Exponentially many steps for finding a NE in a bimatrix game Rahul Savani, Bernhard von Stengel...

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Exponentially many steps for finding a NE in a bimatrix game Rahul Savani, Bernhard von Stengel (2004) Presentation: Angelina Vidali
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Page 1: Exponentially many steps for finding a NE in a bimatrix game Rahul Savani, Bernhard von Stengel (2004) Presentation: Angelina Vidali.

Exponentially many steps for finding a NE in a bimatrix game

Rahul Savani, Bernhard von Stengel (2004)

Presentation: Angelina Vidali

Page 2: 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.

Page 3: Exponentially many steps for finding a NE in a bimatrix game Rahul Savani, Bernhard von Stengel (2004) Presentation: Angelina Vidali.

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.

Page 4: Exponentially many steps for finding a NE in a bimatrix game Rahul Savani, Bernhard von Stengel (2004) Presentation: Angelina Vidali.

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!

Page 5: Exponentially many steps for finding a NE in a bimatrix game Rahul Savani, Bernhard von Stengel (2004) Presentation: Angelina Vidali.

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.

Page 6: Exponentially many steps for finding a NE in a bimatrix game Rahul Savani, Bernhard von Stengel (2004) Presentation: Angelina Vidali.

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)

Page 7: Exponentially many steps for finding a NE in a bimatrix game Rahul Savani, Bernhard von Stengel (2004) Presentation: Angelina Vidali.

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

Page 8: Exponentially many steps for finding a NE in a bimatrix game Rahul Savani, Bernhard von Stengel (2004) Presentation: Angelina Vidali.

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

Page 9: Exponentially many steps for finding a NE in a bimatrix game Rahul Savani, Bernhard von Stengel (2004) Presentation: Angelina Vidali.

Lemke-Howson on dual cyclic polytopesdropping label 1

Page 10: Exponentially many steps for finding a NE in a bimatrix game Rahul Savani, Bernhard von Stengel (2004) Presentation: Angelina Vidali.

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:

Page 11: Exponentially many steps for finding a NE in a bimatrix game Rahul Savani, Bernhard von Stengel (2004) Presentation: Angelina Vidali.

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)

Page 12: Exponentially many steps for finding a NE in a bimatrix game Rahul Savani, Bernhard von Stengel (2004) Presentation: Angelina Vidali.

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

Page 13: Exponentially many steps for finding a NE in a bimatrix game Rahul Savani, Bernhard von Stengel (2004) Presentation: Angelina Vidali.

Symmetries of π(d,2d)

Remaining path: B(d),

excluding the zero columns

is point symmetric around

Page 14: Exponentially many steps for finding a NE in a bimatrix game Rahul Savani, Bernhard von Stengel (2004) Presentation: Angelina Vidali.

A(4)=π(4,1)Label 1 is dropped

Page 15: Exponentially many steps for finding a NE in a bimatrix game Rahul Savani, Bernhard von Stengel (2004) Presentation: Angelina Vidali.

B(6)=π(d=6,12) drop label 12

Page 16: Exponentially many steps for finding a NE in a bimatrix game Rahul Savani, Bernhard von Stengel (2004) Presentation: Angelina Vidali.

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

Page 17: Exponentially many steps for finding a NE in a bimatrix game Rahul Savani, Bernhard von Stengel (2004) Presentation: Angelina Vidali.

A(6)

Page 18: Exponentially many steps for finding a NE in a bimatrix game Rahul Savani, Bernhard von Stengel (2004) Presentation: Angelina Vidali.

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)

Page 19: Exponentially many steps for finding a NE in a bimatrix game Rahul Savani, Bernhard von Stengel (2004) Presentation: Angelina Vidali.

A(6)

A(4)

B(6)

C(6)=A(4)+

Page 20: Exponentially many steps for finding a NE in a bimatrix game Rahul Savani, Bernhard von Stengel (2004) Presentation: Angelina Vidali.

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.

Page 21: Exponentially many steps for finding a NE in a bimatrix game Rahul Savani, Bernhard von Stengel (2004) Presentation: Angelina Vidali.

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 )

Page 22: Exponentially many steps for finding a NE in a bimatrix game Rahul Savani, Bernhard von Stengel (2004) Presentation: Angelina Vidali.

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

Page 23: Exponentially many steps for finding a NE in a bimatrix game Rahul Savani, Bernhard von Stengel (2004) Presentation: Angelina Vidali.

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.

Page 24: Exponentially many steps for finding a NE in a bimatrix game Rahul Savani, Bernhard von Stengel (2004) Presentation: Angelina Vidali.

It is the end!