6.896: Topics in Algorithmic Game Theory
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6.896: Topics in Algorithmic Game Theory
Audiovisual Supplement toLecture 5
Constantinos Daskalakis
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On 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.
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Brouwer’ s Fixed Point Theorem
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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.
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Brouwer’s fixed point theorem
fixed point
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Brouwer’s fixed point theorem
fixed point
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Brouwer’s fixed point theorem
fixed point
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Nash’s Proof
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: [0,1]2[0,1]2, continuoussuch that
fixed points Nash eq.
Kick Dive Left Right
Left 1 , -1 -1 , 1
Right -1 , 1 1, -1
Visualizing Nash’s Construction
Penalty Shot Game
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Kick Dive Left Right
Left 1 , -1 -1 , 1
Right -1 , 1 1, -1
Visualizing Nash’s Construction
Penalty Shot Game
0 10
1
Pr[Right]
Pr[R
ight
]
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Kick Dive Left Right
Left 1 , -1 -1 , 1
Right -1 , 1 1, -1
Visualizing Nash’s Construction
Penalty Shot Game
0 10
1
Pr[Right]
Pr[R
ight
]
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Kick Dive Left Right
Left 1 , -1 -1 , 1
Right -1 , 1 1, -1
Visualizing Nash’s Construction
Penalty Shot Game
0 10
1
Pr[Right]
Pr[R
ight
]
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: [0,1]2[0,1]2, cont.such that
fixed point Nash eq.
Kick Dive Left Right
Left 1 , -1 -1 , 1
Right -1 , 1 1, -1
Visualizing Nash’s Construction
Penalty Shot Game
0 10
1
Pr[Right]
Pr[R
ight
]fixed point
½½
½
½
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Sperner’s Lemma
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Sperner’s Lemma
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Sperner’s Lemma
no red
no blue
no yellow
Lemma: Color the boundary using three colors in a legal way.
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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
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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.
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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.
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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.
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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.
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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…
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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.
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2D-Brouwer on the SquareSuppose : [0,1]2[0,1]2, continuous
must be uniformly continuous (by the Heine-Cantor theorem)
1
0 10
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2D-Brouwer on the SquareSuppose : [0,1]2[0,1]2, continuous
must be uniformly continuous (by the Heine-Cantor theorem)
choose some and triangulate so that the diameter of cells is at most
1
0 10
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2D-Brouwer on the SquareSuppose : [0,1]2[0,1]2, continuous
must be uniformly continuous (by the Heine-Cantor theorem)
choose some and triangulate so that the diameter of cells is at most
1
0 10
color the nodes of the triangulation according to the direction of
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1
0 10
2D-Brouwer on the SquareSuppose : [0,1]2[0,1]2, continuous
must 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 lemma
find a trichromatic triangle, guaranteed by Sperner
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2D-Brouwer on the SquareSuppose : [0,1]2[0,1]2, continuous
must be uniformly continuous (by the Heine-Cantor theorem)
Claim: If zY is the yellow corner of a trichromatic triangle, then
1
0 10
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1
0 10
Proof of Claim
Claim: If zY is the yellow corner of a trichromatic triangle, then
Proof: 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:
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2D-Brouwer on the SquareSuppose : [0,1]2[0,1]2, continuous
must be uniformly continuous (by the Heine-Cantor theorem)
Claim: If zY is the yellow corner of a trichromatic triangle, then
1
0 10
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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 subsequencewith 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,