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Algorithmic Game Theory Alexander Skopalik

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Page 1: Alexander Skopalik - Heinz Nixdorf Institut · Sperner’s Lemma Lemma (Sperner’s Lemma) Every Sperner coloring of a subdivided triangle contains a trichromatic triangle. •If

Algorithmic Game Theory

Alexander Skopalik

Page 2: Alexander Skopalik - Heinz Nixdorf Institut · Sperner’s Lemma Lemma (Sperner’s Lemma) Every Sperner coloring of a subdivided triangle contains a trichromatic triangle. •If

Today

1. Strategic games (Normal form games) – Complexity

2. Zero Sum Games – Existence & Complexity

Outlook, next week:

• Network Creation Games Guest lecture by Andreas Cord-Landwehr

Page 3: Alexander Skopalik - Heinz Nixdorf Institut · Sperner’s Lemma Lemma (Sperner’s Lemma) Every Sperner coloring of a subdivided triangle contains a trichromatic triangle. •If

Nash‘s Theorem

Page 4: Alexander Skopalik - Heinz Nixdorf Institut · Sperner’s Lemma Lemma (Sperner’s Lemma) Every Sperner coloring of a subdivided triangle contains a trichromatic triangle. •If

Proof of Nash‘s Theorem

• Idea: Construct a function 𝑓: 𝑋 → 𝑋 (with 𝑋 being set of mixed states) that has a fixpoint 𝑓 𝑥∗ =𝑥∗ if and anly if 𝑥∗is a mixed Nash equilibrium.

• Show: 𝑋 and 𝑓 satisfies the conditions for Brower’s Theorem.

• Proof: a fixpoint 𝑥∗is a mixed Nash equilibrium and vice-versa.

• Then a fixpoint exists and, thus, a mixed equilibrium.

Page 5: Alexander Skopalik - Heinz Nixdorf Institut · Sperner’s Lemma Lemma (Sperner’s Lemma) Every Sperner coloring of a subdivided triangle contains a trichromatic triangle. •If

Complexity of NASH

Theorem: Finding a Nash equilibrium is PPAD-complete. 1. Nash is in PPAD.

It is not harder than for example End-of-a-line. Show that you reduce the problem NASH to End-of-a-line (and then you could use an algorithm for end of the line).

2. Nash is PPAD-hard. It is at least as hard as End-of-a-line. Reduce End-of-a-line to NASH, so you could use an algorithm for NASH to solve End-of-a-line.

We only show 1. The proof of 2. is very involved.

Page 6: Alexander Skopalik - Heinz Nixdorf Institut · Sperner’s Lemma Lemma (Sperner’s Lemma) Every Sperner coloring of a subdivided triangle contains a trichromatic triangle. •If

Nash is in PPAD

Lemma: Finding an (approximate) Nash equilibrum is in PPAD.

Outline of the proof:

1. Reduction: Finding fixed points

2. Subdivide the space into finite number of smaller areas and apply some fancy coloring

3. Use End-of-a-line to find an area close to a fixed point.

Page 7: Alexander Skopalik - Heinz Nixdorf Institut · Sperner’s Lemma Lemma (Sperner’s Lemma) Every Sperner coloring of a subdivided triangle contains a trichromatic triangle. •If

End-of-a-line

Page 8: Alexander Skopalik - Heinz Nixdorf Institut · Sperner’s Lemma Lemma (Sperner’s Lemma) Every Sperner coloring of a subdivided triangle contains a trichromatic triangle. •If

Step 2: Subdivision - Simplifying assumptions.

• For simplicity, we only consider only 𝐷 = ℝ2 and functions 𝑓: 0,1 2 → 0,1 2

• Furthermore, we transform 0,1 2 to a triangle 𝑇:

and consider the corresponding problem on 𝑇.

Page 9: Alexander Skopalik - Heinz Nixdorf Institut · Sperner’s Lemma Lemma (Sperner’s Lemma) Every Sperner coloring of a subdivided triangle contains a trichromatic triangle. •If

Step2: Subdivision

• Divide T into smaller triangles.

• Draw an arrow at the vertices into the direction of f.

These pictures and a nice introduction to PPAD can be found on http://cgi.csc.liv.ac.uk/~pwg/PPADintro/PPADintro.html Thanks to Paul Goldberg.

Page 10: Alexander Skopalik - Heinz Nixdorf Institut · Sperner’s Lemma Lemma (Sperner’s Lemma) Every Sperner coloring of a subdivided triangle contains a trichromatic triangle. •If

Step2: Subdivision

• Give the 3 outermost (extremal) vertices of the triangle the 3 colors red, green and blue

• Color the vertices according to the direction of the arrows.

• Each vertex with an arrow gets colored with a color of an extreme vertex that it is moving away from.

• Note: vertices on the edge between the outermost red and green vertices will be either red or green. Similarly for vertices on edges between other vertices.

Page 11: Alexander Skopalik - Heinz Nixdorf Institut · Sperner’s Lemma Lemma (Sperner’s Lemma) Every Sperner coloring of a subdivided triangle contains a trichromatic triangle. •If

• Intuitively, fixpoints of 𝑓 lie in vicinity of small triangles whose 3 vertices get 3 different colors.

• This is because the points are being dragged in 3 different directions

• By continuity, if we triangulate at a sufficiently fine resolution, we converge to a fixpoint.

• Here: 3 small triangles with vertices of all colors, marked with black spots

Page 12: Alexander Skopalik - Heinz Nixdorf Institut · Sperner’s Lemma Lemma (Sperner’s Lemma) Every Sperner coloring of a subdivided triangle contains a trichromatic triangle. •If

Sperner’s Lemma

Definition: A Sperner coloring of the vertices of a subdivided triangle satisfies: • Each extremal vertex gets a

different color. • A vertex on a side of the

largest triangle gets a color of one of the corresponding endpoints.

• Other vertices are colored arbitrarily.

Lemma (Sperner’s Lemma) Every Sperner coloring of a subdivided triangle contains a trichromatic triangle.

Page 13: Alexander Skopalik - Heinz Nixdorf Institut · Sperner’s Lemma Lemma (Sperner’s Lemma) Every Sperner coloring of a subdivided triangle contains a trichromatic triangle. •If

Sperner’s Lemma

Lemma (Sperner’s Lemma)

Every Sperner coloring of a subdivided triangle contains a trichromatic triangle.

• If we can find a trichromatic triangle, we find an (approximate) Nash equilibrium.

• The proof of Sperner‘s Lemma tells us how to find one using End-of-a-line.

Page 14: Alexander Skopalik - Heinz Nixdorf Institut · Sperner’s Lemma Lemma (Sperner’s Lemma) Every Sperner coloring of a subdivided triangle contains a trichromatic triangle. •If

Sperner’s Lemma

• We start with a valid Sperner Coloring.

Page 15: Alexander Skopalik - Heinz Nixdorf Institut · Sperner’s Lemma Lemma (Sperner’s Lemma) Every Sperner coloring of a subdivided triangle contains a trichromatic triangle. •If

Sperner’s Lemma

• We start with a valid Sperner Coloring

• We extend the triangulation by connecting some extremal vertex to all vertices along one of its incident edges.

Page 16: Alexander Skopalik - Heinz Nixdorf Institut · Sperner’s Lemma Lemma (Sperner’s Lemma) Every Sperner coloring of a subdivided triangle contains a trichromatic triangle. •If

Sperner’s Lemma

• We start with a valid Sperner Coloring

• We extend the triangulation by connecting some extremal vertex to all vertices along one of its incident edges.

• Treat each red-green edge as having a gateway through it.

Page 17: Alexander Skopalik - Heinz Nixdorf Institut · Sperner’s Lemma Lemma (Sperner’s Lemma) Every Sperner coloring of a subdivided triangle contains a trichromatic triangle. •If

Sperner’s Lemma

• We start with a valid Sperner Coloring

• We extend the triangulation by connecting some extremal vertex to all vertices along one of its incident edges.

• Treat each red-green edge as having a gateway through it.

Page 18: Alexander Skopalik - Heinz Nixdorf Institut · Sperner’s Lemma Lemma (Sperner’s Lemma) Every Sperner coloring of a subdivided triangle contains a trichromatic triangle. •If

Sperner’s Lemma

• We can do the same thing with the red-blue edges

Page 19: Alexander Skopalik - Heinz Nixdorf Institut · Sperner’s Lemma Lemma (Sperner’s Lemma) Every Sperner coloring of a subdivided triangle contains a trichromatic triangle. •If

Sperner’s Lemma

• We can do the same thing with the red-blue edges

• And again we find a trichromatic triangle by following the path through the gateways.

Page 20: Alexander Skopalik - Heinz Nixdorf Institut · Sperner’s Lemma Lemma (Sperner’s Lemma) Every Sperner coloring of a subdivided triangle contains a trichromatic triangle. •If

Sperner’s Lemma

We can construct a directed graph:

• vertices are the tiny triangles

• edges are pairs of adjacent triangles connected by a red-blue edge.

Page 21: Alexander Skopalik - Heinz Nixdorf Institut · Sperner’s Lemma Lemma (Sperner’s Lemma) Every Sperner coloring of a subdivided triangle contains a trichromatic triangle. •If

Sperner’s Lemma

• The graph has one known source: The outer plane.

• Each vertex has in indegree and outdegree at most one.

• We can construct simple algorithms/circuits that compute successor and predecessor.

• Now we have an instance of End-of-a-line

Page 22: Alexander Skopalik - Heinz Nixdorf Institut · Sperner’s Lemma Lemma (Sperner’s Lemma) Every Sperner coloring of a subdivided triangle contains a trichromatic triangle. •If

Summary

We have shown that finding an (approximate) Nash equilibrium is in PPAD.

In fact finding a Nash equilibrium is PPAD-complete – even for games with two players.

What does this mean?