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

3. 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.

4. 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.

5. 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!

6. 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.

7. Dual cyclic polytopes space dimension: dd even The vertices of the dual cyclic polytopes are bitstrings (u 1,…,u d, u d+1,…,u 2d ) that: 1.Fullfill the Gale evenness condition 2.Have exactly d ones and d zeros. i.e.: 00001111, 11000011, 11001100,… Each vertex is on exactly d facets. (simple polytope)

8. 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)=d Otherwise: exchanges adjacent numbers: l’ 1,2,3,4,5,6,7,81,3,2,4,6,5,8,7

9. Artificial equilibrium: e 0 =(1 d 0 d,0 d 1 d ) The artificial equilibrium e 0 is a vertex pair (u,v) such that: u has labels 1,…,d v has labels d+1,…,2d (u,v) Completely labeled (u 1,…,u d, u d+1,…,u 2d ) 11 00 Labels1 …dd+1 …2d1 …dd+2 …2d-1 (v 1,…,v d, v d+1,…,v 2d ) 0011

10. Lemke-Howson on dual cyclic polytopes dropping label 1

11. Unique NE: (0 d 1 d,1 d 0 d ) Lemma: Let e 1 =(0 d 1 d,1 d 0 d ) This is the only NE of the Game. Proof outline: let (u,v) be a completely labeled vertex pair (u 0,…,u d,…,v 0,…)(u,v)=e 1 =(0 d 1 d,1 d 0 d ) (u 0,…,u d,…,v 0,…)(u,v)= e 0 =(1 d 0 d,0 d 1 d ) using: Definition of l’ 0 1 01?0 0110 & Gale Evenness:

12. 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)

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

14. Symmetries of π(d,2d) Remaining path: B(d), excluding the zero columns is point symmetric around

15. A(4)=π(4,1) Label 1 is dropped

16. B(6)=π(d=6,12) drop label 12

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

18. A(6)

19. 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)

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

21. 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 21 345589... We know their order of growth is exponential.

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

23. π(4,4) is transformed to π(4,1) using a shift and a reversal π(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

24. 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.

25. It is the end!