Exponentially many steps for finding a NE in a bimatrix game

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

2. 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.

3. Our Result • A family of dxd games with a unique equilibrium for which: the LH algorithm takes anexponential number of steps, forany droppedlabel.

4. 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!

5. 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.

6. Dual cyclic polytopes space dimension:d d even The vertices of thedual cyclic polytopes are bitstrings (u1,…,ud, ud+1,…,u2d) that: • Fullfill theGale evenness condition • Have exactlyd onesandd zeros. i.e.: 00001111, 11000011, 11001100,… Each vertex is on exactly d facets. (simple polytope)

7. l’ • 1,2,3,4,5,6,7,8 1,3,2,4,6,5,8,7 Facet labels =Pemutationsl (for P) and l’ (for Q), of 1,…,2d l : the identity permutation (l(k)=k) Fixed points: l’ (1)=1, l’ (d)=d Otherwise: exchanges adjacent numbers:

8. 1 1 1 1 0 0 0 0 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 Labels 1 … d d+1 … 2d 1 … d d+2 … 2d-1 (u1,…,ud, ud+1,…,u2d) (v1,…,vd, vd+1,…,v2d)

9. Lemke-Howson on dual cyclic polytopesdropping label 1

10. 0 1 0110 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’ & Gale Evenness: 01?0

11. Names for LH paths • π(d,l)=LH path droppinglabel lindim d • π(d,1)=LH path droppinglabel 1indim d=A(d) • π(d,2d)=LH path droppinglabel 2dindim d=B(d)

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

13. Remaining path: B(d), excludingthe zero columns is point symmetric around Symmetries of π(d,2d)

14. Label 1 is dropped A(4)=π(4,1)

15. drop label 12 B(6)=π(d=6,12)

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

17. A(6)

18. C(6) B(6) is preffix of A(6) B(d) is preffix of A(d+2)? -Yes! B(6) A(6)=B(6)+C(6)

19. C(6)=A(4)+ A(6) B(6) A(4)

20. 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 even lengths of B(2)C(2)A(2)B(4)C(4)A(4)B(6)C(6)A(6) . . . are the Fibonacci numbers 235813 21345589. . . We know their order of growth is exponential.

21. Longest paths: drop label 1 or 2d, paths A(d), B(d) path length Ω( q 3d/2 ) Shortest path: drop label 3d/2 We can write it as: B(d/2)+B(d/2+1) path length: Ω( q3d/4 )= (1.434...d)

22. π(4,1) π(4,4) is transformed to π(4,1) using a shift and a reversal The transformation shows that the paths have equal length. dropping label 1 we don’t use cyclic symmetry Here we do π(4,4)

23. Endnote • a construction of dxd games with a unique equilibriumwhich is found by the LH algorithm using an exponential number of steps, for any droppedlabel. • But: it easily guessed since it has full support.

24. It is the end!