Slide 1Slide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12Slide 13Slide 14Sperner’s LemmaSperner’s LemmaSperner’s LemmaSperner’s LemmaSperner’s LemmaProof of Sperner’s LemmaProof of Sperner’s LemmaProof of Sperner’s LemmaSlide 23Slide 24Slide 25Slide 26Slide 27Slide 28Slide 29Slide 30Slide 316.896: Topics in Algorithmic Game TheoryAudiovisual Supplement toLecture 5Constantinos DaskalakisOn the blackboard we defined multi-player games and Nash equilibria, and showed Nash’s theorem that a Nash equilibrium exists in every game.In our proof, we used Brouwer’s fixed point theorem. In this presentation, we explain Brouwer’s theorem, and give an illustration of Nash’s proof.We proceed to prove Brouwer’s Theorem using a combinatorial lemma whose proof we also provide, called Sperner’s Lemma.Brouwer’ s Fixed Point TheoremBrouwer’s fixed point theoremfTheorem: 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 boundedDDBelow we show a few examples, when D is the 2-dimensional disk.Brouwer’s fixed point theoremfixed pointBrouwer’s fixed point theoremfixed pointBrouwer’s fixed point theoremfixed pointNash’s Proof: [0,1]2[0,1]2, continuoussuch thatfixed points  Nash eq. Kick Dive Left RightLeft 1 , -1 -1 , 1Right -1 , 1 1, -1Visualizing Nash’s ConstructionPenalty Shot GameKick Dive Left RightLeft 1 , -1 -1 , 1Right -1 , 1 1, -1Visualizing Nash’s ConstructionPenalty Shot Game0 101Pr[Right]Pr[Right]Kick Dive Left RightLeft 1 , -1 -1 , 1Right -1 , 1 1, -1Visualizing Nash’s ConstructionPenalty Shot Game0 101Pr[Right]Pr[Right]Kick Dive Left RightLeft 1 , -1 -1 , 1Right -1 , 1 1, -1Visualizing Nash’s ConstructionPenalty Shot Game0 101Pr[Right]Pr[Right]: [0,1]2[0,1]2, cont.such thatfixed point  Nash eq. Kick Dive Left RightLeft 1 , -1 -1 , 1Right -1 , 1 1, -1Visualizing Nash’s ConstructionPenalty Shot Game0 101Pr[Right]Pr[Right]fixed point½½½½Sperner’s LemmaSperner’s LemmaSperner’s Lemmano redno blueno yellowLemma: Color the boundary using three colors in a legal way.Sperner’s LemmaLemma: 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 redno blueno yellowSperner’s LemmaLemma: 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.Sperner’s LemmaLemma: 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.Proof of Sperner’s LemmaLemma: 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.Transition Rule: If  red - yellow door cross it with red on your left hand.?Space of Triangles12Proof of Sperner’s LemmaLemma: 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.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…Proof of Brouwer’s Fixed Point TheoremWe show that Sperner’s Lemma implies Brouwer’s Fixed Point Theorem. We start with the 2-dimensional Brouwer problem on the square.2D-Brouwer on the SquareSuppose : [0,1]2[0,1]2, continuousmust be uniformly continuous (by the Heine-Cantor theorem)10 102D-Brouwer on the SquareSuppose : [0,1]2[0,1]2, continuousmust be uniformly continuous (by the Heine-Cantor theorem)choose some and triangulate so that the diameter of cells is at most 10 102D-Brouwer on the SquareSuppose : [0,1]2[0,1]2, continuousmust be uniformly continuous (by the Heine-Cantor theorem)choose some and triangulate so that the diameter of cells is at most 10 10color the nodes of the triangulation according to the direction of10 102D-Brouwer on the SquareSuppose : [0,1]2[0,1]2, continuousmust be uniformly continuous (by the Heine-Cantor theorem)choose some and triangulate so that the diameter of cells is at most 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 lemmafind a trichromatic triangle, guaranteed by Sperner2D-Brouwer on the SquareSuppose : [0,1]2[0,1]2, continuousmust be uniformly continuous (by the Heine-Cantor theorem)Claim: If zY is the yellow corner of a trichromatic triangle, then10 1010 10Proof of ClaimClaim: If zY is the yellow corner of a trichromatic triangle, thenProof: Let zY, zR , zB 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:2D-Brouwer on the SquareSuppose : [0,1]2[0,1]2, continuousmust be uniformly continuous (by the Heine-Cantor theorem)Claim: If zY is the yellow corner of a trichromatic triangle, then10 102D-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 cornerClaim:Finishing the proof of Brouwer’s Theorem:- by compactness, this sequence has a converging subsequencewith limit point Proof: Define the function . Clearly, is continuous since is