1. 6.853: Topics in Algorithmic Game Theory Fall 2011 Constantinos Daskalakis Lecture 6

2. Last time we showed Nash’s theorem that a Nash equilibrium exists in every game. In our proof, we used Brouwer’s fixed point theorem as a Black- box. We proceed to prove Brouwer’s Theorem using a combinatorial lemma, called Sperner’s Lemma, whose proof we also provide. In this lecture, we explain Brouwer’s theorem, and give an illustration of Nash’s proof.

3. Brouwer’ s Fixed Point Theorem

4. Brouwer’s fixed point theorem f Theorem: Let f : D D be a continuous function from a convex and compact subset D of the Euclidean space to itself. Then there exists an x s.t. x = f (x). N.B. All conditions in the statement of the theorem are necessary. closed and bounded D D Below we show a few examples, when D is the 2-dimensional disk.

5. Brouwer’s fixed point theorem fixed point

6. Brouwer’s fixed point theorem fixed point

7. Brouwer’s fixed point theorem fixed point

8. Nash’s Proof

9. Nash’s Function where:

10.  : [0,1] 2  [0,1] 2, continuous such that fixed points  Nash eq. Kick Dive Left Right Left1, -1-1, 1 Right-1, 1 1, -1 Visualizing Nash’s Construction Penalty Shot Game

11. Kick Dive Left Right Left1, -1-1, 1 Right-1, 1 1, -1 Visualizing Nash’s Construction Penalty Shot Game 01 0 1 Pr[Right]

12. Kick Dive Left Right Left1, -1-1, 1 Right-1, 1 1, -1 Visualizing Nash’s Construction Penalty Shot Game 01 0 1 Pr[Right]

13. Kick Dive Left Right Left1, -1-1, 1 Right-1, 1 1, -1 Visualizing Nash’s Construction Penalty Shot Game 01 0 1 Pr[Right]

14.   : [0,1] 2  [0,1] 2, cont. such that fixed point  Nash eq. Kick Dive Left Right Left1, -1-1, 1 Right-1, 1 1, -1 Visualizing Nash’s Construction Penalty Shot Game 01 0 1 Pr[Right] fixed point ½ ½ ½ ½

15. Sperner’s Lemma

17. no red no blue no yellow Lemma: Color the boundary using three colors in a legal way.

18. Sperner’s Lemma Lemma: Color the boundary using three colors in a legal way. No matter how the internal nodes are colored, there exists a tri-chromatic triangle. In fact an odd number of those. no red no blue no yellow

19. Sperner’s Lemma Lemma: Color the boundary using three colors in a legal way. No matter how the internal nodes are colored, there exists a tri-chromatic triangle. In fact an odd number of those.

20. Sperner’s Lemma Lemma: Color the boundary using three colors in a legal way. No matter how the internal nodes are colored, there exists a tri-chromatic triangle. In fact an odd number of those.

21. Proof of Sperner’s Lemma Lemma: Color the boundary using three colors in a legal way. No matter how the internal nodes are colored, there exists a tri-chromatic triangle. In fact an odd number of those. For convenience we introduce an outer boundary, that does not create new tri- chromatic triangles. Next we define a directed walk starting from the bottom-left triangle.

22. Transition Rule: If  red - yellow door cross it with red on your left hand. ? Space of Triangles 1 2 Proof of Sperner’s Lemma Lemma: Color the boundary using three colors in a legal way. No matter how the internal nodes are colored, there exists a tri-chromatic triangle. In fact an odd number of those.

23. Proof of Sperner’s Lemma ! Lemma: Color the boundary using three colors in a legal way. No matter how the internal nodes are colored, there exists a tri-chromatic triangle. In fact an odd number of those. For convenience we introduce an outer boundary, that does not create new tri- chromatic triangles. Next we define a directed walk starting from the bottom-left triangle. Starting from other triangles we do the same going forward or backward. Claim: The walk cannot exit the square, nor can it loop around itself in a rho-shape. Hence, it must stop somewhere inside. This can only happen at tri-chromatic triangle…

24. Proof of Brouwer’s Fixed Point Theorem We show that Sperner’s Lemma implies Brouwer’s Fixed Point Theorem. We start with the 2-dimensional Brouwer problem on the square.

25. 2D-Brouwer on the Square Suppose  : [0,1] 2  [0,1] 2, continuous must be uniformly continuous (by the Heine-Cantor theorem)Heine-Cantor theorem 1 01 0

26. 2D-Brouwer on the Square Suppose  : [0,1] 2  [0,1] 2, continuous must be uniformly continuous (by the Heine-Cantor theorem)Heine-Cantor theorem choose some and triangulate so that the diameter of cells is 1 01 0

27. 2D-Brouwer on the Square Suppose  : [0,1] 2  [0,1] 2, continuous must be uniformly continuous (by the Heine-Cantor theorem)Heine-Cantor theorem 1 01 0 color the nodes of the triangulation according to the direction of choose some and triangulate so that the diameter of cells is

28. 1 01 0 2D-Brouwer on the Square Suppose  : [0,1] 2  [0,1] 2, continuous must be uniformly continuous (by the Heine-Cantor theorem)Heine-Cantor theorem color the nodes of the triangulation according to the direction of tie-break at the boundary angles, so that the resulting coloring respects the boundary conditions required by Sperner’s lemma find a trichromatic triangle, guaranteed by Sperner choose some and triangulate so that the diameter of cells is

29. 2D-Brouwer on the Square Suppose  : [0,1] 2  [0,1] 2, continuous must be uniformly continuous (by the Heine-Cantor theorem)Heine-Cantor theorem Claim: If z Y is the yellow corner of a trichromatic triangle, then 1 01 0

30. 1 01 0 Proof of Claim Claim: If z Y is the yellow corner of a trichromatic triangle, then Proof: Let z Y, z R, z B be the yellow/red/blue corners of a trichromatic triangle. By the definition of the coloring, observe that the product of Hence: Similarly, we can show:

31. 2D-Brouwer on the Square Suppose  : [0,1] 2  [0,1] 2, continuous must be uniformly continuous (by the Heine-Cantor theorem)Heine-Cantor theorem Claim: If z Y is the yellow corner of a trichromatic triangle, then 1 01 0 Choosing

32. 2D-Brouwer on the Square - pick a sequence of epsilons: - define a sequence of triangulations of diameter: - pick a trichromatic triangle in each triangulation, and call its yellow corner Claim: Finishing the proof of Brouwer’s Theorem: - by compactness, this sequence has a converging subsequence with limit point Proof: Define the function. Clearly, is continuous since is continuous and so is. It follows from continuity that But. Hence,. It follows that. Therefore,

33. Sperner’ s Lemma in n dimensions

34. A. Canonical Triangulation of [0,1] n

35. Triangulation High-dimensional analog of triangle? in 2 dimensions: a trianglein n dimensions: an n-simplex i.e. the convex hull of n+1 points in general position

36. Simplicization of [0,1] n ?

37. 1 st Step: Division into Cubelets Divide each dimension into integer multiples of 2 -m, for some integer m.

38. 2 nd Step: 2 nd Step: Simplicization of each Cubelet note that all tetrahedra in this division use the corners 000 and 111 of the cube in 3 dimensions…

39. Generalization to n-dimensions For a permutation of the coordinates, define: 00000… 00001… 00011… 00111… 11111… … Claim 1: The unique integral corners of are the following n+1 points :

40. Simplicization Claim 2: is a simplex.