!